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www.tntextbooks.in, , GOVERNMENT OF TAMIL NADU, , HIGHER SECONDARY SECOND YEAR, , CHEMISTRY, VOLUME - II, , A publication under Free Textbook Programme of Government of Tamil Nadu, , Department of School Education, Untouchability is Inhuman and a Crime, , Introduction Pages.indd 1, , 2/19/2020 5:08:37 PM

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www.tntextbooks.in, , Government of Tamil Nadu, First Edition, , -, , 2019, , Revised Edition, , -, , 2020, , (Published under new syllabus), , NOT FOR SALE, , Content Creation, , The wise, possess all, , State Council of Educational, Research and Training, © SCERT 2019, , Printing & Publishing, , Tamil NaduTextbook and Educational, Services Corporation, www.textbooksonline.tn.nic.in, , II, , Introduction Pages.indd 2, , 2/19/2020 5:08:37 PM

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www.tntextbooks.in, , Key features …, , Scope of Chemistry, Learning objectives, Do you know, Example Problems, , Awareness about higher education avenues in, the field of Chemistry, Describe the specific competency /, performance capability acquired by the learner, Additional information provided to relate the, content to day-to-day life / development in the, field, Model problems worked out for clear-cut, comprehension by the learners, , Evaluate yourself, , To help the students to assess their own, conceptual understanding, , Q.R code, , Quick access to concepts, videos, animations, and tutorials, , ICT, , opens up resources for learning; enables the, learners to access, extend transform ideas /, informations, , Summary, , A glance on the substance of the unit, , Concept map, Evaluation, , Inter relating the concepts for enabling learners, to visualize the essence of the unit, To assess the level of understanding through, multiple choice question,numerical problems, etc…, , Books for Reference List of relevant books for further reading, Key answers, Glossary, , To help the learners confirm the accuracy of, the answers arrived and remedy the gaps in, learning, Important terms are enlisted with equivalent, Tamil words, , III, , Introduction Pages.indd 3, , 2/19/2020 5:08:38 PM

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www.tntextbooks.in, , CONTENTS, CHEMISTRY, S.No., , Topic, , Page.No., , Month, , 8, , Ionic Equilibrium, , 01, , August, , 9, , Electro Chemistry, , 33, , October, , 10, , Surface Chemistry, , 69, , November, , 11, , Hydroxy Compounds and Ethers, , 104, , July, , 12, , Carbonyl Compounds and Carboxylic Acids, , 145, , August upto, page 176., Remaining, september, , 13, , Organic Nitrogen Compounds, , 196, , October, , 14, , Biomolecules, , 237, , November, , 15, , Chemistry in Everyday Life, , 272, , November, , ANSWERS, , 297, , GLOSSARY, , 320, , E-book, , Assessment, , DIGI links, , Let’s use the QR code in the text books!, •, , Download DIKSHA app from the Google Play Store., , •, , Tap the QR code icon to scan QR codes in the textbook., , •, , Point the device and focus on the QR code., , •, , On successful scan, content linked to the QR code gets listed., , Note: For ICT corner, Digi Links QR codes use any other QR scanner., , IV, , Introduction Pages.indd 4, , 2/19/2020 5:08:38 PM

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www.tntextbooks.in, , UNIT, , 8, , IONIC, EQUILIBRIUM, , Learning Objectives, , Peter Joseph William Debye, , After studying this unit, the students, will be able to,  classify the substances into acids and, bases based on Arrhenius, Lowry –, Bronsted and Lewis concepts.,  define, pH scale and establish, relationship between pH and pOH,  describe the equilibrium involved in, the ionisation of water.,  explain Ostwald’s dilution Law and, derive a relationship between the, dissociation constant and degree of, dissociation of a weak electrolyte.,  recognise the concept of common ion, effect and explain buffer action.,  apply Henderson equation for the, preparation of buffer solution,  calculate, solubility product and, understand the relation between, solubility and solubility product.,  solve numerical problems involving, ionic equilibria., , Peter Joseph William Debye was, Dutch-American, physicist, greatly contributed to the theory, of electrolyte solutions. He also, studied the dipole moments, of molecules, Debye won the, Nobel Prize in Chemistry, (1936) for his contributions, to, the, determination, of, molecular structure through, his investigations on dipole, moments and X-rays diffraction., , 1, , XII U8-Ionic equilibrium.indd 1, , 2/19/2020 5:11:18 PM

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www.tntextbooks.in, , INTRODUCTION, We have already learnt the chemical equilibrium in XI standard. In this unit, we discuss, the ionic equilibria, specifically acid – base equilibria. Some of the important processes in our, body involve aqueous equilibria. For example, the carbonic acid – bicarbonate buffer in the, blood., H3O+ (aq)+HCO3- (aq)  H2CO3 (aq)+H2O(l), We have come across many chemical compounds in our daily life among them acids and, bases are the most common. For example, milk contains lactic acid, vinegar acetic acid, tea, tannic acid and antacid tablet aluminium hydroxide / magnesium hydroxide. Acids and bases, have many important industrial applications. For example, sulphuric acid is used in fertilizer, industry and sodium hydroxide in soap industry etc… Hence, it is important to understand, the properties of acids and bases., In this unit we shall learn the definitions of acids and bases and study, their ionisation in, aqueous solution. We learn the pH scale and also apply the principles of chemical equilibrium, to determine the concentration of the species furnished in aqueous solution by acids and, bases., , 8.1 Acids and bases, The term ‘acid’ is derived from the latin word ‘acidus’ meaning sour. We have already, learnt in earlier classes that acid tastes sour, turns the blue litmus to red and reacts with metals, such as zinc and produces hydrogen gas. Similarly base tastes bitter and turns the red litmus, to blue., These classical concepts are not adequate to explain the complete behaviour of acids and, bases. So, the scientists developed the acid – base concept based on their behaviour., Let us, learn the concept developed by scientists Arrhenius, Bronsted and Lowry and, Lewis to describe the properties of acids and bases., 8.1.1 Arrhenius Concept, One of the earliest theories about acids and bases was proposed by swedish chemist, Svante Arrhenius. According to him, an acid is a substance that dissociates to give hydrogen, ions in water. For example, HCl, H2SO4 etc., are acids. Their dissociation in aqueous solution, is expressed as, +, , HCl(g) , ,  H (aq)+Cl (aq), H2 O, , The H+ ion in aqueous solution is highly hydrated and usually represented as H3O+ ,, +, the simplest hydrate of proton [H(H2O)] . We use both H+and H3O+ to mean the same., Similarly a base is a substance that dissociates to give hydroxyl ions in water. For, example, substances like NaOH, Ca(OH)2 etc., are bases., HO, 2+, , Ca(OH)2 , ,  Ca (aq)+2OH (aq), 2, , 2, , XII U8-Ionic equilibrium.indd 2, , 2/19/2020 5:11:19 PM

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www.tntextbooks.in, , Limitations of Arrhenius concept, i. Arrhenius theory does not explain the behaviour of acids and bases in non aqueous solvents, such as acetone, Tetrahydrofuran etc..., ii. This theory does not account for the basicity of the substances like ammonia ( NH3 ) which, do not possess hydroxyl group., Evaluate yourself – 1, Classify the following as acid (or) base using Arrhenius concept, i)HNO3, , ii) Ba(OH)2, , iii) H3PO4 iv) CH 3COOH, , 8.1.2 Lowry – Bronsted Theory (Proton Theory), In 1923, Lowry and Bronsted suggested a more general definition of acids and bases., According to their concept, an acid is defined as a substance that has a tendency to donate, a proton to another substance and base is a substance that has a tendency to accept a proton, from other substance. In other words, an acid is a proton donor and a base is a proton acceptor., When hydrogen chloride is dissolved in water, it donates a proton to the later. Thus,, HCl behaves as an acid and H2O is base. The proton transfer from the acid to base can be, represented as, HCl+H2O  H3O+ +Cl -, , When ammonia is dissolved in water, it accepts a proton from water. In this case, ammonia, ( NH3 ) acts as a base and H2O is acid. The reaction is represented as, H2O+NH3  NH 4+ +OH-, , Let us consider the reverse reaction in the following equilibrium, HCl, , Proton donar (acid), , +, , H2O, , Proton acceptor (Base), , , , H 3 O+, , Proton donar (acid), , +, , Cl -, , Proton acceptor (Base), , H3O+ donates a proton to Cl - to form HCl i.e., the products also behave as acid and base., In general, Lowry – Bronsted (acid – base) reaction is represented as, Acid1 +Base 2  Acid 2 +Base1, , The species that remains after the donation of a proton is a base (Base1 )and is called the, conjugate base of the Bronsted acid ( Acid 1 ) . In other words, chemical species that differ only, by a proton are called conjugate acid – base pairs., 3, , XII U8-Ionic equilibrium.indd 3, , 2/19/2020 5:11:22 PM

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www.tntextbooks.in, , Conjugate acid - base pair, HCl, , Proton donar (acid), , +, , H 2O, , , , Proton acceptor (Base), , H 3O +, , +, , Proton donar (acid), , Cl-, , Proton acceptor (Base), , Conjugate acid - base pair, HCl and Cl- , H 2 O and H3O + are two conjugate acid – base pairs. i.e., Cl - is the conjugate, base of the acid HCl . (or) HCl is conjugate acid of Cl- . Similarly H3O+ is the conjugate acid, , of H2O ., , Limitations of Lowry – Bronsted theory, i. Substances like BF3 , AlCl 3 etc., that do not donate protons are known to behave as acids., Evaluate yourself – 2, Write a balanced equation for the dissociation of the following in water and identify, the conjugate acid –base pairs., i) NH 4 +, , ii) H2SO4, , iii) CH3COOH., , 8.1.3 Lewis concept, In 1923, Gilbert . N. Lewis proposed a more generalised concept of acids and bases. He, considered the electron pair to define a species as an acid (or) a base. According to him, an, acid is a species that accepts an electron pair while base is a species that donates an electron, pair. We call such species as Lewis acids and bases., A Lewis acid is a positive ion (or) an electron deficient molecule and a Lewis base is a, anion (or) neutral molecule with at least one lone pair of electrons., Les us consider the reaction between Boron tri fluoride and ammonia, , F, , B, , +, F, , H, , F, , F, , H, , : NH3, B, , ammonia, , Boron tri fluoride, , N, , F, H, , F, electron deficient - electron pair donar, (Lewis acid), , -, , adduct, , - (Lewis base), , Here, boron has a vacant 2p orbital to accept the lone pair of electrons donated by, ammonia to form a new coordinate covalent bond. We have already learnt that in coordination, compounds, the Ligands act as a Lewis base and the central metal atom or ion that accepts a, pair of electrons from the ligand behaves as a Lewis acid., 4, , XII U8-Ionic equilibrium.indd 4, , 2/19/2020 5:11:25 PM

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www.tntextbooks.in, , Lewis acids, Electron deficient molecules, BF3 ,AlCl 3 ,BeF2 etc..., , Lewis bases, such, , as, , Molecules with one (or) more lone pairs of, electrons., NH3 ,H2O,R-O-H,R-O-R, R - NH2, , All metal ions, , All anions, , Examples: Fe2+ ,Fe3+ ,Cr 3+ ,Cu 2+ etc..., , F- ,Cl - ,CN- ,SCN - ,SO2-4 etc..., , Molecules that contain a polar double bond, , Molecules that contain carbon – carbon, multiple bond, Examples: CH 2 =CH 2 ,CH ≡ CH etc..., , Examples : SO2 ,CO2 ,SO3 etc..., Molecules in which the central atom can, expand its octet due to the availability of, empty d – orbitals, , All metal oxides, CaO,MgO,Na 2O etc..., , Example: SiF4 ,SF4 ,FeCl 3 etc.., Carbonium ion, , Carbanion, , (CH ), , CH3−, , 3 3, , C+, Example, , Identify the Lewis acid and the Lewis base in the following reactions., Cr 3+ + 6 H2O → [Cr(H2O)6 ]3+, In the hydration of ion, each of six water molecules donates a pair of electron to Cr 3+, to from the hydrated cation, hexaaquachromium (III) ion, thus, the Lewis acid is Cr 3+ and, the Lewis base H2O ., Evaluate yourself – 3, Identify the Lewis acid and the Lewis base in the following reactions., i., , CaO+CO2 → CaCO3, CH3, , ii., , CH3, , O, , Cl, O, , CH3 + AlCl3, CH3, , Al, , Cl, Cl, , 5, , XII U8-Ionic equilibrium.indd 5, , 2/19/2020 5:11:29 PM

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www.tntextbooks.in, , Evaluate yourself – 4, H3BO3 accepts hydroxide ion from water as shown below, H3BO3 (aq)+H2O(l)  B(OH)4 - +H+, Predict the nature of H3BO3 using Lewis concept, , 8.2 Strength of Acids and Bases, The strength of acids and bases can be determined by the concentration of H3O+ (or) OHproduced per mole of the substance dissolved in H2O . Generally we classify the acids / bases, either as strong or weak. A strong acid is the one that is almost completely dissociated in water, while a weak acid is only partially dissociated in water., Let us quantitatively define the strength of an acid (HA) by considering the following, general equilibrium., HA + H2O  H3O+ + Aacid 2 base 1, acid 1 base 2, , The equilibrium constant for the above ionisation is given by the following expression, K=, , [H3O+ ][A - ], .....(8.1), [HA][H2O], , We can omit the concentration of H2O in the above expression since it is present in large, excess and essentially unchanged., [H3O+ ][A - ], Ka =, .....(8.2), [HA], Here, K a is called the ionisation constant or dissociation constant of the acid. It measures, the strength of an acid. Acids such as HCl,HNO3 etc... are almost completely ionised and, hence they have high K a value (K a for HCl at 25o C is 2×106 ) Acids such as formic acid, (K a =1.8 × 10-4at 25oC) , acetic acid (1.8 × 10-5at 25oC) etc.. are partially ionised in solution, and in such cases, there is an equilibrium between the unionised acid molecules and their, dissociated ions. Generally, acids with K a value greater than ten are considered as strong acids, and less than one are considered as weak acids., Let us consider the dissociation of HCl in aqueous solution,, HCl + H - OH  H3O+ + Cl acid 1, , base 2, , acid 2, , base 1, , As discussed earlier, due to the complete dissociation, the equilibrium lies almost 100%, to the right. i.e., the Cl - ion has only a negligible tendency to accept a proton form H3O+ . It, means that the conjugate base of a strong acid is a weak base and vice versa., The following table illustrates the relative strength of conjugate acid – base pairs., 6, , XII U8-Ionic equilibrium.indd 6, , 2/19/2020 5:11:32 PM

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www.tntextbooks.in, , HClO4, HCl, H 2 SO4, , ClO 4 Strong, acids, , ClHSO 4, , HNO3, , NO3-, , H 3 O+, , H 2O, , HNO2, HF, CH 3COOH, , NO 2 -, , Weak, acids, , F-, , H2, , Very weak, bases, , Weak, bases, , CH3 COO NH 2 -, , NH 3, OH-, , -, , Strong, bases, , O 2-, , Very weak, acids, , H-, , 8.3 Ionisation of water, We have learnt that when an acidic or a basic substance is dissolved in water, depending, upon its nature, it can either donate (or) accept a proton. In addition to that the pure water, itself has a little tendency to dissociate. i.e, one water molecule donates a proton to an another, water molecule. This is known as auto ionisation of water and it is represented as below., , .., , HO - H+H 2O  H 3O + + OH acid 1, , base 2, , acid 2, , base 1, , Conjugate acid - base pairs, , In the above ionisation, one water molecule acts as an acid while the another water, molecule acts as a base., The dissociation constant for the above ionisation is given by the following expression, [H3O+ ][OH- ], K=, .....(8.3), [H2O]2, The concentration of pure liquid water is one. i.e, [H 2 O]2 = 1,  K w =[H3O+ ][OH- ] .....(8.4), Here, K w represents the ionic product (ionic product constant) of water, It was experimentally found that the concentration of H3O+ in pure water is 1×10-7 at 25oC ., Since the dissociation of water produces equal number of H3O+ and OH- , the concentration, of OH- is also equal to 1×10-7 at 25o C ., 7, , XII U8-Ionic equilibrium.indd 7, , 2/19/2020 5:11:35 PM

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www.tntextbooks.in, , Therefore, the ionic product of water at 25oC is, K w =[H3O]+[OH- ]......(8.4), K w =(1 × 10-7 )(1 × 10-7 ), = 1 × 10-14 ., , K w values at different temperatures, , Like all equilibrium constants, K w is also are given in the following below, a constant at a particular temperature. The, Temperature ( o C ), Kw, dissociation of water is an endothermic reaction., With the increase in temperature, the concentration, 0, 1.14 × 10−15, of H3O+ and OH - also increases, and hence the, ionic product also increases., −15, 10, In neutral aqueous solution like NaCl, solution, the concentration of H3O+ is always, equal to the concentration of OH- whereas in case, of an aqueous solution of a substance which may, behave as an acid (or) a base, the concentration of, H 3O + will not be equal to [OH ] ., , 2.95 × 10, , 25, , 1.00 × 10−14, , 40, , 2.71 × 10−14, , 50, , 5.30 × 10−14, , We can understand this by considering the aqueous HCl as an example. In addition to, the auto ionisation of water, the following equilibrium due to the dissociation of HCl can also, exist., HCl+H2O  H3O+ +Cl -, , In this case, in addition to the auto ionisation of water, HCl molecules also produces, H3O+ ion by donating a proton to water and hence [H3O+ ]>[OH- ] . It means that the aqueous, HCl solution is acidic. Similarly, in basic solution such as aqueous NH 3 , NaOH etc….., [OH–]>[H3O+]., Example 8.1, Calculate the concentration of OH- in a fruit juice which contains 2 × 10-3 M, H3O+ ion., Identify the nature of the solution., Given that H3O+ =2 × 10-3M, K w =[H3O+ ][OH- ], ∴[OH- ]=, , Kw, 1 × 10-14, =, =5 × 10-12 M, +, 3, [H3O ] 2 × 10, , 2 × 10-3 > > 5 × 10-12, i.e., [H 3O + ]>>[OH- ], hence the juice is acidic in nature, , 8, , XII U8-Ionic equilibrium.indd 8, , 2/19/2020 5:11:50 PM

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www.tntextbooks.in, , Evaluate yourself - 5, At a particular temperature, the K w of a neutral solution was equal to 4 × 10-14 ., Calculate the concentration of [H3O+ ] and [OH- ] ., , 8.4 The pH scale, We usually deal with acid / base solution in the concentration range 10-1 to 10-7 M . To, express the strength of such low concentrations, Sorensen introduced a logarithmic scale, known as the pH scale. The term pH is derived from the French word ‘Purissance de hydrogene’, meaning, the power of hydrogen. pH of a solution is defined as the negative logarithm of base, 10 of the molar concentration of the hydronium ions present in the solution., pH = - log 10[H3O+ ] .....(8.5), The concentration of H 3O + in a solution of known pH can be calculated using the following, expression., -pH, [H3O+ ]=10, (or) [H3O+ ]= antilog of (-pH) , , .....(8.6), , Similarly, pOH can also be defined as follows, pOH = - log 10 [OH- ] , , .....(8.7), , As discussed earlier, in neutral solutions, the concentration of [H3O ] as well as [OH + ] is, equal to 1 × 10-7 M at 25C . The pH of a neutral solution can be calculated by substituting this, H3O+concentration in the expression (8.5), +, , pH = - log 10 [H3O+ ], = - log 10 10, , -7, , = (-7)(-1) log1010= +7 (1)= 7, , [, , log 1010 =1], , Similary, we can calculate the pOH of a neutral solution using the expression (8.7), it is also, equal to 7., The negative sign in the expression (8.5) indicates that when the concentration of [H 3O + ], -7, -5, increases the pH value decreases. For example, if the [H3O+ ] increases from to 10 to10 M ,, the pH value of the solution decreases from 7 to 5. We know that in acidic solution,, [H 3O + ] > [OH - ] , i.e., [H 3O+ ] > 10-7 . Similarly in basic solution [H 3O+ ] < 10-7 . So, we can, conclude that acidic solution should have pH value less than 7 and basic solution should have, pH value greater than 7., 8.4.1 Relation between pH and pOH, A relation between pH and pOH can be established using their following definitions, , +, pH=-log, [H, pH=-log10, [H3O, O+ ]] .....(8.5), 10, 3, pOH, =, log, [OH, pOH = - log10, [OH- ]], 10, , .....(8.7), , Adding equation (8.5) and (8.7), 9, , XII U8-Ionic equilibrium.indd 9, , 2/19/2020 5:12:03 PM

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www.tntextbooks.in, , Example 8.2, Calculate the pH of 0.001M HCl solution, H2O, +, , HCl , ,  H3O + Cl, 0.001 M, , 0.001 M, , 0.001M, , H3O+ from the auto ionisation of H2O (10-7 M) is negligible when compared to the, H3O+ from 10-3M HCl., Hence [H3O+ ] = 0.001 mol dm–3, pH=-log [H O+ ], 10, , 3, , =-log10 (0.001), -3, , =-log10 (10 ) = 3, Note: If the concentration of the acid or base is less than 10–6 M, the concentration of, H3O+ produced due to the auto ionisation of water cannot be negleted and in such cases, [H3O+ ]=10-7 (from water) + [H3O+ ] (from the acid), similarly, [OH- ]=10-7 M (from water) + [OH- ] (from the base), , Example 8.3, -7, , Calculate pH of 10 M HCl, If we do not consider [H3O+ ] from the ionisation of H2O,, then [H3O+ ] =[HCl]=10-7 M, i.e., pH = 7, which is a pH of a neutral solution. We know that HCl solution is acidic, whatever may be the concentration of HCl i.e, the pH value should be less than 7. In this, case the concentration of the acid is very low (10-7 M) Hence, the H3O+ (10-7 M) formed, due to the auto ionisation of water cannot be neglected., so, in this case we should consider [H3O+ ] from ionisation of H 2 O, [H3O+ ] = 10-7 (from HCl) +10-7 (from water), = 10-7 (1+1), , pH=-log10[H3O+ ], , = 2 × 10-7, , =-log10 (2 × 10-7 ) = - log 2 + log 10-7 , =- log 2-(-7).log10 10, =7-log 2, =7-0.3010 = 6.6990, = 6.70, 11, , XII U8-Ionic equilibrium.indd 11, , 2/19/2020 5:12:24 PM

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www.tntextbooks.in, , Evaluate yourself - 6, a) Calculate pH of 10-8 M H2SO4, B) Calculate the concentration of hydrogen ion in moles per litre of a solution whose, pH is 5.4, c) Calculate the pH of an aqueous solution obtained by mixing 50ml of 0.2 M HCl with, 50ml 0.1 M NaOH, , 8.5 Ionisation of weak acids, We have already learnt that weak acids are partially dissociated in water and there is an, equilibrium between the undissociated acid and its dissociated ions., Consider the ionisation of a weak monobasic acid HA in water., HA+H2O  H3O+ +A Applying law of chemical equilibrium, the equilibrium constant K c is given by the, expression, [H3O+ ][A - ], KC =, [HA][H2O] .....(8.9),   , The square brackets, as usual, represent the concentrations of the respective species in, moles per litre., In dilute solutions, water is present in large excess and hence, its concentration may be, taken as constant say K. Further H3O+ indicates that hydrogen ion is hydrated, for simplicity, it may be replaced by H+ . The above equation may then be written as,, [H+ ][A - ], KC =, [HA] × K .....(8.10), , The product of the two constants K C and K gives another constant. Let it be K a, [H+ ][A - ], Ka =, [HA] .....(8.11), The constant K a is called dissociation constant of the acid. Like other equilibrium, constants, K a also varies only with temperature., Similarly, for a weak base, the dissociation constant can be written as below., Kb =, , [B+ ][OH- ], [BOH] ....(8.12), , 8.5.1 Ostwald’s dilution law, Ostwald’s dilution law relates the dissociation constant of the weak acid (K a ) with its, degree of dissociation (α) and the concentration (c). Degree of dissociation (α) is the fraction, of the total number of moles of a substance that dissociates at equilibrium., 12, , XII U8-Ionic equilibrium.indd 12, , 2/19/2020 5:12:36 PM

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www.tntextbooks.in, , α=, , Number of moles dissociated, total number of moles, , We shall derive an expression for ostwald's law by considering a weak acid, i.e. acetic acid, (CH3COOH) . The dissociation of acetic acid can be represented as, CH3COOH  H + + CH3COOThe dissociation constant of acetic acid is,, [H+ ][CH3COO- ], ka =, [CH3COOH] .....(8.13), CH3COOH, , H+, , CH3COO-, , Initial number of, moles, , 1, , -, , -, , Degree of, dissociation of, CH3COOH, , α, , -, , -, , Number of moles at, equilibrium, , 1- α, , α, , α, , Equilibrium, concentration, , (1 - α) C, , αC, , αC, , Substituting the equilibrium concentration in equation (8.13), , (αC)(αC), (1-α)C, α2C, ka =, 1-α, .....(8.14), ka =, , We know that weak acid dissociates only to a very small extent. Compared to one, α is so, small and hence in the denominator (1 - α)  1. The above expression (8.14) now becomes,, K a =α2C, K, ⇒ α2 = a, C, K, α= a, C .....(8.15), Let us consider an acid with K a value 4 × 10-4 and calculate the degree of dissociation, of that acid at two different concentration 1 × 10-2 M and 1 × 10-4 M using the above expression, (8.15), For 1 × 10-2 M ,, 13, , XII U8-Ionic equilibrium.indd 13, , 2/19/2020 5:12:52 PM

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www.tntextbooks.in, , α=, , 4 × 10-4, 10-2, , = 4 × 10-2, = 2 × 10-1, = 0.2, For 1 × 10-4 M acid,, α=, =2, , 4 × 10-4, 10-4, , i.e, When the dilution increases by 100 times, (Concentration decreases from 1 × 10-2 M, to 1 × 10-4 M ), the dissociation increases by 10 times., Thus, we can conclude that, when dilution increases, the degree of dissociation of weak, electrolyte also increases. This statement is known as Ostwald’s dilution Law., The concentration of H+ (H3O+ ) can be calculated using the K a value as below., [ H+ ]= αC , , (Refer table) .....(8.16), , Equilibrium molar concentration of [H ] is equal to αC, +, ,  Ka , ∴[H+ ]= , C,  C, , [ equation (8.15)], , K aC2, =, C, [H+ ] = K a C, , .....(8.17), Similarly, for a weak base, K b = α2C and α =, [OH- ] = αC, (or), , Kb, C, , [OH- ]= K bC, , .....(8.18), , Example 8.4, A solution of 0.10M of a weak electrolyte is found to be dissociated to the extent of, 1.20% at 25oC . Find the dissociation constant of the acid., Given that α=1.20%=, , 1.20, K a =α2 c, =1.2 × 10-2, 100, = (1.2 × 10-2 )2 (0.1)=1.44 × 10-4 × 10-1, = 1.44 × 10-5, 14, , XII U8-Ionic equilibrium.indd 14, , 2/19/2020 5:13:08 PM

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www.tntextbooks.in, , Example 8.5, Calculate the pH of 0.1M CH3COOH solution. Dissociation constant of acetic acid is, 1.8 × 10-5 ., pH=-log[H+ ], For weak acids,, [H+ ]= K a × C, , = 1.8 × 10-5 × 0.1, , =1.34 × 10-3M, =3 - log1.34, = 3 - 0.1271, = 2.8729, , , , pH=-log (1.34 × 10−3 ), , 2.87, , Evaluate yourself - 7, K b for NH 4 OH is 1.8 × 10-5 . Calculate the percentage of ionisation of 0.06M, ammonium hydroxide solution., , 8.6 Common Ion Effect, When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is, suppressed further. For example, the addition of sodium acetate to acetic acid solution leads to, the suppression in the dissociation of acetic acid which is already weakly dissociated. In this, case, CH3COOH and CH3COONa have the common ion, CH 3COOLet us analyse why this happens. Acetic acid is a weak acid. It is not completely dissociated, in aqueous solution and hence the following equilibrium exists., CH 3COOH(aq)  H + (aq)+CH 3COO- (aq), , However, the added salt, sodium acetate, completely dissociates to produce Na + and, CH3COO- ion., CH3COONa(aq) → Na + (aq)+CH3COO- (aq), Hence, the overall concentration of CH 3COO- is increased, and the acid dissociation, equilibrium is disturbed. We know from Le chatelier's principle that when a stress is applied, to a system at equilibrium, the system adjusts itself to nullify the effect produced by that stress., So, inorder to maintain the equilibrium, the excess CH3COO- ions combines with H+ ions to, produce much more unionized CH 3COOH i.e, the equilibrium will shift towards the left. In, other words, the dissociation of CH3COOH is suppressed. Thus, the dissociation of a weak, acid (CH3COOH) is suppressed in the presence of a salt (CH3COONa) containing an ion, common to the weak electrolyte. It is called the common ion effect., 15, , XII U8-Ionic equilibrium.indd 15, , 2/19/2020 5:13:24 PM

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www.tntextbooks.in, , 8.7 Buffer Solution, Do you know that our blood maintains a constant pH, irrespective of a number of cellular, acid – base reactions. Is it possible to maintain a constant hydronium ion concentration in, such reactions? Yes, it is possible due to buffer action., Buffer is a solution which consists of a mixture of a weak acid and its conjugate base (or), a weak base and its conjugate acid. This buffer solution resists drastic changes in its pH upon, addition of a small quantities of acids (or) bases, and this ability is called buffer action. The, buffer containing carbonic acid (H2CO3 ) and its conjugate base HCO-3 is present in our, blood. There are two types of buffer solutions., 1. Acidic buffer solution : a solution containing a weak acid and its salt., Example : solution containing acetic acid and sodium acetate, 2. Basic buffer solution : a solution containing a weak base and its salt., Example : Solution containing NH 4OH and NH 4 Cl, 8.7.1 Buffer action, To resist changes in its pH on the addition of an acid (or) a base, the buffer solution should, contain both acidic as well as basic components so as to neutralize the effect of added acid (or), base and at the same time, these components should not consume each other., Let us explain the buffer action in a solution containing CH3COOH and CH 3COONa ., The dissociation of the buffer components occurs as below., +, CH3 - COO(aq) + H3O (aq), , CH3COOH (aq), , -, , If an acid is added to this mixture, it will be consumed by the conjugate base CH3COO- to, form the undissociated weak acid i.e, the increase in the concentration of H+ does not reduce, the pH significantly., , CH3COO (aq) + H (aq), , CH3COOH (aq), , If a base is added, it will be neutralized by H3O+, and the acetic acid is dissociated to, maintain the equlibrium. Hence the pH is not significantly altered., OH (aq) + H3O+ ((aq), CH3COOH (aq), , H2O (l), H2O (l), , CH3COO (aq) + H3O+ ((aq), CH3COO (aq) + H2O (l), , OH (aq) + CH3COOH (aq), , These neutralization reactions are identical to those reactions that we have already, discussed in common ion effect., 16, , XII U8-Ionic equilibrium.indd 16, , 2/19/2020 5:13:34 PM

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www.tntextbooks.in, , CH3COOH + OH- (aq) → CH3COO- (aq) + H2O (l), ∴[CH3COOH] = 0.8 - α - 0.01 = 0.79 - α, [CH3COO- ]=α+0.8+0.01=0.81+α, , α <<0.8;, , 0.79 - α  0.79 and 0.81 + α  0.81, , ∴ [H+ ] = (1.8 × 10−5 ) ×, [H+ ] = 1.76 × 10−5, , 0.79, 0.81, , ∴ pH = - log ( 1.76 × 10-5 ), = 5 - log 1.76, = 5 - 0.25, pH = 4.75, The addition of a strong base (0.01 M NaOH) increased the pH only slightly ie., from 4.74, to 4.75 . So, the buffer action is verified., Evaluate yourself - 8, a) Explain the buffer action in a basic buffer containing equimolar ammonium hydroxide, and ammonium chloride., b) Calculate the pH of a buffer solution consisting of 0.4M CH3COOH and 0.4M, CH3COONa . What is the change in the pH after adding 0.01 mol of HCl to 500ml of, the above buffer solution. Assume that the addition of HCl causes negligible change in, the volume. Given: ( K a = 1.8 × 10−5. ), 8.7.2 Buffer capacity and buffer index, The buffering ability of a solution can be measured in terms of buffer capacity. Vanslyke, introduced a quantity called buffer index, β , as a quantitative measure of the buffer capacity., It is defined as the number of gram equivalents of acid or base added to 1 litre of the buffer, solution to change its pH by unity., β=, Here,, , dB, ....(8.19), d(pH), , dB = number of gram equivalents of acid / base added to one litre of buffer solution., d(pH) = The change in the pH after the addition of acid / base., 8.7.3 Henderson – Hasselbalch equation, We have already learnt that the concentration of hydronium ion in an acidic buffer solution, depends on the ratio of the concentration of the weak acid to the concentration of its conjugate, base present in the solution i.e.,, [acid]eq, H 3O +  =K a, [base]eq, ....(8.20), 18, , XII U8-Ionic equilibrium.indd 18, , 2/19/2020 5:13:54 PM

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www.tntextbooks.in, , The weak acid is dissociated only to a small extent. Moreover, due to common ion effect,, the dissociation is further suppressed and hence the equilibrium concentration of the acid is, nearly equal to the initial concentration of the unionised acid. Similarly, the concentration of, the conjugate base is nearly equal to the initial concentration of the added salt., [acid], H3O+  =K a, [salt] ....(8.21), Here [acid] and [salt] represent the initial concentration of the acid and salt, respectively, used to prepare the buffer solution, Taking logarithm on both sides of the equation, [acid], log [H 3O+ ]=logK a +log, [salt] ....(8.22), reverse the sign on both sides, [acid], -log [H3O+ ]=-logK a -log, [salt] ....(8.23), , We know that, pH = - log [H 3O+ ] and pK a = - logK a, [acid], ⇒ pH = pK a - log, [salt], ....(8.24), ⇒ pH = pK a + log, , [salt], [acid] ....(8.25), ,    Similarly for a basic buffer, pOH = pK b +log, , [salt], ....(8.26), [base], , Example 8.6, 1. Find the pH of a buffer solution containing 0.20 mole per litre sodium acetate and 0.18, mole per litre acetic acid. K a for acetic acid is 1.8 × 10-5 ., pH = pK a +log, , [salt], [acid], , Given that K a = 1.8 × 10-5, ∴ pK a = −log(1.8 × 10−5 ) = 5 −log 1.8, = 5 -0.26, = 4.74, 0.20, 0.18, 10, = 4.74 + log, = 4.74 + log 10 - log 9, 9, = 4.74 + 1 - 0.95 = 5.74 - 0..95, , ∴pH = 4.74 + log, , = 4.79, , 19, , XII U8-Ionic equilibrium.indd 19, , 2/19/2020 5:14:06 PM

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www.tntextbooks.in, , Example 8.7, What is the pH of an aqueous solution obtained by mixing 6 gram of acetic acid and, 8.2 gram of sodium acetate making the volume equal to 500 ml. (Given: K a for acetic acid, is 1.8 × 10-5 ), According to Henderson – Hasselbalch equation,, [salt], [acid], P K =-logK a =-log(1.8 × 10-5 )= 4.74 (Refer previous example), pH = pK a +log, a, , Number of moles of sodium acetate, Volume of the solution ( litre ), mass of sodium acetate, Number of moles of sodium acetate =, molar mass of sodium acetate, 8.2, = 0.1, =, 82, 0.1 mole, ∴ [Salt ] =, = 0. 2 M, 1 Litre, 2, , [Salt ]=, , [acid ]=, , , mass of CH3COOH ,  molar mass of CH COOH , 3, , Volume of solution in litre,  6,  60 , =, 1, 2, = 0.2M, , ∴pH = 4.74 + log, , (0.2), (0.2), , pH = 4.74 + log 1, pH = 4.74 + 0 = 4.74, Evaluate yourself - 9, , a) How can you prepare a buffer solution of pH 9. You are provided with 0.1M NH 4OH, solution and ammonium chloride crystals. (Given: pK b for NH 4OH is 4.7 at 25C ., b) What volume of 0.6M sodium formate solution is required to prepare a buffer, solution of pH 4.0 by mixing it with 100ml of 0.8M formic acid. (Given: pK a for formic, acid is 3.75. ), , 20, , XII U8-Ionic equilibrium.indd 20, , 2/19/2020 5:14:14 PM

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www.tntextbooks.in, , 8.8 Salt Hydrolysis, , When an acid reacts with a base, a salt and water are formed and the reaction is called, neutralization. Salts completely dissociate in aqueous solutions to give their constituent ions., The ions so produced are hydrated in water. In certain cases, the cation, anion or both react, with water and the reaction is called salt hydrolysis. Hence, salt hydrolysis is the reverse of, neutralization reaction., 8.8.1 Salts of strong acid and a strong base, Let us consider the reaction between NaOH and nitric acid to give sodium nitrate and, water., NaOH(aq)+HNO3 (aq) → NaNO3 (aq)+H 2 O(l), , The salt NaNO3 completely dissociates in water to produce Na + and NO3− ions., NaNO3 (aq) → Na + (aq)+NO3- (aq), Water dissociates to a small extent as, H2O(l)  H+ (aq)+OH- (aq), Since [H+ ]=[OH- ], water is neutral, NO3- ion is the conjugate base of the strong acid HNO3 and hence it has no tendency to, react with H+ ., Similarly, Na + is the conjugate acid of the strong base NaOH and it has no tendency to, react with OH- ., It means that there is no hydrolysis. In such cases [H+ ]=[OH- ] pH is maintained and,, therefore, the solution is neutral., 8.8.2 Hydrolysis of Salt of strong base and weak acid (Anionic Hydrolysis)., Let us consider the reactions between sodium hydroxide and acetic acid to give sodium, acetate and water., NaOH (aq) + CH3COOH(aq)  CH3COONa(aq)+H2O(l), In aqueous solution, CH3COONa is completely dissociated as below, CH3COONa (aq) , → CH3COO- (aq)+Na + (aq), , CH3COO- is a conjugate base of the weak acid CH3COOH and it has a tendency to react, with H+ from water to produce unionised acid ., , There is no such tendency for Na + to react with OH- ., +, CH3COO- (aq) + H2O(l)  CH3COOH (aq) + OH - (aq) and therefore [OH ]>[H ] , in, such cases, the solution is basic due to hydrolysis and the pH is greater than 7., , Let us find a relation between the equilibrium constant for the hydrolysis reaction, (hydrolysis constant) and the dissociation constant of the acid., 21, , XII U8-Ionic equilibrium.indd 21, , 2/19/2020 5:14:26 PM

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www.tntextbooks.in, , Kh =, , [CH3COOH][OH- ], [CH3COO- ][H2O], , [CH3COOH][OH- ], Kh =, [CH3COO- ], .....(1), CH3COOH (aq)  CH3COO- (aq) + H+ (aq), [CH3COO- ][H+ ], Ka =, [CH3COOH] .....(2), , (1) × (2), , ⇒ K h .K a =[H+ ][OH- ], we know that [H+ ][OH- ]=K w, K h .K a =K w, K h value in terms of degree of hydrolysis (h) and the concentration of salt (C) for the, equilibrium can be obtained as in the case of ostwald's dilution law. K h = h 2C. and, , i.e [OH- ]= K h .C, , pH of salt solution in terms of Ka and the concentration of the electrolyte., pH + pOH = 14, , pH = 14 - p OH = 14 - {-log [OH- ]}, = 14 + log [OH- ], ∴ pH = 14 + log (K h C),  K C, pH = 14 + log  w ,  K , , (, , 1, 1, , 2, , 2, , a, , pH = 14 + 1 log K w + 1 log C - 1 log K a, 2, 2, 2, 1, 1, pK a, pH = 14 - 7 +, log C +, 2, 2, pH = 7 + 1 pK a + 1 log C., 2, 2, , ), , [ K w =10-14 ., 1 log K = 1 × log10-14 = -14 (1) = -7., w, 2, 2, 2, - log K a =pK a ], , 8.8.3 Hydrolysis of salt of strong acid and weak base (Cationic Hydrolysis), Let us consider the reactions between a strong acid, HCl, and a weak base, NH 4OH , to, produce a salt, NH 4Cl , and water, HCl (aq) + NH 4OH (aq)  NH 4Cl(aq)+H2O(l), NH 4Cl(aq) → NH 4 + +Cl - (aq), NH 4+ is a strong conjugate acid of the weak base NH 4OH and it has a tendency to react, with OH- from water to produce unionised NH 4OH shown below., NH 4+ (aq) + H2O(l)  NH 4OH (aq) +H+ (aq), 22, , XII U8-Ionic equilibrium.indd 22, , 2/19/2020 5:14:39 PM

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www.tntextbooks.in, , There is no such tendency shown by Cl - and therefore [H + ] > [OH - ]; the solution is acidic, and the pH is less than 7., As discussed in the salt hydrolysis of strong base and weak acid. In this case also, we can, establish a relationship between the K h and K b as, K h .K b =K w, Let us calculate the K h value in terms of degree of hydrolysis (h) and the concentration, of salt, K h = h 2C, , and [H+ ]= K h .C, [H+ ]=, , Kw, .C, Kb, , pH = - log [H+ ],  K .C , = - log  w ,  K , , 1, , 2, , b, , = - 1 log K w − 1 log C + 1 log K b, 2, 2, 2, 1, 1, pK b −, log C., pH = 7 2, 2, 8.8.4 Hydrolysis of Salt of weak acid and weak base (Anionic & Cationic Hydrolysis)., Let us consider the hydrolysis of ammonium acetate., CH 3COONH 4 (aq) → CH 3COO- (aq) + NH 4 + (aq), , In this case, both the cation (NH 4+ ) and anion (CH3COO−) have the tendency to react, with water, CH 3COO−+H 2 O  CH 3COOH+OH −, NH 4 + +H 2 O  NH 4 OH+H +, , The nature of the solution depends on the strength of acid (or) base i.e, if K a >K b ; then, the solution is acidic and pH < 7, if K a < K b ; then the solution is basic and pH > 7, if K a =K b ;, then the solution is neutral., The relation between the dissociation constant (K a ,K b ) and the hydrolysis constant is, given by the following expression., K a .K b .K h =K w, pH of the solution, pH of the solution can be calculated using the following expression,, pH = 7 + 1 pK a − 1 pK b ., 2, 2, 23, , XII U8-Ionic equilibrium.indd 23, , 2/19/2020 5:14:51 PM

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www.tntextbooks.in, , Example 8.8, Calculate i) degree of hydrolysis, ii) the hydrolysis constant and, CH3COONa solution ( pK a for CH3COOH is 4.74)., , iii) pH of 0.1M, , Solution (a) CH3COONa is a salt of weak acid (CH 3COOH) and a strong base (NaOH)., Hence, the solutions is alkaline due to hydrolysis., CH 3COO- (aq) + H 2 O(aq)  CH 3COOH (aq) + OH- (aq), , Give that pKa = 4.74, pK a = -log K a, i.e., K a = antilog of (-pK a ), 1 × 10-14, =, = antilog of (-4.74), 1.8 × 10-5 × 0.1 , = antilog of (-5 + 0.26), h = 7.5 × 10-5, , i) h=, , Kw, Ka × C, , = 10-5 × 1.8, [antilog of 0.26 = 1.82  1.8], ii) K h =, , K w 1 × 10-14, =, K a 1.8 × 10-5, , = 5.56 × 10-10, , pK a logC, +, 2, 2, 4.74 log 0.1, +, = 7 + 2.37 −0.5, =7+, 2, 2, = 8.87, , iii) pH = 7 +, , Evaluate yourself - 10, Calculate the i) hydrolysis constant, ii) degree of hydrolysis and iii) pH of 0.05M, sodium carbonate solution ( pK a for HCO3− is 10.26)., , 8.9 Solubility Product, We have come across many precipitation reactions in inorganic qualitative analysis. For, example, dil HCl is used to precipitate Pb2+ ions as PbCl2 which is sparingly soluble in water., Kidney stones are developed over a period of time due to the precipitation of Ca 2+ (as calcium, oxalate etc…). To understand the precipitation, let us consider the solubility equilibria that, exist between the undissociated sparingly soluble salt and its constituent ions in solution., For a general salt X m Yn ,, H2O, , , → mX n+ (aq) + nY m- (aq), X m Yn (s) ←, , , The equilibrium constant for the above is, [X n+ ]m [Y m- ]n, K=, [X m Ym ], In solubility equilibria, the equilibrium constant is referred as solubility product constant, (or) Solubility product., 24, , XII U8-Ionic equilibrium.indd 24, , 2/19/2020 5:15:06 PM

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www.tntextbooks.in, , In such heterogeneous equilibria, the concentration of the solid is a constant and is omitted, in the above expression, K sp = [X n+ ]m [Y m- ]n, The solubility product of a compound is defined as the product of the molar concentration, of the constituent ions, each raised to the power of its stoichiometric co – efficient in a balanced, equilibrium equation., Solubility product finds useful to decide whether an ionic compound gets precipitated, when solution that contains the constituent ions are mixed., When the product of molar concentration of the constituent ions i.e., ionic product,, exceeds the solubility product then the compound gets precipitated., The expression for the solubility product and the ionic product appears to be the same, but in the solubility product expression, the molar concentration represents the equilibrium, concentration and in ionic product, the initial concentration (or) concentration at a given, time ‘t’ is used., In general we can summarise as,, Ionic product > K sp , precipitation will occur and the solution is super saturated., Ionic product < K sp , no precipitation and the solution is unsaturated., Ionic product = K sp , equilibrium exist and the solution is saturated., Example 8.9, Indicate find out whether lead chloride gets precipitated or not when 1 mL of 0.1M, lead nitrate and 0.5 mL of 0.2 M NaCl solution are mixed? K sp of PbCl 2 is 1.2 × 10-5 ., 2+, , PbCl 2(s) , ,  Pb (aq)+2Cl (aq), H2 O, , Ionic product = [Pb2+ ][Cl - ]2, Total volume = 1.5 mL, 2+, −, , Pb ( NO3 )2 , ,  Pb +2NO3, 0. 1 M, , 0. 1 M, , No of moles of Pb2+ = Molarity × volume of the solution in litre,  , = 0.1×1×10–3 = 10–4, [Pb, , 2+, , ]=, , number of moles of Pb 2+, 10−4, =, = 6.7 × 10-2 M, Volume of the solution in L 1.5 × 10−3 mL, , NaCl , → Na + + Cl 0. 2 M, , 0. 2 M, , 0. 2 M, , No of moles of Cl– = 0.2 × 0.5 × 10–3= 10–4, 10−4 moles, = 6.7 × 10-2 M, -3, 1.5 × 10 L, Ionic product = (6.7 × 10-2 )(6.7 × 10−2 )2 = 3.01 × 10-4, [Cl - ] =, , Since, the ionic product 3.01 × 10-4 is greater than the solubility product (1.2 × 10-5 ) ,, PbCl 2 will get precipitated., 25, , XII U8-Ionic equilibrium.indd 25, , 2/19/2020 5:15:19 PM

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www.tntextbooks.in, , 8.9.1 Determination of solubility product from molar solubility, Solubility product can be calculated from the molar solubility i.e., the maximum number, of moles of solute that can be dissolved in one litre of the solution., For a solute X m Yn ,, X m Yn (s)  mX n+ (aq) +nY m- (aq), From the above stoichiometrically balanced equation we have come to know that 1 mole, of X m Yn (s) dissociated to furnish ‘m’ moles of Xn+ and ‘n’ moles of Y m- if ‘s’ is molar solubility, of X m Yn , then, [X n+ ]=ms, , and, , [ Y m- ]=ns, , ∴ K sp = [X n+ ]m [Y m- ]n, K sp =(ms)m (ns)n, K sp =(m)m (n)n (s)m+n, Example 8.10, •, , Establish a relationship between the solubility product and molar solubility for the, following, a) BaSO4, , BaSO4 (s), , b) Ag 2 (CrO 4 ), H2 O, , K sp =[Ba 2+ ][SO4 2- ], , Ba 2+ (aq)+SO 4 2- (aq), , = (s) (s), K sp = S 2, , , → 2Ag + (aq)+CrO4 2- (aq), Ag 2CrO4 (s) ←, H2 O, , K sp =[Ag + ]2[CrO4 2- ], = (2s)2 (s), K sp =4s 3, , 26, , XII U8-Ionic equilibrium.indd 26, , 2/19/2020 5:15:24 PM

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www.tntextbooks.in, , Summary, „„ According to Arrhenius, an acid is, , litre of the buffer solution to change its, pH by unity., , a substance that dissociates to give, hydrogen ions in water., , β=, , „„ According to Lowry and Bronsted, , concept, an acid is defined as a substance, that has a tendency to donate a proton, to another substance and base is a, substance that has a tendency to accept, a proton from other substance., , „„ Henderson – Hasselbalch equation, , For Acid buffer, ⇒ pH = pK a + log, , „„ According to Gilbert . N. Lewis , an acid, , [salt], [acid], , For Basic buffer, , is a species that accepts an electron pair, while base is a species that donates an, electron pair., , ⇒ pOH = pK b + log, , „„ ionic product (ionic product constant), , [salt], [base], , „„ Hydrolysis of Salt of strong base and, , of water (Kw)=[H3O ][OH ], +, , dB, d(pH), , weak acid, , –, , K h .K a =K w, , „„ pH of a solution is defined as the, , negative logarithm of base 10 of the, molar concentration of the hydronium, ions present in the solution., , pH = 7 + 1 pK a + 1 log C., 2, 2, „„ Hydrolysis of salt of strong acid and, , pH = –log 10 [H3O+], , weak base, , „„ when dilution increases, the degree of, ,   K h .K b =K w, , dissociation of weak electrolyte also, increases. This statement is known as, Ostwald’s dilution Law., , pH = 7 - 1 pK b − 1 log C., 2, 2, „„ Hydrolysis of Salt of weak acid and weak, , „„ When a salt of a weak acid is added to, , base, , the acid itself, the dissociation of the, weak acid is suppressed further this is, known as common ion effect, , K a .K b .K h =K w, pH = 7 + 1 pK a − 1 pK b ., 2, 2, , „„ Buffer is a solution which consists of a, , mixture of a weak acid and its conjugate, base (or) a weak base and its conjugate, acid., , „„ The solubility product of a compound, , is defined as the product of the molar, concentration of the constituent, ions, each raised to the power of, its stoichiometric co – efficient in a, balanced equilibrium equation., , „„ Buffer, , capacity and buffer index, is defined as the number of gram, equivalents of acid or base added to 1, 27, , XII U8-Ionic equilibrium.indd 27, , 2/19/2020 5:15:29 PM

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www.tntextbooks.in, , 7. Which of the following fluro compounds is most likely to behave as a Lewis base?, , (NEET – 2016), b) PF3, c) CF4, d) SiF4, a) BF3, 8. Which of these is not likely to act as Lewis base?, a) BF3, b) PF3, c) CO, , d) F–, , 9. The aqueous solutions of sodium formate, anilinium chloride and potassium cyanide are, respectively, a) acidic, acidic, basic , , b) basic, acidic, basic, , c) basic, neutral, basic , , d) none of these, , 10. The percentage of pyridine (C 5H5N) that forms pyridinium ion (C 5H5 NH) in a 0.10M, aqueous pyridine solution (K b for C5 H 5 N= 1.7 × 10-9 ) is, a) 0.006%, , b) 0.013%, , c) 0.77%, , d) 1.6%, , 11. Equal volumes of three acid solutions of pH 1,2 and 3 are mixed in a vessel. What will be, the H+ ion concentration in the mixture?, a) 3.7 × 10-2, , b) 10-6, , c) 0.111, , d) none of these, , 12. The solubility of AgCl (s) with solubility product 1.6 × 10-10 in 0.1M NaCl solution would, be, a) 1.26 × 10-5M, , b) 1.6 × 10-9 M, , c) 1.6 × 10-11M, , d) Zero, , 13. If the solubility product of lead iodide is 3.2 × 10-8 , its solubility will be, a) 2×10-3M, , b) 4 × 10-4 M, , c) 1.6 × 10-5M, , d) 1.8 × 10-5M, , 14. MY and NY3 , are insoluble salts and have the same K sp values of 6.2 × 10-13 at room, temperature. Which statement would be true with regard to MY and NY3?, a) The salts MY and NY3 are more soluble in 0.5M KY than in pure water, b) The addition of the salt of KY to the suspension of MY and NY3 will have no effect, on their solubility’s, c) The molar solubilities of MY and NY3 in water are identical, d) The molar solubility of MY in water is less than that of NY3, 15. What is the pH of the resulting solution when equal volumes of 0.1M NaOH and 0.01M, HCl are mixed?, a) 2.0, , b) 3, , c) 7.0, , d) 12.65, , 16. The dissociation constant of a weak acid is 1 × 10-3 . In order to prepare a buffer solution, with a pH = 4, the [Acid], , [Salt], , a) 4:3, , b) 3:4, , ratio should be, c) 10:1, , d) 1:10, , 29, , XII U8-Ionic equilibrium.indd 29, , 2/19/2020 5:16:15 PM

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www.tntextbooks.in, , 17. The pH of 10-5M KOH solution will be, a) 9, , b) 5, , c) 19, , d) none of these, , c) H3PO4, , d) HPO 4 2-, , 18. H2 PO4- the conjugate base of, a) PO4 3−, , b) P2O5, , 19. Which of the following can act as Lowry – Bronsted acid as well as base?, a) HCl, , b) SO4 2−, , c) HPO4 2−, , d) Br -, , 20. The pH of an aqueous solution is Zero. The solution is, a) slightly acidic    b) strongly acidic   c) neutral    d) basic, 21. The hydrogen ion concentration of a buffer solution consisting of a weak acid and its salts, is given by, K [acid], K [salt], +, a) [H ]= a, b) [H+ ]= K a [salt]   c) [H+ ]=K a [acid]   d) [H+ ]= a, [salt], [acid], 22. Which of the following relation is correct for degree of hydrolysis of ammonium, acetate?, a) h =, , Kh, C, , b) h =, , Ka, Kb, , c) h =, , Kw, K a .K b, , d) h =, , K a .K b, Kw, , 23. Dissociation constant of NH 4OH is 1.8 × 10-5 the hydrolysis constant of NH 4Cl would, be, a) 1.8 × 10-19, , b) 5.55 × 10-10, , c) 5.55 × 10-5, , d) 1.80 × 10-5, , Answer the following questions:, 1. What are Lewis acids and bases? Give two example for each., 2. Discuss the Lowry – Bronsted concept of acids and bases., 3. Indentify the conjugate acid base pair for the following reaction in aqueous solution, i)HS- (aq) + HF  F- (aq) + H2S(aq), , ii) HPO4 2- + SO32-  PO4 3- + HSO3-, , iii)NH 4 + + CO32-  NH3 + HCO34. Account for the acidic nature of HClO4 in terms of Bronsted – Lowry theory, identify its, conjugate base., 5. When aqueous ammonia is added to CuSO4 solution, the solution turns, deep blue due to the formation of tetramminecopper (II) complex,, 2+, 2+, [Cu(H2O)4 ](aq), +4NH3 (aq)  [Cu(NH3 )4 ](aq), , among H2O and NH3 Which is stronger, Lewis base., 6. The concentration of hydroxide ion in a water sample is found to be 2.5 × 10-6 M . Identify, the nature of the solution., 30, , XII U8-Ionic equilibrium.indd 30, , 2/19/2020 5:16:28 PM

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www.tntextbooks.in, , 7. A lab assistant prepared a solution by adding a calculated quantity of HCl gas at 25oC to, get a solution with [H3O+ ]= 4 × 10-5M . Is the solution neutral (or) acidic (or) basic., 8. Calculate the pH of 0.04 M HNO3 Solution., 9. Define solubility product, 10. Define ionic product of water. Give its value at room temperature., 11. Explain common ion effect with an example, 12. Derive an expression for Ostwald’s dilution law, 13. Define pH, 14. Calculate the pH of 1.5×10-3M solution of Ba (OH)2, 15. 50ml of 0.05M HNO3 is added to 50ml of 0.025M KOH . Calculate the pH of the resultant, solution., 16. The K a value for HCN is 10-9 . What is the pH of 0.4M HCN solution?, 17. Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that, K a =K b =1.8 × 10-5, 18. Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of strong, acid and weak base, 19. Solubility product of Ag 2CrO4 is 1 × 10-12 . What is the solubility of Ag 2CrO4 in 0.01M, AgNO3 solution?, 20. Write the expression for the solubility product of Ca 3 (PO4 )2, 21. A saturated solution, prepared by dissolving CaF2 (s) in water, has [Ca 2+ ]=3.3 × 10-4 M What, is the K sp of CaF2 ?, 22. K sp of AgCl is 1.8 × 10−10 . Calculate molar solubility in 1 M AgNO3, 23. A particular saturated solution of silver chromate Ag 2CrO4 has [Ag + ]=5 × 10-5 and, [CrO4 ]2- =4.4 × 10-4 M. What is the value of K sp for Ag 2CrO4 ?, 24. Write the expression for the solubility product of Hg 2Cl 2 ., 25. K sp of Ag 2CrO4 is 1.1 × 10-12 . what is solubility of Ag 2CrO4 in 0.1M K 2CrO4 ., 26. Will a precipitate be formed when 0.150 L of 0.1M Pb(NO3 )2 and 0.100L of 0.2 M NaCl, are mixed? K sp (PbCl 2 )=1.2 × 10-5 ., 27. K sp of Al(OH)3 is 1 × 10-15M . At what pH does 1.0 × 10-3M Al 3+ precipitate on the addition, of buffer of NH 4Cl and NH 4OH solution?, 31, , XII U8-Ionic equilibrium.indd 31, , 2/19/2020 5:16:56 PM

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www.tntextbooks.in, , ICT Corner, Buffers and pH, By using this tool you can, simulate the prepara�on of, a buffer and measure its pH, values, , Please go to the URL, h�p://pages.uoregon.edu/, tgreenbo/pHbuffer20.html, (or), Scan the QR code on the right, side, , Step – 1, Open the Browser and type the URL given (or) Scan the QR Code. You can see a webpage as shown in the, figure., , Step – 2, Now you can select a combination of an acid/base (Box 1) and its corresponding salt (Box 2) from the given, choices and also select the desired concentrations (Box 3) and volume (Box 4) of these for the buffer., , Step – 3, In order to measure the pH of the made-up buffer click the ‘Insert Probe’ (Box 5) on the pH meter. Now the, pH meter shows the pH. After measuring you need to remove the probe by clicking ‘Remove Probe” (Box 5), to make any changes in the composition., , Step – 4, Now you can vary the concentration and volume of the components and see how the pH changes., , 32, , XII U8-Ionic equilibrium.indd 32, , 2/19/2020 5:16:58 PM

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www.tntextbooks.in, , ELECTRO, CHEMISTRY, , UNIT, , 9, , Learning Objectives, After learning this unit, the students will be, able to,  Recognise, the, conductivity, of, electrolylic solution,  Define the terms resistivity, conductivity, equivalent and molar conductivity,  Explain the variation of conductivity, with concentration,  Apply Kohlrausch law to calculate the, conductivity of weak electrolyte at, infinite dilution.,  Describe an electrochemical cell,  Differentiate, between, an, electrochemical and electolylic cell,  Represent a galvanic cell using IUPAC, cell notation,  Derive Nernst equation and apply it to, calculate Ecell.,  Define Faraday’s Law of electrolysis,  Describe the construction of batteries, , Walther Hermann Nernst, Walther Hermann Nernst,, , was, , a German chemist known for his, work in thermodynamics, physical, chemistry, electrochemistry, and solid, state, , physics., , His, , formulation, , of, , the Nernst heat theorem helped to, pave the way for the third law of, thermodynamics, for which he won the, 1920 Nobel Prize in Chemistry. He is, also known for developing the Nernst, equation in 1887. He also derived, the Nernst equation for the electrical, potential, , generated, , by, , unequal, , concentrations of an ion separated by, a membrane that is permeable to the, ion. His equation is widely used in cell, physiology and neurobiology., , , , Explain corrosion as an electrochemical, process., , 33, , XII U9 Electro Chemistry.indd 33, , 2/19/2020 5:11:40 PM

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www.tntextbooks.in, , INTRODUCTION, We have come across many materials in our life, and they can be broadly classified, into conductors, semiconductors and insulators based on their electrical conductivity. You, might have noticed that conducting materials such as copper, aluminium etc., are used to, transport electrical energy from one place to another place, and the insulating materials such, as PVC, Bakelite etc., in switches, circuit boards etc., Do you know how the electrical energy is, generated? We know from first Law of thermodynamics that energy can neither be created nor, be destroyed, but one form of energy can be converted into another form. It is not possible to, create electrical energy but we can generate electrical energy in many ways i.e., by converting, solar energy, wind energy, tidal energy etc.... one such a way is converting chemical energy, into electrical energy as in the case of batteries. We cannot imagine a modern technological, world without batteries. Hence it is important to know the principles behind this type of, energy conversion. The branch of chemistry that deals with the study of electrical energy, transport and the inter conversion of electrical and chemical energy is called electrochemistry., Electrochemical reactions are redox reactions and they involve the transfer of electron from, one substance to another., In this unit, we will learn about the electrical conduction, construction of batteries and, the thermodynamic principles involved in electro chemical reactions., , 9.1 Conductivity of electrolytic solution, We have already learnt that when an electrolyte such as, sodium chloride, potassium chloride etc... is dissolved in a, solvent like water, the electrolyte is completely dissociated, to give its constituent ions (namely cations and anions)., When an electric field is applied to such an electrolytic, solution, the ions present in the solution carry charge, from one electrode to another electrode and thereby they, conduct electricity. The conductivity of the electrolytic, solution is measured using a conductivity cell. (Fig 9.1), A conductivity cell consists of two electrodes, immersed in an electrolytic solution. It obeys Ohm's law, like metallic conductor. i.e., at a constant temperature, the, current flowing through the cell (I) is directly proportional, to the voltage across the cell (V)., V, ⇒ V = IR .....(9.1), R, Where ‘R’ is the resistance of the solution in ohm (Ω), Here the resistance is the opposition that a cell, offers to the flow of electric current through it., , Platinium, electrode, (Area-A), , i.e., I α V (or) I =, , Electrolytic, solution, , Figure 9.1 conductivity cell, , 34, , XII U9 Electro Chemistry.indd 34, , 2/19/2020 5:11:41 PM

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www.tntextbooks.in, , Resistivity (ρ), Let us consider a conductivity cell in which the electrolytic solution is confined between, the two electrodes having cross sectional area (A) and are separated by a distance ‘l ’ . Like the, metallic conductor, the resistance of such an electrolytic solution is also directly proportional, to the length (l ) and inversely proportional to the cross sectional area (A)., l, Rα, A, l, R=ρ, A (9.2), Where ρ (rho) is called the specific resistance or resistivity, which depends on the nature, of the electrolyte., l, If = 1 m-1 , then, ρ = R . Hence the resistivity is defind as the resistance of an electrolyte, A, confined between two electrodes having unit cross sectional area and are separated by a unit,  l, distance. The ratio   is called the cell constant, Unit of resistivity is ohm metre (Ωm).,  A, Conductivity, It is more convenient to use conductance rather than resistance. The reciprocal of the, resistance 1, gives the conductance of an, R, electrolytic solution. The SI unit of conductance is, , ( ), , Siemen (S) ., , 1m, , 1, ..... (9.3), R, Substitute (R) from (9.2) in (9.3), 1 A, ⇒ i.e., C = . .....(9.4), ρ l, C=, , The reciprocal of the specific resistance, , Electrolytic, Solution, 1m, 1m, , Fig 9.2 conductivity of a cube, of an electrolytic solution, , ( 1ρ) is called the specific conductance (or), , conductivity. It is represented by the symbol κappa (κ) ., 1, = κ in equation (9.4) and rearranging, ρ,  l, ......(9.5), ⇒ κ = C.  ,  A, , Substitute, , Unit of κ, 1 l  1, m, ., . 2, , R A  ohm m , = ohm -1 m -1 = mho m -1 (or) Sm -1, κ=, , If A = 1m 2 and l = 1m ; then κ = C., The specific conductance is defined as the conductance of a cube of an electrolytic solution, of unit dimensions(Fig 9.2). The SI unit of specific conductance is Sm-1 ., Example, A conductivity cell has two platinum electrodes separated by a distance 1.5 cm and the, cross sectional area of each electrode is 4.5 sq cm. Using this cell, the resistance of 0.5 N, 35, , XII U9 Electro Chemistry.indd 35, , 2/19/2020 5:11:50 PM

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www.tntextbooks.in, , electrolytic solution was measured as 15 Ω . Find the specific conductance of the solution., Solution, κ=, , 1 l, R  A , , l = 1.5 cm = 1.5 × 10-2m, , 1, 1.5 × 10-2 m, ×, 15Ω 4.5 × 10-4 m2, = 2.22 Sm -1, , κ=, , A = 4.5 cm2 = 4.5 × (10-4 )m2, R = 15Ω, , 9.1.1 Molar conductivity (Λ m ), Solutions of different concentrations have different number of electrolytic ions in a given, volume of solution and hence they have different specific conductance. Therefore a new, quantity called molar conductance (Λ m ) was introduced., Let us imagine a conductivity cell in which the electrodes are separated by 1m and having, V m 3 of electrolytic solution which contains 1 mole of electrolyte. The conductance of such a, system is called the molar conductance (Λ m ), We have just learnt that the conductance of 1 m 3 electrolytic solution is called the specific, conductance (κ) . Therefore, the conductance of the above mentioned V m 3 solution (Λ m ) is, given by the following expression., (Λ m ) = κ × V , We know that, molarity (M) =, , Number of moles of solute (n), Volume of the solution (V in dm3 ), , Therefore, Volume of the solution containing one mole of so, olute =, , 10-3, ∴ Volume per m (V) =, mol -1m 3 ), (, M, Substitute (9.7) in (9.6), , .....(9.6), , 1, mol -1 L ), (, M, , 3, , κ (Sm -1 ) × 10-3, (9.6) ⇒ Λ m =, mol -1m -3 .....(9.8), M, The above relation defines the molar conductance in terms of the specific conductance, and the concentration of the electrolyte., Example, Calculate the molar conductance of 0.025M aqueous solution of calcium chloride at, 25 C. The specific conductance of calcium chloride is 12.04 × 10-2 Sm-1., , , κ (Sm -1 ) × 10-3, Molar conductance = Λ m =, mol -1m 3, M, -2, (12.04 × 10 Sm-1 ) × 10-3 ( mol -1m 3 ), =, 0.025, -5, 2, = 481.6 × 10 Sm mol -1, 36, , XII U9 Electro Chemistry.indd 36, , 2/19/2020 5:12:04 PM

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www.tntextbooks.in, , Evaluate yourself : 1, Calculate the molar conductance of 0.01M aqueous KCl solution at 25 C . The specific, , conductance of KCl at 25 C is 14.114 × 10-2 Sm-1 ., 9.1.2 Equivalent conductance (Λ ), Equivalent conductance is defined as the conductance of 'V' m3 of electrolytic solution, containing one gram equivalent of electrolyte in a conductivity cell in which the electrodes are, one metre apart., The relation between the equivalent conductance and the specific conductance is given, below., κ (Sm -1 ) × 10-3 (gram equivalent)-1 m3, Λ=, .....(9..9), N, Where κ the specific conductance and N is the concentration of the electrolytic solution, expressed in normality., Evaluate yourself : 2, The resistance of 0.15N solution of an electrolyte is 50 Ω. The specific conductance of, the solution is 2.4 Sm -1 . The resistance of 0.5 N solution of the same electrolyte measured, using the same conductivity cell is 480 Ω. Find the equivalent conductivity of 0.5 N solution, of the electrolyte., Given that, R1 =, 50 Ω, , R2 =, 480 Ω, , κ1 = 2.4 Sm -1, , κ2 = ?, , N1 = 0.15 N, , N 2 = 0.5 N, , κ2 (Sm -1 ) × 10-3 (gram equivalent)-1 m3, Λ=, N, -3, 0.25 × 10 S (gram equivalent)-1 m2, =, 0. 5, Λ = 5 × 10-4 Sm2 gram equivalent-1, , we know that, Cell constant, R, κ, R, ∴ 2 = 1, κ1, R2, R, κ2 = κ1 × 1, R2, 50 Ω, = 2.4 Sm-1 ×, 480 Ω, -1, = 0.25 Sm, κ=, , 9.1.3 Factors affecting electrolytic conductance, If the interionic attraction between the oppositely charged ions of solutes increases, the, conductance will decrease., 37, , XII U9 Electro Chemistry.indd 37, , 2/19/2020 5:12:10 PM

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www.tntextbooks.in, , zz, zz, , zz, , zz, , Solvent of higher dielectric constant show high conductance in solution., Conductance is inversely proportional to the Viscosity of the medium. i.e., conductivity, increases with the decrease in viscosity., If the temperature of the electrolytic solution increases, conductance also increases. Increase, in temperature increases the kinetic energy of the ions and decreases the attractive force, between the oppositely charged ions and hence conductivity increases., Molar conductance of a solution increases with increase in dilution. This is because, for, a strong electrolyte, interionic forces of attraction decrease with dilution. For a weak, electrolyte, degree of dissociation increases with dilution., , 9.1.4 Measurement of conductivity of ionic solutions, We have already learnt to measure the, B, specific resistance of a metallic wire using, a metre bridge in your physics practical, P, Q, experiment. We know that it works on the, principle of wheatstone bridge. Similarly,, the conductivity of an electrolytic solution, E, C, A, is determined by using a wheatstone, bridge arrangement in which one, S, resistance is replaced by a conductivity, cell filled with the electrolytic solution of, D, unknown conductivity., Conductivity cell, In the measurement of specific, Water, resistance of a metallic wire, a DC power, thermostat, supply is used. Here, if we apply DC, Electrolytic, solution, current through the conductivity cell, it, (Resistance = R), will lead to the electrolysis of the solution, taken in the cell. So, AC current is used for, this measurement to prevent electrolysis., A wheatstone bridge is constituted, AC Source, using known resistances P, Q, a variable, Fig 9.3 Schematic diagram of a conductivity, resistance S and conductivity cell (Let, cell in a wheatstone bridge circuit, the resistance of the electrolytic solution, taken in it be R) as shown in the figure 9.3. An AC source (550 Hz to 5 KHz) is connected, between the junctions A and C. Connect a suitable detector E (Such as the telephone ear piece, detector) between the junctions ‘B’ and ‘D’., The variable resistance ‘S’ is adjusted until the bridge is balanced and in this conditions, there is no current flow through the detector., Under balanced condition,, P =R, Q, S, ∴R = P × S, Q, .....(9.10), 38, , XII U9 Electro Chemistry.indd 38, , 2/19/2020 5:12:12 PM

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www.tntextbooks.in, , The resistance of the electrolytic solution (R) is calculated from the known resistance values, P, Q and the measured ‘S’ value under balanced condition using the above expression (9.10)., Conductivity calculation, Specific conductance (or) conductivity of an electrolyte can be calculated from the, resistance value using the following expression., 1  l, κ=, [  equation 9.5], R  A , The value of the cell constant l is usually provided by the cell manufacturer. Alternatively, A, the cell constant may be determined using KCl solution whose concentration and specific, conductance are known., Example, The resistance of a conductivity cell is measured as 190 Ω using 0.1M KCl solution, (specific conductance of 0.1M KCl is 1.3 Sm -1 ) . When the same cell is filled with 0.003M, sodium chloride solution, the measured resistance is 6.3KΩ. Both these measurements are, made at a particular temperature. Calculate the specific and molar conductance of NaCl, solution., Given that, κ = 1.3 Sm-1 (for 0.1M KCl solution), R = 190 Ω,  l, -1,  A  = κ . R = (1.3 Sm ) (190 Ω ), = 247 m-1, , κ(NaCl ) =, , 1, R (NaCl ), ,  l,  A , , 1, (247 m-1 ), 6.3 KΩ, = 39.2 × 10-3 Sm -1, , 6.3KΩ = 6.3 × 103 Ω, , =, , Λm =, =, , κ × 10-3 mol-1 m3, M, 39.2 × 10-3 (Sm-1 ) × 10-3 (mol-1m3 ), 0.003, , Λ m = 13.04 × 10-3 Sm 2 mol-1, , 9.2 Variation of molar conductivity with concentration, Friedrich Kohlraush studied the molar conductance of different electrolytes at different, concentrations. He observed that, increase of the molar conductance of an electrolytic solution, with the increase in the dilution. One such experimental results is given in the following table, for better understanding., 39, , XII U9 Electro Chemistry.indd 39, , 2/19/2020 5:12:17 PM

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www.tntextbooks.in, , Concentration, (M), , Molar conductance ( × 10-3Sm 2 mol -1 ), NaCl, , KCl, , HCl, , 0.1, , 10.674, , 12.896, , 39.132, , 0.01, , 11.851, , 14.127, , 41.20, , 0.0001, , 12.374, , 14.695, , 42.136, , Based on the above such results, Kohlraush deduced the following empirical relationship, between the molar conductance (Λ m ) and the concentration of the electrolyte (C)., , Λm = Λm, - k C .....(9.11), The above equation represents a straight line of the form y = mx + c . Hence, the plot of, Λ m Vs C gives a straight line with a negative slope of –k and the y intercept, Λ m . Where, Λ m is called the limiting molar conductivity. i.e., the molar conductance approaches a limiting, value in very dilute solutions., , 0.04, Weak electrolyte (CH3COOH), , m (S m2 mol–1), , For strong electrolytes such, as KCl, NaCl etc., the plot, Λ m, Vs C , gives a straight line as, shown in the graph (9.4). It is, also observed that the plot is not, a linear one for weak electrolytes., , For a strong electrolyte, at, high concentration, the number, 0.02, of constituent ions of the, electrolyte in a given volume is, Strong electrolyte (KCI), high and hence the attractive force, between the oppositely charged, ions is also high. Moreover the, ions also experience a viscous, drag due to greater solvation., 0, 0.2, 0.4, These factors attribute for the, c (mol L-1)1/2, low molar conductivity at high, concentration. When the dilution, Figure 9.4 Variation of molar, increases, the ions are far apart, conductance with concentration, and the attractive forces decrease., At infinite dilution the ions are so far apart, the interaction between them becomes insignificant, and hence, the molar conductivity increases and reaches a maximum value at infinite dilution., For a weak electrolyte, at high concentration, the plot is almost parallel to concentration, axis with slight increase in conductivity as the dilution increases. When the concentration, 40, , XII U9 Electro Chemistry.indd 40, , 2/19/2020 5:12:26 PM

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www.tntextbooks.in, , approaches zero, there is a sudden increase in the molar conductance and the curve is almost, parallel to Λ m axis. This is due to the fact that the dissociation of the weak electrolyte increases, with the increase in dilution (Ostwald dilution law). Λ 0m values for strong electrolytes can be, obtained by extrapolating the straight line, as shown in figure (9.4). But the same procedure, is not applicable for weak electrolytes, as the plot is not a linear one, Λ 0m values of the weak, electrolytes can be determined using Kohlraush's law., 9.2.1 Debye - Huckel and Onsager equation, We have learnt that at infinite dilution, the interaction between the ions in the electrolyte, solution is negligible. Except this condition, electrostatic interaction between the ions alters, the properties of the solution from those expected from the free – ions value. The influence, of ion-ion interactions on the conductivity of strong electrolytes was studied by Debye and, Huckel. They considered that each ion is surrounded by an ionic atmosphere of opposite, sign, and derived an expression relating the molar conductance of strong electrolytes with the, concentration by assuming complete dissociation. Later, the equation was further developed, by Onsager. For a uni – univalent electrolyte the Debye Huckel and Onsager equation is given, below., , (, , Λ m = Λ 0m − A + B Λ 0m, , ), , C .....(9.12), , Where A and B are the constants which depend only on the nature of the solvent and, temperature. The expression for A and B are, A=, , 82.4, , DT η, , ;, , B=, , 8.20 × 105, 3, , DT, , Here, D is the dielectric constant of the medium, η the viscosity of the medium and T the, temperature in Kelvin., 9.2.2 Kohlraush's law, The limiting molar conductance Λ 0m is the basis for kohlraush law. At infinite dilution,, the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar, conductivities of its constituent ions. i.e., the molar conductivity is due to the independent, migration of cations in one direction and anions in the opposite direction., For a uni – univalent electrolyte such as NaCl, the Kohlraush's law is expressed as, , ( Λ 0m )NaCl = (λ 0m )Na+ + (λ 0m )Cl-, , In general, according to Kohlraush's law, the molar conductivity at infinite dilution for a, electrolyte represented by the formula A x B y , is given below., , ( Λ 0m )AxBy = x (λ0m )Ay+ + y (λ0m )Bx-, , .....(9.13), , Kohlraush arrived the above mentioned relationship based on the experimental, observations such as the one as shown in the table. These result show that at infinite dilution, each constituent ion of the electrolyte makes a definite contribution towards the molar, conductance of the electrolyte irrespective of nature of other ion with which it is associated, 41, , XII U9 Electro Chemistry.indd 41, , 2/19/2020 5:12:31 PM

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www.tntextbooks.in, , Λ 2m C, Ka =, (Λ  −Λ ), Λ m 2 m  m, Λm, Λ 2mC, ⇒ Ka = , .....(9.16), Λ m (Λ m −Λ m ), 3., , Calculation of solubility of sparingly soluble salts, , Substances like AgCl, PbSO 4 etc., are sparingly soluble in water. The solubility product of, such substances can be determined using conductivity measurements., Let us consider AgCl as an example, AgCl (s)  Ag + + Cl −, K sp = [Ag + ][Cl - ], Let the concentration of [Ag + ] be ‘C’ molL–1., As per the stoichiometry, if [Ag + ] = C, then [Cl −]also equal to ‘C’ mol L−1 ., K sp =C.C, ⇒ K sp = C 2, We know that the concentration (in mol dm-3) is related to the molar and specific, conductance by the following expressions, Λo =, (or), , κ × 10-3, C (in mol L-1 ), , κ × 10-3, C=, Λ, , Substitute the concentration value in the relation K sp = C 2,  κ × 10-3 , K sp = , , Λ  , , 2, , .....(9.17), , 9.3 Electrochemical Cell, Electrochemical cell is a device which converts chemical energy into electrical energy, and vice versa. It consists of two separate electrodes which are in contact with an electrolyte, solution. Electrochemical cells are mainly classified into the following two types., 1. Galvanic Cell ( Voltaic cell) : It is a device in which a spontaneous chemical reaction, generates an electric current i.e., it converts chemical energy into electrical energy. It is, commonly known as a battery., 2. Electrolytic cell : It is a device in which an electric current from an external source drives, a nonspontaneous reaction i.e., it converts electrical energy into chemical energy., , 43, , XII U9 Electro Chemistry.indd 43, , 2/19/2020 5:12:52 PM

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www.tntextbooks.in, , 9.3.1 Galvanic cell, We have already learnt in XI standard that when a zinc metal strip is placed in a copper, sulphate solution, the blue colour of the solution fades and the copper is deposited on the zinc, strip as red – brown crust due to the following spontaneous chemical reaction., Zn(s) + CuSO 4 (aq) → ZnSO4 (aq) + Cu (s), The energy produced in the above reaction is lost to the surroundings as heat., In the above redox reaction, Zinc is oxidised to Zn2+ ions and the Cu2+ ions are reduced, to metallic copper. The half reactions are represented as below., Zn(s) → Zn2+ (aq) + 2e-, , (oxidation), , Cu 2+ (aq) + 2e- → Cu (s), , (reduction), , If we perform the above two half reactions separately in an apparatus as shown in, figure 9.5, some of the energy produced in the reaction will be converted into electrical energy., Let us understand the function of a galvanic cell by considering Daniel cell as an example. It, uses the above reaction for generation of electrical energy., The separation of half reaction is the basis for the construction of Daniel cell. It consists, of two half cells., Oxidation half cell, A metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker,, as shown in Figure 9.5., Reduction half cell, A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker, as, shown in Figure 9.5., Joining the half cells, The zinc and copper strips are externally connected using a wire through a switch (k), and a load (example: volt meter). The electrolytic solution present in the cathodic and anodic, compartment are connected using an inverted U tube containing a agar-agar gel mixed with an, inert electrolytes such as KCl, Na 2SO4 etc., The ions of inert electrolyte do not react with other, ions present in the half cells and they are not either oxidised (or) reduced at the electrodes. The, solution in the salt bridge cannot get poured out, but through which the ions can move into, (or) out of the half cells., When the switch (k) closes the circuit, the electrons flows from zinc strip to copper strip., This is due to the following redox reactions which are taking place at the respective electrodes., 44, , XII U9 Electro Chemistry.indd 44, , 2/19/2020 5:13:00 PM

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www.tntextbooks.in, , Digital, voltmeter, 1.10, , e−, , Zinc, (anode), , 2SO4, , +, , Copper, (cathode), , 2(Na+), 2e-, , Na2SO4(aq), , 2e-, , Zn, , Salt bridge, , e-, , 2+, Cu, , Zn, , ZnSO4 (aq), , Cu, , (Zn2++SO42-), , 2+, , CuSO4(aq), (Cu2++SO42-), , 2-, , SO4, , 2e, , 2e, , Zn, Zn2, , Cu, Cu2, , Zn2, , Cu2, , Cu2+(aq) + 2e− Cu(s), , Zn(s) Zn2+(aq) + 2e−, , Figure 9.5 : Daniel cell, Anodic oxidation, The electrode at which the oxidation occurs is called the anode. In Daniel cell, the oxidation, take place at zinc electrode, i.e., zinc is oxidised to Zn2+ ions by loosing its electrons. The Zn2+, ions enter the solution and the electrons enter the zinc metal, then flow through the external, wire and then enter the copper strip. Electrons are liberated at zinc electrode and hence it is, negative ( - ve)., Zn(s) → Zn2+ (aq) + 2e-, , (loss of electron-oxidation), , Cathodic reduction, As discussed earlier, the electrons flow through the circuit from zinc to copper, where, the Cu 2+ ions in the solution accept the electrons, get reduced to copper and the same get, deposited on the electrode. Here, the electrons are consumed and hence it is positive (+ve)., Cu 2+ (aq) + 2e- → Cu (s), , (gain of electron-reduction), 45, , XII U9 Electro Chemistry.indd 45, , 2/19/2020 5:13:05 PM

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www.tntextbooks.in, , Salt bridge, The electrolytes present in two half cells are connected using a salt bridge. We have learnt, that the anodic oxidation of zinc electrodes results in the increase in concentration of Zn2+, in solution. i.e., the solution contains more number of Zn2+ ions as compared to SO4 2- and, hence the solution in the anodic compartment would become positively charged. Similarly,, the solution in the cathodic compartment would become negatively charged as the Cu2+ ions, are reduced to copper i.e., the cathodic solution contain more number of SO4 2- ions compared, to Cu2+ ., To maintain the electrical neutrality in both the compartments, the non reactive anions, Cl (from KCl taken in the salt bridge) move from the salt bridge and enter into the anodic, compartment, at the same time some of the K+ ions move from the salt bridge into the cathodic, compartment., -, , Completion of circuit, Electrons flow from the negatively charged zinc anode into the positively charged copper, cathode through the external wire, at the same time, anions move towards anode and cations, are move towards the cathode compartment. This completes the circuit., Consumption of Electrodes, As the Daniel cell operates, the mass of zinc electrode gradually decreases while the mass, of the copper electrode increases and hence the cell will function until the entire metallic zinc, electrode is converted in to Zn2+ or the entire Cu2+ ions are converted in to metallic copper., Unlike Daniel cell, in certain cases, the reactants (or) products cannot serve as electrodes, and in such cases inert electrode such as graphite (or) platinum is used which conducts current, in the external circuit., 9.3.2 Galvanic cell notation, The galvanic cell is represented by a cell diagram, for example, Daniel cell is represented as, Zn (s) Zn2+ (aq) Cu 2+ (aq) Cu (s), In the above notation, a single vertical bar (|) represents a phase boundary and the double, vertical bar (||) represents the salt bridge., The anode half cell is written on the left side of the salt bridge and the cathode half cell on, the right side., The anode and cathode are written on the extreme left and extreme right, respectively., The emf of the cell is written on the right side after cell diagram., , 46, , XII U9 Electro Chemistry.indd 46, , 2/19/2020 5:13:12 PM

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www.tntextbooks.in, , Phase boundary, , Phase boundary, , Zn (s) Zn2+ (aq), , Cu2+ (aq) Cu (s), , Anode, (extreme Left), Anode half, cell, , =, , 1.1V, , Cathode, Standard emf of the cell, (extreme right), Left, , Salt, bridge, , Right, , Cathode, Half cell, , Example, The net redox reaction of a galvanic cell is given below, 2 Cr (s) + 3Cu2+ (aq) → 2Cr 3+ (aq) + 3Cu (s), Write the half reactions and describe the cell using cell notation., Anodic oxidation 2Cr (s) → 2Cr 3+ (aq) + 6e- .....(1), Cathodic reduction : 3Cu 2+ (aq) + 6e- → 3 Cu (s) .....(2), Cell Notation is, Cr (s) Cr 3+ (aq) Cu 2+ (aq) Cu(s), 9.3.3 emf of a Cell, We have learnt that when two half cells of a Daniel cell are connected, a spontaneous, redox reaction will take place which results in the flow of electrons from anode to cathode. The, force that pushes the electrons away from the anode and pulls them towards cathode is called, the electromotive force (emf) (or) the cell potential. The SI unit of cell potential is the volt (v)., When there is one volt difference in electrical potential between the anode and cathode,, one joule of energy is released for each columb of charge that moves between them., i.e.,, , 1J = 1C × 1V , , .....(9.18), , The cell voltage depends on the nature of the electrodes, the concentration of the, electrolytes and the temperature at which the cell is operated. For example, At, 25 C , The emf of the below mentioned Daniel cell is 1.107 Volts, Zn (s) Zn 2+ (aq,1M) Cu 2+ (aq,1M) Cu ( s ), , E 0 = 1.107 V, , 47, , XII U9 Electro Chemistry.indd 47, , 2/19/2020 5:13:19 PM

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www.tntextbooks.in, , 9.3.4 Measurement of electrode potential, The overall redox reaction can be considered as the sum, of two half reactions i.e., oxidation and reduction. Similarly,, the emf of a cell can be considered as the sum of the electrode, potentials at the cathode and anode,, Ecell = ( E ox )anode + ( Ered )cathode .....(9.19), Here, ( E ox )anode represents the oxidation potential at, anode and ( Ered )cathode represents the reduction potential, at cathode. It is impossible to measure the emf of a single, electrode, but we can measure the potential difference, between the two electrodes ( Ecell ) using a voltmeter. If we, know the emf of any one of the electrodes which constitute, the cell, we can calculate the emf of the other electrode from, the measured emf of the cell using the expression (9.19)., Hence, we need a reference electrode whose emf is known, For that purpose, Standard Hydrogen Electrode (SHE) is, used as the reference electrode. It has been assigned an arbitrary, emf of exactly zero volt. It consists of a platinum electrode in, contact with 1M HCl solution and 1 atm hydrogen gas. The, hydrogen gas is bubbled through the solution at 25 C as shown, in the figure 9.6. SHE can act as a cathode as well as an anode., The Half cell reactions are given below., If SHE is used as a cathode, the reduction reaction is, 2H+ (aq,1M)+2e- → H2 (g, 1 atm), , Eo = 0 volt, , If SHE is used as an anode, the oxidation reaction is, H2 (g,1 atm) → 2H+ (aq, 1M) + 2eEo = 0 volt, Illustration, , H2 out, , H2 in, , H2gas, Pt, electrode, , (SHE), Figure 9.6 Standard, Hydrogen electrode, , Let us calculate the reduction potential of zinc electrode dipped in zinc sulphate solution, using SHE., Step : 1 The following galvanic cell is constructed using SHE, Zn (s), , Zn2+ (aq, 1M) H+ (aq, 1M) H2 (g, 1atm) Pt (s), , Step : 2 The emf of the above galvanic cell is measured using a volt meter. In this case, the, measured emf of the above galvanic cell is 0.76V., Calculation, We know that,, o, , ( ), o, , Ecell = Eox, o, , ( ), , [ From equation (9.19)], , o, , Zn Zn2+, , + Ered, , SHE, , o, , Ecell = 0.76 and (E red )SHE = 0V . Substitute these values in the above equation, 48, , XII U9 Electro Chemistry.indd 48, , 2/19/2020 5:13:28 PM

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www.tntextbooks.in, , Digital voltmeter, 0.76, , e-, , Salt bridge, , −, , Zinc, (anode), , 2-, , SO4, , e+, , 2(Na+), H2(g) (1 atm), , Na2SO4(aq), , 2eZn, , ZnSO4(aq), , H2, , 2+, , Zn, , H+, (1 M), , 2SO4, , Standard hydrogen, electrode (cathode), , e–, , 2H+ (aq) + 2e- H2(g), , Figure 9.7 emf measurement (Zn | Zn2+ electrode), o, , ⇒ 0.76V = (Eox )zn Zn + 0V, 2+, , o, , ⇒ (Eox )zn Zn = 0.76V, 2+, , This oxidation potential corresponds to the below mentioned half cell reaction which, takes place at the cathode., Zn → Zn2+ + 2e- (Oxidation), The emf for the reverse reaction will give the reduction potential, ∴ (Ered )Zn Zn = -0.76V., Zn2+ +2e- → Zn ; E = - 0.76V , o, , o, , 2+, , IUPAC definition, Electrode potential (E), Electromotive force of a cell in which the electrode on the left is a standard hydrogen, electrode and the electrode on the right is the electrode in question., Standard electrode potential, E, The value of the standard emf of a cell in which molecular hydrogen under standard, pressure is oxidised to solvated protons at the left hand electrode., , , Evaluate yourself, 1. The emf of the following cell at 25 C is equal to 0.34v. Calculate the reduction potential, of copper electrode., Pt (s), , H 2 (g, 1atm), , H + (aq, 1M) Cu 2+ (aq, 1M) Cu (s), , 2. Using the calculated emf value of zinc and copper electrode, calculate the emf of the, following cell at 25 C ., Zn (s) Zn2+ (aq, 1M) Cu 2+ (aq, 1M) Cu (s), 49, , XII U9 Electro Chemistry.indd 49, , 2/19/2020 5:13:38 PM

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www.tntextbooks.in, , Evaluate yourself, Write the overall redox reaction which takes place in the galvanic cell,, Pt(s) Fe2+ (aq),Fe 3+ (aq) MnO-4 (aq), H+ (aq),Mn2+ (aq) Pt(s), , 9.4 Thermodynamics of cell reactions, We have just learnt that in a galvanic cell, the chemical energy is converted into electrical, energy. The electrical energy produced by the cell is equal to the product of the total charge of, electrons and the emf of the cell which drives these electrons between the electrodes., If ‘n’ is the number of moles of electrons exchanged between the oxidising and reducing, agent in the overall cell reaction, then the electrical energy produced by the cell is given as, below., Electrical energy = Charge of ’n’ mole of electrons × Ecell, ......(9.20), Charge of 1 mole of electrons = one Faraday (1F), ∴ Charge of ’n’ mole of electrons = nF, .......(9.21), Equation (9.20) ⇒ Electrical energy = nFEcell, Charge of one elctron = 1.602 × 10-19 C, −19, , ∴ Charge one mole of elctron = 6.023 × 1023 × 1.602 × 10 C, = 96488 C, i.e., 1F  96500 C, , This energy is used to do the electric work. Therefore the maximum work that can be, obtained from a galvanic cell is, (Wmax )cell = - nFEcell .....(9.22), Here the (-) sign is introduced to indicate that the work is done by the system on the, surroundings., We know from the Second Law of thermodynamics that the maximum work done by the, system is equal to the change in the Gibbs free energy of the system., i.e., Wmax = ∆ G , , .....(9.23), , From (9.22) and (9.23),, DG = - nFEcell .....(9.24), For a spontaneous cell reaction, the DG should be negative. The above expression (9.24), indicates that Ecell should be positive to get a negative DG value., When all the cell components are in their standard state, the equation (9.24) becomes, DG o = - nFEocell .....(9.25), We know that the standard free energy change is related to the equilibrium constant as, per the following expression., 50, , XII U9 Electro Chemistry.indd 50, , 2/19/2020 5:13:44 PM

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www.tntextbooks.in, , DG o = - RT lnK eq .....(9.26), Comparing (9.25) and (9.26),, nFEcell = RT lnK eq, 2.303 RT, ⇒ Ecell =, log K eq .....(9.27), nF, , 9.4.1 Nernst equation, Nernst equation is the one which relates the cell potential and the concentration of the, species involved in an electrochemical reaction. Let us consider an electrochemical cell for, which the overall redox reaction is,, xA + yB  lC + mD, The reaction quotient Q for the above reaction is given below, [C]l [D]m, Q=, [A]x [B]y .....(9.28), We have already learnt that,, DG = DG + RT lnQ .....(9.29), The Gibbs free energy can be related to the cell emf as follows, [∴ equation (9.24) and (9.25)], DG = - nFE cell ; DG o = - nFEocell, Substitute these values and Q from (9.28) in the equation (9.29), (9.29) ⇒ - nFEcell = - nFEocell + RT ln, , [C]l [D]m, .....(9.30), [A]x [B]y, , Divide the whole equation (9.30) by (-nF), (9.25) ⇒ Ecell = E, (or) Ecell = E, , , cell, , , cell, , RT [C]l [D]m, ln, nF [A]x [B]y, , 2.303RT, [C]l [D]m, log, .....(9.31), nF, [A]x [B]y, , The above equation (9.31) is called the Nernst equation, At 25o C (298K), the above equation (9.31) becomes,, o, Ecell = Ecell, -, , 2.303 × 8.314 × 298, [C]l [D]m, log x, n(96500), [A] [B]y, , Ecell = Ecell -, , [C] [D], 0.0591, log, [A]x [B]y, n, l, , ∴ R = 8.314 JK -1 mol -1  .....(9.32), , ,  T = 298 K., , 1 F = 96500 C mol -1 , , , , m, , Let us calculate the emf of the following cell at 25oC using Nernst equation., 51, , XII U9 Electro Chemistry.indd 51, , 2/19/2020 5:13:53 PM

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www.tntextbooks.in, , Cu (s) Cu 2+ (0.25 aq, M) Fe3+ (0.005 aq M) Fe2+ (0.1 aq M) Pt (s), , Given : ( Eo, , ), , Fe3+ Fe2+, , = 0.77V and ( Eo ) Cu, , 2+, , Cu, , = 0.34 V, , Half reactions are, Cu (s) → Cu 2+ (aq) + 2e-, , ...... (1), , 2 Fe (aq)+2e → 2Fe (aq), , ..... (2), , 3+, , -, , 2+, , the overall reaction is, Cu (s) + 2 Fe3+ (aq) → Cu 2+ (aq) + 2 Fe 2+ (aq), and n = 2, Apply Nernst equation at . 25oC ., Ecell =Eocell −, , [Cu 2+ ][Fe2+ ]2, 0.0591, log, 2, [Fe3+ ]2, , , Eocell = ( Eox )Cu Cu + (Ered, )Fe, 2+, , 3+, , [ [Cu (s)] = 1], , Fe2+, , Given standard reduction potential of Cu 2+ Cu is 0.34V, ∴ (Eoox )Cu Cu = -0.34V, 2+, , o, (Ered, )Fe, , 3+, , Fe2+, , = 0.77V, , ∴ Ecell = - 0.34 + 0.77, Ecell = 0.43V, ∴ Ecell, , 0.0591, (0.25)(0.1)2, = 0.43 × log, , 2, (0.005)2, 0.0591, ×2, 2, = 0.43 - 0.0591, = 0.3709V., = 0.43 -, , (0.25)(0.1)2, (0.005)2, 25 × 10-2 × 1 × 10-2, = log, 25 × 10−6, = log 102, = 2 log10 10, = log, , = 2., , Evaluate yourself, The electrochemical cell reaction of the Daniel cell is, Zn (s) + Cu 2+ (aq) → Zn2+ (aq)+Cu (s), What is the change in the cell voltage on increasing the ion concentration in the anode, compartment by a factor 10?, , 52, , XII U9 Electro Chemistry.indd 52, , 2/19/2020 5:14:06 PM

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www.tntextbooks.in, , Electrolytic cell and electrolysis, Electrolysis is a process in which the electrical energy is used to cause a non-spontaneous, chemical reaction to occur; the energy is often used to decompose a compound into its, elements. The device which is used to carry out the electrolysis is called the electrolytic cell. The, electrochemical process occurring in the electrolytic cell and galvanic cell are the reverse of, each other. Let us understand the function of a electrolytic cell by considering the electrolysis, of molten sodium chloride., The electrolytic, Cl2 gas, cell consists of two, iron, electrodes, NaCl, dipped in molten, sodium, chloride, and, they, are, connected to an, external DC power, Molten Na, supply via a key as, Molten Na, shown in the figure, Cylindrical steel cathode, (9.8). The electrode, which is attached to, 2 Na(l), 2 Na+(l) + 2 e−, the negative end of, the power supply is, called the cathode,, and the one which, Graphite anode, Molten NaCl, attached to the, Cl2( g) + 2 e−, 2 Cl−(l), and CaCl2, +, positive end is called, Figure, 9.8, Electrolysis, of molten NaCl, the anode. Once the, key is closed, the, external DC power supply drives the electrons to the cathode and at the same time pull the, electrons from the anode., Cell reactions, Na + ions are attracted towards cathode, where they combine with the electrons and, reduced to liquid sodium., Cathode (reduction), Na + (l )+e- → Na (l ), , E o = -2.71V, , Similarly, Cl– ions are attracted towards anode where they lose their electrons and oxidised, to chlorine gas., Anode (oxidation), 2Cl - (l ) → Cl 2 (g) + 2e -, , E o = -1.36V, 53, , XII U9 Electro Chemistry.indd 53, , 2/19/2020 5:14:10 PM

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www.tntextbooks.in, , The overall reaction is,, 2Na + (l ) + 2Cl - (l ) → 2Na(l ) + Cl 2 ( g ), , E o = - 4.07V, , The negative Eo value shows that the above reaction is a non spontaneous one. Hence, we, have to supply a voltage greater than 4.07V to cause the electrolysis of molten NaCl., In electrolytic cell, oxidation occurs at the anode and reduction occur at the cathode as in, a galvanic cell, but the sign of the electrodes is the reverse i.e., in the electrolytic cell cathode, is –ve and anode is +ve., Faraday’s Laws of electrolysis, First Law, The mass of the substance (m) liberated at an electrode during electrolysis is directly, proportional to the quantity of charge (Q) passed through the cell., i.e m α Q, We know that the charge is related to the current by the equation I =, ∴ m α It, , Q, ⇒ Q = It, t, , (or), m = Z It, ..... (9.33), Where is Z is known as the electro chemical equivalent of the substance produced of the, electrode., When, I = 1A and t = 1sec,, , Q = 1C , in such case the equation (9.32) becomes, (9.33), , ⇒ m = Z .....(9.34), Thus, the electrochemical equivalent is defined as the amount of substance deposited or, liberated at the electrode by a charge of 1 coulomb., Electro chemical equivalent and molar mass, Consider the following general electrochemical redox reaction, Mn+ (aq)+ne- → M(s), , We can infer from the above equation that ‘n’ moles of electrons are required to precipitate, 1 mole of Mn+ as M(s)., The quantity of charge required to, = Charge of 'n' moles of electrons, precipitate one mole of Mn+, = nF, In other words, the mass of substance deposited by one coulomb of charge, Electrochemical equivalent of Mn+ =, , Z=, , Molarmass of M, n (96500), (or), Equivalent mass, 96500, , , .....(9.35), , 54, , XII U9 Electro Chemistry.indd 54, , 2/19/2020 5:14:17 PM

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www.tntextbooks.in, , Second Law, , Key, , Rh, +, , –, , NiSO4 (aq), , – A +, , +, , –, , CuSO4 (aq), , +, , –, , CoSO4 (aq), , Figure 9.9 Electrolysis of various electrolytes using same quantity of charge, When the same quantity of charge is passed through the solutions of different electrolytes,, the amount of substances liberated at the respective electrodes are directly proportional to, their electrochemical equivalents., Let us consider three electrolytic cells connected in series to the same DC electrical source, as shown in the figure 9.9. Each cell is filled with a different electrolytes namely NiSO4, CuSO4, and CoSO4, respectively., When Q coulomb charge is passed through the electrolytic cells the masses of Nickel,, copper and cobalt deposited at the respective electrodes be m Ni , m Cu and mCo , respectively., According to Faraday’s second Law,, , m Ni α Z Ni , m Cu α Z Cu and mCo α Z Co, (or), m Ni m Cu m Co, =, =, Z Ni Z Cu Z Co, , .....(9.36), , Example, A solution of silver nitrate is electrolysed for 20 minutes with a current of 2 amperes., Calculate the mass of silver deposited at the cathode., Electrochemical reaction at cathode is Ag + +e- → Ag (reduction), molarmass of Ag, 108, m = ZIt, Z=, =, (96500), 1 × 96500, -1, 108 gmol, I = 2A, m=, × 2400C, 96500 C mol -1, m = 2.68 g., t = 20 × 60S = 1200 S, , It = 2A × 1200S = 2400C, 55, , XII U9 Electro Chemistry.indd 55, , 2/19/2020 5:14:20 PM

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www.tntextbooks.in, , Evaluate yourself A solution of a salt of metal was electrolysed for 15 minutes with, a current of 0.15 amperes. The mass of the metal deposited at the cathode is 0.783g., calculate the equivalent mass of the metal., Batteries, Batteries are indispensable in the modern electronic world. For example, Li – ion batteries, are used in cell phones, dry cell in flashlight etc…. These batteries are used as a source of, direct current at a constant voltage. We can classify them into primary batteries (non –, rechargeable) and secondary batteries (rechargeable). In this section, we will briefly discuss, the electrochemistry of some batteries., Leclanche cell, Anode, : Zinc container, Cathode, : Graphite rod in contact with MnO2, Electrolyte : ammonium chloride and zinc chloride in water, Emf of the cell is about 1.5V, Cell reaction, Oxidation at anode, Zn (s) → Zn2+ (aq)+2e- .....(1), Reduction at cathode, 2 NH 4+ (aq) + 2e- → 2NH3 (aq) + H2 (g) .....(2), The hydrogen gas is oxidised to water by MnO2, H2 (g) + 2 MnO2 (s) → Mn2O3 (s) + H2O (l ) .....(3), Equation (1) + (2)+(3) gives the overall redox reaction, Electrons (e-), , Dry cell battery, Electrons (e-), Carbon (cathode), Light bulb, (load), , Electrolyte, paste, Separator, , Carbon and, manganse, doxide mixture, , Electrons (e-), , Zinc (anode), , Electrons (e-), , Figure 9.10 Leclanche cell, 56, , XII U9 Electro Chemistry.indd 56, , 2/19/2020 5:14:23 PM

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www.tntextbooks.in, , Zn (s) + 2NH4+ (aq) + 2 MnO2 (s) → Zn2+ (aq) + Mn 2O3 (s) + H2O (l )+2NH3, , ...... (4), , Ammonia produced at the cathode combines with Zn to form a complex ion, 2+, , [Zn (NH ) ], , 2+, , 3 4, , (aq) . As the reaction proceeds, the concentration of NH 4 + will decrease and, , the aqueous NH3 will increase which lead to the decrease in the emf of cell., Mercury button cell, Anode, : zinc amalgamated with mercury, Cathode, : HgO mixed with graphite, Electrolyte, : Paste of KOH and ZnO, Oxidation occurs at anode, , :, , 0, , +2, , Zn(s) + 2OH- (aq) → ZnO(s) + H2O (l ) + 2efrom KOH, , +2, , 0, , Reduction occurs at cathode, , :, , HgO(s) + H2O (l) + 2e - → Hg (l) + 2OH- (aq), , Overall reaction, Cell emf, , :, :, , Zn (s) + HgO (s) → ZnO (s) + Hg (l ), about 1.35V., , Uses, , :, , It has higher capacity and longer life. Used in, pacemakers, electronic watches, cameras etc…, Sealing and, insulation Zn (anode), gasket, , Tin-plated, inner top, , KOH saturated, with ZnO in, absorbent material, (electrolyte), HgO mixed with, graphite (cathode), , Steel outer, top, Inner steel case, , Outer steel, case, Gas vent, , Barrier, , Figure 9.11 Mercury button cell, Secondary batteries, We have already learnt that the electrochemical reactions which take place in a galvanic, cell may be reversed by applying a potential slightly greater than the emf generated by the, cell. This principle is used in secondary batteries to regenerate the original reactants. Let us, understand the function of secondary cell by considering the lead storage battery as an example, 57, , XII U9 Electro Chemistry.indd 57, , 2/19/2020 5:14:29 PM

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www.tntextbooks.in, , Lead storage battery, Anode, : spongy lead, Cathode, : lead plate bearing PbO2, Electrolyte : 38% by mass of H2SO4 with density 1.2g / mL., Oxidation occurs at the anode, Pb(s) → Pb2+ (aq)+2e- .....(1), The Pb2+ ions combine with SO4 -2 to from PbSO4 precipitate., Pb2+ (aq) + SO 4 2−(aq) → PbSO 4 (s) .....(2), Reduction occurs at the cathode, PbO2 (s) + 4 H+ (aq) + 2e - → Pb2+ (aq) + 2H 2O (l ) ......(3), The Pb 2+ ions also combine with SO4 2- ions from sulphuric acid to form PbSO4 precipitate., Pb 2+ (aq) + SO 4 2− (aq) → PbSO4 (s).....(4), The Overall reactions is, Equation (1) + (2) + (3) + (4), Pb (s) + PbO 2 (s) + 4H + (aq) + 2SO 4 2− (aq) → 2 PbSO 4 (s) + 2H 2 O (l ), , The emf of a single cell is about 2V . Usually six such cells are combined in series to, produce 12volt, The emf of the cell depends on the concentration of H 2SO 4 . As the cell reaction uses SO4 2−, ions, the concentration H2SO4 decreases. When the cell potential falls to about 1.8V, the cell, has to be recharged., Recharge of the cell, As said earlier, a potential greater than 2V is applied across the electrodes, the cell reactions, that take place during the discharge process are reversed. During recharge process, the role of, anode and cathode is reversed and H2SO4 is regenerated., Oxidation occurs at the cathode ( now act as anode), +2, +4, PbSO4 (s) + 2H2O (l ) → PbO2 (s) + 4 H+ (aq) + SO4 2- (aq) + 2e-, , Reduction occurs at the anode (now act as cathode) PbSO 4 (s) + 2e- → Pb(s) + SO 4 2- (aq), Overall reaction, 2PbSO 4 (s) + 2H 2 O (l ) → Pb (s) + PbO 2 (s) + 4H + (aq) + 2SO 4 2- (aq)., , Thus, the overall cell reaction is exactly the, reverse of the redox reaction which takes place, while discharging ., Uses:, Used in automobiles, trains, inverters etc…, The lithium – ion Battery, Anode, : Porus graphite, Cathode, : transition metal oxide, such as CoO2., , Figure 9.12 lithium – ion Battery, , 58, , XII U9 Electro Chemistry.indd 58, , 2/19/2020 5:14:45 PM

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www.tntextbooks.in, , Electrolyte : Lithium salt in an organic solvent, At the anode oxidation occurs, Li (s) → Li+ (aq) + e-, , Positive electrode, , Negative electrode, , At the cathode reduction occurs, Li+ + CoO2 (s) + e- → Li CoO2 (s), Overall reactions, Li (s) + CoO2 → LiCoO2 (s), Both electrodes allow Li+ ions to, move in and out of their structures., , Charge, , Li, , Co, O, Li, , During discharge, the Li+ ions, produced at the anode move towards, cathode through the non – aqueous, electrolyte. When a potential greater than, the emf produced by the cell is applied, across the electrode, the cell reaction is, reversed and now the Li+ ions move from, cathode to anode where they become, embedded on the porous graphite, electrode. This is known as intercalation., , Li, Discharge, , LiCoO2, , Specialty carbon, , Figure 9.13 Li-ion battery, , Uses :, Used in cellular phones, laptops, computers, digital cameras, etc…, Fuel cell, The, galvanic, cell, in which the energy of, combustion of fuels is, directly converted into, electrical energy is called, the fuel cell. It requires, a continuous supply of, reactant to keep functioning., The general representation, of a fuel cell is follows, , Porous gas diffusion layer, Flow field Catalyst layer, , Membrane, , H2, , O2, , e, , e, H, , Cathode, , Anode, , Fuel | Electrode |, Electrolyte | Electrode |, Oxidant, , H 2O, , H2, , Let us understand the, function of fuel cell by, , 2H, , 2e, , 1, , /2O2, , 2H, , 2e, , Figure 9.14 H2O2 fuel cell, , H 2O, , 59, , XII U9 Electro Chemistry.indd 59, , 2/19/2020 5:14:51 PM

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www.tntextbooks.in, , considering hydrogen – oxygen fuel cell. In this case, hydrogen act as a fuel and oxygen as, an oxidant and the electrolyte is aqueous KOH maintained at 200oC and 20 – 40 atm. Porous, graphite electrode containing Ni and NiO serves as the inert electrodes., Hydrogen and oxygen gases are bubbled through the anode and cathode, respectively., Oxidation occurs at the anode:, 2H2 ( g ) + 4 OH− (aq) → 4 H2O (l ) + 4 e−, Reduction occurs at the cathode O2 (g) + 2 H2O (l) + 4e − → 4 OH− (aq), The overall reaction is 2H2 (g) + O2 (g) → 2H2O (l), The above reaction is the same as the hydrogen combustion reaction, however, they do, not react directly ie., the oxidation and reduction reactions take place separately at the anode, and cathode respectively like H2 - O2 fuel cell. Other fuel cells like propane –O2 and methane, O2 have also been developed., Corrosion, We are familiar with the rusting of iron. Have you ever noticed a green film formed on, copper and brass vessels?. In both, the metal is oxidised by oxygen in presence of moisture., This redox process which causes the deterioration of metal is called corrosion. As the corrosion, of iron causes damages to our buildings, bridges etc....it is important to know the chemistry of, rusting and how to prevent it. Rusting of iron is an electrochemical process., Electrochemical mechanism of corrosion, The formation of rust requires both oxygen and water. Since it is an electrochemical redox, process, it requires an anode and cathode in different places on the surface of iron. The iron, surface and a droplet of water on the surface as shown in figure (9.15) form a tiny galvanic, cell. The region enclosed by water is exposed to low amount of oxygen and it acts as the anode., The remaining area has high amount of oxygen and it acts as cathode. So based on the oxygen, content, an electro chemical cell is formed. corrosion occurs at the anode i,e,. in the region, enclosed by the water as discussed below., O2, , water, , Fe2+, Rust, , H2O, , Fe2O3• x H2O, iron, , anodic, site, , O2, cathodic, site, , e–, , Fe(s) → Fe2+(aq)+2e–, , O2(g) + 4H+(aq)+4e– → 2H2O(I), , Figure 9.15 Rusting of iron, 60, , XII U9 Electro Chemistry.indd 60, , 2/19/2020 5:14:57 PM

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www.tntextbooks.in, , At anode (oxidation): Iron dissolves in the anode region, 2Fe(s) → 2Fe2+ (aq)+4e-, , E = 0.44V., , The electrons move through the iron metal from the anode to the cathode area where the, oxygen dissolved in water, is reduced to water., At Cathode (reduction), The reaction of atmospheric carbon dioxide with water gives carbonic acid which furnishes, the H + ions for reduction., E = 1.23V, , O2 (g)+4H + (aq)+4e- → 2H 2 O (l ), , The electrical circuit is completed by the migration of ions through water droplet., The overall redox reactions is,, 2Fe(s)+O2 (g)+4H + (aq) → 2Fe2+ (aq) + 2H 2O(l ), , E = 0.444 + 1.23 = 1.67V, , The positive emf value indicates that the reaction is spontaneous., 3+, Fe2+ ions are further oxidised to Fe , which on further reaction with oxygen to form rust., 4Fe2+ (aq)+O2 (g)+4H + (aq) → 4Fe3+ (aq)+2H 2O(l), 2Fe3+ (aq)+4H 2O(l) → Fe2O3.H 2O(s) + 6H + (aq), , Other metals such as aluminium, copper and silver also undergo corrosion, but at a slower, rate than iron. For example, let us consider the reduction of aluminium,, Al(s) → Al3+ (aq)+3e−, Al3+ , which reacts with oxygen in air to forms a protective coating of Al2O3 . This coating act as, a protective film for the inner surface. So,further corrosion is prevented., Protection of metals form corrosion, This can be achieved by the following methods., i. Coating metal surface by paint., ii. Galvanizing - by coating with another metal such as zinc. zinc is stronger reducing agent, than iron and hence it can be more easily corroded than iron. i.e., instead of iron, the zinc is, oxidised., iii. Cathodic protection - In this technique, unlike galvanising the entire surface of the metal to, be protected need not be covered with a protecting metal. Instead, metals such as Mg or zinc, which is corroded more easily than iron can be used as a sacrificial anode and the iron material, acts as a cathode. So iron is protected, but Mg or Zn is corroded., Passivation - The metal is treated with strong oxidising agents such as concentrated HNO3 . As a, result, a protective oxide layer is formed on the surface of metal., Alloy formation - The oxidising tendency of iron can be reduced by forming its alloy with other more, anodic metals., Example, stainless steel - an alloy of Fe and Cr ., , 61, , XII U9 Electro Chemistry.indd 61, , 2/19/2020 5:15:11 PM

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www.tntextbooks.in, , Electrochemical series, , The standard reduction potential, , (E ) is a measure of the oxidising, tendency of the species. The greater the, E value, greater is the tendency shown, by the species to accept electrons and, undergo reduction. So higher the (E∞), Value, lesser is the tendency to undergo, corrosion, , Half Reaction, , Stronger oxidizing agent, , This series is called electrochemical, series., , Standard, Potential (V), , F2, Pb4+, , + 2e-, , 2F-, , +2.87, , Pb2+, , +1.67, , 2Cl, 2H2O, Ag, Fe2+, Cu, H2, Pb, , Al3+, , + 2e+ 2e+ 4e+ 1e+ 1e+ 2e+ 2e+ 2e+ 2e+ 2e+ 3e-, , +1.36, +1.23, +0.80, +0.77, +0.34, 0.00, -0.13, -0.44, -0.76, -1.66, , Mg2+, , + 2e-, , Li+, , + 1e-, , Cl2, O2+4H+, Ag+, Fe3+, Cu2+, 2H+, Pb2+, Fe2+, Zn2+, , -, , Fe, Zn, Al, Mg, Li, , Stronger reducing agent, , We have already learnt that the, standard single electrode potentials, are measured using standard hydrogen, electrode. The standard electrode, potential at 298K for various metal metal ion electrodes are arranged in, the decreasing order of their standard, reduction potential values as shown in, the figure., , -2.36, -3.05, , EVALUATION, Choose the correct answer:, 1. The number of electrons that have a total charge of 9650 coulombs is, a) 6.22 × 1023 b) 6.022 × 1024, c) 6.022 × 1022 c) 6.022 × 10−34, 2. Consider the following half cell reactions:, Mn2+ + 2e− → Mn E = -1.18V, Mn2+ → Mn3+ + e- E = -1.51V, The E for the reaction 3Mn2+ → Mn+2Mn3+ , and the possibility of the forward reaction, are respectively., a) 2.69V and spontaneous , b) -2.69 and non spontaneous, c) 0.33V and Spontaneous , d) 4.18V and non spontaneous, 3. The button cell used in watches function as follows, 2+, −, (aq) the half cell potentials, Zn (s) + Ag 2O (s) + H2O (l )−  2 Ag (s) + Zn − (aq) + 2OH, , 0.34V The cell potential will, are Ag 2O (s) + H2O (l ) + 2e → 2Ag (s) + 2 OH (aq) E =, be, , 62, , XII U9 Electro Chemistry.indd 62, , 2/19/2020 5:15:19 PM

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www.tntextbooks.in, , a) 0.84V, , b) 1.34V, , c) 1.10V, , d) 0.42V, , 4. The molar conductivity of a 0.5 mol dm-3 solution of AgNO3 with electrolytic conductivity, of 5.76 × 10−3 S cm −1 at 298 K is, 2, -1, a) 2.88 S cm mol b) 11.52 S cm 2 mol -1, , c) 0.086 S cm 2 mol -1 d) 28.8 S cm 2 mol-1, 5., Electrolyte, , KCl, , KNO3, , HCl, , NaOAC, , NaCl, , Λ–, , 149.9, , 145.0, , 426.2, , 91.0, , 126.5, , (S cm 2 mol −1 ), , Calculate Λ HOAC using appropriate molar conductances of the electrolytes listed above at, infinite dilution in water at 25oC ., a) 517.2, , b) 552.7, , c) 390.7, , d) 217.5, , 6. Faradays constant is defined as, a) charge carried by 1 electron, b) charge carried by one mole of electrons, c) charge required to deposit one mole of substance, d) charge carried by 6.22 × 1010 electrons., 7. H,  ow many faradays of electricity are required for the following reaction to occur, MnO-4 → Mn2+, a) 5F, , b) 3F, , c) 1F, , d) 7F, , 8. A current strength of 3.86 A was passed through molten Calcium oxide for 41minutes and, 40 seconds. The mass of Calcium in grams deposited at the cathode is (atomic mass of Ca, is 40g / mol and 1F = 96500C)., a) 4, , b) 2, , c) 8, , d) 6, , 9. During electrolysis of molten sodium chloride, the time required to produce 0.1mole of, chlorine gas using a current of 3A is, a) 55 minutes, , b) 107.2 minutes, , c) 220 minutes, , d) 330 minutes, , 10. The number of electrons delivered at the cathode during electrolysis by a current of 1A in, 60 seconds is (charge of electron = 1.6 × 10−19 C ), a) 6.22 × 1023, , b) 6.022 × 1020, , c) 3.75 × 1020, , d) 7.48 × 1023, , 11. Which of the following electrolytic solution has the least specific conductance, a) 2N, , b) 0.002N, , c) 0.02N, , d) 0.2N, , 12. While charging lead storage battery, a) PbSO 4 on cathode is reduced to Pb, , b) PbSO 4 on anode is oxidised to PbO2, 63, , XII U9 Electro Chemistry.indd 63, , 2/19/2020 5:15:33 PM

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www.tntextbooks.in, , c) PbSO4 on anode is reduced to Pb, , d) PbSO 4 on cathode is oxidised to Pb, , 13. Among the following cells, I) Leclanche cell, II) Nickel – Cadmium cell, III) Lead storage battery, IV) Mercury cell, Primary cells are, a) I and IV, , b) I and III, , c) III and IV, , d) II and III, , 14. Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is, because, a) Zinc is lighter than iron, b) Zinc has lower melting point than iron, c) Zinc has lower negative electrode potential than iron, d) Zinc has higher negative electrode potential than iron, 15. Assertion : pure iron when heated in dry air is converted with a layer of rust., Reason : Rust has the composition Fe3O4, a) if both assertion and reason are true and reason is the correct explanation of assertion., b) if both assertion and reason are true but reason is not the correct explanation of, assertion., c) assertion is true but reason is false, d) both assertion and reason are false., 16. In H2 -O2 fuel cell the reaction occurs at cathode is, a) O2 (g) + 2H2O (l ) + 4e − → 4OH− (aq), b) H+ (aq) + OH − (aq) → H2O (l), c) 2H2 (g) + O2 (g) → 2H2O (g), d) H + + e− → 1 H 2, 2, , 17. The equivalent conductance of M solution of a weak monobasic acid is 6 mho cm2, 36, equivalent –1 and at infinite dilution is 400 mho cm2 equivalent –1. The dissociation constant of, this acid is, a) 1.25 × 10−6, , c) 1.25 × 10−4, , b) 6.25 × 10−6, , d) 6.25 × 10−5, , 18. A conductivity cell has been calibrated with a 0.01M, 1:1 electrolytic solution (specific, conductance ( κ =1.25 × 10−3 S cm−1 ) in the cell and the measured resistance was 800 W at, 25 C . The cell constant is,, a) 10−1 c m −1   b) 101 c m−1   c) 1 c m −1   d) 5.7 × 10−12, 64, , XII U9 Electro Chemistry.indd 64, , 2/19/2020 5:15:42 PM

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www.tntextbooks.in, , 19. Conductivity of a saturated solution of a sparingly soluble salt AB (1:1 electrolyte) at 298K is, 1.85 × 10−5 S m −1 . Solubility product of the salt AB at 298K ( Λ m ) AB = 14 × 10−3 S m 2 mol −1 ., a) 5.7 × 10−12   b) 1.32 × 10−12   c) 7.5 × 10−12   d) 1.74 × 10−12, 20. In the electrochemical cell: Zn ZnSO4 (0.01M) CuSO4 (1.0M) Cu , the emf of this Daniel, cell is E1. When the concentration of, is changed to 1.0M and that CuSO4 changed, to 0.01M, the emf changes to E2. From the above, which one is the relationship between E1, and E2?, a) E1 < E2, , c) E2 E1, , b) E1 > E2, , d) E1 = E2, , 21. Consider the change in oxidation state of Bromine corresponding to different emf values, as shown in the diagram below:, BrO-4 1.82V, , → BrO-3 1.5V, → HBrO 1.595V, , → Br2 1.0652V, → Br Then the species undergoing disproportionation is, a) Br2, , b) BrO−4, , c) BrO-3, , d) HBrO, , 22. For the cell reaction, 2Fe3+ (aq) + 2l −(aq) → 2Fe2+ (aq) + l 2 (aq), Eocell = 0.24V at 298K. The standard Gibbs energy (∆, G∞) of the cell reactions is :, a) -46.32 KJ mol −1, , b) -23.16 KJ mol −1, , c) 46.32 KJ mol −1, , d) 23.16 KJ mol −1, , 23. A certain current liberated 0.504gm of hydrogen in 2 hours. How many grams of copper, can be liberated by the same current flowing for the same time through copper sulphate, solution, a) 31.75, , b) 15.8, , c) 7.5, , d) 63.5, , 24. A gas X at 1 atm is bubbled through a solution containing a mixture of 1MY- and 1MZ- at, 25oC . If the reduction potential of Z>Y>X, then, a) Y will oxidize X and not Z, , b) Y will oxidize Z and not X, , d) Y will oxidize both X and Z, , d) Y will reduce both X and Z, , 25. Cell equation : A + 2B- → A 2+ +2B;, A 2+ + 2e − → A E = + 0.34 V and log10 K = 15.6 at 300K for cell reactions find E for, B+ + e − → B (AIIMS – 2018), a) 0.80, , b) 1.26, , c) -0.54, , d) -10.94, , Short Answer Questions, 1. Define anode and cathode, 2. Why does conductivity of a solution decrease on dilution of the solution, 3. State Kohlrausch Law. How is it useful to determine the molar conductivity of weak, electrolyte at infinite dilution., 65, , XII U9 Electro Chemistry.indd 65, , 2/19/2020 5:16:07 PM

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www.tntextbooks.in, , 4. Describe the electrolysis of molten NaCl using inert electrodes, 5. State Faraday’s Laws of electrolysis, 6. Describe the construction of Daniel cell. Write the cell reaction., 7. Why is anode in galvanic cell considered to be negative and cathode positive electrode?, 8. The conductivity of a 0.01M solution of a 1 :1 weak electrolyte at 298K is 1.5 × 10-4 S cm −1 ., i) molar conductivity of the solution, ii) degree of dissociation and the dissociation constant of the weak electrolyte, Given that, λcation = 248.2 S cm 2 mol −1, λanion = 51.8 S cm 2 mol −1, , 0, , 9. Which of 0.1M HCl and 0.1 M KCl do you expect to have greater Λ m and why?, 10. Arrange the following solutions in the decreasing order of specific conductance., i) 0.01M KCl, , ii) 0.005M KCl, , iv) 0.25 M KCl, , v) 0.5 M KCl, , iii) 0.1M KCl, , 11. Why is AC current used instead of DC in measuring the electrolytic conductance?, 12. 0.1M NaCl solution is placed in two different cells having cell constant 0.5 and 0.25cm-1, respectively. Which of the two will have greater value of specific conductance., 13. A current of 1.608A is passed through 250 mL of 0.5M solution of copper sulphate for 50, minutes. Calculate the strength of Cu 2+ after electrolysis assuming volume to be constant, and the current efficiency is 100%., 14. Can Fe3+ oxidises bromide to bromine under standard conditions?, Given: EFe, EBr, , 2, , Br −, , 3+, , Fe2+, , = 0.771, , = 1.09V., , 15. Is it possible to store copper sulphate in an iron vessel for a long time?, Given : ECu, , 2+, , Cu, , = 0.34 V and EFe, , 2+, , Fe, , = −0.44 V ., , 16. Two metals M1 and M2 have reduction potential values of -xV and +yV respectively., Which will liberate H2 and H2SO4., 17. Reduction potential of two metals M1 and M2 are EM, , 2+, 1, , M1, , = − 2.3V and EM, , Predict which one is better for coating the surface of iron. Given : EFe, , 2+, 1, , 2+, , Fe, , M2, , = 0. 2 V, , = − 0.44 V, , 18. Calculate the standard emf of the cell: Cd Cd 2+ Cu 2+ Cu and determine the cell reaction., The standard reduction potentials of Cu 2+ Cu and Cd 2+ Cd are 0.34V and -0.40 volts, respectively. Predict the feasibility of the cell reaction., 66, , XII U9 Electro Chemistry.indd 66, , 2/19/2020 5:16:15 PM

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www.tntextbooks.in, , 19. In fuel cell H2 and O2 react to produce electricity. In the process, H2 gas is oxidised, at the anode and O2 at cathode. If 44.8 litre of H2 at 25oC and 1atm pressure reacts in, 10 minutes, what is average current produced? If the entire current is used for electro, deposited?, deposition of Cu from Cu 2+ , how many grams of, 20. The same amount of electricity was passed through two separate electrolytic cells containing, solutions of nickel nitrate and chromium nitrate respectively. If 2.935g of Ni was deposited, in the first cell. The amount of Cr deposited in the another cell? Give : molar mass of Nickel, and chromium are 58.74 and 52gm-1 respectively., 21. A copper electrode is dipped in 0.1M copper sulphate solution at 25oC . Calculate the, electrode potential of copper. [Given: E Cu Cu = 0.34 V ]., 2+, , 22. For the cell Mg (s) Mg 2+ (aq) Ag + (aq) Ag (s),, , calculate the equilibrium constant, , at 25oC and maximum work that can be obtained during operation of cell. Given :, EMg Mg = − 2.37 V and EAg Ag = 0.80V., 2+, , +, , 23. 8.2 × 1012 litres of water is available in a lake. A power reactor using the electrolysis of, water in the lake produces electricity at the rate of 2 × 106 Cs −1 at an appropriate voltage., How many years would it like to completely electrolyse the water in the lake. Assume that, there is no loss of water except due to electrolysis., 24. Derive an expression for Nernst equation, 25. Write a note on sacrificial protection., 26. Explain the function of H2 - O2 fuel cell., 27. Ionic conductance at infinite dilution of Al 3+ and SO4 2- are 189 and 160 mho cm2 equiv-1., Calculate the equivalent and molar conductance of the electrolyte Al 2 (SO4 )3 at infinite, dilution., , 67, , XII U9 Electro Chemistry.indd 67, , 2/19/2020 5:16:23 PM

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www.tntextbooks.in, , ICT Corner, Simulating an Voltaic Cell, By using this tool you can, construct an electrochemical, cell with using Ag/Cu/Zn, electrodes and measure the, emf of the cell. You can also, learn how the concentra�on, affects the emf value of the, cell., , Please go to the URL h�ps://, pages.uoregon.edu/tgreenbo/, voltaicCellEMF.html, (or), Scan the QR codeon the right, side, , Step – 1, Open the Browser and type the URL given (or) Scan the QR Code. You will see the webpage as shown in, the figure., Step – 2, Choose the metal electrode and appropriate electrolytic solution for both cathode and anode following the, on screen instructions. Now switch on the volt meter by clicking the red power switch. Now you can see, the flow of electrons and the emf value on the screen., Step – 3, The above steps can be repeated by varying the concentrations of electrolytic solutions of cathode and, anode by selecting appropriate concentration from the list., , 68, , XII U9 Electro Chemistry.indd 68, , 2/19/2020 5:16:25 PM

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www.tntextbooks.in, , SURFACE, CHEMISTRY, , UNIT, , 10, , Learning Objectives, , Irving Langmuir, , After studying this unit the student will be, able to, ’ classify adsorption., , Irving Langmuir was an American, Chemist and Physicist. He was, , ’, , awarded Nobel Prize in the year, 1932 in Chemistry for his works, , ’, , in Surface Chemistry. He outlined, , ’, , the concentric theory of atomic, structure. He invented the gas filled, , ’, , incandescent lamp and hydrogen, welding technique. The Langmuir, , ’, , Laboratory, , ’, , for, , Atmospheric, , Research near Socorro, New Mexico, , ’, , was named in his honor. Langmuir, and Tonks discovered electron, , ’, , density waves in plasmas that are, , distinguish between absorption and, adsorption, explain Freundlich adsorption isotherm, understand, catalysis, and, characteristics of catalysts., , the, , explain the theories of catalysis and, enzyme catalysis., classify colloids., explain the methods of preparation and, purification of colloids., discuss the properties of colloidal, solution., explain the role of colloids and, emulsions in daily life., , now known as Langmuir waves., , 69, , XII U10-Surface Chemistry.indd 69, , 2/19/2020 5:11:50 PM

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www.tntextbooks.in, , INTRODUCTION, Surface chemistry is the branch of chemistry that deals with the processes occurring at, interfaces between phases for example, solid and liquid, solid and gas and liquid and liquid., This topic is of immense importance to our everyday life and to numerous industries, from, materials and paints to medicine and biotechnology. Surfaces play a key role in heterogeneous, catalysis, formation and stability of colloids and electrode reactions. Surfaces of solids are, inherently different from their bulk portion. The bonding between the atoms at the mere, surface is different from that in the bulk. Hydrogen that exists in the interstellar space are, formed on the surfaces of grains and dust particles. Mosquitoes and other small insects can, walk on the surface of water but they will drown into the water when soaps are added in, the neighbourhood. We are fascinated by the spherical shape of water droplets and mercury, droplets. We are also impressed by the non-sticky wings of butterfly and leaves of plants. Blue, colour of the sky and red colour of the sunset strongly attract us. In all the above only the surface, of matter is important. Many of creams, lotions and other personal care products are complex, emulsions. Food companies are interested in developing healthy, tasty and longlasting food, products. All these are based on the principles of colloids and surface chemistry. So, surface, Chemistry is an exciting topic to learn., , 10.1 Adsorption and Absorption, Solid surfaces have the ability to attract the contacting species due to free valency or, residual force on them., charcoal adsorbs ammonia, silica gel adsorbs water., charcoal adsorbs, For example: , colorants from sugar., These examples prove that adsorption is a surface phenomenon. In contrast to adsorption,, absorption is a bulk phenomenon i.e. the adsorbate molecules are distributed throughout the, adsorbent., ’’, , Adsorbent is the material on which adsorption takes place., , ’’, , Adsorbed substance is called an adsorbate., , ’’, , ’’, , ’’, ’’, , ’’, , The surface of separation of the two phases where the concentration of adsorbed molecule, is high is known as interface., In adsorption, if the concentration of a substance in the interface is high, then it is called, positive adsorption. If it is less, then it is called negative adsorption., The process of removing an adsorbed substance from the surface is called desorption., The gaseous molecules like He, Ne,O2 ,N 2 ,SO2 and NH3 and solutions of NaCl or KCl, can be adsorbed by suitable adsorbents. These are referred as adsorbates., Silica gel and metals like Ni,Cu, Pt, Ag and Pd and certain colloids can act as adsorbents., , 70, , XII U10-Surface Chemistry.indd 70, , 2/19/2020 5:11:51 PM

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www.tntextbooks.in, , Characteristics of adsorption, 1. Adsorption can occur in all interfacial surfaces i.e. the adsorption can occur in between, gas-solid, liquid solid, liquid-liquid, solid- solid and gas-liquid., 2. Adsorption is a spontaneous process and it is always accompanied by decrease in free, energy. When DG reaches zero, the equilibrium is attained., We know, DG = DH - T DS where DG is Change in Free energy., DH is Change in enthalpy and DS = Change in entropy., , 3. When molecules are adsorbed, there is always a decrease in randomness of the molecules., ie., ∆S<0, and T∆S is negative. Hence, adsorption is exothermic., Adsorption is a quick process whereas absorption is a slow process., M.C. Bain introduced a term ‘sorption’ to represent the simultaneous, adsorption and absorption. T. Graham used a term occlusion for sorption of, gases on metal surfaces., 10.1.1 Types of adsorption, Adsorption is classified as physical adsorption and chemical adsorption, depending on, the nature of forces acting between adsorbent and adsorbate. In chemical adsorption, gas, molecules are held to the surface by formation of chemical bonds. Since strong bond is formed,, nearly 400 KJ / mole is given out as heat of adsorption., Examples, • Adsorption of O2 on tungsten, Adsorption of H2 on nickel, Adsorption of ethyl, alcohol vapours on nickel., In physical adsorption, physical forces such as van der waals force of attraction, dipole, - dipole interaction, dispersion forces etc., exist between adsorbent and adsorbate. As these, forces are weak, heat of adsorption is low, hence physical adsorption occurs at low temperatures., Examples, (a) Adsorption of N 2 on mica., (b) Adsorption of gases on charcoal., The following table 10.1 illustrates the distinction between chemical and physical, adsorption., Table 10.1 Distinction between chemical and physical adsorption, Chemical adsorption or Chemisorption or Physical adsorption or van der waals, Activated adsorption, adsorption or Physisorption, 1. It is very slow, , 1. It is instantaneous, , 2. It is very specific depends on nature of 2. It is non-specific, adsorbent and adsorbate., 71, , XII U10-Surface Chemistry.indd 71, , 2/19/2020 5:11:58 PM

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www.tntextbooks.in, , 3. Chemical adsorption is fast with increase 3. In Physisorption, when pressure increases, pressure, it can not alter the amount., the extent of adsorption increases., 4. When, temperature, is, raised 4. Physisorption decreases with increase in, chemisorption first increases and then, temperature., decreases., 5. Chemisorption involves transfer of 5. No transfer of electrons, electrons between the adsorbent and, adsorbate., 6. Heat of adsorption is high i.e., from 40- 6. Heat of adsorption is low in the order of, 400kJ/mole., 40kJ/mole., 7. Monolayer of the adsorbate is formed., , 7. Multilayer of the adsorbate is formed on, the adsorbent., , 8. Adsorption occurs at fixed sites called 8. It occurs on all sides., active centres. It depends on surface area, 9. Chemisorption involves the formation 9. Activation energy is insignificant., of activated complex with appreciable, activation energy., 10.1.2 Factors affecting adsorption, The adsorption is well understood by considering the various factors affecting it., Qualitatively, the extent of surface adsorption depends on the following factors, (i), Nature of adsorbent, (ii), Nature of adsorbate, (iii) Pressure, (iv) Concentration at a given temperature., 1., , Surface area of adsorbent:, , As the adsorption is a surface phenomenon it depends on the surface area of adsorbent., i.e., higher the surface area, higher is the amount adsorbed., 2., , Nature of adsorbate, , The nature of adsorbate can influence the adsorption. Gases like SO2 ,NH3 ,HCl and CO2, are easily liquefiable as they have greater van der waal’s force of attraction. On the other hand,, permanent gases like H 2 ,N 2 and O2 can not be liquefied easily. These permanent gases are, having low critical temperature and adsorbed slowly, while gases with high critical temperature, are adsorbed readily., , 72, , XII U10-Surface Chemistry.indd 72, , 2/19/2020 5:11:59 PM

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www.tntextbooks.in, , 3., , Effect of temperature, , When temperature is raised chemisorption first increases and then decreases. whereas, physisorption decreases with increase in temperature., 4., , Effect of pressure:, , chemical adsorption is fast with increase in pressure, it can not alter the amount of, adsorption. In Physisorption the extent of adsorption increases with increase in pressure., 10.1.3 Adsorption isotherms and isobars., Adsorption isotherms represents the variation of adsorption at constant temperature., When amount of adsorption is plotted versus temperature at constant pressure it is called, adsorption isobar., Adsorption isobars of physisorption and chemisorption are different as represented in the, graphs., Figure 10.1 (a) Physical Adsorption, , x, , Figure 10.1 (b) Chemical Adsorption, , x, , m, , Temp, , m, , Temp, , x is the amount of adsorbate adsorbed on ‘m’ g of adsorbent., x, In physical adsorption,, decreases with increase in Temprature, But in chemical, m, x, adsorption,, increases with rise in temperature and then decreases. The increase illustrates, m, the requirement of activation of the surface for adsorption is due to fact that formation of, activated complex requires certain energy., The decrease at high temperature is due to desorption, as the kinetic energy of the, adsorbate increases., 10.1.3.1 Adsorption isotherms, Adsorption isotherm can be studied quantitatively. A plot between the amount of, adsorbate adsorbed and pressure (or concentration of adsorbate) at constant temperature is, called adsorption isotherms., In order to explain these isotherms various equations were suggested as follows:, 73, , XII U10-Surface Chemistry.indd 73, , 2/19/2020 5:12:02 PM

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www.tntextbooks.in, , (i) Freundlich adsorption isotherm., According to Freundlich,, 1, x, = kp n, m, , where x is the amount of adsorbate, adsorbed on ‘m’ gm of adsorbent at a pressure of p. K, and n are constant introduced by freundlich., Value n is always less than unity., This equation is applicable for adsorption of gases on solid surfaces. The same equation, 1, x, n, becomes =K c , when used for adsorption in solutions with c as concentration., m, This equation quantitively predicts the effect of pressure(or concentration) on the, adsorption of gases(or adsorbates) at constant temperature., 1, Taking log on both sides of equation x =Kp n, , m, , log, x, 1, m p, , x, 0, m p, log (x/m), , x/m, , x, 1, = logK + log p, m, n, , x, 1/n, m p, p, , 1, slope = n, , log p, , Freundhich adsorption, isothermi plot of x/m against p, , Figure 10.2 log, , Plot of log x/m against log p for, the adsorption of a gason a solid, , x, vs log p graph, m, , Hence the intercept represents the value of log k and the slope gives, , 1, ., n, , x, This equation explains the increase of, with increase in pressure. But experimental, m, values show the deviation at low pressure., Limitations, This equation is purely empirical and valid over a limited pressure range., The values of constants k and n also found vary with temperature. No theoretical, explanations were given., 74, , XII U10-Surface Chemistry.indd 74, , 2/19/2020 5:12:08 PM

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www.tntextbooks.in, , 10.1.4 Applications of adsorption, Though we have innumerable applications for adsorption, we consider few of them, 1. Gas masks: During world war I charcoal gas mask was employed by both the British and, American. Activated charcoal was found to be one of the best adsorbents., 2. To create high vacuum in vessels, Tail and Dewar used activated charcoal.For dehydration, and also purification of gases like CO2, N2, Cl2, O2 and He, alumina and silica are employed., In the blast furnace silica gel is also used for drying air., , 3. One of the highly important use of adsorption is the softening of hardwater. Permutit is, employed for this process which adsorbs Ca2+ and Mg2+ ions in its surface, there is an ion, exchange as shown below it occurs on the surface., Na 2 Al 2Si 4O12 +CaCl 2 → CaAl 2Si 4O12 + 2NaCl, , Exhausted permutit is regenerated by adding a solution of common salt., CaAl 2Si 4O12 +2NaCl → Na 2 Al 2Si 4O12 +CaCl 2, , 4., , Ion exchange resins, , Ion exchange resins are working only based on the process of adsorption. Ion exchange, resins are used to demineralise water. This process is carried out by passing water through two, columns of cation and anion exchange resins., 2 RSO3H+Ca 2+ (Mg 2+ ) → (RSO3 )2Ca(Mg)+2H+, resin, , minerals, in water, , resin with, mineral, , R1, 2+, R2 N Cl ( SO4 ) + OH, R3 R, , R1, 2+, R2 N OH +Cl (SO4 ), R3 R resin, minerals, 4, , 5., , 4, resin with mineral, , in water, , Petroleum refining and refining of vegetable oil, Fuller’s earth and silica gel are used for refining process., , 6., , Decolourisation of Sugar:, , Sugar prepared from molasses is decolourised to remove coloured impurities by adding, animal charcoal which acts as decolourising material., 7., , Chromatography, , The chromatographic technique is applied for separation of components in a mixture. It is, mainly based on adsorption of components on the surface of adsorbents. This method is very, effective and used for identification, detection and estimation of many substances even if they, are contained in micro quantities., 75, , XII U10-Surface Chemistry.indd 75, , 2/19/2020 5:12:09 PM

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www.tntextbooks.in, , 8., , Catalysed reaction, , Catalysis is an important branch of surface chemistry which is based on the phenomenon, of adsorption of materials on the catalyst surface., Examples:, In the Haber’s process, ammonia is manufactured from N2 and H2 as shown by the, following reactions., N 2 +3H2 → 2NH3, , In this process, Fe is the catalyst and Mo is a promoter. The surface of the Fe catalyses the, reaction., In the hydrogenation of oils to obtain vanaspathi, Nickel is used as a catalyst. Nickel, surface catalyses the reaction., catalyst, vegetable oil+H Ni, → vanaspathi, 473K, 2, , 9., , Qualitative analysis, , When blue litmus solution is added to Al 3+ ion, a red coloration is seen due to the acidic, nature of the solution. Addition of ammonium hydroxide to it gives a blue lake. This is due to, the adsorption of blue colour litmus compound on the surface of Al (OH)3 Which is formed, during the addition of NH4OH, 10., , Medicine:, Drugs cure diseases by adsorption on body tissues., , Concentration of Ores of metals, Sulphides ores are concentrated by a process called froth flotation in which light ore, particles are wetted by pine oil., , 11., , Mordants and Dyes, Most of the dyes are adsorbed on the surface of the fabrics. Mordants are the substances, used for fixing dyes onto the fabric., , 12., , 13., , Adsorption indicators, , In the precipitation titrations, the end point is indicated by an external indicator which, changes its colour after getting adsorbed on precipitate. It is used to indicate the end point of, the titration., , 10.2 Catalysis, In 1836 Berzelius identified certain substances loosen the bond in the reacting molecules, and increased the rate of the reaction. But he also found these substances didn’t undergo any, change chemically. In order to indicate the property, he gave them the name catalyst. (In greek,, kata-wholly, lein-to loosen)., 76, , XII U10-Surface Chemistry.indd 76, , 2/19/2020 5:12:11 PM

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www.tntextbooks.in, , Later it was identified that there were many substances which retarded the speed of a, reaction., Hence a catalyst is defined as a substance which alters the rate of chemical reaction, without itself undergoing chemical change. The phenomenon which involves the action of a, catalyst is called catalysis., Positive and negative catalysis:, In positive catalysis, the rate of a reaction is increased by the presence of catalyst but in, negative catalysis, the rate of reaction is decreased by the presence of a catalyst., The two main types of catalysis (i) Homogeneous catalysis and (ii) Heterogeneous catalysis, Homogeneous catalysis, In a homogeneous catalysed reaction, the reactants, products and catalyst are present in, the same phase., Illustration (1):, 2SO2 +O2 +[NO]  2SO3 +[NO], (g), , (g), , (g), , (g), , (g), , In this reaction the catalyst NO, reactants, SO2 and O2 , and product, SO3 are present in, the gaseous form., Illustration (2):, In the decomposition of acetaldehyde by I2 catalyst, the reactants and products are all, present in the vapour phase., CH3CHO( g ) +[I2 ]( g ) → CH 4( g ) +CO( g ) +[I2 ]( g ), , Let us consider some examples in which the reactants, products and catalyst are present, in aqueous solution., (1), , Hydrolysis of cane sugar with a mineral acid as catalyst, H2SO4(l), → C 6 H12O6 + C 6 H12O6, C12 H22O11 +H2O(l), (l), (l), (l), Sucrose, Glucose Fructtose, , (2) Ester hydrolysis with acid or alkali as catalyst, , H2SO4(l), CH3COOC 2 H5 + H2O , → CH3COOH + C 2 H5OH, , Heterogeneous catalysis, , Ethyl acetate, , (l), , (l), , acetic acid, , (l), , ethyl alcohol (l), , In a reaction, the catalyst is present in a different phase i.e. it is not present in the same, phase as that of reactants or products. This is generally referred as contact catalysis and the, catalyst present is in the form of finely divided metal or as gauze, Illustration, (i) In the manufacture of sulphuric acid by contact process SO3 is prepared by the action of, SO2 and O2 in the presence of Pt or V2O5 as a catalyst., 77, , XII U10-Surface Chemistry.indd 77, , 2/19/2020 5:12:14 PM

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www.tntextbooks.in, , Pt(s) or V2O5(s), 2SO2( g ) +O2( g ) → 2SO3( g ), , ii) In the Haber’s process for the manufacture of ammonia, iron is used as a catalyst for the, reaction between Hydrogen and Nitrogen., Fe(s), N 2( g ) +3H2( g ) , → 2NH3( g ), , iii) Oxidation of ammonia is carried out in presence of platinum gauze, , 4NH3(g ) +5O2(g ) Pt→ 4NO(g ) +6H2O( g ), (s), , iv) The hydrogenation of unsaturated organic compounds is carried out using finely divided, nickel as a catalyst., Ni(s), , CH2 =CH2(g ) +H2(g )  , → CH3 -CH3(g ), , v) Decomposition of H2O2 occurs in the presence of the Pt catalyst, Pt(s) 2H2 O(l)+O2 (g), 2 H2 O2(l) , , vi) In the presence of anhydrous AlCl3, benzene reacts with ethanoyl chloride to produce, acetophenone, O, (l), , + CH3, , C, , O, anhydrous AlCl 3(s), , →, Cl(l)  , , C, , CH3 (l)+ HCl(l), , acetophenone, , 10.2.1 Characteristics of catalysts, 1. For a chemical reaction, catalyst is needed in very small quantity. Generally, a pinch of, catalyst is enough for a reaction in bulk., 2. There may be some physical changes, but the catalyst remains unchanged in mass and, chemical composition in a chemical reaction., 3. A catalyst itself cannot initiate a reaction. It means it can not start a reaction which is not, taking place. But, if the reaction is taking place in a slow rate it can increase its rate., 4. A solid catalyst will be more effective if it is taken in a finely divided form., 5. A catalyst can catalyse a particular type of reaction, hence they are said to be specific in, nature., 6. In an equilibrium reaction, presence of catalyst reduces the time for attainment of, equilibrium and hence it does not affect the position of equilibrium and the value of, equilibrium constant., 7. A catalyst is highly effective at a particular temperature called as optimum temperature., 8. Presence of a catalyst generally does not change the nature of products, For example. 2SO2 +O2 → 2SO3, This reaction is slow in the absence of a catalyst, but fast in the presence of Pt catalyst, Promoters and catalyst poison, 78, , XII U10-Surface Chemistry.indd 78, , 2/19/2020 5:12:21 PM

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www.tntextbooks.in, , In a catalysed reaction the presence of a certain substance increases the activity of a, catalyst. Such a substance is called a promoter., For example in the Haber’s process of manufacture of ammonia, the activity of the iron, catalyst is increased by the presence of molybdenum. Hence molybdenum is called a promoter., In the same way Al2O3 can also be used as a promoter to increase the activity of the iron, catalyst., On the other hand, certain substances when added to a catalysed reaction decreases or, completely destroys the activity of catalyst and they are often known as catalytic poisons., Few examples,, In the reaction, 2SO2 +O2 → 2SO3 with a Pt catalyst, the poison is As 2O3, i.e., As 2O3 destroys the activity of Pt . As 2O3 blocks the activity of the catalyst., So, the activity is lost., In the Haber’s process of the manufacture of ammonia, the Fe catalyst is poisoned by the, presence of H2S ., In the reaction, 2H2 +O2 → 2H2O ,, CO acts as a catalytic poison for Pt - catalyst, Auto catalysis, In certain reactions one of the products formed acts as a catalyst to the reaction. Initially, the rate of reaction will be very slow but with the increase in time the rate of reaction, increases., Auto catalysis is observed in the following reactions., CH3COOC 2 H5 +H2O → CH3COOH+C 2 H5OH, , Acetic acid acts as the autocatalyst, 2AsH3 → 2As+3H2, , Arsenic acts as an autocatalyst, Negative Catalysis, In certain reactions, presence of certain substances, decreases the rate of the reaction., Ethanol is a negative catalyst for the following reaction., (i) 4CHCl 3 +3O2 → 4COCl 2 +2H2O+2Cl 2, Ethanol decreases the rate of the reaction, (ii) 2H 2O 2 → 2H 2O+O 2, In the decomposition of hydrogen peroxide, dilute acid or glycerol acts as a negative, catalyst., 10.2.2 Theories of Catalysis, 79, , XII U10-Surface Chemistry.indd 79, , 2/19/2020 5:12:28 PM

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www.tntextbooks.in, , For a chemical reaction to occur, the reactants are to be activated to form the activated, complex. The energy required for the reactants to reach the activated complex is called, the activation energy. The activation energy can be decreased by increasing the reaction, temperature. In the presence of a catalyst, the reactants are activated at reduced temperatures, in otherwords, the activation energy is lowered. The catalyst adsorbs the reactants activates, them by weakening the bonds and allows them to react to form the products., As activation energy is lowered in presence of a catalyst, more molecules take part in the, reaction and hence the rate of the reaction increases., The action of catalysis in chemical reactions is explained mainly by two important theories., They are, (i), the intermediate compound formation theory, (ii), the adsorption theory., The intermediate compound formation theory, A catalyst acts by providing a new path with low energy of activation. In homogeneous, catalysed reactions a catalyst may combine with one or more reactant to form an intermediate, which reacts with other reactant or decompose to give products and the catalyst is regenerated., Consider the reactions:, A+B → AB , , (1), , A+C → AC (intermediate), , (2), , C is the catalyst, AC+B → AB+C , , (3), , Activation energies for the reactions (2) and (3) are lowered compared to that of (1). Hence, the formation and decomposition of the intermediate accelerate the rate of the reaction., Example 1, The mechanism of Fridel crafts reaction is given below, anhydrous, AlCl 3, C 6 H6 +CH3Cl →, C 6 H5CH3 +HCl, , The action of catalyst is explained as follows, , CH3Cl+AlCl 3 → [CH3 ]+[AlCl 4 ]-, , It is an intermediate., , C 6 H6 +[CH3+ ][AlCl 4 ]- → C 6 H5CH3 +AlCl 3 +HCl, , Example 2, Thermal decomposition of KClO3 in presence of MnO 2 proceeds as follows., Steps in the reaction 2KClO3 → 2KCl+3O 2 can be given as, 80, , XII U10-Surface Chemistry.indd 80, , 2/19/2020 5:12:33 PM

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www.tntextbooks.in, , 2KClO3 +6MnO2 → 6MnO3 +2KCl, , It is an intermediate., , 6MnO3 → 6MnO2 +3O2, , Example 3:, Formation of water due to the reaction of H2 and O2 in the presence of Cu proceeds as, +½O23 → 6MnO, H2O can, be2 given as, follows. Steps in the reaction H6MnO, 2 +3O, 2, 1, 2Cu+ O2 → Cu 2O, 2, , It is an intermediate., , Cu 2O+H2 → H2O+2Cu, , Example 4:, Oxidation of HCl by air in presence of CuCl2 proceeds as follows. Steps in the reaction, 6MnO32 → 2H, 6MnO, O+2Cl, can be given as, 4HCl+O, 2 +3O, 2, 22, 2CuCl 2 → Cl 2 +Cu 2Cl 2, , 2Cu 2Cl 2 +O2 → 2Cu 2OCl 2, , It is an intermediate., 2Cu 2OCl 2 +4HCl → 2H2O+4CuCl 2, , This theory describes, (i) the specificity of a catalyst and, (ii) the increase in the rate of the reaction with increase in the concentration of a catalyst., Limitations, (i) Th,  e intermediate compound theory fails to explain the action of catalytic poison and, activators (promoters)., (ii) This theory is unable to explain the mechanism of heterogeneous catalysed reactions., 2. Adsorption theory, Langmuir explained the action of catalyst in heterogeneous catalysed reactions based on, adsorption. The reactant molecules are adsorbed on the catalyst surfaces, so this can also be, called as contact catalysis., According to this theory, the reactants are adsorbed on the catalyst surface to form an, activated complex which subsequently decomposes and gives the product., The various steps involved in a heterogeneous catalysed reaction are given as follows:, 1. Reactant molecules diffuse from bulk to the catalyst surface., 2. The reactant molecules are adsorbed on the surface of the catalyst., 3. The adsorbed reactant molecules are activated and form activated complex which is, decomposed to form the products., 4. The product molecules are desorbed., 5. The product diffuse away from the surface of the catalyst., 81, , XII U10-Surface Chemistry.indd 81, , 2/19/2020 5:12:36 PM

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www.tntextbooks.in, , Ethylene, H, , Ni surface, H, , H, , H, , H, , H, H, , H, , H, , H, H, , H, , H, , H, , H, , H H, , H, , H, , H H, H, , H, , H, , H H, , H, , H, , H, , H, , H, , H, , Ethylene absorbed on, surface breaking  bonds, , Figure 10.3 Hydrogenation of ethylene in presence of a nickel catalyst., Active centres, The surface of a catalyst is not smooth. It bears steps, cracks and corners. Hence the atoms, on such locations of the surface are co-ordinatively unsaturated. So, they have much residual, force of attraction. Such sites are called active centres. So, the surface carries high surface free, energy., The presence of such active centres increases the rate of reaction by adsorbing and, activating the reactants., The adsorption theory explains the following, i. I ncrease in the surface area of metals and metal oxides by reducing the particle size, increases acting of the catalyst and hence the rate of the reaction., , Figure 10.4 Finely divided catalyst is more effective due to increase, in the number of active centres., ii. Th,  e action of catalytic poison occurs when the poison blocks the active centres of the, catalyst., iii. A promoter or activator increases the number of active centres on the surfaces., , 82, , XII U10-Surface Chemistry.indd 82, , 2/19/2020 5:12:38 PM

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www.tntextbooks.in, , 10.3 Enzyme Catalysis, , Enzymes are complex protein molecules with three dimensional structures. They catalyse, the chemical reaction in living organism. They are often present in colloidal state and extremely, specific in catalytic action. Each enzyme produced in a particular living cell can catalyse a, particular reaction in the cell., Some common examples for enzyme catalysis, 1) The peptide glycyl L-glutamyl L-tyrosin is hydrolysed by an enzyme called pepsin., 2) The enzyme diastase hydrolyses starch into maltose, 2(C 6 H10O5 )n +nH2O → nC12 H22O11, , 3) The yeast contains the enzyme zymase which converts glucose into ethanol., , C 6 H12O6 → 2C 2 H5OH+2CO2, , 4) The enzyme micoderma aceti oxidises alcohol into acetic acid., C 2 H5OH+O2 → CH3COOH+H2O, , 5) The enzyme urease present in soya beans hydrolyses the urea., NH2 -CO-NH2 +H2O → 2NH3 +CO2, , 10.3.1 Mechanism of enzyme catalysed reaction, The following mechanism is proposed for the enzyme catalysis, , E+S  ES → P+E, Where E is the enzyme, S the substrate (reactant), ES represents activated complex and, P the products., , Figure 10.5 Mechanism of Enzyme Catalysis, Enzyme catalysed reaction show certain general special characteristics., (i) Effective and efficient conversion is the special characteristic of enzyme catalysed reactions., An enzyme may transform a million molecules of reactant into product in a minute., For eg. 2H2O2 → 2H2O+O2, For this reaction, the activation energy is 18k cal/mole without a catalyst., With colloidal platinum as a catalyst the activation energy is 11.7kcal /mole, 83, , XII U10-Surface Chemistry.indd 83, , 2/19/2020 5:12:42 PM

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www.tntextbooks.in, , Rate of reaction, , But with the enzyme catalyst the activation energy of this reaction is less than 2kcal/mole., (ii) Enzyme catalysis is highly specific in nature., H 2 N-CO-NH 2 +H 2 O → 2NH 3 +HCO, 2 O2, The enzyme urease which catalyses the reaction of urea does not catalyse the following, reaction of methyl urea, H2 N-CO-NH-CH3 +H2O → No reaction, (3) Enzyme catalysed reaction has, Y, Optimum temperature is, maximum rate at optimum temperature., usually around 40 ºC, At first rate of reaction increases with, the increase of temperature, but above, a particular temperature the activity of, enzyme is destroyed. The rate may even, drop to zero. The temperature at which, enzymic activity is high or maximum is, called as optimum temperature., X, 0, , 10, , 20, , 30, , 40, , 50, , 60, , Rate of reaction, , For example:, Temperature ºC, • Enzymes involved in human body, At high temperatures enzymes denature,, rate falls rapidly, have an optimum temperature, Figure 10.6 Rate vs Temperature, 37∞C /98∞F, • During high fever, as body, temperature rises the enzymatic, activity may collapse and lead to, 100, danger., 80, 4. The rate of enzyme catalysed reactions, varies with the pH of the system., 60, The rate is maximum at a pH called, 40, optimum pH., 20, pHopt, 5. Enzymes can be inhibited i.e. poisoned., 0, Activity of an enzyme is decreased and, 7, 3, 4, 5, 6, pH, destroyed by a poison., Figure 10.7 Rate vs pH, The physiological action of drugs is, related to their inhibiting action., Example: Sulpha drugs. Penicillin inhibits the action of bacteria and used for curing, diseases like pneumonia, dysentery, cholera and other infectious diseases., 6. Catalytic activity of enzymes is increased by coenzymes or activators., A small non protein (vitamin) called a coenzyme promotes the catalytic activity of enzyme., , 10.4 Zeolite Catalysis:, , The details of heterogeneous catalysis will be incomplete, if zeolites are not discussed., Zeolites are microporous, crystalline, hydrated, alumino silicates, made of silicon and, aluminium tetrahedron. There are about 50 natural zeolites and 150 synthetic zeolites. As, 84, , XII U10-Surface Chemistry.indd 84, , 2/19/2020 5:12:45 PM

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www.tntextbooks.in, , silicon is tetravalent and aluminium is trivalent, the zeolite matrix carries extra negative charge., To balance the negative charge, there are extra framework cations for example, H + or Na +, ions. Zeolites carrying protons are used as solid acid catalysts and they are extensively used in, the petrochemical industry for cracking heavy hydrocarbon fractions into gasoline, diesel,etc.,, Zeolites carrying Na + ions are used as basic catalysts., One of the most important applications of zeolites is their shape selectivity. In zeolites,, the active sites namely protons are lying inside their pores. So, reactions occur only inside the, pores of zeolites., Reactant selectivity:, When bulkier molecules in a reactant mixture are prevented from reaching the active sites, within the zeolite crystal, this selectivity is called reactant shape selectivity., Transition state selectivity:, If the transition state of a reaction is large compared to the pore size of the zeolite, then, no product will be formed., Product selectivity:, It is encountered when certain product molecules are too big to diffuse out of the zeolite, pores., Phase Transfer catalysis:, Suppose the reactant of a reaction is present in one solvent and the other reactant is present, in an another solvent. The reaction between them is very slow, if the solvents are immiscible., As the solvents form separate phases, the reactants have to migrate across the boundary to, react. But migration of reactants across the boundary is not easy. For such situations a third, solvent is added which is miscible with both. So, the phase boundary is eliminated, reactants, freely mix and react fast. But for large scale production of any product, use of a third solvent is, not convenient as it may be expensive. For such problems phase transfer catalysis provides a, simple solution, which avoids the use of solvents. It directs the use a phase transfer catalyst (a, phase transfer reagent) to facilitate transport of a reactant in one solvent to the other solvent, where the second reactant is present. As the reactants are now brought together, they rapidly, react and form the product., Example:, Substitution of Cl - and CN– in the following reaction., R-Cl, + NaCN, → R-CN, + NaCl, organic phase aqueous phase organic phase aqu, ueous phase, , R-Cl=1-chlorooctane, R-CN=1-cyanooctane, By direct heating of two phase mixture of organic 1-chlorooctane with aqueous sodium, cyanide for several days, 1-cyanooctane is not obtained. However, if a small amount of, quaternary ammonium salt like tetraalkylammoniumchloride is added, a rapid transition, 85, , XII U10-Surface Chemistry.indd 85, , 2/19/2020 5:12:49 PM

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www.tntextbooks.in, , of 1-cyanooctane occurs in about 100% yield after 1 or 2 hours. In this reaction, the, tetraalkylammonium cation, which has hydrophobic and hydrophilic ends, transports CN from the aqueous phase to the organic phase using its hydrophilic end and facilitates the, reaction with 1-chloroocatne as shown below:, NaCN, + R 4 N +Cl R 4 N + CN - + NaCl, , , aqueous phase, R 4 N +CN -, , +, , It moves to organic phas e, , , R-Cl, , Both in organic phase, , R-CN, , +, , R 4 N+Cl -, , organic phase It moves to aqueous phase, releases Cl again pick s up CN - and transports it., , So phase transfer catalyst, speeds up the reaction by transporting one reactant from one, phase to another., Nano Catalysis:, Nano materials such a metallic nano particles, metal oxides, etc., are used as catalyst in, many chemical transformation, Nanocatalysts carry the advantages of both homogeneous and, heterogeneous catalyses. Like homogeneous catalysts, the nanocatalysts give 100% selective, transformations and excellent yield and show extremely high activity. Like the heterogeneous, catalysts, nanocatalysts can be recovered and recycled. Nanocatalysts are actually soluble, heterogeneous catalysts. An example for nanoparticles catalysed reaction is given below, Cl, Cl, , Cl, , (i), Cl, , Cl, , Fe0/Pd0, + 6 HCl, , H2O, , Cl, Lindane, , cyclohexane, , Fe0/Pd0 : Nanobimetallic catalyst (Zerovalent state), CH3, 10.5 Colloid, Dispersion phase and dispersion medium, CH, , CH OH, , 3, C=O, Origin of study of colloid, starts with Thomas Graham who observed, diffusion of that a, solution of sugar, urea or sodium chloride, a membrane but not glue, gelatine or gum., Co/Fethrough, 3O4 80°C, He called the(ii), former substances as crystalloids and the latter as colloids ( In Greek, kola as, tert-butylhydroperoxide, gum, eidos-like)., Later it was realised that any substance can be converted into a colloid by reducing its, particle size to 1-200nm., Co/Fe3O4: Magnetite supported Co metal nano catalyst, Hence, colloid is a homogeneous mixture of two substances in which one substance, (smaller proportion) is dispersed in another substance( large proportion)., In a colloid, the substance present, in larger, amount is called dispersing medium and the, Cu-NP, 5, (iii) in lessCH, substance present, amount, is called dispersed phase.HCOOH + H2, 2O, NaOh, 25°C, , 86, Cu-NP5 : Copper metal nano particles, XII U10-Surface Chemistry.indd 86, , 2/19/2020 5:12:50 PM

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www.tntextbooks.in, , 10.5.1 Classifications of Colloidal solution, Probably the most important colloidal systems have dispersed phase as solid and the, dispersion medium as liquid., If the dispersion medium considered is water, then the colloids are referred as hydrosols, or aquasols., If the dispersion medium is an alcohol, the colloid is termed as alcosol, and if benzene is, the dispersion medium, it is called as benzosol., One more type of classification is based on the forces acting between the dispersal phase, and dispersion medium., In lyophillic colloids definite attractive force or affinity exists between dispersion medium, and dispersed phase. Examples: sols of protein and starch. They are more stable and will not get, precipitated easily. They can be brought back to colloidal solution even after the precipitation, by addition of the dispersion medium., In a lyophobic colloids, no attractive force exists between the dispersed phase and, dispersion medium. They are less stable and precipitated readily, but can not be produced, again by just adding the dispersion medium. They themselves undergo coagulation after a, span of characteristic life time., They are called irreversible sols, examples: sols of gold, silver, platinum and copper., The following table lists the types of colloids based on the physical states of dispersed, phase and dispersion medium., Classification of colloids based on the physical state of dispersed phase and dispersion, medium., S.No., , Dispersion, medium, , Dispersed phase, , Name of the, colloid, , Examples, , 1., , Gas, , Liquid, , Liquid Aerosol, , Fog, Aerosol spray, , Solid Aerosol, , Smoke, Air, pollutants like, fumes, dust., , 2., , Gas, , Solid, , 3., , Liquid, , Gas, , Foam, , Whipped cream,, Shaving cream,, Soda water, Froth., , 4., , Liquid, , Liquid, , Emulsion, , Milk, Cream,, Mayonnaise, , 5., , Liquid, , Solid, , Sol, , Inks, Paints,, colloidal gold., , 87, , XII U10-Surface Chemistry.indd 87, , 2/19/2020 5:12:50 PM

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www.tntextbooks.in, , 6., , Solid, , Gas, , Solid foam, , Pumice stone, Foam, rubber bread., , 7., , Solid, , Liquid, , Gel, , Butter, cheese, , Solid sol, , Pearls, opals, coloured glass, alloys colloidal, dispersed eutetics., , 8., , Solid, , Solid, , 10.5.2 Preparation of Colloids, Many lyophillic substances are made in their colloidal form by warming with water., Rubber forms colloidal solution with benzene. Soap spontaneously forms a colloidal solution, by just mixing with water., In general, colloidal are prepared by the following methods., i. Dispersion methods: In this method larger particles are broken to colloidal dimension.,  ondensation method: In this method, smaller atom or molecules are converted into, ii. C, larger colloidal sized particles., 1) Dispersion methods, (i) Mechanical Dispersion:, Using a colloid mill, the solid is ground to colloidal dimension. The colloid mill, consists of two metal plates rotating in opposite direction at very high speed of nearly, 7000 revolution / minute., Coarse dispersion, , Colloid mill, , Figure 10.8 Colloid mill, The colloidal particles of required colloidal size is obtained by adjusting the distance, between two plates., By this method, colloidal solutions of ink and graphite are prepared., 88, , XII U10-Surface Chemistry.indd 88, , 2/19/2020 5:12:52 PM

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www.tntextbooks.in, , (ii) Electro Dispersion:, Metal electrodes, A brown colloidal solution, of platinum was first prepared by, –, +, George Bredig in 1898. An electrical, arc is struck between electrodes, dispersed in water surrounded by, Water, ice. When a current of 1 amp /100 V, Arc, is passed an arc produced forms, vapours of metal which immediately, Ice, condense to form colloidal solution., By this method colloidal solution of, many metals like copper, silver, gold,, platinum, etc. can be prepared Alkali, hydroxide is added as an stabilising, agent for the colloidal solution., Svedberg, modified, this, Figure 10.9 Bredig’s are method, method for the preparation of non, aqueous inflammable liquids like, pentane, ether and benzene, etc using high frequency alternating current which prevents the, decomposition of liquid., (iii), , Ultrasonic dispersion, Sound waves of frequency more than 20kHz (audible limit) could cause transformation of, coarse suspension to colloidal dimensions., , Water, Oil, Mercury, , Electrical oscillator, , Quartz generator, Figure : 10.10 Ultrasonic dispersion, , Claus obtained mercury sol by subjecting mercury to sufficiently high frequency ultrasonic, vibrations., 89, , XII U10-Surface Chemistry.indd 89, , 2/19/2020 5:13:08 PM

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www.tntextbooks.in, , The ultrasonic vibrations produced by generator spread the oil and transfer the vibration, to the vessel with mercury in water., (iv) Peptisation:, By addition of suitable electrolytes, precipitated particles can be brought into colloidal, state. The process is termed as peptisation and the electrolyte added is called peptising or, dispersing agent., AgCl HCl, , → AgCl, Precipitate, Colloid, 2) Condensation Methods:, When the substance for colloidal particle is present as small sized particle, molecule or, ion, they are brought to the colloidal dimension by condensation methods. Here care should, be taken to produce the particle with colloidal size otherwise precipitation will occur. Various, chemical methods for the formation of colloidal particles., (i) Oxidation:, Sols of some non metals are prepared by this method., (a) When hydroiodic acid is treated with iodic acid, I2 sol is obtained., HIO3 +5HI → 3H2O+I2 (Sol), , (b), , When O2 is passed through H2Se, a sol of selenium is obtained., H2Se+O2 → 2H2O+Se(sol), , (ii) Reduction:, Many organic reagents like phenyl hydrazine, formaldehyde, etc are used for the formation, of sols. For example: Gold sol is prepared by reduction of auric chloride using formaldehyde., 2AuCl 3 +3HCHO+3H2O → 2Au(sol)+6HCl+3HCOOH, , (iii) Hydrolysis, Sols of hydroxides of metals like chromium and aluminium can be produced by this, method., For Example,, , FeCl 3 +3H2O → Fe(OH)3 +3HCl, , (iv) Double decomposition, For the preparation of water insoluble sols this method can be used., When hydrogen sulphide gas is passed through a solution of arsenic oxide, a yellow, coloured arsenic sulphide is obtained as a colloidal solution., As 2O3 +3H2S → As 2S3 +3H2O, , (v) Decomposition, When few drops of an acid is added to a dilute solution of sodium thiosulphate, the, insoluble free sulphur produced by decomposition of sodium thiosulphate accumulates into, small, clusters which impart various colours blue, yellow and even red to the system depending, on their growth within the size of colloidal dimensions., S2O32- +2H+  S+H2O+SO2, sol, , 90, , XII U10-Surface Chemistry.indd 90, , 2/19/2020 5:13:14 PM

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www.tntextbooks.in, , 3) By exchange of solvent:, Colloidal solution of few substances like phosphorous or sulphur is obtained by preparing, the solutions in alcohol and pouring them into water. As they are insoluble in water, they form, colloidal solution., P in alcohol + water→Psol., 10.5.3 Purification of colloids, The colloidal solutions due to their different methods of preparation may contain impurities., If they are not removed, they may destablise and precipitate the colloidal solution. This is called, coagulation. Hence the impurities mainly electrolytes should be removed to increase the stabilisation, of colloid. Purification of colloidal solution can be done by the following methods., (i), Dialysis, (ii) Electrodialysis , (iii) Ultrafilteration., (i) Dialysis, In 1861, T. Graham separated the electrolyte from a colloid using a semipermeable, membrane (dialyser). In this method, the colloidal solution is taken in a bag made up of, semipermeable membrane. It is suspended in a trough of flowing water, the electrolytes diffuse, out of the membrane and they are carried away by water., Do you Know? Kidney malfunction results in the building up of electrolyte concentration, within the blood to toxic levels., In the Dialysis, recycling of patient’s blood is done through considerable length of, seimpermeable tube in an isotonic saline solution., ii) Electrodialysis, The presence of electric field increases the, speed of removal of electrolytes from colloidal, solution. The colloidal solution containing, an electrolyte as impurity is placed between, two dialysing membranes enclosed into two, compartments filled with water. When current, is passed, the impurities pass into water, compartment and get removed periodically., This process is faster than dialysis, as the rate, of diffusion of electrolytes is increased by the, application of electricity., , Cation-transfer, membrane, Cathode, , Anode, , The pores of ordinary filter papers permit, the passage of colloidal solutions. In ultra, filtrations, the membranes are made by using, collodion cellophane or visiking. When a, colloidal solution is filtered using such a filter,, , Pure water, , Solution, , Solution, , Cl–, , Na+, Na+, , –, –, –, –, –, , iii) Ultrafiltration, , Anion-transer, membrane, , Na, , Cl–, , Cl–, , Cl–, , Cl–, Na+, , Cl–, , –, , Na+, , Cl–, , Cl–, , Na+, , +, , Na, , +, , Cl–, , Na+, , Na+, , +, , Cl–, , +, , +, , Na, , Cl–, , +, +, , Na, , Cl–, Na+, , +, +, +, , Cl–, Na+, , Cl–, , Figure 10.11 Electro Dialysis, , 91, , XII U10-Surface Chemistry.indd 91, , 2/19/2020 5:13:16 PM

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www.tntextbooks.in, , colloidal particles are separated on the filter and the impurities are removed as washings., This process is quickened by application of pressure. The separation of sol particles from, electrolyte by filteration through an ultrafilter is called ultrafiltration. Collodion is 4% solution, of nitrocellulose in a mixture of alcohol and water., 10.5.4 Properties of Colloids, 1) Colour:, The colour of a sol is not always the same as the colour of the substance in the bulk., For example bluish tinge is given by diluted milk in reflected light and reddish tinge in, transmitted light., Colour of the sol, generally depends on the following factors., (i), Method of preparation, (ii), Wavelength of source of light., (iii) Size and shape of colloidal particle, (iv) whether the observer views the reflected light or transmitted light., 2) Size:, The size of colloidal particles ranges from 1nm (10-9m) to 1000 nm (10-6m) diameter., 3) Colloidal solutions are heterogeneous in nature having two distinct phases., Though experiments like dialysis, ultrafiltration and ultracentrifuging clearly show the, heterogeneous nature in the recent times colloidal solution are considered as border line, cases., 4) Filtrability:, As the size of pores in ordinary filter paper are large the colloidal particles easily pass, through the ordinary filter papers., 5) Non-Setting nature, Colloidal solutions are quite stable i.e. they are not affected by gravity., 6), , Concentration and density, When the colloidal solution is dilute, it is stable. When the volume of medium is decreased, coagulation occurs. Generally, density of sol decreases with decrease in the concentration., , 7), , Diffusability, Unlike true solution, colloids diffuse less readily through membranes., , 8), , Colligative properties, The colloidal solutions show colligative properties i.e. elevation of boiling point, depression, in freezing point and osmotic pressure. Measurements of osmotic pressure is used to find, molecular weight of colloidal particle., , 9), , Shape of colloidal particles, It is very interesting to know the various shapes of colloidal particles. Here are some, examples, 92, , XII U10-Surface Chemistry.indd 92, , 2/19/2020 5:13:16 PM

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www.tntextbooks.in, , Colloidal Particles, , Shapes, , As 2S3, , Spherical, , Fe (OH)3sol (blue gold sol), , Disc or plate like, , W3O5 sol (tungstic acid sol), , Rod like, , 10) Optical property, Colloids have optical property. When a homogeneous solution is seen in the direction of, light, it appears clear but it appears dark, in a perpendicular direction., , Figure 10.12 Tyndall effect, But when light passes through colloidal solution, it is scattered in all directions. This effect, was first observed by Faraday, but investigations are made by Tyndall in detail, hence called as, Tyndall effect., The colloidal particles absorb a portion of light and the remaining portion is scattered, from the surface of the colloid. Hence the path of light is made clear., 11), , Kinetic property, , Robert Brown observed that when the pollen grains suspended in water were viewed, through ultra microscope, they showed a random, zigzag ceaseless motion., This is called Brownian movement of colloidal particles., This can be explained as follows, The colloidal sol particles are continuously bombarded with the molecules of the, dispersion medium and hence they follow a zigzag, random, continuous movement., 93, , XII U10-Surface Chemistry.indd 93, , 2/19/2020 5:13:21 PM

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www.tntextbooks.in, , Brownian movement enables us,, I. to calculate Avogadro, number., II. to confirm kinetic theory, which considers the ceaseless, rapid movement of molecules, that increases with increase in, temperature., III. to understand the stability of, colloids: As the particles in, continuous rapid movement, they do not come close and, hence not get condensed. That, is Brownian movement does, not allow the particles to be, acted on by force of gravity., 12) Electrical property, (1) Helmholtz double layer, The surface of colloidal, particle adsorbs one type of ion, due to preferential adsorption., This layer attracts the oppositely, charged ions in the medium, and hence at the boundary, separating the two electrical, double layers are setup. This is, called as Helmholtz electrical, double layer., , Figure 10. 13 Brownian movement, , +, , –, , +, +, , +, , –, , +, , + –, + + + +, +, +, + +, +, +, + +, + +, Negative charged + –, – +, +, particle, +, +, +, +, +, +, +, +, +, +, +, – +, + + +, –, –, +, , +, , +, , +, , Stern layer, Slipping plane, , +, , ζ Potential, , Figure 10.14 Electrical double layer, , As the particles nearby are having similar charges, they cannot come close and condense., Hence this helps to explain the stability of a colloid., (ii) Electrophoresis:, When electric potential is applied across two platinum electrodes dipped in a, hydrophilic sol, the dispersed particles move toward one or other electrode., This migration of sol particles under the influence of electric field is called electrophoresis, or cataphoresis. If the sol particles migrate to the cathode, then they posses positive (+) charges,, 94, , XII U10-Surface Chemistry.indd 94, , 2/19/2020 5:13:27 PM

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www.tntextbooks.in, , and if the sol particles migrate, to the anode then they, have negative charges(-)., Thus from the direction of, migration of sol particles we, can determine the charge, of the sol particles. Hence, electrophoresis is used for, detection of presence of, charges on the sol particles., , Electrophoresis of a sol, , +, , –, , +, , –, , Platinum, electrode, , Deionized, water, , Sol, , Before, , After, , Figure 10.15 Electrophoresis, , Few examples of charges of sols detected by electrophoresis are given below:, Positively charge colloids, , Negatively charge colloids, , Ferric hydroxide, , Ag, Au & Pt, , Aluminium hydroxide, , Arsenic sulphide, , Basic dyes, , Clay, , Haemoglobin, , Starch, , (iii) Electro osmosis, , Electro-osmosis, , Original level, of water, , Original level, of water, Water, , Sol, , Water, , +, , –, , Semipermeable, membrane, , Figure 10.16 Electro osmosis, A sol is electrically neutral. Hence the medium carries an equal but opposite charge to that of, dispersed particles. When sol particles are prevented from moving, under the influence of electric, field the medium moves in a direction opposite to that of the sol particles. This movement of, dispersion medium under the influence of electric potential is called electro osmosis., 95, , XII U10-Surface Chemistry.indd 95, , 2/19/2020 5:13:31 PM

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www.tntextbooks.in, , 13. Coagulation or precipitation, The flocculation and settling down of the sol particles is called coagulation., Various method of coagulation are given below:, (i), Addition of electrolytes, (ii), Electrophoresis, (iii) Mixing oppositively charged sols., (iv) Boiling, (1) Addition of electrolytes, A negative ion causes the precipitation of positively charged sol and vice versa., When the valency of ion is high, the precipitation power is increased. For example, the, precipitation power of some cations and anions varies in the following order, Al 3+ >Ba 2+ >Na + , Similarly [Fe(CN)6 ] > SO4 2− > Cl 3−, , The precipitation power of electrolyte is determined by finding the minimum, concentration (millimoles/lit) required to cause precipitation of a sol in 2hours. This value is, called flocculation value. The smaller the flocculation value greater will be precipitation., (ii), , Electrophoresis:, , In the electrophoresis, charged particles migrate to the electrode of opposite sign. It is, due to neutralization of the charge of the colloids. The particles are discharged and so they get, precipitated., (iii), , By mixing two oppositively charged sols, , When colloidal sols with opposite charges are mixed mutual coagulation takes place. It is, due to migration of ions from the surface of the particles., (iv) By boiling, When boiled due to increased collisions, the sol particles combine and settle down., 14. Protective action, Generally, lyophobic sols are precipitated, readily even with small amount of electrolytes. But, they are stabilised by addition of a small amount of, lyophillic colloid., A small amount of gelatine sol is added to gold, sol to protect the gold sol., , Colloid, , Gold number, , Gelatin, , 0.005-0.01, , Egg albumin, , 0.08-0.10, , Gum Arabic, , 0.1-0.15, , Potato starch, , 25, , Zsigmondy introduced the term ‘gold number’ as a measure of protecting power of a, colloid. Gold number is defined as the number of milligrams of hydrophilic colloid that will, just prevent the precipitation of 10ml of gold sol on the addition of 1ml of 10% NaCl solution., Smaller the gold number greater the protective power., 96, , XII U10-Surface Chemistry.indd 96, , 2/19/2020 5:13:32 PM

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www.tntextbooks.in, , 10.6 Emulsions, Emulsions are colloidal solution in which a liquid is dispersed in an another liquid., Generally there are two types of emulsions., (i) Oil in water (O/W) , (ii) Water in oil (W/O), Example:, Milk is example of the oil in water type emulsion., Stiff greases are emulsion of water in oil i.e. water dispersed in lubricating oil., The process of preparation of emulsion by the dispersal of one liquid in another liquid is, called Emulsification., A colloid mill can be used as a homogeniser to mix the two liquid. To have a stable, emulsion a small amount of emulsifier or emulsification agent is added., Several types of emulsifiers are known., i. Most of the lyophillic colloids also act as emulsifiers. Example: glue, gelatine., ii. Long chain compounds with polar groups like soap and sulphonic acids., iii. Insoluble powders like clay and lamp black also act as emulsifiers., Identification of types of emulsion, The two types of emulsions can be identified by the following tests., (i) Dye test:, A small amount of dye soluble in oil is added to the emulsion. The emulsion is shaken, well. The aqueous emulsion will not take the colour whereas oily emulsion will take up the, colour of the dye., (ii)Viscosity test, Viscosity of the emulsion is determined by experiments. Oily emulsions will have higher, value than aqueous emulsion., (iii) Conductivity test, Conductivity of aqueous emulsions are always higher than oily emulsions., (iv) Spreading test, Oily emulsions spread readily than aqueous emulsion when spread on an oily surface., 10.5.1 Deemulsification:, Emulsion can be separated into two separate layers. The process is called Deemulsification., Various deemulsification techniques are given below, 1. Distilling of one component, 2. Adding an electrolyte to destroy the charge., 3. Destroying the emulsifier using chemical methods., 97, , XII U10-Surface Chemistry.indd 97, , 2/19/2020 5:13:32 PM

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www.tntextbooks.in, , 4., 5., 6., 7., 8., 9., , Using solvent extraction to remove one component., By freezing one of the components., By applying centrifugal force., Adding dehydrating agents for water in oil (W/O) type., Using ultrasonic waves., Heating at high pressures., , Inversion of Phase:, The change of W/O emulsion into O/W emulsion is called inversion of phases., For example:, An oil in water emulsion containing potassium soap as emulsifying agent can be converted, into water in oil emulsion by adding CaCl2 or AlCl3. The mechanism of inversion is in the, recent developments of research., , 10.7 Various application of colloids, In every path of life, colloids play a great role. Human body contains the numerous, colloidal solutions. The blood in our body, protoplasma of plant and animal cell, and fats in, our intestines are in the form of emulsions. Synthetic polymers like polystyrene silicones and, PVC are colloids., Food, Food stuffs like milk cream, butter, etc are present in colloidal form., Medicines, Antibodies such as penicillin and streptomycin are produced in colloidal form for suitable, injections. Colloidal gold and colloidal calcium are used as tonics. Milk of magnesia is used, for stomach troubles. Silver sol protected by gelatine known as Argyrol is used as eye lotion., In Industry, Colloids find many applications in industries., (i) Water purification:, Purification of drinking water is activated by coagulation of suspended impurities in, water using alums containing Al3+, (ii) In washing:, The cleansing action of soap is due to the formation of emulsion of soap molecules with, dirt and grease., (iii) Tanning of leather, Skin and hides are protein containing positively charged particles which are coagulated, by adding tannin to give hardened leather for further application. Chromium salts are used for, the purpose. Chrome tanning can produce soft and polishable leather., , 98, , XII U10-Surface Chemistry.indd 98, , 2/19/2020 5:13:32 PM

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www.tntextbooks.in, , (iv) Rubber industry:, Latex is the emulsion of natural rubber with negative particles. By heating rubber with, sulphur, vulcanized rubbers are produced for tyres, tubes, etc., (v) Sewage disposal, Sewage contains dirt, mud and wastes dispersed in water. The passage of electric current, deposits the wastes materials which can be used as a manure., Primary treatment, , Secondary treatment, Disinfectant, , Aeration tank, , Raw sewage, , Secondary clarifier, , Screens, , Primary, effluent, , Secondary, effluent, , Comminutor, Primary clarifier, , Air compressor, Return sludge, , Activated, sludge, , Discharge, to surface water or, tertiary treatment, , Return sludge pump, Grit chamber, , Raw of primary sludge, , Grit disposal, Sludge treatment and disposal, , Figure 10.17 Sewage disposal, Vi ) Cortrell’s precipitator, Carbon dust in air is solidified by cortrell’s precipitator. In it, a high potential difference of, about 50,000V is used. The charge on carbon is neutralized and solidified. Thus the air is free, from carbon particles., Vii) The blue colour of the sky in nature is due to Tyndall effect of air particles., Viii) Formation of delta:, The electrolyte in sea and river water coagulates the solid particles in river water at their, intersection. So, the earth becomes a fertile land., Ix) Analytical application, Qualitative and quantitative analysis are based on the various properties of colloids., Hence we can conclude that in our life, there is hardly any field which is not including the, applications of colloids., 99, , XII U10-Surface Chemistry.indd 99, , 2/19/2020 5:13:34 PM

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www.tntextbooks.in, , Natural honey is a colloidal sol. It is distinguished from artificial one by, adding ammoniacal AgNO3, In case of natural honey a metallic silver is produced, assumes a reddish, yellow color due to traces of albumin or ethereal oil which acts as a protective, colloid. In case of artificial honey a dark yellow or greenish yellow precipitate, is formed., , EVALUATION, Choose the correct answer:, x, is plotted against log p. The slope of the line and, m, its y – axis intercept respectively corresponds to, , 1. For Freundlich isotherm a graph of log, , b) log 1 , k, c) 1 , log k, a) 1 , k, n, n, n, 2. Which of the following is incorrect for physisorption?, , d) log 1 , log k, n, , a) reversible , , b) increases with increase in temperature, , c) low heat of adsorption, , d) increases with increase in surface area, , 3. Which one of the following characteristics are associated with adsorption? (NEET), a) DG and DH are negative but DS is positive, b) DG and ∆S are negative but ∆H is positive, c) ∆G is negative but ∆H and ∆S are positive, , d)∆G, ∆H and ∆S all are negative., , 4. Fog is colloidal solution of, a) solid in gas, , b) gas in gas, , c) liquid in gas, , d) gas in liquid, , 5. Assertion : Coagulation power of Al 3+ is more than Na ., +, , Reason : greater the valency of the flocculating ion added, greater is its power to cause, precipitation, a) if both assertion and reason are true and reason is the correct explanation of assertion., b) if both assertion and reason are true but reason is not the correct explanation of, assertion., c) assertion is true but reason is false, , d) both assertion and reason are false., , 6. Statement :, To stop bleeding from an injury, ferric chloride can be applied. Which comment about, the statement is justified?, 100, , XII U10-Surface Chemistry.indd 100, , 2/19/2020 5:13:44 PM

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www.tntextbooks.in, , a) It is not true, ferric chloride is a poison., b) It is true, Fe3+ ions coagulate blood which is a negatively charged sol, c) It is not true; ferric chloride is ionic and gets into the blood stream., d) It is true, coagulation takes place because of formation of negatively charged sol with Cl-., 7. Hair cream is, a) gel, , b) emulsion, , c) solid sol, , d) sol., , 8. Which one of the following is correctly matched?, a) Emulsion, , –, , Smoke, , b) Gel, , –, , butter, , c) foam, , –, , Mist, , d) whipped cream, , –, , sol, , 9. The most effective electrolyte for the coagulation of As 2S3Sol is, a) NaCl, , b) Ba(NO3 )2, , c) K 3[Fe(CN)6 ] , , d) Al2(SO4)3, , 10. Which one of the is not a surfactant?, a) CH3, b) CH3, , (CH2)15, (CH2)15, , N (CH3)2 CH2Br, NH2, , c) CH3 ( CH2 )16CH2 OSO2 Na+, d), , OHC, , (CH2)14, , CH2, , COO, , Na+, , 11. The phenomenon observed when a beam of light is passed through a colloidal solution is, a) Cataphoresis    b) Electrophoresis    c) Coagulation   d) Tyndall effect, 12. In an electrical field, the particles of a colloidal system move towards cathode. The, coagulation of the same sol is studied using K 2SO4 (i), Na 3PO4 (ii),K 4[Fe(CN)6 ] (iii), and NaCl (iv) Their coagulating power should be, a) II > I>IV > III, b) III > II > I > IV, c) I > II > III > IV, d) none of these, 13. Collodion is a 4% solution of which one of the following compounds in alcohol – ether, mixture?, a) Nitroglycerine, b) Cellulose acetate c) Glycoldinitrate d) Nitrocellulose, 14. Which one of the following is an example for homogeneous catalysis?, a) manufacture of ammonia by Haber’s process, b) manufacture of sulphuric acid by contact process, c) hydrogenation of oil, d) Hydrolysis of sucrose in presence of dil HCl, , 101, , XII U10-Surface Chemistry.indd 101, , 2/19/2020 5:13:48 PM

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www.tntextbooks.in, , 15. Match the following, A) V2O5, , i) High density polyethylene, , B) Ziegler – Natta, , ii) PAN, , C) Peroxide, , iii) NH3, , D) Finely divided Fe, , iv) H2SO4, , A, B, C, D, a) (iv), (i), (ii), (iii), b) (i), (ii), (iv), (iii), c) (ii), (iii), (iv), (i), d) (iii), (iv), (ii), (i), 16. The coagulation values in millimoles per litre of the electrolytes used for the coagulation of, As 2S3 are given below, (I) (NaCl)=52, , (II) ((BaCl 2 )=0.69, , (III) (MgSO4 )=0.22, , The correct order of their coagulating power is, a) III > II > I, , b) I > II > III, , c) I > III > II, , d) II > III>I, , 17. Adsorption of a gas on solid metal surface is spontaneous and exothermic, then, a) ∆H increases, , b) ∆S increases, , c) ∆G increases, , d) ∆S decreases, , 18. If x is the amount of adsorbate and m is the amount of adsorbent, which of the following, relations is not related to adsorption process?, b) x =f(T) at constant P, a) x =f(P) at constant T, m, m, c) P = f(T) at constant x, d) x = PT, m, m, 19. On which of the following properties does the coagulating power of an ion depend ?, (NEET – 2018), a) Both magnitude and sign of the charge on the ion., b) Size of the ion alone, c) the magnitude of the charge on the ion alone, d) the sign of charge on the ion alone., 20. Match the following, A) Pure nitrogen, , i) Chlorine, , B) Haber process, , ii) Sulphuric acid, , C) Contact process, , iii) Ammonia, , D) Deacons Process, , iv) sodium azide, (or) Barium azide, , Which of the following is the correct option?, A, B, C, D, a) (i), (ii), (iii), (iv), b) (ii), (iv), (i), (iii), c) (iii), (iv), (ii), (i), d) (iv), (iii), (ii), (i), , 102, , XII U10-Surface Chemistry.indd 102, , 2/19/2020 5:13:55 PM

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www.tntextbooks.in, , Short Answer, 1. Give two important characteristics of physiscorption, 2. Differentiate physisorption and chemisorption, 3. In case of chemisorption, why adsorption first increases and then decreases with, temperature?, 4. Which will be adsorbed more readily on the surface of charcoal and why? NH3 or O2 ?, 5. Heat of adsorption is greater for chemisorptions than physisorption. Why?, 6. Peptising agent is added to convert precipitate into colloidal solution. Explain with an, example., 7. What happens when a colloidal sol of Fe(OH)3 and As2S3 are mixed?, 8. What is the difference between a sol and a gel?, 9. Why are lyophillic colloidal sols are more stable than lyophobic colloidal sol., 10. Addition of Alum purifies water. Why?, 11. What are the factors which influence the adsorption of a gas on a solid?, 12. What are enzymes? Write a brief note on the mechanism of enzyme catalysis., 13. What do you mean by activity and selectivity of catalyst?, 14. Describe some feature of catalysis by Zeolites., 15. Give three uses of emulsions., 16. Why does bleeding stop by rubbing moist alum, 17. Why is desorption important for a substance to act as good catalyst?, 18. Comment on the statement: Colloid is not a substance but it is a state of substance., 19. Explain any one method for coagulation, 20. Write a note on electro osmosis, 21. Write a note on catalytic poison, 22. Explain intermediate compound formation theory of catalysis with an example, 23. What is the difference between homogenous and hetrogenous catalysis?, 24. Describe adsorption theory of catalysis., , 103, , XII U10-Surface Chemistry.indd 103, , 2/19/2020 5:13:55 PM

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www.tntextbooks.in, , UNIT, , 11, , HYDROXY COMPOUNDS, AND ETHERS, , Learning Objectives, , Alfred Bernhard Nobel, , After studying this unit the student will be, able to, , Alfred Bernhard Nobel was a, Swedish chemist, engineer, inventor,, and philanthropist. Nobel found, that when nitroglycerine was, incorporated in an inert absorbent, like kieselguhr (diatomaceous, earth) it became safer and more, convenient to handle. He patented, this mixture in 1867 as "dynamite"., Nobel prizes were established in, accordance with his will. The Noble, prizes is considered as one of the, precious awards in the fields of, chemistry, literature, peace activism,, physics, Economics and physiology, or medicine., , , , , , , , , , , , , , describe the important methods of, preparation and reactions of alcohols, explain the mechanism of Nucleophilic, substitution reaction of alcohols and, ethers., explain the elimination reaction of, alcohols., describe the preparation and properties, of phenols, discuss the preparation of ethers and, explain their chemical reactions., recognise the uses of alcohols and ethers, , 104, , XII U11-Hydroxy compounds.indd 104, , 2/19/2020 5:12:06 PM

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www.tntextbooks.in, , INTRODUCTION, We have already learnt in eleventh standard that the hydrolysis of an alkyl halide gives an, alcohol, an organic compound containing hydroxyl (-OH ) functional group. Many organic, compounds containing –OH group play an important role in our body. For example, cholesteryl, alcohol commonly known as cholesterol is an important component in our cell membrane. Retinol,, the storage form of vitamin A, finds application in proper functioning of our eyes. Alcohols also, find application in many areas like medicine, industry, etc., For example, methanol is used as, an industrial solvent, ethyl alcohol an additive to petrol, isopropyl alcohol as a skin cleanser for, injection, etc., The hydroxyl group of alcohol can be converted to many other functional groups., Hence, alcohols are important resource in synthetic organic chemistry. In this unit, we will learn, the preparation, properties and uses of alcohols, phenols and ethers., , 11.1 Classification of alcohols:, Alcohols can be classified based on the number of hydroxyl groups and the nature of the, carbon to which the functional group (–OH) is attached., Alcohols, , (Monohydric alcohols), containing only one - OH group, CH3 - CH2 - OH(ethanol), , -OH group attached to, sp2hybridised carbon, eg: vinyl alcohol, CH2 = CH - OH, ethenol, , -OH group attached to, sp3 hybridised carbon, , –OH group attached to, an alkyl group, , (Polyhydric alcohols), containing more than, one -OH group, , –OH group attached to, an allyl group, , examples:, i) Dihydric alcohol, HO - CH2 - CH2 - OH, Glycol(ethan-1,2-diol), ii) Trihydric alcohol, HO-CH2-CHOH-CH2-OH, Glycerol(propan-1,2,3-triol), iii) a hexahydric alcohol, HO-CH2-(CHOH)4-CH2-OH, Sorbitol, (hexan-1,2,3,4,5,6-hexol), , –OH group attached to, a benzyl group, , H, 1 alcohol, , 2, , 3, , H, ethanol, CH3, 2 alcohol, , 1CH, , 3, , CH3 - C - OH, , CH2 = CH - CH - OH, 3, , 4, , H, propan-2-ol, 4, , CH3 - C - OH, CH3, 2-methylpropan-2-ol, , 3, , C6H5 - CH2 - OH, Phenylmethanol, , 1, , C6H5 - CH - OH, CH3, , 2, , but -3-en-2-ol, , CH3, , 3alcohol, , 1, , CH2 = CH - CH2 - OH, Prop-2-en-1-ol, , CH3 - C - OH, , 1 CH3, 2, , CH2 = CH - C - OH, CH3, 2-methylbut-3-en-2-ol, , 2, , 1-phenylethanol, 1CH3, 2, , C6H5 - C - OH, 3 CH3, , 2-phenylpropan-2-ol, , 105, , XII U11-Hydroxy compounds.indd 105, , 2/19/2020 5:12:08 PM

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www.tntextbooks.in, , Preparation of glycol, We have already learnt that the hydroxylation of ethylene using cold alkaline solution of, potassium permanganate (Baeyer’s reagent) gives ethylene glycol., Cold alkaline, KMnO4, CH2, CH2, CH2 = CH2 + H2O, [O], ethene, OH, OH, ethane-1,2-diol, Preparation of glycerol, Glycerol occurs in many natural fats and it is also found in long chain fatty acids in, the form of glyceryl esters (Triglycerides). The alkaline hydrolysis of these fats gives glycerol, and the reaction is known as saponification., O, CH2, , O, , C, , (CH2)14, , CH3, , CH2, , O, , O, CH, , O, , C, , OH, , (CH2)14, , CH3, , + 3NaOH, , D, , CH, , OH + 3NaO, , O, CH2, , O, , C, , C, , (CH2)14, , CH3, , Sodium palmitate, (CH2)14, , CH2, , CH3, , OH, , Sodiumhexadeconoate, , Glycerol, (propane-1,2,3-triol), , Glycerylpalmitate, (a t riglyceride), , Evaluate Yourself?, 1. Suggest a suitable carbonyl compound for the preparation of pent-2-en-1-ol using, LiAlH 4 ., 2SO4 /H2O, 2. 2-methylpropene H, →?, 3. How will you prepare the following using Grignard reagent., i) t-butyl alcohol, ii) allyl alcohol, , Methods to differentiate primary, secondary and tertiary alcohols., The following tests are used to distinguish between 1°, 2° and 3° alcohols., a) Lucas test:, When alcohols are treated with Lucas agent (a mixture of concentrated HCl and anhydrous, ZnCl 2 ) at room temperature, tertiary alcohols react immediately to form a turbidity due to the, formation of alkyl chloride which is insoluble in the medium. Secondary alcohols react within, 10 minutes to form a turbidity of alkyl chloride where primary alcohols do not react at room, temperature., 110, , XII U11-Hydroxy compounds.indd 110, , 2/19/2020 5:12:43 PM

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www.tntextbooks.in, , CHO, CHOH, , COOH, [O], , [O], , CHOH, , Glyceraldehyde, , Tart ronic acid, , (2,3-dihydroxypropanoic, acid), , COOH, , CH2OH, O, , COOH, , Glyceric acid, , (2,3-dihydroxypropanal), , C, , CHOH, , CH2OH, , CH2OH, Glycerol, , [O], , CHOH, , CH2OH, CH2OH, , COOH, , [O], , C, , (2 hydroxypropane1,3-dioic acid), , [O], , O, COOH, , [O], , CH2OH, Dihydroxyacetone, , COOH, Mesooxalic acid, , (1,3-dihydroxypropan, -2-one), , (2-Oxopropane, -1,3 dioic acid), , COOH, Oxalic acid, (ethan-1,2-dioic acid), , Uses of alcohols, Uses of methanol :, 1. Methanol is used as a solvent for paints, varnishes, shellac, gums, cement, etc., 2. In the manufacture of dyes, drugs, perfumes and formaldehyde., Uses of ethanol:, 1. It is also used in the preparation of, a) Paints and varnishes., b) Organic compounds like ether, chloroform, iodoform, etc.,, c) Dyes, transparent soaps., 2. As a substitute for petrol under the name power alcohol used as fuel for aeroplane, 3. It is used as a preservative for biological specimens., Uses of ethylene glycol:, 1. Ethylene glycol is used as an antifreeze in automobile radiator, 2. Its dinitrate is used as an explosive with TNG., Uses of glycerol, 1. Glycerol is used as a sweetening agent in confectionary and beverages., 2. It is used in the manufacture of cosmetics and transparent soaps., 3. It is used in making printing inks and stamp pad ink and lubricant for watches and clocks., 4. It is used in the manufacture of explosive like dynamite and cordite by mixing it with, china clay, , 122, , XII U11-Hydroxy compounds.indd 122, , 2/19/2020 5:13:28 PM

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www.tntextbooks.in, , Acidity of alcohols, According to Bronsted theory, an acid is defined as a proton donor and the acid strength, is the tendency to give up a proton. Alcohols are weakly acidic and their acidity is comparable, with water. Except methanol, all other alcohols are weaker acid than water. The K a value for, water is 1.8 × 10−16 where as for alcohols, the K a value in the order 10−18 to 10−16 ., Alcohols react with active metals such as sodium, aluminium etc… to form the, corresponding alkoxides with the liberation of hydrogen gas and similar reaction to give, alkoxide is not observed in the reaction of alcohol with NaOH., 2C2 H 5 - OH + 2Na → 2C2 H 5ONa + H 2 ↑, , The above reaction explains the acidic nature of alcohols., Comparison of acidity of 1∞, 2∞and 3∞ alcohols, The acidic nature of the alcohol is due to the polar nature of O –H bond. When an electron, withdrawing -I groups such as -Cl, - F etc… is attached to the carbon bearing the OH group,, it withdraws the electron density towards itself and thereby facilitating the proton donation., In contrast, the electron releasing group such as alkyl group increases the electron density on, oxygen and decreases the polar nature of O – H bond, Hence it results in the decrease in acidity., on moving from primary to secondary and tertiary alcohols, the number of alkyl groups which, attached to the carbon bearing -OH group increases, which results in the following order of, acidity., 1∞ alcohol > 2∞ alcohol > 3∞ alcohol, , For example, H3C, , H3C, CH3, , CH2, , CH, , OH, , OH, , H3C, , H3C, K a  1.3 1016, , CH, , OH, , H3C, K a  3.2 1017, , K a  11018, , Alcohols can also act as a Bronsted bases. It is due to the presence of unshared electron, pairs on oxygen which make them proton acceptors., , CH3, , CH2, , .., , O, , H +H, , O, , H, , CH3, , CH2, , O, , H + OH, , H, , 123, , XII U11-Hydroxy compounds.indd 123, , 2/19/2020 5:13:36 PM

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www.tntextbooks.in, , Acidity of Phenol, Phenol is more acidic than aliphatic alcohols. Unlike alcohols it reacts with bases like, sodium hydroxide to form sodium phenoxide. This explains the acidic behaviour of phenol.let, us consider the aqueous solution of phenol in which the following equilibrium exists., C6H5, , C6H5, , OH + H.OH, , O + H3O, , K a value for the above equilibrium is 1×10-10 at 25o C . This Ka value indicates that it is, , more acidic than aliphatic alcohols. This increased acidic behaviour can be explained on the, basis of the stability of phenoxide ion. We have already learnt in XI standard that the phenoxide, is more stabilised by resonance than phenol., , In substituted phenols, the electron withdrawing groups such as -NO 2 ,-Cl enhances the, acidic nature of phenol especially when they are present at ortho and para positions. In such, cases,there is a possibility for the extended delocalisation of negative charge on the phenoxide, ion. On the other hand the alkyl substitued phenols show a decreased acidity due to the, electron releasing +I effect of alkyl group., Table:, , pKaValues of some alcohols and phenols, , S.No., , Compound, , pK a Value, , 1, , methanol, , 15.5, , 2, , ethanol, , 15.9, , 3, , propan – 2- ol, , 16.5, , 4, , 2 – methyl propan 2 - ol, , 18.0, , 5, , Cyclohexanol, , 18.0, , 6, , Phenol, , 10.0, , 7, , o – nitrophenol, , 7.2, , 8, , p – nitrophenol, , 7.1, , 9, , m - nitrophenol, , 8.3, , 10, , o – cresol, , 10.2, , 11, , m – cresol, , 10.1, , 12, , p – cresol, , 10.2, 124, , XII U11-Hydroxy compounds.indd 124, , 2/19/2020 5:13:41 PM

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www.tntextbooks.in, , vii) Phthalein reaction:, On heating phenol with phthalic anhydride in presence of con.H2SO4 , phenolphthalein, is obtained., O, O, , OH, , O, , Con H2SO4, , O, , 2, , +, O, , Phenol, , HO, , phthalic, anhydride, , OH, , Phenolphthalein, , viii) Coupling reaction:, Phenol couples with benzene diazonium chloride in an alkaline solution to form p-hydroxy, azobenzene(a red orange dye)., N2 Cl +, , OH, , NaOH, , N, , 273-278K, Benzene, diazonium, chloride, , Phenol, , N, , OH, , p-hydroxy azobenzene, , Test to differentiate alcohol and phenols, i) Phenol react with benzene diazonium chloride to form a red orange dye, but ethanol, has no reaction with it., ii) Phenol gives purple colouration with neutral ferric chloride solution, alcohols do not, give such coloration with FeCl 3 ., iii) Phenol reacts with NaOH to give sodium phenoxide. Ethyl alcohol does not react with, NaOH ., Uses of phenol, 1) About half of world production of phenol is used for making phenol formaldehyde, resin. (Bakelite)., 2) Phenol is a starting material for the preparation of, i), , drugs such as phenacetin, Salol, aspirin, etc., , ii) phenolphthalein indicator., iii) explosive like picric acid., 3) It is used as an antiseptic-carbolic lotion and carbolic soaps., 131, , XII U11-Hydroxy compounds.indd 131, , 2/19/2020 5:14:13 PM

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www.tntextbooks.in, , Evaluate Yourself:, 1. Which of the following reaction will give 1-methoxy-4-nitrobenzene., a) 4-nitro-1-bromobenzene + sodium methoxide., b) 4-nitrosodium phenoxide+bromomethane, 2. Arrange the following compounds in the increasing order of their acid strength. propan1-ol, 2,4,6-trinitrophenol, 3-nitrophenol, 3,5-dinitrophenol, phenol, 4-methylphenol., Physical Properties:, Ethers are polar in nature. The dipolemoment of ether is the vector sum of two polar, C-O bonds with significant contribution from two lone pairs of electrons. For example, the, dipole moment of diethyl ether is 1.18D. Boiling point of ethers are slightly higher than that of, alkanes and lower than that of alcohols of comparable masses., Compound, CH3 (CH2)5, , Molar Mass, , Boiling point, , 100.21, , 371K, , 102.17, , 373K, , 102.16, , 430K, , CH3, , n-heptane, CH3 O, , (CH2)4 CH3, , 1-methoxypentane, CH3, , (CH2)5, , OH, , hexan-1-ol, , Oxygen of ether can also form Hydrogen bond with water and hence they are miscible, with water. Ethers dissolve wide range of polar and non-polar substances., , O, , O, H, , H, , H, , H, , O, R, , R, Chemical Properties of ethers:, 1., , Nucleophilic substitution reactions of ethers., , Ethers can undergo nucleophilic substitution reactions with HBr or HI . HI is more reactive, than HBr ., CH 3 -O-CH 2 -CH 3 + HI D, → CH 3 I, methoxy ethane, C6 H 5 -O-CH 3, , iodo methane, →, + HI , , methoxy benzene, , + CH 3 -CH 2 -OH, ethanol, , C6 H 5 -OH + CH 3 I, Phenol, , iodomethane, , 136, , XII U11-Hydroxy compounds.indd 136, , 2/19/2020 5:14:32 PM

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www.tntextbooks.in, , 14. Isopropylbenzene on air oxidation in the presence of dilute acid gives, a) C 6 H5COOH   b) C 6 H5COCH3   c) C 6 H5COC 6 H5   d) C 6 H5 - OH, 15. Assertion : Phenol is more reactive than benzene towards electrophilic substitution reaction, Reason : In the case of phenol, the intermediate arenium ion is more stabilized by, resonance., a) if both assertion and reason are true and reason is the correct explanation of assertion., b) if both assertion and reason are true but reason is not the correct explanation of, assertion., c) assertion is true but reason is false, d) both assertion and reason are false., 16. HO CH2 CH2 – OH on heating with periodic acid gives, a) methanoic acid, , b) Glyoxal, , c) methanal, , d) CO2, , 17. Which of the following compound can be used as artifreeze in automobile radiators?, a) methanol, , b) ethanol, , c) Neopentyl alcohol d) ethan -1, 2-diol, , 18. The reactions, OH, , O, i) NaOH, , CH2, , ii) CH2I2, , O, , OH, , a) Wurtz reaction, , b) cyclic reaction, , is an example of, , c) Williamson reaction, , d) Kolbe reactions, , 19. One mole of an organic compound (A) with the formula C 3H8O reacts completely with, two moles of HI to form X and Y. When Y is boiled with aqueous alkali it forms Z. Z, answers the iodoform test. The compound (A) is, a) propan – 2-ol, , b) propan -1-ol, , c) ethoxy ethane, , d) methoxy ehane, , 20. Among the following ethers which one will produce methyl alcohol on treatment with hot, HI?, a) ( H3C ) C O CH3, b) (CH3 ) CH CH2 O, CH3, 3, , 2, , c) CH3 (CH2 ) O CH3, 3, , d) CH3, , CH2, , CH, , O, , CH3, , CH3, 21. Williamson synthesis of preparing dimethyl ether is a / an /, a) SN1 reactions , , b) SN2 reaction, , c) electrophilic addition, , d) electrophilic substitution, , 22. On reacting with neutral ferric chloride, phenol gives, a) red colour, , b) violet colour, , c) dark green colour d) no colouration., 141, , XII U11-Hydroxy compounds.indd 141, , 2/19/2020 5:14:53 PM

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www.tntextbooks.in, , Short Answer Questions, 1. Identify the product (s) is / are formed when 1 – methoxy propane is heated with excess, HI. Name the mechanism involved in the reaction, 2. Draw the major product formed when 1-ethoxyprop-1-ene is heated with one equivalent, of HI, 3. Suggest a suitable reagent to prepare secondary alcohol with identical group using Grignard, reagent., 4. What is the major product obtained when two moles of ethyl magnesium bromide is, treated with methyl benzoate followed by acid hydrolysis., 5. Predict the major product, when 2-methyl but -2-ene is converted into an alcohol in each, of the following methods., (i.), , Acid catalysed hydration, , (ii.), , Hydroboration, , (iii.) Hydroxylation using bayers reagent, 6. Arrange the following in the increasing order of their boiling point and give a reason for, your ordering, (i.) Butan – 2- ol, Butan -1-ol, 2 –methylpropan -2-ol, (ii.)  Propan -1-ol, propan -1,2,3-triol, propan -1,3 – diol, propan -2-ol, 7. Can we use nucelophiles such as NH3 ,CH3O- for the Nucleophilic substitution of alcohols, 8. Is it possible to oxidise t – butyl alcohol using acidified dichromate to form a carbonyl, compound., 9. What happens when 1-phenyl ethanol is treated with acidified KMnO4., 10. Write the mechanism of acid catalysed dehydration of ethanol to give ethene., 11. How is phenol prepared from, i) chloro benzene , , ii) isopropyl benzene, , 12. Explain Kolbe’s reaction, 13. Write the chemical equation for Williamson synthesis of 2-ethoxy – 2- methyl pentane, starting from ethanol and 2 – methyl pentan -2-ol, 14. Write the structure of the aldehyde, carboxylic acid and ester that yield 4- methylpent, -2-en-1-ol., 15. What is metamerism? Give the structure and IUPAC name of metamers of 2-methyoxy, propane, 16. How are the following conversions effected, i) benzylchloride to benzylalcohol, , ii) benzyl alcohol to benzoic acid, , 17. Complete the following reactions, 142, , XII U11-Hydroxy compounds.indd 142, , 2/19/2020 5:14:54 PM

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www.tntextbooks.in, , Br, i) CH3 - CH2 - OH P, , → A aq.NaOH, → B Na, →C, 3, , CH Cl, KMnO, Zn dust, , → B acid, ,  → C, ii) C6 H 5 - OH → A Anhydrs, AlCl, 3, , 4, , 3, , Cl 2 /FeCL3, , → A  , → B  → C, iii) Anisole  , AlCl, t-butylchloride, , HBr, , 3, , CHOHCH3, , iv), , H+, , i) O3, , A, , B, , ii) H2O, , CH3, , 18. 0.44g of a monohydric alcohol when added to methyl magnesium iodide in ether liberates, at STP 112 cm3 of methane with PCC the same alcohol form a carbonyl compound that, answers silver mirror test. Identify the compound., 19. Complete the following reactions, i), , OH, , Nitration, , C6H5COCl, , B, , A, , -, , OH, , (major product), SO, ii) C 6 H5 -CHCH(OH)CH(CH 3 )2 ConH, , →, 2, , 4, , 20. Phenol is distilled with Zn dust followed by friedel – crafts alkylation with propyl chloride, to give a compound B, B on oxidation gives (c) Identify A,B and C., 21., , H3O+, , CH3MgBr+, , HBr, , A, , B, , Mg / ether, , C, , HCHO / H3O+, , D, , O, Identify A,B,C,D and write the complete equation, 22. W,  hat will be the product (X and A)for the following reaction, CH MgBr, K Cr O, acetylchloride i), → X acid, , →A, ii) H O, 3, , 3, , 2, , +, , 2, , 7, , 23. How will you convert acetylene into n-butyl alcohol., 24. Predict the product A,B,X and Y in the following sequence of reaction, , butan - 2- ol, , SOCl2, , A, , Mg, ether, , B, , Cu / 573K, , X, Y, , X, , 25. 3 ,3 – dimethylbutan -2-ol on treatment with conc. H2SO4 to give tetramethyl ethylene as, a major product. Suggest a suitable mechanism, , 143, , XII U11-Hydroxy compounds.indd 143, , 2/19/2020 5:14:59 PM

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www.tntextbooks.in, , UNIT, , 12, , CARBONYL, COMPOUNDS AND, CARBOXYLIC ACIDS, , Learning Objectives, , Adolf von Baeyer, , After studying this unit the student will be, able to, , Adolf, Von, Baeyer,, German, research, chemist, who, synthesized indigo (1880) and, formulated its structure (1883). He, was awarded the Nobel Prize for, Chemistry in 1905. Notable among, Baeyer’s many achievements were, the discovery of the phthalein, dyes and his investigations of uric, acid derivatives, polyacetylenes, and, oxonium salts. One derivative of uric, acid that he discovered was barbituric, acid, the parent compound of the, sedative-hypnotic drugs known as, barbiturates., , , , , , , , , , describes the important methods of, preparation and reactions of Carbonyl, compounds, explains the mechanism of Nucleophilic, addition, reaction, of, carbonyl, compounds, describes the preparation and chemical, reactions of carboxylic acids and its, derivatives, list the uses of aldehydes, ketones and, carboxylic acids, , 145, , XII U12-Carbonyl Compounds.indd 145, , 2/19/2020 5:14:06 PM

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www.tntextbooks.in, , Evaluate yourself, i) Write the IUPAC name for the following compound, , i), , ii) (CH3)2 C = CHCOCH3, , CHO, , CH3, , iv ) (CH3)2 C (OH) CH2 CHO, , iii), , O, ii) Wrtie all possible structural isomers and position isomers for the ketone represented by, the molecular formula C 5H10O ., , 12.2 Structure of carbonyl group, C, , The carbonyl carbon, , is sp 2 hybridised and the carbon – oxygen bond is, , O, similar to carbon – carbon double bond in alkenes. The carbonyl carbon forms three σ bonds, using their three sp2 hybridised orbital. One of the sigma bond is formed with oxygen and, the other two with hydrogen and carbon (in aldehydes) or with two carbons (in ketones). All, the three ' σ ' bonded atoms are lying on the same plane as shown in the fig (12.1). The fourth, valence electron of carbon remains in its unhybridised ‘2p’ orbital which lies perpendicular, to the plane and it overlaps with 2p orbital of oxygen to form a carbon – oxygen p bond., The oxygen atom has two nonbonding pairs of electrons, which occupy its remaining two, p-orbitals. Oxygen, the second most electro negative atom attracts the shaired pair of electron, between the carbon and oxygen towards itself and hence the bond is polar. This polarisation, contributes to the reactivity of aldehydes and ketones., , R, , R, C, , C+, , O, , R, , O:, , R, , , , 120°, , , , C, , , , 120°, , O, , C, , O, , 120°, , C, , O, 120°, , , Fig 12.1 structure of carbonyl group, 148, , XII U12-Carbonyl Compounds.indd 148, , 2/19/2020 5:14:18 PM

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www.tntextbooks.in, , Solubility, Lower members of aldehydes and ketones like formaldehyde, acetaldehyde and acetone, are miscible with water in all proportions because they form hydrogen bond with water., Solubility of aldehydes and ketones decreases rapidly on increasing the length of alkyl, chain., , 3., , O, , R, , C= O, , R, , H, , R, O = C, , H, , R, , Dipolemoment:, The carbonyl group of aldehydes and ketones contains a double bond between carbon and, oxygen. Oxygen is more electronegative than carbon and it attracts the shared pair of electron, which makes the carbonyl group as polar and hence aldehydes and ketones have high dipole, moments., , 4., , , C, , , O, , 12.5 Chemical properties of aldehydes and ketones, A) Nucleophilic addition reactions, This reaction is the most common reactions of aldehydes and ketones. The carbonyl carbon, carries a small degree of positive charge. Nucleophile such as CN − can attack the carbonyl, carbon and uses its lone pair to form a new carbon – nucleophile ' σ ' bond, at the same time, two electrons from the carbon – oxygen double bond move to the most electronegative oxygen, atom. This results in the formation of an alkoxide ion. In this process, the hybridisation of, carbon changes from sp2 to sp3 ., Nu, , R, , C, , R, , Nu, , O, , C, , O, , R, R, , sp3 hybridised carbon, tetrahedral alkoxide ion, , sp2hybridised carbon, , The tetrahedral intermediate can be protonated by water or an acid to form an alcohol., Nu, , Nu, , H, C, , C, , O, , R, , OH, , R, , R, , R, , In general, aldehydes are more reactive than ketones towards nucleophilic addition, reactions due to +I and steric effect of alkyl groups., 154, , XII U12-Carbonyl Compounds.indd 154, , 2/19/2020 5:14:32 PM

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www.tntextbooks.in, , CH3COO- + Cu2O, , CH3 CHO + 2Cu2+ + 5OH(blue), , + 3H2O, , (red), , iii) Benedict's solution Test:, Benedicts solution is a mixture of CuSO4 + sodium citrate + NaOH. Cu2+ is reduced by, aldehyde to give red precipitate of cuprous oxide., CH3COO- + Cu2O, , CH3 CHO + 2Cu2+ + 5OH(blue), , + 3H2O, , (red), , iv) Schiffs’ reagent Test, Dilute solution of aldehydes when added to schiffs’ reagent (Rosaniline hydrochloride dissolved, in water and its red colour decolourised by passing SO2) yields its red colour. This is known as Schiffs’, test for aldehydes . Ketones do not give this test. Acetone however gives a positive test but slowly., , 12.7 Uses of Aldehydes and Ketones, Formaldehyde, (i), , 40% aqueous solution of formaldehyde is called formalin. It is used for preserving, biological specimens., , (ii), , Formalin has hardening effect, hence it is used for tanning., , (iii), , Formalin is used in the production of thermo setting plastic known as bakelite, which, is obtained by heating phenol with formalin., , Acetaldehye, (i), , Acetaldehyde is used for silvering of mirrors, , (ii), , Paraldehyde is used in medicine as a hypnotic., , (iii), , Acetaldehyde is used in the commercial preparation of number of organic compounds, like acetic acid, ethyl acetate etc.,, , Acetone, (i), , Acetone is used as a solvent, in the manufacture of smokeless gun powder (cordite), , (ii), , It is used as a nail polish remover., , (iii), , It is used in the preparation of sulphonal, a hypnotic., , (iv), , It is used in the manufacture of thermosoftening plastic Perspex., , Benzaldehyde is used, (i), , as a flavoring agent   (ii) in perfumes   (iii) in dye intermediates, , (iv), , as starting material for the synthesis of several other organic compounds like, cinnamaldehyde, cinnamic acid, benzoyl chloride etc., , Aromatic Ketones, (i), , Acetophenone has been used in perfumery and as a hypnotic under the name hypnone., , (ii), , Benzophenone is used in perfumery and in the preparation of benzhydrol eye drop., 167, , XII U12-Carbonyl Compounds.indd 167, , 2/19/2020 5:15:11 PM

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www.tntextbooks.in, , O, R, , C, , O, , H O, H, , C, , R, , O, , Inter molecular hydrogen bonding, In fact, most of the carboxylic acids exist as dimer in its vapour phase., iii), , Lower aliphatic carboxylic acids (up to four carbon) are miscible with water due to the, formation of hydrogen bonds with water. Higher carboxylic acid are insoluble in water, due to increased hydrophobic interaction of hydrocarbon part. The simplest aromatic, carboxylic acid, benzoic acid is insoluble in water., Vinegar is 6 to 8% solution of acetic acid in water. Pure acetic acid is called glacial acetic, acid. Because it forms ice like crystal when cooled. When aqueous acetic acid is cooled, at 289.5 K, acetic acid solidifies and forms ice like crystals, where as water remains in, liquid state and removed by filtration. This process is repeated to obtain glacial acetic, acid., , iv), , 12.12 Chemical properties of carboxylic acids., Carboxylic acid do not give the characteristic reaction of carbonyl group, , C = O as given, , by the aldehydes and ketones. as the carbonyl group of carboxylic acid is involved in resonance:, The reactions of carboxylic acids can be classified as follows:, A), , Reactions involving cleavage of O – H bond., , B), , Reactions involving cleavage of C – OH bond., , C), , Reactions involving – COOH group., , D), , Substitution reactions involving hydrocarbon part., , A) Reactions involving cleavage of O – H bond., 1) Reactions with metals, Carboxylic acid react with active metals like Na, Mg, Zn etc to form corresponding salts, with the liberation of hydrogen., Example, O, 2 CH3 C, , O, OH + 2Na, , Acetic acid, , 2 CH3 C, , ONa + H2, , Sodium acetate, , 2) Reaction with alkalies, Carboxylic acid reacts with alkalies to neutralise them and form salts., 172, , XII U12-Carbonyl Compounds.indd 172, , 2/19/2020 5:15:21 PM

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www.tntextbooks.in, , iv), , Benzoic acid does not undergo friedal craft's reaction. This is due to the strong, deactivating nature of the carboxyl group., E) Reducing action of Formic acid, Formic acid contains both an aldehyde as well as an acid group. Hence, like other aldehydes,, formic acid can easily be oxidised and therefore acts as a strong reducing agent, O, H, , C, , O, OH, , H, , Formic acid reduces Tollens reagent (ammonical silver nitrate solution) to metallic, silver., , HCOO - + 2Ag+ + 3OH-, , 2 Ag, , (Tollens reagent), , ii), , OH, , Carboxylic acid group, , Aldehyde group, , i), , C, , +, , CO32 -, , + 2H2O, , Silver mirror, , Formic acid reduces Fehlings solution. It reduces blue coloured cupric ions to red, coloured cuprous ions., HCOO- + 2Cu2+ + 5 OH Cu2O, + CO32 - + 3 H2O, (Fehlings solution), , red precipitate, , Tests for carboxylic acid group, i), In aqueous solution carboxylic acid turn blue litmus red., ii), Carboxylic acids give brisk effervescence with sodium bicarbonate due to the evolution, of carbon-di -oxide., iii), When carboxylic acid is warmed with alcohol and Con H2SO4 it forms an ester, which, is detected by its fruity odour., , 12.13 Acidity of Carboxylic acids, Carboxylic acids undergo ionisation to produce H+ and carboxylate ions in aqueous, solution. The carboxylate anion is stabilised by resonance which make the Carboxylic acid to, donate the proton easily., O, , O, R, , C, , R, , OH, , C, , O + H+, , Carboxylate ion, , Carboxylic acid, , The resonance structure of carboxylate ion are given below., , The strength of carboxylic acid can be expressed in terms of the dissociation constant(Ka):, 177, , XII U12-Carbonyl Compounds.indd 177, , 2/19/2020 5:15:35 PM

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www.tntextbooks.in, , Relative reactivity of Acid derivatives, The reactivity of the acid derivatives follows the order, O, , O, R, , C, , Cl >, , O, , O, , O, , R C O C R, , >, , R C OR', , >, , R C NH2, , The above order of reactivity can be explained in terms of, i) Basicity of the leaving group    ii) Resonance effect, (i) Basicity of the leaving group, Weaker bases are good leaving groups. Hence acyl derivatives with weaker bases as, leaving groups (L) can easily rupture the bond and are more reactive. The correct order of the, basicity of the leaving group is H2N : > : OR > RCOO : > : Cl Hence the reverse is the order, of reactivity., Resonance effect, Lesser the electronegativity of the group, greater would, O, O, be the resonance stabilization as shown below., R C, This effect makes the molecule more stable and reduces the R C, G, reactivity of the acyl compound. The order of electronegativity, G, of the leaving groups follows the order – Cl > - OCOR > - OR > - NH2, Hence the order of reactivity of the acid derivatives with nucleophilic reagent, follows the order, acid halide > acid anhydride > esters > acid amides, , (ii), , 12.14.1 Nomenclature, Compound, (common name, Structural formula,, IUPAC Name), , IUPAC Name, Prefix with position, number, , Root used, , Primary, suffix, , Secondary, Suffix, , –, , eth, , ane/, , oyl, chloride, , –, , prop, , ane/, , oyl, chloride, , –, , Benz, , ane/, , oyl, chloride, , Acetyl chloride, CH3, , C, , Cl, , O, , Ethanoylchloride, Propionyl chloride, C2H5, , C, , Cl, , O, , Propanoylchloride, Benzoyl chloride, C6H5, , C, , Cl, , O, , Benzoylchloride, 180, , XII U12-Carbonyl Compounds.indd 180, , 2/19/2020 5:16:00 PM

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www.tntextbooks.in, , Propionamide, C2H5, , NH2, , C, O, , –, , prop, , ane/, , amide, , –, , benz, , –, , amide, , Propanamide, Benzamide, C6H5, , NH2, , C, O, , Benzamide, 12. 14. 2. Acid Halides:, Methods of Preparation of acid chloride:, Acid chlorides are prepared from carboxylic acid by treating it with anyone of the, chlorinating agent such as SOCl2, PCl5, or PCl3, 1) By reaction with thionyl Chloride (SOCl2), O, , O, CH3, , C, , OH + SOCl2, , CH3, , C, , Cl + HCl + SO2, , Acetyl chloride, , Acetic acid, , This method is superior to others as the by products being gases escape leaving the acid, chloride in the pure state., Physical properties:, • They emit pale fumes of hydrogen chloride when exposed to air on account of their, reaction with water vapour., • They are insoluble in water but slowly begins to dissolve due to hydrolysis., Chemical properties:, They react with weak nucleophiles such as water, alcohols, ammonia and amines to, produce the corresponding acid, ester, amide or substituted amides., 1) Hydrolysis. Acyl halides undergo hydrolysis to form corresponding carboxylic acids, O, , O, CH3, , CH3, , C, Cl + HOH, Acetyl chloride, , C, , OH + HCl, , Acetic acid, , 182, , XII U12-Carbonyl Compounds.indd 182, , 2/19/2020 5:16:05 PM

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www.tntextbooks.in, , 12.14.4 Esters, Methods of preparation, 1. Esterification, We have already learnt that treatment of alcohols with carboxylic acids in presence of, mineral acid gives esters. The reaction is carried to completion by using an excess of reactant, or by removing the water from the reaction mixture., 2. Alcoholysis of Acid chloride or Acid anhydrides, ii) Treatment of acid chloride or acid anhydride with alcohol also gives esters, Physical Properties, Esters are colour less liquids or solids with characteristic fruity smell. Flavours of some of, the esters are given below., S.No, , Ester, , Flavour, , 1, , Amyl acetate, , Banana, , 2, , Ethyl butyrate, , Pineapple, , 3, , Octyl acetate, , Orange, , 4, , Isobutyl formate, , Raspberry, , 5, , Amyl butyrate, , Apricot, , Chemical Properties, 1., , Hydrolysis, We have already learnt that hydrolysis of esters gives alcohol and carboxylic acid., 2. Reaction with alcohol ( Transesterification), Esters of an alcohol can react with another alcohol in the presence of a mineral acid to, give the ester of second alcohol. The interchange of alcohol portions of the esters is termed, transesterification, O, CH3, , C, , H+, , OC2H5 + HOC3H7, , Ethyl acetate, , O, C, , CH3, , OC3H7+ C2H5OH, , Propyl acetate, , Propyl alcohol, , Ethyl alcohol, , The reaction is generally used for the preparation of the esters of a higher alcohol from, that of a lower alcohol., 3. Reaction with ammonia (Ammonolysis), Esters react slowly with ammonia to form amides and alcohol., O, CH3, , C, , O, OC2H5 + H, , CH3, , NH2, , C, , Acetamide, , Ethyl acetate, , NH2 + C2H5OH, Ethyl alcohol, , 185, , XII U12-Carbonyl Compounds.indd 185, , 2/19/2020 5:16:12 PM

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www.tntextbooks.in, , 4) Hoff mann’s degradation, Amides reacts with bromine in the presence of caustic alkali to form a primary amine, carrying one carbon less than the parent amide., O, CH3, , C, , NH2 + Br2 + 4 KOH, , CH3NH2, , Acetamide, , + K2CO3 + 2KBr + 2H2O, , Methyl amine, , 5) Reduction, Amides on reduction with LiAlH4or Sodium and ethyl alcohol to form corresponding amines., O, CH3, , C, , NH2 + 4 (H), , LiAIH4, , CH3, , CH2, , NH2 + H2O, , Ethyl amine, , Acetamide, , 12.15 Uses of carboxylic acids and its derivatives, , Formic acid, It is used, i), for the dehydration of hides., ii), as a coagulating agent for rubber latex, iii), in medicine for treatment of gout, iv), as an antiseptic in the preservation of fruit juice., Acetic acid, It is used, i), ii), iii), , as table vinegar, for coagulating rubber latex, for manufacture of cellulose acetate and poly vinylacetate, , Benzoic acid, It is used, i), as food preservative either in the pure form or in the form of sodium benzoate, ii), in medicine as an urinary antiseptic, iii), for manufacture of dyes, Acetyl Chloride, It is used, i), as acetylating agent in organic synthesis, ii), in detection and estimation of – OH, - NH2 groups in organic compounds, Acetic anhydride, It is used, i), acetylating agent, ii), in the preparation of medicine like asprin and phenacetin, iii), for the manufacture plastics like cellulose acetate and poly vinyl acetate., Ethyl acetate is used, i), in the preparation of artificial fruit essences., ii), as a solvent for lacquers., iii), in the preparation of organic synthetic reagent like ethyl acetoacetate., 188, , XII U12-Carbonyl Compounds.indd 188, , 2/19/2020 5:16:19 PM

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www.tntextbooks.in, , a) but – 3- enoicacid , , b) but – 1- ene-4-oicacid, , c) but – 2- ene-1-oic acid, , d) but -3-ene-1-oicacid, O, , 15., , C, , Identify the product formed in the reaction, , N2H4, , CH3, , C2H5 ONa, , a), , b), NH2, , C), , d), , O, C, O - C2H5, , 16. In which case chiral carbon is not generated by reaction with HCN, O, , OH, , a), , b), , O, O, , O, , C), , d), Ph, , Ph, ,   , , OH, , 17. Assertion : p – N, N – dimethyl aminobenzaldehyde undergoes benzoin condensation, Reason : The aldehydic (-CHO) group is meta directing, a) if both assertion and reason are true and reason is the correct explanation of assertion., b) if both assertion and reason are true but reason is not the correct explanation of assertion., c) assertion is true but reason is false , d) both assertion and reason are false., 18. Which one of the following reaction is an example of disproportionation reaction, a) Aldol condensation , b) cannizaro reaction, c) Benzoin condensation, d) none of these, 19. Which one of the following undergoes reaction with 50% sodium hydroxide solution to, give the corresponding alcohol and acid, a) Phenylmethanal, , b) ethanal, , c) ethanol, , d) methanol, , 191, , XII U12-Carbonyl Compounds.indd 191, , 2/19/2020 5:16:27 PM

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www.tntextbooks.in, , 20. The reagent used to distinguish between acetaldehyde and benzaldehyde is, a) Tollens reagent , , b) Fehling’s solution, , c) 2,4 – dinitrophenyl hydrazine, , d) semicarbazide, , 21. Phenyl methanal is reacted with concentrated NaOH to give two products X and Y. X, reacts with metallic sodium to liberate hydrogen X and Y are, a) sodiumbenzoate and phenol, , b) Sodium benzoate and phenyl methanol, , c) phenyl methanol and sodium benzoate, , d) none of these, , 22. In which of the following reactions new carbon – carbon bond is not formed?, a) Aldol condensation , , b) Friedel craft reaction, , c) Kolbe’s reaction , , d) Wolf kishner reduction, , 23. An alkene “A” on reaction with O3 and Zn - H2O gives propanone and ethanol in equimolar, ratio. Addition of HCl to alkene “A” gives “B” as the major product. The structure of product “B” is, CH2Cl, , CH3, a) Cl CH2, , b) H3C CH2, , CH2 CH, , CH, , CH3, , CH3, CH3, , CH3, c) H3C, , CH2 C, , d) H3C CH CH, , CH3, , Cl, , Cl, , 24. Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of, comparable molecular mass. It is due to their, (NEET), a) more extensive association of carboxylic acid via van der Waals force of attraction, b) formation of carboxylate ion, c) formation of intramolecular H-bonding, d) formation of intermolecular H – bonding, Short Answer Questions, 1. How is propanoic acid is prepared starting from, (a) an alcohol, (b) an alkylhalide, , (c) an alkene, , 2. A Compound (A) with molecular formula C 2 H3 N on acid hydrolysis gives(B) which, reacts with thionylchloride to give compound(C). Benzene reacts with compound (C) in, presence of anhydrous AlCl 3 to give compound(D). Compound (D) on reduction with, Zn/Hg and Conc.HCl gives (E). Identify (A), (B), (C), (D) and (E). Write the equations., 3. Identify X and Y., +, , MgBr, O, CH3COCH2CH2COOC 2 H5 CH, , → X H, →Y, 3, , 3, , 192, , XII U12-Carbonyl Compounds.indd 192, , 2/19/2020 5:16:31 PM

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www.tntextbooks.in, , UNIT, , 13, , ORGANIC NITROGEN, COMPOUNDS, , Learning Objectives, , Donald James Cram, Donald, an, , James, , American, , shared, , the, , Cram, , was, , chemist, , who, , 1987, , After studying this unit the student will be, able to, , , Nobel, , Prize in Chemistry with Jean-Marie, Lehn and Charles J. Pedersen "for their, , , , development and use of molecules with, structure-specific interactions of high, , , , selectivity." They were the founders of, the field of host–guest chemistry Cram, , , , expanded upon Charles Pedersen's, ground-breaking synthesis of crown, ethers,, , two-dimensional, , , , organic, , compounds that are able to recognize, , , , and selectively combine with the ions, of certain metal elements. He also did, , , , work in stereochemistry and Cram's, rule of asymmetric induction is named, , , , after him., , understand isomerism in organic nitro, compounds, describe the preparation and properties, of nitro compounds, classify amines as primary, secondary, and tertiary, describe the methods of preparation of, amines, explain the properties of amines, distinguish between primary, secondary, and tertiary amines, describe the method of preparation of, diazonium salts, explain the preparation and properties, of cyanides, , 196, , XII U13-Organic Nitrogen Compounds.indd 196, , 2/19/2020 5:16:48 PM

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www.tntextbooks.in, , H, R, , N, , H, , N, , R, H, , N, , H, , .., , .., , NH2 > (CH3)2 NH, , CH3, , H, , R, , H, , The boiling point of various amines follows, the order,, , 1°, , >, , 2°, , .., , > (CH3)3 N, >, , 3°, , Amines have lower boiling point than alcohols because nitrogen has lower electronegative, value than oxygen and hence the N-H bond is less polar than -OH bond., Table Boiling points of amines, alcohols and alkanes of comparable molecular weight., S.NO., , Compound, , Molecular mass, , Boiling point (K), , 1., , CH3 (CH2 )2 NH2, , 59, , 321, , 2., , C 2 H5 -NH-CH3, , 59, , 308, , 3., , (CH3 )3 N, , 59, , 277, , 4., , CH3CH(OH)CH3, , 60, , 355, , 5., , CH3CH2CH2CH3, , 58, , 272.5, , 3) Solubility, Lower aliphatic amines are soluble in water, because they can form hydrogen bonds with, water molecules. However, solubility decreases with increase in molecular mass of amines due, to increase in size of the hydrophobic alkyl group. Amines are insoluble in water but readily, soluble in organic solvents like benzene, ether etc., 13.2.5 Chemical properties, The lone pair of electrons on nitrogen atom in amines makes them basic as well as, nucleophilic. They react with acids to form salts and also react with electrophiles., They form salts with mineral acids, Example:, C6H5, , .., , C6H5, , NH2 + HCl, , NH3Cl-, , Anilinium chloride, , Aniline, , Expression for basic strength of amines, In the aqueous solutions, the following equilibrium exists and it lies far to the left, hence, amines are weak bases compared to NaOH ., 211, , XII U13-Organic Nitrogen Compounds.indd 211, , 2/19/2020 5:17:28 PM

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www.tntextbooks.in, , R, , .., , R, , NH2 + H - OH, , NH3 + OH -, ,  +, , R-N H3  OH , basicity constant Kb =, [R-NH2 ], The basicity constant Kb gives a measure of the extent to which the amine accepts the, hydrogen ion (H+ ) from water,, we know that,, Larger the value of Kb or smaller the value of pKb , stronger is the base., Table : pKb values of Amines in Aqueous solution. (pKb for NH3 is 4.74), Amines, .., , CH3 NH2, , .., (CH3)2 NH, , .., , (CH3)3 N, , pKb, , Amines, , pKb, , 3.38, , C2H5NH2, , 3.29, , C6H5CH2, , 3.28, , (C2H5)2 NH, , 3.00, , C6H5, , 4.22, , (C2H5)3 N, , 3.25, , .., , .., , Amines, , .., , NH2, , 4.70, , NH CH3, , 9.30, , .., , .., , pKb, , C6H5N (CH3)2, , 8.92, , Influence of structure on basic character of amines, The factors which increase the availability of electron pair on nitrogen for sharing with, an acid will increase the basic character of an amine. When a +I group like an alkyl group is, attached to the nitrogen increase the electron density on nitrogen which makes the electron, pair readily available for protonation., a) Hence alkyl amines are stronger bases than ammonia., .., , Consider the reaction of an alkyl amine (R- N H 2 ) with a proton, H, H, R > N+, , R > N: + H+, Acid, H, Base, , H, Alkylammonium ion, , H, H, , H, , H, , N: + H+, Acid, H, Base, , H, , N, , + H, Acid, , H, ammonium ion, 212, , XII U13-Organic Nitrogen Compounds.indd 212, , 2/19/2020 5:17:30 PM

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www.tntextbooks.in, , The electron – releasing alkyl group R pushes electron towards nitrogen in the amine, , .., , (R-N H2 ) and provide unshared electron pair more available for sharing with proton., Therefore, the expected order of basicity of aliphatic amines (in gas phase) is, .., .., .., R3N > R2NH > R-NH2, (30), (20), (10), , The above order is not regular in their aqueous solution as evident by their pKb values, given in the table., To compare the basicity of amines, the inductive effect, solvation effect, steric hindrance,, etc., should be taken into consideration., Solvation effect, In the aqueous solution, the substituted ammonium cations get stabilized not only by, electron releasing (+I) effect of the alkyl group but also by solvation with water molecules. The, greater the size of the ion, lesser will be the solvation. The order of stability of the protonated, amines is greater the size of the ion, lesser is the solvation and lesser is the stability. In case of, secondary and tertiary amines, due to steric hindrance, the alkyl groups decrease the number, of water molecules that can approach the protonated amine. Therefore the order of basicity is,, 1˚ > 2˚ > 3˚, OH2, , H, R, , N, , H, , R, , H, , R, , OH2, , 1, , H, , OH2, , R, , OH2, , H, , N, , R, , >, , N, , >, , OH2, , R, , OH2, , H, , 2, , 3, , Based on these effects we can conclude that the order of basic strength in case of alkyl, substituted amines in aqueous solution is, , .., , .., , .., , .., , (CH3)2NH > CH3 - NH2 > (CH3)3N > NH3, , .., , .., , .., , .., , (C2H5)2 NH > (C2H5)3 N > C2H5 NH2 > NH3, The resultant of +I effect, steric effect and hydration effect cause the 20 amine, more basic., Basic strength of aniline, In aniline, the NH 2 group is directly attached to the benzene ring. The lone pair of electron, on nitrogen atom in aniline gets delocalised over the benzene ring and hence it is less available, for protonation makes the, aromatic amines (aniline) less basic than NH3 ., In case of substituted aniline, electron releasing groups like -CH3 ,-OCH3 ,-NH2 increase, the basic strength and electron withdrawing group like - NO2 ,-X,-COOH decrease the basic, strength., , 213, , XII U13-Organic Nitrogen Compounds.indd 213, , 2/19/2020 5:17:32 PM

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www.tntextbooks.in, , Example, N, , N : Cl :, , H3C, , Benzenediazonium, Chloride, , N : Br :, , N, , O2N, , p - Toluenediazonium, Bromide, , N : BF4, , N, , p- Nitrobenzenediazonium, tetra fluoroborate, , 13.3.2 Resonance structure, The stability of arene diazonium salt is due to the dispersal of the positive charge over the, benzene ring., , N, , .., , .., , N:, , N, , N:, , N, , N:, , .., N, , N:, , .., N, , N:, , Resonating structures of benzenediazonium salt, 13.3.3 Method of preparation of Diazonium salts, We have already learnt that benzene diazonium chloride is prepared by the reaction of, aniline with nitrous acid (Which is produced by the reaction of NaNO2 and HCl) at 273 –, 278K, 13.3.4 Physical properties, • Benzene diazonium chloride is a colourless, crystalline solid., • These are readily soluble in water and stable in cold water. However it reacts with warm, water., • Their aqueous solutions are neutral to litmus and conduct electricity due to the presence to, ions., • Benzenediazonium tetrafluoro borate is soluble in water and stable at room temperature., 13.3.5 Chemical reactions, Benzene diazoniumchloride gives two types of chemical reactions, A. Replacement reactions involving loss of nitrogen, In these reactions diazonium group is replaced by nucleophiles such as X - ,CN - ,H- ,OHetc.,, B. Reactions involving retention of diazogroup., Coupling reaction., , 219, , XII U13-Organic Nitrogen Compounds.indd 219, , 2/19/2020 5:17:40 PM

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www.tntextbooks.in, , 13.4.9, , Uses of organic nitrogen compounds, , nitroalkanes, 1. Nitromethane is used as a fuel for cars, 2. Chloropicrin (CCl 3 NO2 ) is used as an insecticide, 3. Nitroethane is used as a fuel additive and precursor to explosive and they are good solvents, for polymers, cellulose ester, synthetic rubber and dyes etc.,, 4. 4% solution of ethylnitrite in alcohol is known as sweet spirit of nitre and is used as diuretic., nitrobenzene, 1 Nitrobenzene is used to produce lubricating oils in motors and machinery., 2 I t is used in the manufacture of dyes, drugs, pesticides, synthetic rubber, aniline and, explosives like TNT, TNB., cyanides and isocyanides, 1. Alkyl cyanides are important intermediates in the organic synthesis of larger number of, compounds like acids, amides, esters, amines etc., 2 N,  itriles are used in textile industry in the manufacture of nitrile rubber and also as a solvent, particularly in perfume industry., Cancer Drug, Mitomycin C, and anticancer agent used to treat stomach and colon, cancer, contains an aziridine ring. The aziridine functional group participates, in the drug’s degradation by DNA, resulting in the death of cancerous cells., O, , O, , CH2O, , NH2, , C, , H2N, OCH3, , N, , NH, , O, Mitomycin, , 228, , XII U13-Organic Nitrogen Compounds.indd 228, , 2/19/2020 5:17:48 PM

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www.tntextbooks.in, , 6. Write short notes on the following, i., , Hofmann’s bromide reaction, , ii., , Ammonolysis, , iii., , Gabriel phthalimide synthesis, , iv., , Schotten – Baumann reaction, , v., , Carbylamine reaction, , vi., , Mustard oil reaction, , vii., , Coupling reaction, , viii., , Diazotisation, , ix., , Gomberg reaction, , 7. How will you distinguish between primary secondary and tertiary alphatic amines., 8. Account for the following, i. Aniline does not undergo Friedel – Crafts reaction, ii. Diazonium salts of aromatic amines are more stable than those of aliphatic amines, iii. pK b of aniline is more than that of methylamine, iv. Gabriel phthalimide synthesis is preferred for synthesising primary amines., v. Ethylamine is soluble in water whereas aniline is not, vi. Amines are more basic than amides,  lthough amino group is o – and p – directing in aromatic electrophilic substitution, vii. A, reactions, aniline on nitration gives a substantial amount of m – nitroaniline., 9. Arrange the following, i. In increasing order of solubility in water, C 6 H5 NH2 ,(C 2 H5 )2 NH,C 2 H5 NH2, ii. In increasing order of basic strength, a) aniline, p- toludine and p – nitroaniline, b) C 6 H5 NH2 ,C 6 H5 NHCH3 ,C 6 H5 NH2 ,p-Cl-C 6 H 4 -NH2, iii. In decreasing order of basic strength in gas phase, , (C 2 H5 )NH2 ,(C 2 H5 )NH, ( C 2 H5 )3 N and NH3, iv. In increasing order of boiling point, C6H5OH, (CH3 )2 NH, C2H5NH2, , v. In decreasing order of the pK b values, , C 2 H5 NH2 , C 6 H5 NHCH3 ,(C 2 H5 )2 NH and CH 3 NH2, vi. Increasing order of basic strength, C 6 H5 NH2 ,C 6 H5 N(CH3 )2 ,(C 2 H5 )2 NH and CH3 NH2, 234, , XII U13-Organic Nitrogen Compounds.indd 234, , 2/19/2020 5:17:59 PM

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www.tntextbooks.in, , UNIT, , 14, , BIOMOLECULES, , Learning Objectives, , G.N. Ramachandran, , After studying this unit, students will be, able to,  Describe, the, importance, of, carbohydrates and their classification, based on structures/functions.,  Explain the structure of glucose and, fructose and their elucidation.,  List the twenty amino acids and explain, the peptide bond formation,  Explain the four levels of structure of, proteins,  Outline the mechanism of enzyme, catalysis,  Summarise the sources and deficiency, diseases of vitamins,  Outline the composition and the, structure of nucleic acids.,  Differentiate RNA from DNA and, explain DNA finger printing,  Appreciate, the, importance, of, biomolecules in our life, , Dr. G.N. Ramachandran received, Master's Degree in Physics from, Madras University. In 1954, he, identified and published the Triple, helical structure of Collagen using, X-ray diffraction. He pioneered the, field of protein structure validation, through the study of available, crystal structures of peptides. From, his studies, in 1962, he developed, the Ramachandran Plot which is, used even today for stereochemical, validation of protein structures., , 237, , XII U14-Biomolecules_New.indd 237, , 2/19/2020 5:15:11 PM

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www.tntextbooks.in, , INTRODUCTION, All living things are made up of many biomolecules such as carbohydrates, proteins, lipids, and nucleic acids etc... The major elements present in the human body are carbon, hydrogen,, oxygen, nitrogen and phosphorous, and they combine to form a variety of biomolecules. These, biomolecules are used as fuel to provide the necessary energy for the various functions of, living systems in addition to many other biological functions. The field of studying about the, chemistry behind the biological processes is called ‘Biochemistry’. In this unit, we will learn, about some essential informations of the, biomolecules, their structure and their importance., CHO, CHO, , 14.1 Carbohydrates:, , HO, C H, H C, OH, Carbohydrates are the most abundant organic compounds in every living organism. They, CHfrom, are also known, as saccharides (derived, 2OH Greek word ‘sakcharon’ which means sugar), CH2OH, as many of them are sweet. They are considered as hydrates of carbon, containing hydrogen, andD-Glyceraldehyde, oxygen in the same ratio L-Glyceraldehyde, as in water. Chemically, they are defined as polyhydroxy, aldehydes or ketones with a general formula Cn(H2O)n. Some common examples are glucose, (monosaccharide), sucrose (disaccharide) and starch (polysaccharide), CH2OH, , D - Glucose, , C, , O, , HO, , C, , H, , H, , C, , OH, , H, , C, , OH, , Sucrose, (α-D-glucopyranosyl-β-D-fructofuranoside), AMYLOSE, , CH2OH, L - Glucose, AMYLOPECTIN, , D - Fructose, , Figure 14.1. Structure of carbohydrates, , Starch, , Carbohydrates are synthesised by green leaves during photo synthesis, a complex process, in which sun light provides the energy to convert carbon dioxide and water into glucose and, oxygen. Glucose is then converted into other carbohydrates and is consumed by animals., , 6CO2 + 6 H2O, , Sun light, , C6H12O6 + 6O2, , 14.1.1 Configuration of carbohydrates:, Almost all carbohydrates are optically active as they have one or more chiral carbons., The number of optical isomers depends on the number of chiral carbons (2n isomers, where n, is the total number of chiral carbons). We have already learnt in XI standard to represent an, 238, , XII U14-Biomolecules_New.indd 238, , 2/19/2020 5:15:14 PM

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www.tntextbooks.in, , organic compound using Fischer projection formula. Fischer has devised a projection formula, to relate the structure of a carbohydrate to one of the two enantiomeric forms of glyceraldehyde, (Figure 14.2). Based on these structures, carbohydrates are named as D or L.The carbohydrates, are usually named with two prefixes namely D or L and followed by sign either (+) or (-)., Carbohydrates are assigned the notation (D/L) by comparing the configuration of the carbon, that is attached to -CH2OH group with that of glyceraldehyde.For example D-glucose is so, named because the H and OH on C5 carbon are in the same configuration as the H and OH, on C2 carbon in D-Glyceraldehyde., There + and – sign indicates the dextro rotatory and levo rotatory respectively. Dextro, rotatory compounds rotate the plane of plane polarised light in clockwise direction while the, levo rotatory compounds rotate in anticlockwise direction. The D or L isomers can either be, dextro or levo rotatory compounds. Dextro rotatory compounds are represented as D-(+) or, L-(+) and the levo rotatory compounds as D-(–) or L-(–), CHO, CHO, CHO, CHO, H, , OH, , H, , H, , OH, , HO, , OH, H, , H, , OH, , H, , OH, , H, , OH, , H, , OH, , H, , OH, , H, , OH, , CH2OH, , CH2OH, , CH2OH, , D-Erythrose, , D-Glyceraldehyde, , CH2OH, , D-Glucose, , D-Ribose, , CHO, HO, , CHO, CHO, CHO, HO, , H, CH2OH, , L-Glyceraldehyde, , HO, , H, , H, , H, OH, , HO, , H, , HO, , H, , HO, , H, , HO, , H, , HO, , H, , HO, , H, , CH2OH, , L-Erythrose, , CH2OH, , CH2OH, , L-Ribose, , L-Glucose, , Figure 14.2 Configuration of carbohydrates, 14.1.2 Classification of carbohydrates:, Carbohydrates can be classified into three major groups based on their product of, hydrolysis, namely monosaccharides, oligosaccharides and polysaccharides., 239, , XII U14-Biomolecules_New.indd 239, , 2/19/2020 5:15:15 PM

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www.tntextbooks.in, , Monosaccharides:Monosaccharides are carbohydrates that cannot be hydrolysed further, and are also called simple sugars. Monosaccharides have general formula Cn(H2O)n. While, there are many monosaccharides known only about 20 of them occur in nature. Some common, examples are glucose, fructose, ribose, erythrose, Monosaccharides are further classified based on the functional group present (aldoses or, ketoses) and the number of carbon present in the chain (trioses, tetroses, pentoses, hexoses etc...)., If the carbonyl group is an aldehyde, the sugar is an aldose. If the carbonyl group is a ketone, the, sugar is a ketose. The most common monosaccharides have three to eight carbon atoms., Table 14.1 Different types of monosaccharides:, No. of carbon atoms, in the chain, , Functional group, present, , Type of sugar, , Example, , 3, 3, 4, 4, 5, 5, 6, 6, , Aldehyde, Ketone, Aldehyde, Ketone, Aldehyde, Ketone, Aldehyde, Ketone, , Aldotriose, Ketotriose, Aldotetrose, Ketotetrose, Aldopentose, Ketopentose, Aldohexose, Ketohexose, , Glyceraldehyde, Dihydroxy acetone, Erythrose, Erythrulose, Ribose, Ribulose, Glucose, Fructose, , 14.1.3 Glucose, Glucose is a simple sugar, which serves as a major energy, source for us. It is the most, important and most abundant, sugar. It is present in honey,, sweet fruits such as grapes, and mangoes etc... Human, blood contains about 100 mg/, dL of glucose, hence it is also, known as blood sugar. In the, combined form it is present in, sucrose, starch, cellulose etc.,, , H2C, CHO, CH, , C, OH, , CH2OH, , OH, O, , CH OH, , n, , CH2OH, , ALDOSE, n=1,2,3,4, , n, , KETOSE, n=0,1,2,3,4, , Figure 14.3 Structure of aldoses and ketoses, , Preparation of glucose, 1. When sucrose (cane sugar) is boiled with dilute H 2SO 4 in alcoholic solution, it undergoes, hydrolysis and give glucose and fructose., , C12H22O11 + H2O, , H+, , Sucrose, , C6H12O6 + C6H12O6, Glucose, , Fructose, , 240, , XII U14-Biomolecules_New.indd 240, , 2/19/2020 5:15:18 PM

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www.tntextbooks.in, , 4. Glucose gets oxidized to gluconic acid with mild oxidizing agents like bromine water, suggesting that the carbonyl group is an aldehyde group and it occupies one end of the carbon, chain. When oxidised using strong oxidising agent such as conc. nitric acid gives glucaric, acid (saccharic acid) suggesting the other end is occupied by a primary alcohol group., O, C, H, , O, OH, , HO, , H, , H, , C, , OH, Con, HNO3, , HO, H, , OH, COOH, Glucoric acid, , H, , H, , C, , H, OH, , H, , OH, , O, , H, OH, , Br2, H 2O, , OH, CH2OH, D-Glucose, , OH, OH, , H, HO, , H, , H, , OH, , H, , OH, CH2OH, Gluconic acid, , 5. Glucose is oxidised to gluconic acid with ammonical silver nitrate (Tollen’s reagent) and, alkaline copper sulphate (Fehling’s solution). Tollen’s reagent is reduced to metallic silver, and Fehling’s solution to cuprous oxide which appears as red precipitate. These reactions, further confirm the presence of an aldehyde group, CHO, OH, , H, HO, , H, , H, , OH, , H, , OH, CH2OH, , COOH, H, , Ag(NH3)+2 -OH, (Tollens reagent), , HO, , OH, H, , H, , OH, , H, , OH, CH2OH, , Glucose, , Gluconic acid, , CHO, , COOH, , OH, , H, HO, , H, , H, , OH, , H, , OH, CH2OH, , H, , CuO, , HO, , (Fehlings reagent), , Glucose, , +, , Ag, , , , Silver Mirror, , OH, H, , H, , OH, , H, , OH, CH2OH, , + Cu2O, , , , Red Precipitate, , Gluconic acid, , 6. Glucose forms penta acetate with acetic anhydride suggesting the presence of five alcohol, groups., 242, , XII U14-Biomolecules_New.indd 242, , 2/19/2020 5:15:22 PM

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www.tntextbooks.in, , CHO, , O, , O, , (CHOH)4 + CH 3–C–O–C–CH3, CH2OH, , CHO, Pyridine, , Acetic anhydride, , O, , O, , (CH–O–C-CH3)4 + CH 3–C–OH, O, Acetic Acid, CH2–O–C–CH3, , Glucose, , Glucose Penta Acetate, , 7. Glucose is a stable compound and does not undergo dehydration easily. It indicates that, not more than one hydroxyl group is bonded to a single carbon atom. Thus the five the, hydroxyl groups are attached to five different carbon atoms and the sixth carbon is an, aldehyde group., 8. The exact spacial arrangement of -OH groups was given by Emil Fischer as shown in, Figure 14.4. The glucose is referred to as D(+) glucose as it has D configuration and is, dextrorotatory., Cyclic structure of glucose, Fischer identified that the open chain penta hydroxyl aldehyde structure of glucose, that he, proposed, did not completely explain its chemical behaviour. Unlike simple aldehydes, glucose, did not form crystalline bisulphite compound with sodium bisulphite. Glucose does not give, Schiff ’s test and the penta acetate derivative of glucose was not oxidized by Tollen’s reagent or, Fehling’s solution. This behaviour could not be explained by the open chain structure., , α, (α−D-Glucopyranose), , β, (Open Chain) (β−D-Glucopyranose), , Figure 14.5 Cyclic structures of glucose, In addition, glucose is found to crystallise in two different forms depending upon the, crystallisation conditions with different melting points (419 and 423 K). In order to explain, these it was proposed that one of the hydroxyl group reacts with the aldehyde group to form a, cyclic structure (hemiacetal form) as shown in figure 14.5. This also results in the conversion, of the achiral aldehyde carbon into a chiral one leading to the possibility of two isomers., These two isomers differ only in the configuration of C1 carbon. These isomers are called, anomers. The two anomeric forms of glucose are called a and β-forms. This cyclic structure, of glucose is similar to pyran, a cyclic compound with 5 carbon and one oxygen atom, and, hence is called pyranose form. The specific rotation of pure α- and β-(D) glucose are 112º, & 18.7º respectively. However, when a pure form of any one of these sugars is dissolved in, 243, , XII U14-Biomolecules_New.indd 243, , 2/19/2020 5:15:23 PM

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www.tntextbooks.in, , Structure of fructose:, Fructose is the sweetest of all known sugars. It is readily soluble in water. Fresh solution of, fructose has a specific rotation -1330 which changes to – 920 at equilibrium due to mutarotation., Similar to glucose the structure of fructose is deduced from the following facts., 1. E,  lemental analysis and molecular weight determination of fructose show that it has the, molecular formula C6 H12 O6, 2. F,  ructose on reduction with HI and red phosphorus gives a mixture of n – hexane (major, product) and 2 – iodohexane (minor product). This reaction indicates that the six carbon, atoms in fructose are in a straight chain., , Fructose, , HI / P, Reduction, , CH3 ( CH2 ) CH3 + CH3, 4, , n – hexane, , CH ( CH2 )3 CH3, 2 – iodohexane, , I, 3. Fructose reacts with NH 2 OH and HCN . It shows the presence of a carbonyl groups in the, fructose., 4. F,  ructose reacts with acetic anhydride in the presence of pyridine to form penta acetate., This reaction indicates the presence of five hydroxyl groups in a fructose molecule., 5. F,  ructose is not oxidized by bromine water. This rules out the possibility of absence of an, aldehyde (-CHO) group., 6. P,  artial reduction of fructose with sodium amalgam and water produces mixtures of sorbitol, and mannitol which are epimers at the second carbon. New asymmetric carbon is formed, at C-2. This confirms the presence of a keto group., , CH2OH, C, , O, , HO, , C, , H, , H, , C, , H, , C, , CH2OH, , CH2OH, , Sodium, amalgam H2O, , HO, , C, , H, , + HO, , C, , H, , OH, , H, , C, , OH, , OH, , H, , C, , OH, , H, , C, , OH, , HO, , C, , H, , OH, , H, , C, , OH, , H, , C, , + 4[H], , CH2OH, , CH2OH, Fructose, , Sorbitol, , CH2OH, Mannitol, , 7. O,  n oxidation with nitric acid, it gives glycollic acid and tartaric acids which contain smaller, number of carbon atoms than in fructose., , 245, , XII U14-Biomolecules_New.indd 245, , 2/19/2020 5:15:32 PM

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www.tntextbooks.in, , CH2OH, C, , O, , HO, , C, , H, , H, , C, , OH, , H, , C, , OH, , COOH, + (O), , CH2OH, , Con. HNO3, , CHOH, , +, , COOH, , CHOH, , Glycollic acid, , COOH, Tartaric acid, , CH2OH, Fructose, , This shows that a keto group is present in C-2. It also shows that 1O alcoholic groups are, present at C- 1 and C- 6. Based on these evidences, the following structure is proposed for, fructose (Figure 14-7), CH2-OH, C= O, HO, , *, , H, , *, , H, , *, , C, , H, , C, , OH, , C, , *, , C- asymmetric carbon, , OH, , CH2-OH, , Figure 14.7 Structure of D(+) Fructose, Cyclic structure of fructose, Like glucose, fructose also forms cyclic structure. Unlike glucose it forms a five membered, ring similar to furan. Hence it is called furanose form. When fructose is a component of a, saccharide as in sucrose, it usually occurs in furanose form., , (α−D-Fructofuranose), , (Open Chain), , (β−D-Fructofuranose), , Figure 14.8 Cyclic Structures of fructose, 246, , XII U14-Biomolecules_New.indd 246, , 2/19/2020 5:15:33 PM

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www.tntextbooks.in, , 14.1.5 Disaccharides, Disaccharides are sugars that yield two molecules of monosaccharides on hydrolysis. This, reaction is usually catalysed by dilute acid or enzyme. Disaccharides have general formula, Cn(H2O)n-1.In disaccharides two monosaccharide’s are linked by oxide linkage called ‘glycosidic, linkage’, which is formed by the reaction of the anomeric carbon of one monosaccharide reacts, with a hydroxyl group of another monosaccharide., Example: Sucrose, Lactose, Maltose, Sucrose: Sucrose, commonly, known as table sugar is the most, abundant disaccharide. It is obtained, mainly from the juice of sugar cane, and sugar beets. Insects such as honey, bees have the enzyme called invertases, that catalyzes the hydrolysis of sucrose, to a glucose and fructose mixture., Honey in fact, is primarily a mixture, of glucose, fructose and sucrose., , Sucrose, (α-D-glucopyranosyl-β-D-fructofuranoside), , Figure 14.9 Structure of sucrose, , On hydrolysis sucrose yields equal, amount of glucose and fructose units., Sucrose Invertase, → Glucose+Fructose, , Sucrose (+66.6°) and glucose (+52.5°) are dextrorotatory compounds while fructose is, levo rotatory (-92.4°). During hydrolysis of sucrose the optical rotation of the reaction mixture, changes from dextro to levo. Hence, sucrose is also called as invert sugar., Structure:, In sucrose, C1 of α-D-glucose is joined to C2 of β-D-fructose. The glycosidic bond thus, formed is called α-1,2 glycosidic bond. Since, both the carbonyl carbons (reducing groups) are, involved in the glycosidic bonding, sucrose is a non-reducing sugar., Lactose: Lactose is a disaccharide, found in milk of mammals and, hence it is referred to as milk sugar., On hydrolysis, it yields galactose and, glucose. Here, the β-D–galactose, and β-D–glucose are linked by β-1,4, glycosidic bond as shown in the, figure 14.10 . The aldehyde carbon is, not involved in the glycosidic bond, hence, it retains its reducing property, and is called a reducing sugar., , Lactose, (β-D-galactopyranosyl-(14)β-D-glucopyranose), Figure 14.10 Structure of Lactose, , 247, , XII U14-Biomolecules_New.indd 247, , 2/19/2020 5:15:35 PM

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www.tntextbooks.in, , Maltose: Maltose derives, its name from malt from, which it is extracted. It, is commonly called as, malt sugar. Malt from, sprouting barley is the, major source of maltose., Maltose is produced, during digestion of, starch by the enzyme, a-amylase., , H, , OH, , Maltose, (α-D-glucopyranosyl-(14) α-D-glucopyranose), Figure 14.11 Structure of Maltose, , Maltose consists two molecules of a-D-glucose units linked by an a-1,4 glycosidic bond, between anomeric carbon of one unit and C-4 of the other unit. Since one of the glucose has, the carbonyl group intact, it also acts as a reducing sugar., 14.1.6 Polysaccharides:, Polysaccharides consist of large number of monosaccharide units bonded together by, glycosidic bonds and are the most common form of carbohydrates. Since, they do not have, sweet taste polysaccharides are called as non-sugars. They form linear and branched chain, molecules., Polysaccharides are classified into two types, namely, homopolysaccharides, and heteropolysaccharides depending upon the constituent monosaccharides., Homopolysaccharides are composed of only one type of monosaccharides while the, heteropolysaccharides are composed of more than one. Example: starch, cellulose and, glycogen (homopolysaccharides); hyaluronic acid and heparin (heteropolysaccharides)., STARCH, Starch is used for energy storage in plants. Potatoes, corn, wheat and rice are the rich, sources of starch. It is a polymer of glucose in which glucose molecules are lined by a(1,4), glycosidic bonds. Starch can be separated into two fractions namely, water soluble amylose, and water insoluble amylopectin. Starch contains about 20 % of amylose and about 80% of, amylocpectin., Amylose is composed of unbranched chains upto 4000 a-D-glucose molecules joined, by a(1,4)glycosidic bonds. Amylopetin contains chains upto 10000 a-D-glucose molecules, linked by a(1,4)glycosidic bonds. In addition, there is a branching from linear chain. At branch, points, new chains of 24 to 30 glucose molecules are linked by a(1,6)glycosidic bonds. With, iodine solution amylose gives blue colour while amylopectin gives a purple colour., 248, , XII U14-Biomolecules_New.indd 248, , 2/19/2020 5:15:35 PM

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www.tntextbooks.in, , AMYLOSE, , AMYLOPECTIN, , Figure 14.12 Structure of Starch (Amylose & Amylopectin), , Cellulose, Cellulose is the major constituent of plant cell walls. Cotton is almost pure cellulose. On, hydrolysis cellulose yields D-glucose molecules. Cellulose is a straight chain polysaccharide., The glucose molecules are linked by β(1,4)glycosidic bond., , H, O, , CH2OH, O, H, H, OH, H, , O, H, , OH, , H, , CH2OH, O, H, H, OH, H, , H, O, H, , OH, , CH2OH, O, H, H, OH, H, , O, H, , OH, , -D-glucose, , n, Figure 14.13 Structure of Cellulose, Cellulose is used extensively in the manufacturing paper, cellulose fibres, rayon explosive,, (Gun cotton – Nitrated ester of cellulose) and so on. Human cannot use cellulose as food, because our digestive systems do not contain the necessary enzymes (glycosidases or cellulases), that can hydrolyse the cellulose., 249, , XII U14-Biomolecules_New.indd 249, , 2/19/2020 5:15:35 PM

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www.tntextbooks.in, , Glycogen: Glycogen is the storage polysaccharide of animals. It is present in the liver and, muscles of animals. Glycogen is also called as animal starch. On hydrolysis it gives glucose, molecules. Structurally, glycogen resembles amylopectin with more branching. In glycogen, the branching occurs every 8-14 glucose units opposed to 24-30 units in amylopectin. The, excessive glucose in the body is stored in the form of glycogen., 14.1.7 Importance of carbohydrates, 1. C,  arbohydrates, widely distributed in plants and animals, act mainly as energy sources, and structural polymers., 2. Carbohydrate is stored in the body as glycogen and in plant as starch., 3. C,  arbohydrates such as cellulose which is the primary components of plant cell wall, is, used to make paper, furniture (wood) and cloths (cotton), 4. Simple sugar glucose serves as an instant source of energy., 5. Ribose sugars are one of the components of nucleic acids., 6. M,  odified carbohydrates such as hyaluronate (glycosaminoglycans) act as shock, absorber and lubricant., , 14.2 Proteins, Proteins are most abundant biomolecules in all living organisms. The term protein is, derived from Greek word ‘Proteious’ meaning primary or holding first place. They are main, functional units for the living things. They are involved in every function of the cell including, respiration. Proteins are polymers of α-amino acids., 14.2.1 Amino acids, Amino acids are compounds which contain an amino group and a carboxylic acid group., The protein molecules are made up α-amino acids which can be represented by the following, general formula., H, R, , C*, , COOH, , NH2, , There are 20 α-amino acids commonly found in the protein molecules. Each amino acid, is given a trivial name, a three letter code and a one letter code. In writing the amino acid, sequence of a protein, generally either one letter or three letter codes are used., 14.2.2 Classification of α-amino acids, The amino acids are classified based on the nature of their R groups commonly known as, side chain. They can be classified as acidic, basic and neutral amino acids. They can also be, classified as polar and non-polar (hydrophobic) amino acids., 250, , XII U14-Biomolecules_New.indd 250, , 2/19/2020 5:15:36 PM

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www.tntextbooks.in, , Amino acids can also be classified as essential and non-essential amino acids based on the, ability to be synthesise by the human. The amino acids that can be synthesised by us are called, non-essential amino acids (Gly, Ala, Glu, Asp, Gln, Asn, Ser, Cys, Tyr & Pro) and those needs, to be obtained through diet are called essential amino acids (Phe, Val, Thr, Trp, Ile, Met, His,, Arg, Leu and Lys). These ten essential amino acids can be memorised by mnemonic called, PVT TIM HALL., Although the vast majority of plant and animal proteins are formed by these 20 α- amino, acids, many other amino acids are also found in the cells. These amino acids are called as, non–protein amino acids. Example: ornithine and citrulline (components of urea cycle where, ammonia is converted into urea), 14.2.3 Properties of amino acid, Amino acids are colourless, water soluble crystalline solids. Since they have both carboxyl, group and amino group their properties differ from regular amines and carboxylic acids. The, carboxyl group can lose a proton and become negatively charged or the amino group can, accept a proton to become positively charged depending upon the pH of the solution. At a, specific pH the net charge of an amino acid is neutral and this pH is called isoelectric point., At a pH above the isoelectric point the amino acid will be negatively charged and positively, charged at pH values below the isoelectric point., In aqueous solution the proton from carboxyl group can be transferred to the amino group of, an amino acid leaving these groups with opposite charges. Despite having both positive and negative, charges this molecule is neutral and has amphoteric behaviour. These ions are called zwitter ions., , COOH, +H, , 3N, , CH, , H+, OH-, , R, Acidic pH, Net charge = +1, , COO+H, , 3N, , CH, , R, Zwitter Ion, Neutral pH, Net charge = 0, (Isoelectric point), , OHH+, , COOH2N, , CH, R, , Alkaline pH, Net charge = -1, , Except glycine all other amino acids have at least one chiral carbon atom and hence are optically, active. They exist in two forms namely D and L amino acids. However, L-amino acids are used, predominantly by the living organism for synthesising proteins. Presence of D-amino acids, has been observed rarely in certain organisms., 14.2.4 Peptide bond formation, The amino acids are linked covalently by peptide bonds. The carboxyl group of the first amino, acid react with the amino group of the second amino acid to give an amide linkage between, these amino acids. This amide linkage is called peptide bond. The resulting compound is, called a dipeptide. Addition an another amino acid to this dipeptide a second peptide bond, results in tripeptide. Thus we can generate tetra peptide, penta peptide etc… When you have, 252, , XII U14-Biomolecules_New.indd 252, , 2/19/2020 5:15:37 PM

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www.tntextbooks.in, , more number of amino acids linked this way you get a polypeptide. If the number of amino, acids are less it is called as a polypeptide, if it has large number of amino acids (and preferably, has a function) then it is called a protein., Peptide Bond, , CH3, , Glycine, , O, , Alanine, , H, , —, , H2N — CH2 — C — N —CH — COOH, —, , -H2O, , —, —, , —, , H2N — CH2 — COOH + H2N — CH — COOH, , CH3, , Glycyl alanine - Dipeptide, , The amino end of the peptide is known as N-terminal or amino terminal while the carboxy, end is called C-terminal or carboxy terminal. In general protein sequences are written from, N-Terminal to C-Terminal. The atoms other than the side chains (R-groups) are called main, chain or the back bone of the polypeptide., 14.2.5 Classification of proteins, Proteins are classified based on their structure (overall shape) into two major types. They, are fibrous proteins and globular proteins., Fibrous proteins, Fibrous proteins are linear molecules similar to fibres. These are generally insoluble in, water and are held together by disulphide bridges and weak intermolecular hydrogen bonds., The proteins are often used as structural proteins. Example: Keratin, Collagen etc…, Globular proteins, Globular proteins have an overall spherical shape. The polypeptide chain is folded into a, spherical shape. These proteins are usually soluble in water and have many functions including, catalysis (enzyme). Example: myoglobin, insuline, , Figure 14.15 (a) Structure of fibrous proteins, , Figure 14.15 (b) Structure of globular proteins, , 253, , XII U14-Biomolecules_New.indd 253, , 2/19/2020 5:15:41 PM

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www.tntextbooks.in, , 14.2.6 Structure of proteins, Proteins are polymers of amino acids. Their three dimensional structure depends mainly on, the sequence of amino acids (residues). The protein structure can be described at four hierarchal, levels called primary, secondary, tertiary and quaternary structures as shown in the figure 14.16, 1. Primary structure of proteins:, Proteins are polypeptide chains, made up of amino acids are connected through peptide, bonds. The relative arrangement of the amino acids in the polypeptide chain is called the, primary structure of the protein. Knowledge of this is essential as even small changes have, potential to alter the overall structure and function of a protein., , H2N, , Gly, , Met, , Phe, , Cys, , Arg, , Asp, , COOH, , 2. Secondary structure of proteins:, The amino acids in the polypeptide chain forms highly regular shapes (sub-structures) through, the hydrogen bond between the carbonyl oxygen ( C=O) and the neighbouring amine, hydrogen (-NH)of the main chain.α-Helix and β-strands or sheets are two most, common sub-structures formed by proteins., α-Helix, In the α-helix sub-structure, the amino acids are arranged in a right handed helical (spiral), structure and are stabilised by the hydrogen bond between the carbonyl oxygen of one amino, acid (nth residue) with amino hydrogen of the fifth residue (n+4th residue). The side chains of, the residues protrude outside of the helix. Each turn of an α-helix contains about 3.6 residues, and is about 5.4 Ao long. The amino acid proline produces a kink in the helical structure and, often called as a helix breaker due to its rigid cyclic structure., β-Strand, β-Strands are extended peptide chain rather than coiled. The hydrogen bonds occur, between main chain carbonyl group one such strand and the amino group of the adjacent, strand resulting in the formation of a sheet like structure. This arrangement is called β-sheets., , β-sheet, α-Helix, , Figure 14.16 Secondary structure of proteins, , 254, , XII U14-Biomolecules_New.indd 254, , 2/19/2020 5:15:43 PM

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www.tntextbooks.in, , 3. Tertiary structure:, The secondary structure elements (α-helix & β-sheets) further folds to form the three, dimensional arrangement. This structure is called tertiary structure of the polypeptide, (protein). Tertiary structure of proteins are stabilised by the interactions between the side, chains of the amino acids. These interactions include the disulphide bridges between cysteine, residues, electrostatic, hydrophobic, hydrogen bonds and van der Waals interactions., 4. Quaternary Structure, Some proteins are made up of more than one polypeptide chains. For example, the oxygen, transporting protein, haemoglobin contains four polypeptide chains while DNA polymerase, enzyme that make copies of DNA, has ten polypeptide chains. In these proteins the individual, polypeptide chains (subunits) interacts with each other to form the multimeric structure which, are known as quaternary structure. The interactions that stabilises the tertiary structures also, stabilises the quaternary structures., , Figure 14.17 Four levels of protein structure, 255, , XII U14-Biomolecules_New.indd 255, , 2/19/2020 5:16:32 PM

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www.tntextbooks.in, , 14.2.7 Denaturation of proteins, Each protein has a unique three-dimensional structure formed by interactions such, as disulphide bond, hydrogen bond, hydrophobic and electrostatic interactions. These, interactions can be disturbed when the protein is exposed to a higher temperature, by adding, certain chemicals such as urea, alteration of pH and ionic strength etc., It leads to the loss, of the three-dimensional structure partially or completely. The process of a losing its higher, order structure without losing the primary structure, is called denaturation. When a protein, denatures, its biological function is also lost., Since the primary structure is intact, this process can be reversed in certain proteins. This, can happen spontaneously upon restoring the original conditions or with the help of special, enzymes called cheperons (proteins that help proteins to fold correctly)., Example: coagulation of egg white by action of heat., , Figure 14.18 Denaturation of proteins, 14.2.8 Importance of proteins, Proteins are the functional units of living things and play vital role in all biological processes, 1. All biochemical reactions occur in the living systems are catalysed by the catalytic, proteins called enzymes., 2. Proteins such as keratin, collagen act as structural back bones., 3. P,  roteins are used for transporting molecules (Haemoglobin), organelles (Kinesins) in, the cell and control the movement of molecules in and out of the cells (Transporters)., 4. Antibodies help the body to fight various diseases., 5. P,  roteins are used as messengers to coordinate many functions. Insulin and glucagon, control the glucose level in the blood., 6. P,  roteins act as receptors that detect presence of certain signal molecules and activate, the proper response., 7. Proteins are also used to store metals such as iron (Ferritin) etc., 256, , XII U14-Biomolecules_New.indd 256, , 2/19/2020 5:16:33 PM

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www.tntextbooks.in, , 14.2.9 Enzymes:, There are many biochemical reactions that occur in our living cells. Digestion of food, and harvesting the energy from them, and synthesis of necessary molecules required for, various cellular functions are examples for such reactions. All these reactions are catalysed by, special proteins called enzymes. These biocatalysts accelerate the reaction rate in the orders, of 105 and also make them highly specific. The high specificity is followed allowing many, reactions to occur within the cell. For example, the Carbonic anhydrase enzyme catalyses the, interconversion of carbonic acid to water and carbon dioxide. Sucrase enzyme catalyses the, hydrolysis of sucrose to fructose and glucose. Lactase enzyme hydrolyses the lactose into its, constituent monosaccharides, glucose and galactose., 14.2.10 Mechanism of enzyme action:, Enzymes are biocatalysts that catalyse a specific biochemical reaction. They generally, activate the reaction by reducing the activation energy by stabilising the transition state. In, a typical reaction enzyme (E) binds with the substrate (S) molecule reversibly to produce an, enzyme-substrate complex (ES). During this stage the substrate is converted into product and, the enzyme becomes free, and ready to bind to another substrate molecule. More detailed, mechanism is discussed in the unit XI surface chemistry., E + S, , Enzyme, , Substate, , [ES], Enzyme - substrate complex, , [ES] → E+P, , Figure 14.19 Mechanism of enzyme action (lock and key model), , 14.3 Lipids:, Lipids are organic molecules that are soluble in organic solvents such as chloroform and, methanol and insoluble in water. The word lipid is derived from the Greek work ‘lipos’ meaning, fat. They are the principal components of cell membranes. In addition, they also act as energy, source for living systems. Fat provide 2-3 fold higher energy compared to carbohydrates /, proteins., 257, , XII U14-Biomolecules_New.indd 257, , 2/19/2020 5:16:36 PM

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www.tntextbooks.in, , 14.3.1 Classification of lipids:, Based on their structures lipids can be classified as simple lipids, compound lipids and, derived lipids. Simple lipids can be further classified into fats, which are esters of long chain, fatty acids with glycerol (triglycerides) and waxes which are the esters of fatty acids with long, chain monohydric alcohols (Bees wax)., Compound lipids are the esters of simple fatty acid with glycerol which contain additional, groups. Based on the groups attached, they are further classified into phospholipids, glycolipids, and lipoproteins. Phospholipids contain a phospho-ester linkage while the glycolipids contain, a sugar molecule attached. The lipoproteins are complexes of lipid with proteins., 14.3.2 Biological importance of lipids, 1. Lipids are the integral component of cell membrane. They are necessary of structural, , integrity of the cell., 2. Th,  e main function of triglycerides in animals is as an energy reserve. They yield more, energy than carbohydrates and proteins., 3. They act as protective coating in aquatic organisms., 4. Lipids of connective tissue give protection to internal organs., 5. Lipids help in the absorption and transport of fat soluble vitamins., 6. They are essential for activation of enzymes such as lipases., 7. Lipids act as emulsifier in fat metabolism., , 14.4 Vitamins:, Vitamins are small organic compounds that cannot be synthesised by our body but are, essential for certain functions. Hence, they must be obtained through diet. The requirements, of these compounds are not high, but their deficiency or excess can cause diseases. Each, vitamin has a specific function in the living system, mostly as co enzymes. They are not served, as energy sources like carbohydrates, lipids, etc.,, The name ‘Vitamin’ is derived from ‘vital amines’, referring to the vitamins earlier identified, amino compounds. Vitamins are essential for the normal growth and maintenance of our, health., 14.4.1 Classification of vitamins, Vitamins are classified into two groups based on their solubility either in water or in fat., Fat soluble vitamins: These vitamins absorbed best when taken with fatty food and are stored, in fatty tissues and livers. These vitamins do not dissolve in water. Hence they are called fat, soluble vitamins. Vitamin A, D, E & K are fat-soluble vitamins., Water soluble vitamins: Vitamins B (B1, B2, B3, B5, B6, B7, B9 and B12) and C are readily soluble, in water. On the contrary to fat soluble vitamins, these can’t be stored. The excess vitamins, present will be excreted through urine and are not stored in our body. Hence, these two, 258, , XII U14-Biomolecules_New.indd 258, , 2/19/2020 5:16:37 PM

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www.tntextbooks.in, , vitamins should be supplied regularly to our body. The missing numbers in B vitamins are, once considered as vitamins but no longer considered as such, and the numbers that were, assigned to them now form the gaps., Table 14.2: Vitamins, their Sources, Functions and their Deficiency disease, Vitamin, Vitamin A, (Retinol), , Vitamin B1, (Thiamine), , Vitamin B2, (Riboflavin), , Vitamin B3, (Niacin), Vitamin B5, (Pantothenic acid), Vitamin B6, (Pyridoxine), Vitamin B7, (Biotin), Vitamin B9, (Folic acid), , Sources, , Functions, , Liver oil, Fish,, Carrot, Milk,, spinach and fruits, such as Papaya and, mango, , Vision and growth, , Co – enzyme in the, Yeast, Milk, Cereals,, form of Thiamine pyro, Green vegetables,, phosphate (TPP) in, Liver, Pork, glycolysis, , Deficiency, Disease, Night blindness,, Xerophthalmia, Keratinisation of, skin, Beri – Beri, (peripheral nerve, damage), , Soybean, Green, vegetable Yeast, Egg, white, Milk, Liver, kidney, , Co enzyme in the, form of FMN (Flavin, mono nucleotide)and, FAD (Flavin adenine, dinucleotide) in redox, reactions, , Cereals, Green leafy, vegetables, Liver,, Kidney, , Co enzyme in the form, of NAD and NADP+ in, redox reactions., , Pellagra (photo, sensitive, dermatitis), , Mushroom,, Avocado, Egg yolk,, Sunflower oil, , Part of coenzyme A in, carbohydrate protein, and Fat metabolism, , Inadequate, growth, , Co enzyme in amino, Meat, Cereals, Milk, acid metabolism,, Whole grains, Egg. formation of Heme in, Hemoglobin, , Cheilosis, (lesions of corner, of mouth, lips and, tongue), , Convulsions, , Liver, kidney,, Milk, Egg yolk,, Vegetables, Grains, , Co enzyme in fatty acid Depression, Hair, Biosynthesis, loss, muscle pain., , Egg, Meat, Beet root,, Leafy vegetables,, Cereals, Yeast, , Nucleic acid synthesis,, maturation of red, blood cells, , Megaloblastic, anaemia, , 259, , XII U14-Biomolecules_New.indd 259, , 2/19/2020 5:16:38 PM

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www.tntextbooks.in, , Vitamin, Vitamin B12, (Cobalamin), Vitamin C, (Ascorbic acid), , Vitamin D, Cholecalciferol(D3),, Ergocalciferol (D2), , Vitamin E, (Tocopherols), , Vitamin K, (Phylloquinone&, Menaquinones), , Sources, , Functions, , Deficiency, Disease, , Egg, Meat, Fish, , Co-enzyme in amino, acid metabolism, Red, blood cells maturation, , Pernicious, Anaemia, , Citrus fruits, (Orange, Lemon, etc…),Tomato,, Amla, Leafy, Vegetables, , Coenzyme in, Antioxidant, building, of collagen, , Scurvy (bleeding, gums), , Fish liver oil, Milk,, Egg yolk, (exposure, to sunlight), , Absorption and, maintenance of calcium, , Rickets (children),, Osteomalacia, (adults), , Cotton seed oil, Sun, flower oil, wheat, Antioxidant, germ oil, Vegetable, oils, , muscular, dystrophy, (muscular, weakness) and, neurological, dysfunction, , Green leafy, vegetable, soybean, oil, tomato, , Increased blood, clotting time,, Haemorrhagic, diseases, , Blood clotting, , 14.5 Nucleic acids, The inherent characteristics of each and every species are transmitted from one, generation to the next. It has been observed that the particles in nucleus of the cell are, responsible for the transmission of these characteristics. They are called chromosomes, and are made up of proteins and another type of biomolecules called nucleic acids. There, are mainly two types nucleic acids, the deoxyribonucleic acid (DNA) and ribonucleic, acid (RNA). They are the molecular repositories that carry genetic information in every, organism., 14.5.1 Composition and structure of nucleic acids, Nucleic acids are biopolymers of nucleotides. Controlled hydrolysis of DNA and RNA, yields three components namely a nitrogenous base, a pentose sugar and phosphate group., Nitrogen base, These are nitrogen containing organic compounds which are derivatives of two parent, 260, , XII U14-Biomolecules_New.indd 260, , 2/19/2020 5:16:38 PM

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www.tntextbooks.in, , compounds, pyrimidine and purine. Both DNA and RNA have two major purine bases,, adenine (A) and guanine (G). In both DNA and RNA, one of the pyrimidines is cytosine, (C), but the second pyrimidine is thymine (T) in DNA and uracil (U) in RNA., , 6, , 1, N, , 7, N, , 5, , 8, , 2, , N9, H, , 4, , N, 3, , Purine, O, , NH2, N, , N, , N, H, , H, , N, , N, , N, H, , N, , N, , NH2, , Guanine, , Adenine, H, C, , 4, , N, , 5, CH, , HC, 2, , CH, 6, , 3, , N, 1, Pyrimidine, , H, O, , H, O, , N, , N, , N, , NH2, , O, , N, , H, , CH3, , O, , N, , H, , H, , O, Thymine (T), , N, , Cytosine (C), , H, , H, , H, , H, , Uracil (U), , 261, , XII U14-Biomolecules_New.indd 261, , 2/19/2020 5:16:40 PM

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www.tntextbooks.in, , Pentose sugar:, Nucleic acids have two types of pentoses. The recurring deoxyribonucleotide units of, DNA contain 2’-deoxy-D-ribose and the ribonucleotide units of RNA contain D-ribose. In, nucleotides, both types of pentoses are in their β-furanose (closed five membered rings) form., , 5' CH2OH, 4', , O, 2', , 3', , 5' CH2OH, , OH, , 4', , 1', , OH, , OH, , 3', OH, , OH, , O, 2', , 1', , H, Deoxyribose, , Ribose, O, O, , P, , O, , O, Phosphate group, Phosphoric acid forms phospho diester bond between nucleotides. Based on the number, of phosphate group present in the nucleotides, they are classified into mono nucleotide,, dinucleotide and trinucleotide., Nucleosides and nucleotides:, The molecule without the phosphate group is called a nucleoside. A nucleotide is derived, from a nucleoside by the addition of a molecule of phosphoric acid. Phosphorylation occurs, generally in the 5’ OH group of the sugar. Nucleotides are linked in DNA and RNA by phospho, diester bond between 5’ OH group of one nucleotide and 3’ OH group on another nucleotide., Sugar + Base, , Nucleoside, , Nucleoside + Phosphate, nNucleotide, , Nucleotide, , Polynucleotide, (Nucleic Acid), , 14.5.2 Double strand helix structure of DNA, In early 1950s, Rosalind Franklin and Maurice Wilkins, used X-ray diffraction to unravel the structure of DNA. The, DNA fibers produced a characteristic diffraction pattern., The central X shaped pattern indicates a helix, whereas, the heavy black arcs at the top and bottom of the diffraction, pattern reveal the spacing of the stacked bases., , Figure 14.20 DNA X-ray, diffraction, , 262, , XII U14-Biomolecules_New.indd 262, , 2/19/2020 5:16:43 PM

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www.tntextbooks.in, , The structure elucidation of DNA by Watson, and Crick in 1953 was a momentous event in science., They postulated a 3-dimensional model of DNA, structure which consisted of two antiparallel helical, DNA chains wound around the same axis to form a, right-handed double helix., The hydrophilic backbones of alternating, deoxyribose and phosphate groups are on the outside, of the double helix, facing the surrounding water., The purine and pyrimidine bases of both strands, are stacked inside the double helix, with their, hydrophobic and ring structures very close together, and perpendicular to the long axis, thereby reducing, the repulsions between the charged phosphate, groups. The offset pairing of the two strands creates, a major groove and minor groove on the surface of, the duplex., , H3C, , H-bonds, , O, H, , N, , N, , H, N, , H, , H, N, , N, , O, Thymidine, , N, , N, , T=A, , Adenine, , H, N, , The model revealed that, there are, 10.5 base pairs (36 Å) per turn of the, helix and 3.4 Å between the stacked, bases. They also found that each base is, hydrogen bonded to a base in opposite, strand to form a planar base pair., , H-bonds, H, O, , N, , N, , Figure 14.21 DNA, Double Helix, , H, O, H, , Cytosine, , C∫G, , N, H, , N, , N, N, , N, Guanine, , Two hydrogen bonds are formed, between adenine and thymine and three, hydrogen bonds are formed between, guanine and cytosine. Other pairing, tends to destabilize the double helical, structure. This specific association of the, two chains of the double helix is known, , 263, , XII U14-Biomolecules_New.indd 263, , 2/19/2020 5:16:45 PM

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www.tntextbooks.in, , as complementary base pairing. The DNA double helix or duplex is held together by two forces,, a) Hydrogen bonding between complementary base pairs, b) Base-stacking interactions, The complementary between the DNA strands is attributable to the hydrogen bonding, between base pairs but the base stacking interactions are largely non-specific, make the major, contribution to the stability of the double helix., 14.5.3 Types of RNA molecules, Ribonucleic acids are similar to DNA. Cells contain up to eight times high quantity of, RNA than DNA. RNA is found in large amount in the cytoplasm and a lesser amount in the, nucleus. In the cytoplasm it is mainly found in ribosomes and in the nucleus, it is found in, nucleolus., RNA molecules are classified according to their structure and function into three major, types, i. Ribosomal RNA (rRNA)      ii. Messenger RNA (mRNA), iii. Transfer RNA (tRNA), rRNA, rRNA is mainly found in cytoplasm and in ribosomes, which contain 60% RNA and 40%, protein. Ribosomes are the sites at which protein synthesis takes place., tRNA, tRNA molecules have lowest molecular weight of all nucleic acids. They consist of 73 –, 94 nucleotides in a single chain. The function of tRNA is to carry amino acids to the sites of, protein synthesis on ribosomes., mRNA, mRNA is present in small quantity and very short lived. They are single stranded, and, their synthesis takes place on DNA. The synthesis of mRNA from DNA strand is called, transcription. mRNA carries genetic information from DNA to the ribosomes for protein, synthesis. This process is known as translation, Table 14.3 Difference between DNA and RNA, DNA, It is mainly present in nucleus,, mitochondria and chloroplast, , RNA, It is mainly present in cytoplasm, nucleolus, and ribosomes, , It contains deoxyribose sugar, , It contains ribose sugar, , Base pair A = T. G ≡ C, , Base pair A = U. C ≡ G, , Double stranded molecules, , Single stranded molecules, , It's life time is high, , It is Short lived, , It is stable and not hydrolysed easily by, alkalis, , It is unstable and hydrolyzed easily by alkalis, , It can replicate itself, , It cannot replicate itself. It is formed from, DNA., 264, , XII U14-Biomolecules_New.indd 264, , 2/19/2020 5:16:45 PM

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www.tntextbooks.in, , More to know, DNA finger printing, Traditionally, one of the most accurate methods for placing an individual at, the scene of a crime has been a fingerprint. With the advent of recombinant DNA, technology, a more powerful tool is now available: DNA fingerprinting is (also called, DNA typing or DNA profiling). It was first invented by Professor Sir Alec Jeffrey sin, 1984. The DNA finger print is unique for every person and can be extracted from, traces of samples from blood, saliva, hair etc…By using this method we can detect the, individual specific variation in human DNA., In this method, the extracted DNA is cut at specific points along the strand with, restriction of enzymes resulting in the formation of DNA fragments of varying lengths, which were analysed by technique called gel electrophoresis. This method separates, the fragments based on their size. The gel containing the DNA fragments is then, transferred to a nylon sheet using a technique called blotting. Then, the fragments, will undergo autoradiography in which they were exposed to DNA probes (pieces of, synthetic DNA that were made radioactive and that bound to the fragments). A piece, of X-ray film was then exposed to the fragments, and a dark mark was produced at any, point where a radioactive probe had become attached. The resultant pattern of marks, could then be compared with other samples. DNA fingerprinting is based on slight, sequence differences (usually single base-pair changes) between individuals. These, methods are proving decisive in court cases worldwide., Chromosomal DNA, (e.g., Suspect 1), Cleave with restriction, endonuclease., DNA fragments, , DN, A, DN mar, k, Su A ma ers, sp r, Ev ect 1 kers, ide, Vic nce, Su tim, sp, DN ect 2, A, DN ma, A m rker, ark s, ers, , Separate fragments by agarose, gel electrophoresis (unlabeled)., , Denature, DNA, and, transfer to, nylon, membrane., , Incubate, with, probe,, then, wash., , Su, sp, Ev ect 1, ide, n, Vic ce, tim, Su, spe, ct, 2, , Radiolabeled, DNA probe, , Expose, X-ray, film, to, membrane., , Figure 14.22 DNA finger printing, 265, , XII U14-Biomolecules_New.indd 265, , 2/19/2020 5:16:47 PM

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www.tntextbooks.in, , 14.5.4 Biological functions of nucleic acids, In addition to their roles as the subunits of nucleic acids, nucleotides have a variety of, other functions in every cell such as,, i. Energy carriers (ATP), H, , N, , N, , O, , P, , N, , H, , OO, , N, , CH2, , O-, , H, , Nitrogenous, base, (adenine), , H, , H, , Phosphate, group, , N, , O, , H, , H, , H, OH H, Sugar, , ii. Components of enzyme cofactors (Example: Coenzyme A, NAD+, FAD), H, , H, , O, , H CH3, , O, , N, , NH2, N, , 5', , N, HS CH2 CH2 N C CH2 CH2 N C C C CH2 O P O P O CH2 O, O, O OHCH3, O, O, 4', H H 1', Pantothenic acid, H, H, -Mercaptoethylamine, 3', , O, , N, , 2', , OH, , O P O, O, , Coenzyme A, , 3'-Phosphoadenosine diphosphate, (3'-P-ADP), , iii. Chemical messengers (Example: Cyclic AMP, cAMP), O, , 5', , CH2, H, , O, , P, , Adenine, , O, , H, 3', , O, , H, , H, , OH, , O, , Adenosine 3', 5'-cyclic monophosphate, (cyclic AMP; cAMP), , 14.6 HORMONES, Hormone is an organic substance (e.g. a peptide or a steroid) that is secreted by one, tissue. it limits the blood stream and induces a physiological response (e.g. growth and, metabolism) in other tissues. It is an intercellular signalling molecule. Virtually every, 266, , XII U14-Biomolecules_New.indd 266, , 2/19/2020 5:16:51 PM

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www.tntextbooks.in, , process in a complex organism is regulated by one or more hormones: maintenance of, blood pressure, blood volume and electrolyte balance, embryogenesis, hunger, eating, behaviour, digestion - to name but a few. Endocrine glands, which are special groups, of cells, make hormones. The major endocrine glands are the pituitary, pineal, thymus,, thyroid, adrenal glands, and pancreas. In addition, men produce hormones in their testes, and women produce them in their ovary. Chemically, hormones may be classified as, either protein (e.g. insulin, epinephrine) or steroids (e.g. estrogen, androgen).Hormones, are classified according to the distance over which they act as, endocrine, paracrine and, autocrine hormones, Endocrine hormones act on cells distant from the site of their release. Example: insulin, and epinephrine are synthesized and released in the bloodstream by specialized ductless, endocrine glands., Paracrine hormones (alternatively, local mediators) act only on cells close to the cell that, released them. For example, interleukin-1 (IL-1), Autocrine hormones act on the same cell that released them. For example, protein growth, factor interleukin-2 (IL-2)., , (b) Paracrine, , (a) Endocrine, , Hormone, molecules, , Endocrine, cells, , (c) Autocrine, , Target, cells, , Bloodstream, , Figure 14.22 Endocrine Paracrine and Autocrine hormones, Only those cells with a specific receptor for a given hormone will respond to its presence, even though nearly all cells in the body may be exposed to the hormone. Hormonal messages, are therefore quite specifically addressed., , 267, , XII U14-Biomolecules_New.indd 267, , 2/19/2020 5:16:52 PM

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www.tntextbooks.in, , EVALUATION, Choose the correct answer:, 1. Which one of the following rotates the plane polarized light towards left?, , (NEET Phase – II), a) D(+) Glucose, , (b) L(+) Glucose, , (c)D(-) Fructose, , d) D(+) Galactose, , 2. The correct corresponding order of names of four aldoses with configuration given below, Respectively is, (NEET Phase – I)1551, a) L-Erythrose, L-Threose, L-Erythrose, D-Threose, b)D-Threose,D-Erythrose, L-Threose, L-Erythrose,, c)L-Erythrose, L-Threose, D-Erythrose, D-Threose, d) D-Erythrose, D-Threose, L-Erythrose, L-Threose, 3. Which one given below is a non-reducing sugar? (NEET Phase – I), a) Glucose, 4. Glucose, , (HCN), , a) Heptanoic acid, , b) Sucrose, Product, , (hydrolysis), , c) maltose, Product, , b) 2-Iodohexane, , (HI + Heat), , d) Lactose., A, the compound A is, , c) Heptane, , d) Heptanol, , 5. Assertion: A solution of sucrose in water is dextrorotatory. But on hydrolysis in the, presence of little hydrochloric acid, it becomes levorotatory. (AIIMS), Reason: Sucrose hydrolysis gives equal amounts of glucose and fructose. As a result of, this change in sign of rotation is observed., a)If both accretion and reason are true and reason is the correct explanation of assertion, b) If both assertion and reason are true but reason is not the correct explanation of, assertion, c) If assertion is true but reason is false., d) if both assertion and reason are false., 6. The central dogma of molecular genetics states that the genetic information flows from, (NEET Phase – II), a) Amino acids, , Protein, , DNA, , b) DNA, , Carbohydrates, , Proteins, , c) DNA, , RNA, , Proteins, , d) DNA, , RNA, , Carbohydrates, , 7. In a protein, various amino acids linked together by (NEET Phase – I), a) Peptide bond , , b) Dative bond, , c) α - Glycosidic bond , , d) β - Glycosidic bond, 268, , XII U14-Biomolecules_New.indd 268, , 2/19/2020 5:16:53 PM

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www.tntextbooks.in, , 17. Which of the following vitamins is water soluble?, a) Vitamin E, , b) Vitamin K, , c) Vitamin A, , d) Vitamin B, , c) D-Ribose, , d) D-Glucose, , 18. Complete hydrolysis of cellulose gives, a) L-Glucose, , b) D-Fructose, , 19. Which of the following statement is not correct?, a) Ovalbumin is a simple food reserve in egg-white, b) Blood proteins thrombin and fibrinogen are involved in blood clotting, c) Denaturation makes protein more active, d) Insulin maintains the sugar level of in the human body., 20. Glucose is an aldose. Which one of the following reactions is not expected with glucose?, a) It does not form oxime, b) It does not react with Grignard reagent, c) It does not form osazones, d) It does not reduce tollens reagent, 21. If one strand of the DNA has the sequence ‘ATGCTTGA’, then the sequence of, complementary strand would be, a) TACGAACT, , b) TCCGAACT, , c) TACGTACT, , d) TACGRAGT, , c) Protein, , d) Carbohydrates, , 22. Insulin, a hormone chemically is, a) Fat, , b) Steroid, , 23. α-D (+) Glucose and β-D (+) glucose are, a) Epimers , , b) Anomers, , c) Enantiomers , , d) Conformational isomers, , 24. Which of the following are epimers, a) D(+)-Glucose and D(+)-Galactose, , (b) D(+)-Glucose and D(+)-Mannose, , c) Neither (a) nor (b) , , (d) Both (a) and (b), , 25. Which of the following amino acids are achiral?, a) Alanine, , b) Leucine, , c) Proline, , d) Glycine, , Short Answer Questions, 1. What type of linkages hold together monomers of DNA?, 2. Give the differences between primary and secondary structure of proteins., 3. Name the Vitamins whose deficiency cause i) rickets ii) scurvy, 4. Write the Zwitter ion structure of alanine, 270, , XII U14-Biomolecules_New.indd 270, , 2/19/2020 5:16:56 PM

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www.tntextbooks.in, , 5. Give any three difference between DNA and RNA, 6. Write a short note on peptide bond, 7. Give two difference between Hormones and vitamins, 8. Write a note on denaturation of proteins, 9. What are reducing and non – reducing sugars?, 10. Why carbohydrates are generally optically active., 11. Classify the following into monosaccharides, oligosaccharides and polysaccharides., i) Starch, , ii) fructose, , iv) lactose, , iv) maltose, , iii) sucrose, , 12. How are vitamins classified, 13. What are harmones? Give examples, 14. Write the structure of all possible dipeptides which can be obtained form glycine and, alanine, 15. Define enzymes, 16. Write the structure of α-D (+) glucophyranose, 17. What are different types of RNA which are found in cell, 18. Write a note on formation of, , α-helix ., , 19. What are the functions of lipids in living organism., 20. Is the following sugar, D – sugar or L – sugar?, CHO, OH, , H, , OH, , H, , OH, , H, CH2 - OH, , 271, , XII U14-Biomolecules_New.indd 271, , 2/19/2020 5:16:57 PM

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www.tntextbooks.in, , UNIT, , 15, , CHEMISTRY IN, EVERYDAY LIFE, , Learning Objectives, After studying this unit, the students will, be able to, , Vladimir Prelog, , Prof. Vladimir Prelog was a, Swiss Chemist who shared 1975, Nobel Prize for Chemistry with, John W Cornforth for his work, on Stereo Chemistry. He has, done wide ranging research on, alkaloids, antibiotics, enzymes, and other natural compounds., He was distinguished for his, contribution to the development, of modern stereo chemistry., Prelog, synthesized, many, natural products and worked on, problems of stereo chemistry like, adamenline, boromycin analoids, and rifamycins, , , , , , , , , , , , , , , , , , , , recognize the, chemotherapy, , term, , drug, , and, , classify the drugs based on their, properties, describe the drug-target interaction., discuss some important classes of, drugs., explain the chemistry of cleansing, agents, describe the chemicals in food, explain the important terms in polymer, chemistry., describe the preparation of some, important synthetic polymers, appreciate the importance of polymers, in today life, , 272, , XII U15 Chemistry in Action.indd 272, , 2/19/2020 5:16:54 PM

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www.tntextbooks.in, , INTRODUCTION, Chemistry touches every aspect of our lives. The three-basic requirement of our life: food,, clothes, shelter are all basically chemical compounds. Infact, life itself is a complicated system, of interrelated chemical process. In this unit, we will learn the chemistry involved in the field, of medicines, food materials, cleansing agents and polymers., , 15.1 Drug, The word drug is derived from the French word “drogue” meaning “dry herb”. A drug is, a substance that is used to modify or explore physiological systems or pathological states for, the benefit of the recipient. It is used for the purpose of diagnosis, prevention, cure/relief of, a disease.The drug which interacts with macromolecular targets such as proteins to produce, a therapeutic and useful biological response is called medicine. The specific treatment of a, disease using medicine is known as chemotherapy. An ideal drug is the one which is nontoxic, bio-compatible and bio-degradable, and it should not have any side effects. Generally,, most of the drug molecules that are used now a days have the above properties at lower, concentrations. However, at higher concentrations, they have side effects and become toxic., The medicinal value of a drug is measured in terms of its therapeutic index, which is defined, as the ratio between the maximum tolerated dose of a drug (above which it become toxic), and the minimum curative dose (below which the drug is ineffective). Higher the value of, therapeutic index, safer is the drug., 15.1.1 Classification of drugs:, Drugs are classified based on their properties such as chemical structure, pharmacological, effect, target system, site of action etc. We will discuss some general classifications here., Classification based on the chemical structure:, In this classification, drugs with a common chemical skeleton are classified into a single, group. For example, ampicillin, amoxicillin, methicillin etc.. all have similar structure and are, classified into a single group called penicillin. Similarly, we have other group of drugs such, as opiates, steroids, catecholamines etc. Compounds having similar chemical structure are, expected to have similar chemical properties. However, their biological actions are not always, similar. For example, all drugs belonging to penicillin group have same biological action, while, groups such as barbiturates, steroids etc.. have different biological action., Penicillins, R group-Drug Name, H, N, , R, , H, S, , C, O, , N, , H3CO, , CH3, CH3, , CH2, , CH2, , CH, , O, , O, , NH2, , OH, O, , Penicillin G, , Penicillin V, , Ampicillin, , CH, , OH, , NH2, , Amoxicillin, , H3CO, , methicillin, , 273, , XII U15 Chemistry in Action.indd 273, , 2/19/2020 5:16:58 PM

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www.tntextbooks.in, , Classification based on Pharmacological effect:, In this classification, the drugs are grouped based on their biological effect that they, produce on the recipient. For example, the medicines that have the ability to kill the pathogenic, bacteria are grouped as antibiotics. This kind of grouping will provide the full range of drugs, that can be used for a particular condition (disease). The physician has to carefully choose a, suitable medicine from the available drugs based on the clinical condition of the recipient., Examples:, Antibiotic drugs:, tetracycline etc.., , amoxicillin,, , ampicillin,cefixime,, , cefpodoxime,, , erythromycin,, , Antihypertensive drugs: propranolol, atenolol, metoprolol succinate, amlodipine etc…, Classification based on the target system (drug action):, In this classification, the drugs are grouped based on the biological system/process, that, they target in the recipient. This classification is more specific than the pharmacological, classification.For example, the antibiotics streptomycin and erythromycin inhibit the protein, synthesis (target process) in bacteria and are classified in a same group. However, their mode of, action is different. Streptomycin inhibits the initiation of protein synthesis, while erythromycin, prevents the incorporation of new amino acids to the protein., Classification based on the site of action (molecular target):, The drug molecule interacts with biomolecules such as enzymes, receptors etc,, which are, referred as drug targets. We can classify the drug based on the drug target with which it binds., This classification is highly specific compared to the others. These compounds often have a, common mechanism of action, as the target is the same., 15.1.2 Drug–target Interaction:, The biochemical processes such as metabolism (which is responsible for breaking down, the food molecules and harvest energy in the form of ATP and biosynthesis of necessary, biomolecules from the available precursor molecules using many enzymes),cell-signaling, (senses any change in the environment using the receptor molecules and send signals to various, processes to elicit an appropriate response) etc… are essential for the normal functioning of our, body. These routine processes may be disturbed by any external factors such as microorganism,, chemicals etc.. or by a disorder in the system itself. Under such conditions we may have to take, medicines to restore the normal functioning of the body., These drug molecules interact with biomolecules such as proteins, lipids, etc..that are, responsible for different functions of the body. For example, proteins which act as biological, catalysts are called enzymes and those which are important for communication systems are, called receptors. The drug interacts with these molecules and modify the normal biochemical, reactions either by modifying the enzyme activity or by stimulating/suppressing certain, receptors., 274, , XII U15 Chemistry in Action.indd 274, , 2/19/2020 5:16:58 PM

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www.tntextbooks.in, , Enzymes as drug targets:, In all living systems, the biochemical reactions are catalysed by enzymes. Hence, these, enzyme actions are highly essential for the normal functioning of the system. If their normal, enzyme activity is inhibited, then the system will be affected. This principle is usually applied, to kill many pathogens., We have already learnt that in enzyme catalysed reactions, the substrate molecule binds, to the active site of the enzyme by means of the weak interaction such as hydrogen bonding,, van der Waals force etc… between the amino acids present in the active site and the substrate., When a drug molecule that has a similar geometry (shape)as the substrate is administered,, it can also bind to the enzyme and inhibit its activity.In other words, the drug acts as an, inhibitor to the enzyme catalyst. These type of inhibitors are often called competitive inhibitors., For example the antibiotic sulphanilamide, which is structurally similar to p-aminobenzoic, acid (PABA) inhibits the bacterial growth. Many bacteria need PABA in order to produce an, important coenzyme, folic acid. When the antibiotic sulphanilamide is administered, it acts as, a competitive inhibitor to the enzyme dihydropteroate synthase (DHPS) in the biosynthetic, pathway of converting PABA into folic acid in the bacteria. It leads to the folic acid deficiency, which retards the growth of the bacteria and can eventually kill them., O, H2N, , S, , H, , O, , NH2, , N, HO, , O, , Sulphanilamide, , H, , p-aminobenzoic acid, , H, , O, , O, N, HO, , H, , H2N, , S, , p-aminobenzoic acid, , NH2, , O, , Sulphanilamide, , In certain enzymes, the inhibitor molecule binds to a different binding site, which is, commonly referred to as allosteric site, and causes a change in its active site geometry (shape)., As a result, the substrate cannot bind to the enzyme. This type of inhibitors are called allosteric, inhibitors., 275, , XII U15 Chemistry in Action.indd 275, , 2/19/2020 5:17:01 PM

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www.tntextbooks.in, , Receptor as drug targets:, Many drugs exert their physiological effects by binding to a specific molecule called a, receptor whose role is to trigger a response in a cell. Most of the receptors are integrated with, the cell membranes in such a way that their active site is exposed to outside region of the cell, membrane. The chemical messengers, the compounds that carry messages to cells, bind to, the active site of these receptors. This brings about the transfer of message into the cell. These, receptors show high selectivity for one chemical messenger over the others. If we want to block, a message, a drug that binds to the receptor site should inhibit its natural function. Such drugs, are called antagonists. In contrast, there are drugs which mimic the natural messenger by, switching on the receptor. These type of drugs are called agonists and are used when there is, lack of chemical messenger., , Outer surface of, cell membrane, , Binding site of receptor, , Receptor, protein, , Small part of, the cell membrane, (plasma membrane), , Interior of cell, , Inner surface, of cell membrane, , Animal cell, , Cell membrane, , Chemical, messenger, , Binding, site, , Outer, surface, of cell, membrane, , Binding, site, , Induced fit, , Interior, of cell, , Cell, membrane, Message, , For example, when adenosine binds to the adenosine receptors, it induces sleepiness. On, the other hand, the antagonist drug caffeine binds to the adenosine receptor and makes it, inactive. This results in the reduced sleepiness (wakefulness)., The agonist drug, morphine, which is used as a pain killer, binds to the opioid receptors, and activates them. This supress the neuro transmitters that causes pain., Most receptors are chiral and hence different enantiomers of a drug can have, different effect, 276, , XII U15 Chemistry in Action.indd 276, , 2/19/2020 5:17:06 PM

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www.tntextbooks.in, , CH3, , N, , N, , N, , N, , O, H2N, , H3C, , N, , N, , N, , OH, , O, , OH, , N, , CH3, , HO, , O, , ADENOSINE, , Therapeutic action of Different classes of Drugs:, The developments in the field of biology allowed us to understand various biological, process and their mechanism in detail. This enabled to develop new safer efficient drugs., For example, to treat acidity, we have been using weak bases such as aluminium and, magnesium hydroxides. But these can make the stomach alkaline and trigger the production, of much acid. Moreover, This treatment only relives the symptoms and does not control the, cause. Detailed studies reveal that histamines stimulate the secretion of HCl by activating, the receptor in the stomach wall. This findings lead to the design of new drugs such as, cimetidine, ranitidine etc.. which binds the receptor and inactivate them. These drugs are, structurally similar to histamine.In this section, we shall discuss the therapeutic action of a, few important classes of drugs., Class of Drugs, , 1) Tranquilizers, They are, neurologically active, drugs., i) Major, tranquilizers:, Haloperidol,, clozapine, ii) Minor, tranquilizers:, Diazepam (Valium),, alprazolam, , Chemical structure of some important, structures, , Mode of action, , O, , Acts on the central, nervous system, by blocking the, neurotransmitter, dopamine in the brain, , N, , N, , Cl, N, N, N, , Cl, , Uses, , N, , Valium, , alprazolam, Cl, , Treatment of stress,, anxiety, depression, sleep, disorders and severe, mental diseases like, schizophrenia, , OH, O, N, , F, , Haloperidol, , 277, , XII U15 Chemistry in Action.indd 277, , 2/19/2020 5:17:07 PM

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www.tntextbooks.in, , 2) Analgesics (Non, – narcotic), Analgesics reduce, the pain without, causing impairment, of consciousness., , They alleviate pain, by reducing local, inflammatory responses, , Uses, , O, , Used for short-term pain, relief and for modest, i) Antiinflammatory drugs painlike headache,, muscle strain, bruising,, Example, or arthritis., Acetaminophen or, These drugs have, paracetamol,, many other effects, Ibuprofen, Asprin., such as reducing fever, ii) Antipyretics, (antipyretic) and, Example, preventing platelet, coagulation. Due to this, Salicylates, property, aspirin finds, Acetylsalicylic acid, useful in the prevention, (aspirin),, of heart attacks, Acetaminophen or, Reduces fever by causing, Paracetamol, the hypothalamus, iii) Nonsteroidal, to override a, anti-inflammatory prostaglandin-induced, drugs (NSAIDs), increase in temperature.., Ibuprofen, , 3) Opioids, (Narcotic, Analgesics), Examples, Morphine, codeine, , Relive pain and produce, sleep. These drugs are, addictive. In poisonous, dose,these produces, coma and ultimately, death., , O, , CH3, COOH, , Acetylsalicylic acid (aspirin), OH, O, , H3C, , N, H, , Paracetamol, , HOOC, , Ibuprofen, , HO, , H, , O, , N, , H, , CH3, , HO, , Morphine, , Uses, Used for either shortterm or long-term relief, of severe pain. Mainly, used for post operative, pain, pain of terminal, cancer., , C, , O, H3C, , O, , H, , N, , H, , CH3, , HO, , codeine, , 278, , XII U15 Chemistry in Action.indd 278, , 2/19/2020 5:17:09 PM

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www.tntextbooks.in, , 4) Anaesthetics, i) Local anaesthetics, Examples, Ester-linked local, anaesthetic Procaine, Amide-linked, local anaesthetic Lidocaine, , It causes loss of, sensation, in the, area in which it is, applied without losing, consciousness. They, block pain perception, that is transmitted via, peripheral nerve fibres to, the brain, , O, N, O, , NH2, , Procaine, CH3, H2, C, , H, N, N, , Uses, They are often used, during minor surgical, procedures., , O, , CH3, , H2C, , CH3, , CH3, , Lidocaine, , ii) General, anaesthetics, , Cause a controlled, and reversible loss, of consciousness by, Example, affecting central nervous, Intravenous general, system, anaesthetics–, Propofol, , CH3, , OH, , CH, , CH3, CH, , H3C, , CH3, , Inhalational general Uses, anaestheticsThey are often used, Isoflurane, for major surgical, procedures., , Propofol, , 5) Antacids, Examples, , Neutralize the acid in, the stomach that causes, acidity., , Milk of Magnesia,, Sodium bicarbonate,, calcium bicarbonate, Uses, aluminium, To relieve symptoms, hydroxide, such as burning, Ranitidine,, sensation in the chest/, Cemitidine, throat area (heart burns), caused by acid reflux., Omeprazole,, rabeprazole, , OH, , HO, Al, OH, , aluminium hydroxide, , 279, , XII U15 Chemistry in Action.indd 279, , 2/19/2020 5:17:10 PM

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www.tntextbooks.in, , 6) Antihistamines, Examples, Cetirizine,, levocetirizine,, desloratadine,, brompheniramine, Terfenadine, , Block histamine release, from histamine-1, receptors, N, N, , Uses, , Examples, Penicillins,, ampicillin,, cephalosporins,, carbapenems, and, monobactams, , O, O, , To provide relief from, the allergic effects, , 7) Antimicrobials, i) Beta-Lactams, , OH, , Cl, , Cetirizine, H, , H, N, , R, , S, , C, , Inhibits bacterial cell, wall biosynthesis, , CH3, , N, , O, , CH3, , O, OH, , Uses, , O, , To treat skin infections,, dental infections, ear, infections, respiratory, tract infections,, pneumonia, urinary, tract infections, and, gonorrhoea, , Penicillins, NH2, H, , H, N, , CH3, , S, , O, , N, , CH3, , O, , OH, O, , Ampicillin, ii) Macrolides, Examples, Erythromycin,, azithromycin, , Targets bacterial, ribosomes and prevent, protein production, , O, , HO, OH, , O, , Uses, , O, , OH, O, , N, OH, , To treat respiratory, tract infections, genital,, gastrointestinal tractand, skin infections, , O, , O, , O, , O, OH, , Erythromycin, , 280, , XII U15 Chemistry in Action.indd 280, , 2/19/2020 5:17:11 PM

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www.tntextbooks.in, , 8) Antiseptics, Examples, Hydrogen peroxide,, povidone-iodine,, benzalkonium, chloride, , Stop or slow down, the growth of, microorganisms –, Applied to living tissue, Uses, To reduce the risk of, infection during surgery, and other procedures, , Povidone-iodine, , H, , 9) Disinfectants, , Stop or slow down, Examples, the growth of, microorganisms –, Chlorine, compounds, alcohol, Generally used on, Hydrogen peroxide. inanimate objects, 10) Antifertility, drugs, Example, Synthetic oestrogen, - Ethynylestradiol,, Menstranol, Synthetic, Progesterone Norethindrone,, Norethynodrel, , O, , O, , H, , Hydrogen peroxide, , These synthetic, hormones that, suppresses ovulation/, fertilisation., Uses, Used in birth control, pills., , Ethynylestradiol, , 15.2 Food additives:, Have you ever noticed the ingredients that is printed on the cover of the packed food, materials such as biscuits, chocolates etc...You might have noticed that emulsifiers such as, 322, 472E, dough conditioners 223 etc… are used in the preparation, in addition to the main, ingredients such as wheat flour, edible oil, sugar, milk solid etc… Do you think that these, substances are necessary? Yes. These substances enhance the nutritive, sensory and practical, value of the food. They also increase the shelf life of food. The substances which are not, naturally a part of the food and added to improve the quality of food are called food additives., 282, , XII U15 Chemistry in Action.indd 282, , 2/19/2020 5:17:12 PM

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www.tntextbooks.in, , 15.2.1 Important categories of food additives, • Aroma compounds, , • Artificial Sweeteners, , • Food colours, , • Antioxidants, , • Preservatives, , • Buffering substances, , • Stabilizers, , • Vitamins and minerals, , Advantages of food additives:, 1. Uses of preservatives reduce the product spoilage and extend the shelf-life of food, 2. Addition of vitamins and minerals reduces the mall nutrient, 3. Flavouring agents enhance the aroma of the food, 4. Antioxidants prevent the formation of potentially toxic oxidation products of lipids, and other food constituents, 15.2.2. Preservatives:, Preservatives are capable of inhibiting, retarding or arresting the process of fermentation,, acidification or other decomposition of food by growth of microorganisms. Organic acids such, as benzoic acid, sorbic acid and their salts are potent inhibitors of a number of fungi, yeast and, bacteria. Alkyl esters of hydroxy benzoic acid are very effective in less acidic conditions. Acetic, acid is used mainly as a preservative for the preparation of pickles and for preserved vegetables., Sodium metasulphite is used as preservatives for fresh vegetables and fruits. Sucrose esters, with palmitic and steric acid are used as emulsifiers. In addition that some organic acids and, their salts are used as preservatives. In addition to chemical treatment, physical methods such, as heat treatment (pasteurisation and sterilisations), cold treatment (chilling and freezing), drying (dehydration) and irradiation are used to preserve food., 15.2.3. Antioxidants:, Antioxidants are substances which retard the oxidative deteriorations of food. Food, containing fats and oils is easily oxidised and turn rancid. To prevent the oxidation of the fats, and oils, chemical BHT(butylhydroxy toluene), BHA(Butylated hydroxy anisole) are added, as food additives. They are generally called antioxidants. These materials readily undergo, oxidation by reacting with free radicals generated by the oxidation of oils, thereby stop the, chain reaction of oxidation of food. Sulphur dioxide and sulphites are also used as food, additives. They act as anti-microbial agents, antioxidants and enzyme inhibitors., 15.2.4 Sugar Substituents:, Those compounds that are used like sugars (glucose, sucrose) for sweetening, but are, metabolised without the influence of insulin are called sugar substituents. Eg. Sorbitol, Xylitol,, Mannitol., 15.2.5 Artificial sweetening agents:, Synthetic compounds which imprint a sweet sensation and possess no or negligible nutritional, value are called artificial sweeteners. Eg. Saccharin, Aspartame, sucralose, alitame etc…, 283, , XII U15 Chemistry in Action.indd 283, , 2/19/2020 5:17:13 PM

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www.tntextbooks.in, , 15.3 Cleansing agents:, , Soaps and detergents are used as cleansing agents. Chemically soap is the sodium or, potassium salt of higher fatty acids. Detergent is sodium salt of alkyl hydrogen sulphates or, alkyl benzene sulphonic acids., 15.3.1 Soaps:, Soaps are made from animal fats or vegetable oils. They contain glyceryl esters of long chain, fatty acids. When the glycerides are heated with a solution of sodium hydroxide they become, soap and glycerol. We have already learnt this reaction under the preparation of glycerol by, saponification. Common salt is added to the reaction mixture to decrease the solubility of, soap and it helps to precipitate out from the aqueous solution. Soap is then mixed with desired, colours, perfumes and chemicals of medicinal importance., Total fatty matter:, The quality of a soap is described in terms of total fatty matter (TFM value). It is defined, as the total amount of fatty matter that can be separated from a sample after splitting with, mineral acids., Higher the TFM quantity in the soap better is its quality., As per BIS standards, Grade-1 soaps should have 76% minimum TFM, while Grade-2, and 3 must have 70 and 60% , minimum respectively. The other quality parameters are lather,, moisture content,mushiness, insoluble matter in alcohol etc.., The cleansing action of soap:, To understand how a soap works as a cleansing agent, let us consider sodium palmitate an, example of a soap. The cleansing action of soap is directly related to the structure of carboxylate, ions (palmitate ion) present in soap. The structure of palmitate exhibit dual polarity. The, hydrocarbon portion is non polar and the carboxyl portion is polar., O, , O-, , The nonpolar portion is hydrophobic, while the polar end is hydrophilic. The, hydrophobic hydro carbon portion is, soluble in oils and greases, but not in, water. The hydrophilic carboxylate group, is soluble in water. The dirt in the cloth, is due to the presence of dust particles, intact or grease which stick. When the, soap is added to an oily or greasy part, of the cloth, the hydrocarbon part of, the soap dissolve in the grease, leaving, the negatively charged carboxylate end, exposed on the grease surface. At the, , Na+, , Na+, , Na+, , Na+, , Na+, , Na+, , Na+, Na+, , Na+, , Na+, , Na+, , Na+, Na+, , Na+, , Na+, Na+, , Na+, , Na+, , Na+, , Na+, Na+, Na+, , Nonpolar, hydrocarbon, end insoluble, in water, , Na+, , Na+, , Na+, , Highly polar, carboxylate, end soluble, in water, , Na+, , Na+, , Na+, , Na+, Na+, , Na+, , Suspended grease, particles floated, away from grease, surface, , Na+, Na+, , Grease, , Na+, , Dirt particles, embedded i a, film of grease, , Orientation of dissolved soap, at waterg–grease interface, , 284, , XII U15 Chemistry in Action.indd 284, , 2/19/2020 5:17:14 PM

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www.tntextbooks.in, , same time the negatively charged carboxylate groups are strongly attracted by water, thus, leading to the formation of small droplets called micelles and grease is floated away from, the solid object. When the water is rinsed away, the grease goes with it. As a result, the cloth, gets free from dirt and the droplets are washed away with water. The micelles do not combine, into large drops because their surfaces are all negatively charged and repel each other. The, cleansing ability of a soap depends upon its tendency to act as a emulsifying agent between, water and water insoluble greases., 15.3.2 Detergents:, Synthetic detergents are formulated products containing either sodium salts of alkyl, hydrogen sulphates or sodium salts of long chain alkyl benzene sulphonic acids. There are, three types of detergents., Detergent Type Example, Sodium Lauryl sulphate (SLS), Anionic, detergent, , O, O, Na+, , S, OO, , n-hexaadecyltrimethyl ammonium chloride, Cationic, detergent, , N+, Cl-, , N,N,N-trimethylhexadecan-1-aminium chloride, , Pentaerythrityl stearate., CH2OH, CH2OH, , Non-ionic, detergent, , O, C, , CH2OH, , O, , 3-hydroxy-2,2-bis(hydroxymethyl)propyl heptanoate, , Detergents are superior to soaps as they can be used even in hard water and in acidic, conditions. The cleansing action of detergents are similar to the cleansing action of soaps., , 15.4 Polymers, The term Polymer is derived from the Greek word ‘polumeres’ meaning “having many, parts”. The constitution of a polymer is described in terms of its structural units called, monomers. Polymers consists of large number of monomer units derived from simple, molecules. For example: PVC(Poly Vinyl Chloride). is a polymer which is obtained from, the monomer vinyl chloride. Polymers can be classified based on the source of availability,, structure, molecular forces and the mode of synthesis. The following chart explain different, classification of polymers., 285, , XII U15 Chemistry in Action.indd 285, , 2/19/2020 5:17:15 PM

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Source, , XII U15 Chemistry in Action.indd 286, , Branched polymers, (one main chain with, small chains as branches), E.g. polypropylene, LDPE., , Cross linked polymers, (linking of chain polymers), E.g. bakelite, melamine,, formaldehyde, , Semisynthetic polymers, (natural polymers-modified, by chemical treatment), E.g. viscose rayon,, cellulose diacetate, , Linear polymers, (long continuous chain), E.g. HDPE, PVC., , Thermosetting, Do not be come soft, on heating but set to an, infusible mass upon, heating., E.g. bakelite, melamine,, formaldehyde, , Condensation Polymer, formed by the, condensation of two, or more, monomers with the, elimination of simple, molecules like H2O,, NH3, etc.,, E.g. Nylon-66,, polyester., , Addition polymers., Formed by polymerisation, of monomers without, the elimination of any, byproduct., E.g. polyethylene,, PVC, teflon., , Mode of synthesis, , Thermoplastic, They become soft, on heating and hard on, cooling. they can be remolded, E.g.polythene,, PVC, polystrene, , Fibers-Polymer chains, forms fibers by, hydrogen bonding, (high tensile strength), E.g.Nylon-6,6,, Terylene, , Elastomers, (soft and stretchy), E.g. neoprene, buna-S,, buna-N, , Molecular foces, , Synthetic polymers, (man made from chemicals), E.g. polythene, PVC, etc.,, , Natural polymers, (obtained from plants/, animals)., E.g. Cellulose, Silk, , Structure, , Polymers, , www.tntextbooks.in, , 15.4.1 Classification of Polymers:, , 286, , 2/19/2020 5:17:15 PM

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www.tntextbooks.in, , 15.4.2 Types of polymerisation, The process of forming a very large, high molecular mass polymer from small structural, units i.e., monomer is called polymerisation. Polymerisation occurs in the following two ways, i. Addition polymerisation or chain growth polymerisation, ii. Condensation polymerisation or step growth polymerisation, Addition polymerisation, Many alkenes undergo polymerisation under suitable conditions. The chain growth, mechanism involves the addition of the reactive end of the growing chain across the double, bond of the monomer. The addition polymerisation can follow any of the following three, mechanisms depending upon the reactive intermediate involved in the process., i. Free radical polymerisation, ii. Cationic polymerisation, iii. Anionic polymerisation, Free radical polymerisation, When alkenes are heated with free radical initiator such as benzyl peroxide, they undergo, polymerisation reaction. For example styrene polymerises to polystyrene when it is heated to, ionic with a peroxide initiator. The mechanism involves the following steps., 1. initiation – formation of free radical, O, , O, , C, , O, , O, , O, , C, , 2, , ., , C, , O, , ., , 2, , phenyl free radical, + 2CO2, , benzoyl peroxide, , 2. Propagation step, H, , C6H5, C, , C, , CH•, H, , H, , Styrene, (monomer), , phenyl free radical, , H, , C6H5, , C, , C•, , H, , H, , Stabilised radical, , The stabilized radical attacks another monomer molecule to give an elongated radical, H, C6H5, , C6H5, , C, , C•, , H, , H, , C, , +, , C6H5, , C6H5, , H, H, , CH CH2 CH, •, growing chain, , C6H5 CH2, , C, H, , C6H5, , C6H5, , ( CH2 CH ), n, •, Polymer cahin, , monomer, , Chain growth will continue with the successive addition of several thousands of monomer units., , 287, , XII U15 Chemistry in Action.indd 287, , 2/19/2020 5:17:16 PM

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www.tntextbooks.in, , Termination, The above chain reaction can be stopped by stopping the supply of monomer or by, coupling of two chains or reaction with an impurity such as oxygen., C 6 H5, , C 6 H5, , 2 ( CH2, , CH ), , •, , ( CH2, , n, , C 6 H5, , CH ) (CH, , CH2 ), , n, , n, , 15.4. 3 Preparation of some important addition polymers, 1. Polythene, It is an addition polymer of ethene. There are two types of polyethylene i) HDPE (High, Density Polyethylene) ii) LDPE (Low Density polyethylene)., LDPE, It is formed by heating ethene at 200o to 300 C under oxygen as a catalyst. The reaction, follows free radical mechanism. The peroxides formed from oxygen acts as a free radical, initiator., 200o - 300 C, , n CH2 = CH2, , 1000 atm, , ( CH2 CH2 ), n, Polythene, , ethene, , It is used as insulation for cables, making toys etc…, HDPE, The polymerization of ethylene is carried out at 373K and 6to7 atm pressure using Zeiglar, – Natta catalyst [TiCl4 +(C2 H5 )3Al ] HDPE has high density and melting point and it is used to, make bottles, pipe etc..,, Preparation of Teflon (PTFE), The monomer is tetrafluroethylene. When the monomer is heated with oxygen (or), ammonium persulphate under high pressure, Teflon is obtained., ( CF2 CF2 ), n CF2 = CF2, n, , It is used for coating articles and preparing non – stick utensils., I. Preparation of Orlon (polyacrylonitrile – PAN), It is prepared by the addition polymerisation of vinylcyanide (acrylonitrile) using a, peroxide initiator., Peroxides, ( CH2 CH ), n (CH2 = CH), n, CN, , CN, , PAN, , Prop - 2-enenitrile, , It is used as a substitute of wool for making blankets, sweaters etc…, 288, , XII U15 Chemistry in Action.indd 288, , 2/19/2020 5:17:19 PM

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www.tntextbooks.in, , Condensation polymerisation, Condensation polymers are formed by the reaction between functional groups an adjacent, monomers with the elimination of simple molecules like H 2 O, NH 3 etc…. Each monomer, must undergo at least two substitution reactions to continue to grow the polymer chain i.e.,, the monomer must be at least bi functional. Examples : Nylon– 6,6, terylene…., Nylon – 6,6, Nylon – 6,6 can be prepared by mixing equimolar adipic acid and hexamethylene – diamine, to form a nylon salt which on heating eliminate a water molecule to form amide bonds., O, , O, , HO C (CH2 ) C, 4, , OH + H2N, , O, , (CH2 ) NH2, 6, , hexan - 1,6- diamine, , hexan - 1,6 - dioicacid, , O, , O, , C, , (CH2 ) C, , O, , 4, , (CH2 ) NH3, , NH3, , 6, , Nylon salt, , O, , O, , ..... C, , (CH2), , C, , NH, , 4, , (CH2 ) NH, 6, , O, , O, , C, , (CH2 ) C, 4, , NH, , (CH, * 2 ) 6 N......, , n, , Poly (hexamethyleneadipamide), Nylon 6,6, , It is used in textiles, manufacture of cards etc…, Nylon – 6, Capro lactam (monomer) on heating at 533K in an inert atmosphere with traces of water, gives Є-v amino carproic acid which polymerises to give nylon – 6, H, N, , O, , O, , 533K, H2O, , H2N, , (CH2) COOH, 5, , -H2O, , NH, , (CH2), , 5, , C, n, , Nylon -6, , It is used in the manufacture of tyrecards fabrics etc…., II. Preparation of terylene (Dacron), The monomers are ethylene glycol and terepathalic acid (or) dimethylterephthalate. When, these monomers are mixed and heated at 500K in the presence of zinc acetate and antimony, trioxide catalyst, terylene is formed., 289, , XII U15 Chemistry in Action.indd 289, , 2/19/2020 5:17:20 PM

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www.tntextbooks.in, , Melamine (Formaldehyde melamine):, The monomers are melamine and formaldehyde. These monomers undergo, condensation polymerisation to form melamine formaldehyde resin., H2N, , N, , NH2, O, , N, , N, , H2N, , H, , + H C, , condensation, , H, N, , N, N, , CH2, , OH, , *, , N, , NH2, , Uses: It is used for making unbreakable crockery, , NH, N, , HN, , NH2, , Methanal, , Melamine, , N, , C H2, , Polymerisation, , N, , CH2, , H, N, , CH2, , n, , Melamine-formaldehyde polymer, , Urea formaldehyde polymer:, It is formed by the condensation polymerisation of the monomers urea and formaldehyde., O, , H, H C=O +, , H2N C NH2, O, urea, , formaldehyde, , -H2O, , HO, , CH2 NH C, , NH2, , H, -H2O, H, , H2 C O, N CH2 N C, O=C, , N, CH2, , N CH2 N C, , H C=O, , H C=O, -H2O, , O, HO, , CH2 NH C, , NH CH2 OH, , N, , H2 C O, , urea formaldehyde, polymer, , 15.4.4 Co-polymers:, A polymer containing two or more different kinds of monomer units is called a copolymer. For example, SBR rubber(Buna-S) contains styrene and butadiene monomer units., Co-polymers have properties quite different from the homopolymers., 15.4.5 Natural and Synthetic rubbers:, Rubber is a naturally occurring polymer. It is obtained from the latex that excludes from, cuts in the bark of rubber tree (Ficus elastic). The monomer unit of natural rubber is cis, isoprene (2-methyl buta-1,3-diene). Thousands of isoprene units are linearly linked together, in natural rubber. Natural rubber is not so strong or elastic. The properties of natural rubber, can be modified by the process called vulcanization., 291, , XII U15 Chemistry in Action.indd 291, , 2/19/2020 5:17:22 PM

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www.tntextbooks.in, , n, cispolyisoprene, , Vulcanization: Cross linking of Rubber, In the year 1839, Charles Good year accidently dropped a mixture of natural rubber and, sulphur onto a hot stove. He was surprised to find that the rubber had become strong and, elastic. This discovery led to the process that Good year called vulcanization., Natural rubber is mixed with 3-5% sulphur and heated at 100-150˚C causes cross linking, of the cis-1,4-polyisoprene chains through disulphide (-S-S-) bonds. The physical properties, of rubber can be altered by controlling the amount of sulphur that is used for vulcanization. In, sulphur rubber, made with about 1 to 3% sulphur is soft and stretchy. When 3 to 10% sulphur, is used the resultant rubber is somewhat harder but flexible., Synthetic rubber:, Polymerisation of certain organic compounds such as buta-1,3-diene or its derivatives, gives rubber like polymer with desirable properties like stretching to a greater extent etc., such, polymers are called synthetic rubbers., Preparation of Neoprene:, The free radical polymeristion of the monomer, 2-chloro buta-1,3-diene(chloroprene), gives neoprene., n CH2 = C, , CH = CH2, , free, radical, Polymerisation, , CH2 C = CH CH2, , Cl, , Cl, , n, , It is superior to rubber and resistant to chemical action., Uses: It is used in the manufacture of chemical containers, conveyer belts., Preparation of Buna-N:, It is a co-polymer of acrylonitrile and buta-1,3-diene., n CH2 = CH, , CH = CH2 + nCH2 = CH, CN, , Na, , CH2 CH = CH CH2 CH CH2, CN, , Vinyl cyanide, , n, , Buna-N, , It is used in the manufacture of hoses and tanklinings., Preparation of Buna-S:, It is a co-polymer. It is obtained by the polymerisation of buta-1,3-diene and styrene in, the ratio 3:1 in the presence of sodium., 292, , XII U15 Chemistry in Action.indd 292, , 2/19/2020 5:17:23 PM

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www.tntextbooks.in, , n H2N, , CH2, , COOH, , +, , n, , H2N, , Glycine, , HN, , (CH2) COOH, 5, , C, , CH2, , Aminocaproic acid, , NH, , (CH2 ) C, 5, , O, , + (2n-1) H2O, , O, , Nylon -2-nylon - 6, , EVALUATION, Choose the correct answer:, 1. Which of the following is an analgesic?, a) Streptomycin, , b) Chloromycetin, , c) Asprin, , d) Penicillin, , 2. Antiseptics and disinfectants either kill or prevent growth of microorganisms. Identify, which of the following statement is not true., a) dilute solutions of boric acid and hydrogen peroxide are strong antiseptics., b) Disinfectants harm the living tissues., c) A 0.2% solution of phenol is an antiseptic while 1% solution acts as a disinfectant., d) Chlorine and iodine are used as strong disinfectants., 3. Drugs that bind to the receptor site and inhibit its natural function are called, a) antagonists, , b) agonists, , c) enzymes, , d) molecular targets, , 4. Aspirin is a/an, a) acetylsalicylic acid b) benzoyl salicylic acid c) chlorobenzoic acid d) anthranilic acid, 5. Which one of the following structures represents nylon 6,6 polymer?, (a), (a), , (b), (b), , (a), , NH2, NH2, , CH3, CH3 66, , NH2, , CH3, , (b), , 66, 66, , NH2, NH2, , NH2, NH2 66, , NH2, , NH2, , 66, 66, , (c), (c), (c), H2N, H2N, , Cl, Cl, , H3C, H3C, , HOOC, HOOC, , 6, 6, , H2N, , Cl, , H3C, , HOOC, , 6, , O, O, , (d), (d), , O, C, C, , (d), , C, , 2, 2, 2, , C, C, C, O, O, O, , N, N, H, N, H, , H, H, N, N, (CH2)6, (CH2)H, 6, N, (CH2)6, , n, n, , H, , n, , 294, , XII U15 Chemistry in Action.indd 294, , 2/19/2020 5:17:24 PM

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www.tntextbooks.in, , 6. Natural rubber has, a) alternate cis- and trans-configuration, b) random cis- and trans-configuration, c) all cis-configuration , d) all trans-configuration, 7. Nylon is an example of, a) polyamide, b) polythene, c) polyester, d) poly saccharide, 8. Terylene is an example of, a) polyamide, b) polythene, c) polyester, d) polysaccharide, 9. Which is the monomer of neoprene in the following?, b) CH2 =CH C, , a) CH2 C CH = CH2, , CH, , Cl, c) CH2 =CH CH =CH2, , d)CH2 = C CH = CH2, CH3, , 10. Which one of the following is a bio-degradable polymer?, , a) HDPE, , b) PVC, , c) Nylon 6, , d) PHBV, , 11. Non stick cook wares generally have a coating of a polymer, whose monomer is, , a) ethane  b) prop-2-enenitrile  c) chloroethene  d) 1,1,2,2-tetrafluoroethane, 12. Assertion: 2-methyl-1,3-butadiene is the monomer of natural rubber, , Reason: Natural rubber is formed through anionic addition polymerisation., a) If both assertion and reason are true and reason is the correct explanation of assertion., b) if both assertion and reason are true but reason is not the correct explanation of, assertion., c) assertion is true but reason is false., , d) both assertion and reason are false., , 13. Which of the following is a co-polymer?, a) Orlon, , b) PVC, , c) Teflon, , d) PHBV, , 14. The polymer used in making blankets (artificial wool) is, a) polystyrene, b) PAN, c) polyester, d) polythene, 15. Regarding cross-linked or network polymers, which of the following statement is incorrect?, (NEET), a) Examples are Bakelite and melamine, b) They are formed from bi and tri-functional monomers, c) They contain covalent bonds between various linear polymer chains, d) They contain strong covalent bonds in their polymer chain, , 295, , XII U15 Chemistry in Action.indd 295, , 2/19/2020 5:17:25 PM

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www.tntextbooks.in, , Short Answer Questions, 1. What are antibiotics?, 2. Name one substance which can act as both analgesic and antiphyretic, 3. Write a note on synthetic detergents, 4. How do antiseptics differ from disinfectants?, 5. What are food preservatives?, 6. What are drugs? How are they classified, 7. How the tranquilizers work in body., 8. Write the structural formula of aspirin., 9. Explain the mechanism of cleansing action of soaps and detergents, 10. Which sweetening agent are used to prepare sweets for a diabetic patient?, 11. What are narcotic and non – narcotic drugs. Give examples, 12. What are anti fertility drugs? Give examples., 13. Write a note on co –polymer, 14. What are bio degradable polymers? Give examples., 15. How is terylene prepared?, 16. Write a note on vulcanization of rubber, 17. Classify the following as linear, branched or cross linked polymers, a) Bakelite, , b) Nylon-6,6, , c) LDPE, , d) HDPE, , 296, , XII U15 Chemistry in Action.indd 296, , 2/19/2020 5:17:25 PM

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www.tntextbooks.in, , ANSWERS, , VOLUME, , II, , UNIT 8, MCQ, 1., , Ag 2C 2O4  2Ag + +C 2O4 2-, , 3., , Ag +  =2.24×10-4 mol L-1, 2.24×10-4, C 2O4 2-  =, mol L-1, 2, =1.12×10-4 mol L-1, , sp, , K sp =(s), =(s)22, K, sp, 2, = 2.42×10, 2.42×10-3-3gg LL-1-1 2, =, , , , = (2.24×10-4 mol L-1 ) (1.12×10-4 mol L-1 ), 2, , -3, -3, , , , =5.619×10-12mol 3L-3, , =(1.038×10 )), =(1.038×10, -10, =1.077×10-10, =1.077×10, , [Option (d)], , M, M, HCl + 25ml NaOH, 5, 5, , -5 2, -5 2, , , , -10, =1.08×10-10, mol22 LL-2-2, =1.08×10, mol, , No of moles of HCl = 0.2×75×10-3 = 15 × 10-3, , [Option (c)], , No of moles of NaOH = 0.2 × 25 × 10-3 = 5 × 10-3, No of moles of HCl after mixing = 15 × 10-3 - 5 × 10-3, = 10 × 10-3, , No of moles of HCl, Vol in litre, 10 × 10−3, =, = 0.1M, 100 × 10−3, , ∴concentration of HCl =, , for (iii) solution, pH of 0.1M HCl = -log, , = 1., , , , 2, -1 2, -1, ,  2.42×10 g L , =  2.42×10 g -1L , =, 233g mol, mol-1 ,  233g, -3 2, -3 2, =, 0.01038×10, , = 0.01038×10, , 2, , K sp = Ag +  C 2O4 2- , , 2. iii) 75 ml, , BaSO4  Ba 2+ +SO4 2K sp =(s), =(s) (s), (s), K, , 10, , (0.1), , 4., , Ca(OH)2  Ca 2+ +2OHGiven that pH=9, pOH=14-9=5, pOH=-log10[OH- ], ∴[OH- ]=10- pOH, [OH- ]=10-5 M, K sp =[Ca 2+ ][OH- ]2, 10-5, =, × (10-5 )2, 2, =0.5 × 10-15, , [Option (d)]., , [Option (a)], , 297, , Answers.indd 297, , 2/19/2020 5:07:50 PM

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www.tntextbooks.in, , Percentage of = 1.7 × 10−4 × 100, dissociation, −2, , =1.3 × 10 = 0.013 %, , [Option (b)], 11. pH=-log10[H+ ], , ∴[H+ ]=10-pH, Let the volume be x mL, V1M1 +V2 M2 +V3M3 =VM, ∴ x mL of 10 M+ x mL, -1, , of 10-2 M + x mL of 10-3 M, = 3x mL of [H ], x[0.1+0.01+0.001], ∴[H+ ]=, 3x, 0.1 + 0.01 + 0.001, =, 3, 0..111, =, 3, = 0.037, +, , = 3.7 × 10−2, , 14. Addition of salt KY (having a common ion Y–) decreases, the solubility of MY and NY3 due to common ion, effect., Option (a) and (b) are wrong., For salt MY , MY  M + +Y -, , K sp =(s)(s), 6.2 × 10-13 =s 2, ∴ s= 6.2 × 10-13  10-7, for salt NY3 ,, NY3  N 3+ +3Y K sp = (s)(3s)3, K sp =27s 4,  6.2 × 10−13 , s= , 27 , , , 1, , 4, , s.  10-4, NY3, , The molar solubility of MY in water is less than of, [Option (d)], , [Option (a)], , 12. AgCl (s)  Ag (aq)+Cl - (aq), +, , 15. x ml of 0.1 M NaOH + x ml of 0.01 M HCl, No of moles of NaOH = 0.1 × x × 10 –3= 0.1x × 10-3, , NaCl , → Na + +Cl -, , No of moles of HCl = 0.01 × x × 10-3 = 0.01x × 10-3, , K sp =1.6 × 10, , No of moles of NaOH after mixing = 0.1x × 10-3 - 0.01x × 10-3, , 0.1 M, , 0.1 M, -10, , 0.1 M, , K sp =[Ag + ][Cl - ], , = 0.09x × 10-3, , 0.09 x × 10−3, Concentration of NaOH=, =0.045, 2 x × 10−3, OH-  = 0.045, , K sp =(s)(s+0.1), 0.1>>>s, ∴ s+0.1  0.1, 1.6 × 10-10, -9, = 1.6 × 10, ∴ S=, 0.1, [Option (b)], 13. PbI2 (s)  Pb (aq)+2I (aq), 2+, , -, , POH = −log ( 4.5 × 10−2 ), = 2 −log 4.5, = 2- 0.65 = 1.35, , pH = 14-1.35=12.65, , K sp =(s)(2s), , 2, , 3.2 × 10-8 =4s 3,  3.2 × 10-8 , s= , , 4, , =(8 × 10-9 ), , 1, , 1, , 3, , 3, , =2 × 10-3M [Option (a)], 299, , Answers.indd 299, , 2/19/2020 5:07:56 PM

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www.tntextbooks.in, , 16. K a =1 × 10, , [Option (c)], , -3, , 20. pH=-log10[H ], +, , pH=4, [salt], =?, [Acid], , ∴[H+ ]=10-pH, =100 =1, , pH = pK a +log, , [Salt], [Acid], , [H+ ]=1M, The solution is strongly acidic, [Option (b)], 21. According to Henderson equation, , [Salt], 4= - log10 (1 × 10-3 )+log, [Acid], [Salt], [Acid], [Salt], 1=log10, [Acid], [Salt], ⇒, =101, [Acid], 4=3+log, , pH=pK a +log, , ie. - log [H + ]=-logK a +log, -log[H + ]=log, , [Acid] 1, i.e.,, =, [Salt] 10, , log, , 10-5m, , +, , 10-5m, -5, , (, , 10-5m, , 22. h=, , pH = 14 - -log OH , =14 + log [OH - ], -, , [salt] 1, ×, [acid] K a, , [acid], [salt], , [option (a)], , + OH-, , [OH ]=10 M., pH= 14 - pOH, -, , [salt], [acid], , [salt] 1, 1, ×, = log, +, [acid] K a, [H ], , ∴[H + ]=K a, , 1:10, [Option (d)], 17. KOH → K, , [salt], [acid], , Kh, K a .K b, , [Option (c)], , ), , 23. K h =, , K w 1×10-14, =, K b 1.8×10-5, , =0.55×10-9, , = 14 + log 10−5, = 14 −5, = 9., , =5.5×10-10, [option (b)], , [Option (a)], 18. H3PO 4 +H − OH  H3O + H2 PO 4, +, , base 1, , acid 1, , acid 2, , -, , base 2, , ∴ H2 PO4 is the conjugate base of H3PO4, −, , [Option (c)], 2−, 19. HPO 4 can have the ability to accept a proton to, −, form H2 PO 4 ., It can also have the ability to donate a proton to, -3, form PO 4, , 300, , Answers.indd 300, , 2/19/2020 5:08:00 PM

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www.tntextbooks.in, , Key answer for short answer question, , 16. Given, , [H3O+ ]=0.04 mol dm -3, , K a =10-9, c=0.4M, , pH=-log[H3O+ ], =-log(0.04), , [H+ ]= K a × c., , 8. Concentration of HNO3 = 0.04M, , pH=-log[H+ ], , =-log(4 × 10-2 ), =2-log4, =2-0.6021, =1.3979 = 1.40, , = 10-9 × 0.4, =2 × 10-5, ∴ pH= - log (2 × 10-5 ), =5-log2, =5-0.3010,    =4.699., , 14. Ba(OH)2 → Ba 2+ +2OH-, , 1.5 × 10-3M, , 2 × 1.5 × 10-3M., , [OH- ] = 3 × 10-3M, [ pH+pOH=14], pH= 14-pOH, , 17. h =, , Kh =, , KW, 1 × 10−14, =, KaK b, 1.8 × 10−5 × 1.8 × 10−5, , 1, × 10−4, 1.8, = 0.7453 × 10−2, , =, , pH= 14- ( −log [OH ]), -, , = 14+ log [OH- ], , pH = 1 2 pK w + 1 2 pK a - 1 2 pK b, Given that K a = K b = 1.8 × 10-5, , =14+log(3 × 10-3 ), =14 + log 3 + log 10 -3, = 14 + 0.4771 −3, =11+0.4771, , if K a = K b , then, pK a = pK b, ∴ pH = 1 2 pK w = 1 2 (14) = 7, , pH=11.48, 15. Number of moles of HNO3 = 0.05 × 50 × 10-3, =2.5 × 10-3, Number of moles of KOH = 0.025 × 50 × 10−3, , = 1.25 × 10-3, , Number of moles of HNO3 after mixing, = 2.5 × 10-3– 1.5 × 10-3, = 1.25 × 10-3, , ∴ concentration of HNO3 =, , Number of moles of HNO3, Volume is litre, , After mixing, total volume = 100 ml = 100×10-3L, , 1.25 × 10-3moles, ∴[H ]=, 100 × 10-3L, =1.25 × 10-2 moles L-1, +, , pH = - log [H + ], pH = − log(1.25 × 10−2 ) = 2 − 0.0969, = 1.9031, 301, , Answers.indd 301, , 2/19/2020 5:08:05 PM

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www.tntextbooks.in, , Evaluate yourself – 2, , Total volume = 0.250L, , Pb(NO3 )2  Pb +2NO3, 2+, , 0.1 M, , 0.1 M, , -, , 0.2 M, , Number of moles, , NH 4 + + H 2O  H 3O + + NH 3, , Pb = molarity × Volume of the solution in lit, 2+, , acid 1, , base 2, , acid 2, , base 1, , = 0.1 × 0.15, , [Pb2+ ]mix =, , 0.1 × 0.15, =0.06 M, 0.25, , NaCl  Na + + Cl0.2 M, , 0.2 M, , 0.2 M, , H 2SO 4 + H 2O  H 3O + + HSO 4 , , No.of moles Cl =0.2 × 0.1, 0.2 × 0.1, =0.08 M, [Cl- ]mix =, 0.25, -, , acid 1, , base 2, , acid 2, , base 1, , Precipitation of PbCl 2 (s) occurs if, , [Pb2+ ][Cl - ]2 >K sp, , CH 3COOH + H 2O  H 3O + + CH3COO -, , [Pb2+ ][Cl - ]2 =(0.06)(0.08)2, , acid 1, , =3.84 × 10-4, , base 2, , acid 2, , base 1, , Since ionic product [Pb ][Cl ] >K sp ,, 2+, , - 2, , PbCl 2 is precipitated, 3+, , 27. Al(OH)3  Al (aq)+3OH (aq), -, , K sp =[Al 3+ ][OH- ]3, Al(OH)3 precipitates when, , Evaluate yourself – 3, , i) CaO - Lewis base ; CO2 -Lewisacid, ii) H3C, , O - Lewis base, H3C, , [Al 3+ ][OH- ]3 >K sp, (1 × 10-3 )[OH- ]3 >1 × 10-15, , [OH]3 >1 × 10-12, , AlCl3 - Lewis acid, , [OH- ]>1 × 10-4 M, , Evaluate yourself – 4, , [OH- ]=1 × 10-4 M, , HO, , POH=-log10[OH- ]=-log(1 × 10-4 )=4, pH = 14-4= 10, Thus, Al (OH)3 precipitates at a pH of 10, , B, , HO, , OH, , HO, , Evaluate yourself – 5, , Evaluate yourself, , Given solution is neutral, , Key, , ∴ [ H3O+ ] = [OH- ], , Evaluate yourself – 1, , K w = [H3O+ ][OH- ], , acid : (i) HNO3 iii) H3PO3, base : ii) Ba (OH)2, , iv) CH3COOH, , ; electron pair acceptor - Lewis acid, , Let [H3O+ ] = x ; then [OH - ] = x, , 4 × 10-14 =x . x, x 2 =4 × 10-14, x = 4 × 10-14 =2 × 10-7, , 303, , Answers.indd 303, , 2/19/2020 5:08:14 PM

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www.tntextbooks.in, , Evaluate yourself – 6, , Evaluate yourself - 7, , a) Answer, , Kb, 1.8 × 10-5, α=, =, C, 6 × 10-2, , H O, 2  2H O+ +SO 2, H SO , ,  3, 2-8 4, 4, -8, -8, 10, , M, , 10, , 2×10, , M, , In this case the concentration of H 2SO 4 is very, +, low and hence [H3O ] from water cannot be neglected, , = 3 × 10-4, = 1.732 × 10-2, , -8, ∴[H3O+ ] = 2 × 10 (from H2SO4 ) + 10-7, , =, , (from water), , 1.732, = 1.732%, 100, , =10-8 (2+10), , Evaluate yourself – 8, , =12×10-8 = 1.2 × 10-7, , a) Answer, Dissociation of buffer components, , pH= - log10[H3O+ ], , NH 4OH (aq)  NH 4 + (aq)+OH- (aq), NH 4Cl → NH 4+ +Cl +, , = - log10 (1.2 × 10 ), =7 - log10 1.2, =7 - 0.0791=6.9209, -7, , Addition of H+, The added H+ ions are neutralized by, NH 4OH and there is no appreciable decrease in, pH., , b) Answer, , pH of the solution = 5.4, , H 3O +  = antilog of (-pH), = anitlog of (-5.4), , NH 4OH(aq) + H+ → NH 4+ (aq) + H2O(l), Addition of OH–, , NH 4 + (aq) + OH- (aq) → NH 4OH (aq), +, The added OH ions react with NH 4 to, produce unionized NH 4OH . Since NH 4OH is, , = antilog of (-6 + 0.6) = 6.6, = 3.981 × 10-6, i.e., 3.98 × 10-6 mol dm -3, , a weak base, there is no appreciable increase in pH, , c) Answer, No of moles of HCl, , b) Answer, pH of buffer, , = 0.2 × 50 × 10 -3 = 10 × 10 -3, , CH3COOH(aq)  CH3COO- (aq)+H+ (aq), α, α, 0. 4 - α, , No of moles of NaOH = 0.1 × 50 × 10 = 5 × 10, -3, , -3, , No of moles of HCl after mixing = 10 × 10-3 - 5 × 10-3, , 0.4, , = 5 × 10-3, after mixing total volume, ∴, , CH 3COONa (aq) , → CH 3COO- (aq) + Na + (aq), 0.4, , 0.4, , K a [CH3COOH], [CH3COO- ], [CH3COOH]=0.4 - α  0.4, [H+ ]=, , = 100mL, , Concentration of HCl in moles per litre =, , 5×10-3 mole, 100 × 10-3 L, , [CH3COO- ] = 0.4 + α  0.4, , K a (0.4), (0.4), +, [H ]=1.8 × 10-5, , ∴ [H+ ]=, H3O+  =5 × 10-2 M, pH = - log (5 × 10-2 ), , ∴ pH = - log (1.8 × 10-5 ) = 4.74, , = 2 - log 5, = 2 - 0.6990, , Addition of 0.01 mol HCl to 500ml of buffer, Added [H ]=, +, , = 1.30, , 0.01 mol 0.01 mol, =, 1 L, 500 mL, 2, , 304, , Answers.indd 304, , 2/19/2020 5:08:21 PM

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www.tntextbooks.in, , 10.70 g ammonium chloride is dissolved in, water and the solution is made up to one litre to, get 0.2M solution. On mixing equal volume of, the given NH 4OH solution and the prepared, NH 4Cl solution will give a buffer solution with, required pH value (pH = 9)., , = 0.02M, CH3COOH(aq)  CH3COO- (aq)+H+ (aq), α, , α, , 0.4 - α, , CH3COONa → CH3COO + Na, 0.4, , -, , +, , 0.4, , 0.4, , CH3COO + HCl → CH3COOH+ Cl -, , 0.02, , (0.02), , 0.02, , 0.02, , b)answer, , ∴ [CH3COOH] = 0.4 - α + 0.02 = 0.42 - α  0.42, [CH3COO ]= 0.4 + α - 0.02 = 0.38 + α  0.38, -, , (1.8 × 10-5 ) (0.42), [H ]=, (0.38), +, [H ] = 1.99 × 10-5, +, , = 0.6 × V × 10−3, , [formic acid] = number of moles of HCOOH, , pH = - log (1.99 × 10-5 ), = 5 - log 1.99, = 5 - 0.30, = 4.70, , Evaluate yourself – 9, a) answer, , pOH = pK b +log, , [salt], [acid], [sodium formate], 4=3.75+log, [formic acid], [Sodium formate] = number of moles of HCOONa, pH = pK a +log, , = 0.8 × 100 × 10−3, = 80 × 10−3, , 4 = 3.75 + log, 0.25=log, , [salt], [base], , 0.6V, 80, , 0.6V, 80, , antilog of 0.25 =, , We know that, , 0.6V = 1.778 × 80, = 1.78 × 80, = 142.4, , pH + pOH = 14, ∴ 9 + pOH = 14, ⇒ pOH = 14 - 9 = 5, [NH 4Cl], 5 = 4.7 + log, [NH 4OH], , V=, , 0.6V, 80, , 142.4 mL, = 237.33mL, 0. 6, , Evaluate yourself - 10, , [NH 4Cl], 0.1, [NH 4Cl], = antilog of (0.3), 0.1, [NH 4Cl] = 0.1M × 1.995, = 0.1995 M, = 0.2M, Amount of NH 4Cl required to, , Sodium carbonate is a salt of weak acid,, , 0.3=log, , H2CO3 and a strong base, NaOH, and hence the, , solution is alkaline due to hydrolysis., , Na 2CO3 (aq) → 2Na + (aq) + CO32- (aq), CO32- (aq)+ H2O (l)  HCO3- +OHi) h=, , Kw, Ka × C, , 1 × 10-14, =, 5.5 × 10-11 × 0.05, h = 6.03 × 10 -2, , prepare 1 litre 0.2M solution = Strength of NH 4Cl, × molar mass of, , NH 4Cl, , = 0.2 × 53.5, = 10.70 g, 305, , Answers.indd 305, , 2/19/2020 5:08:24 PM

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www.tntextbooks.in, , Given that pK a =10.26, pK a = -log K a, i.e., K a =antilog of (-pK a ), = antilog of (-10.26), , 4., , = 10-11 × 5.5, [antilog of 0.74 = 5.49  5.5], , (Option (b)), , K w 1 × 10-14, ii) K h =, =, K a 5.5 × 10-11, , 5., , = 1.8 × 10, pK a logC, iii) pH = 7 +, +, 2, 2, 10.26 log 0.05, =7+, +, = 7 + 5.13 −0.65, 2, 2, = 11.48, -4, , 1. 1F = 96500 C = 1 mole of e = 6.023 × 10, −, , 23, , e, , (Option (b)), , −, , 7., , 7MnO-4 + 5e− → Mn2+ + 4H2O, , 5 moles of electrons i.e., 5F charge is required., (Option (a)), 8., , Option (C), , Mn2+ + 2e − → Mn (Ered ) = -1.18V, , 2 Mn2+ → Mn3+ + e − ( Eoox ) = −1.51V, , m=ZIt, 40 × 3.86 × 2500, =, 2 × 96500, = 2g, , 41min 40sec = 2500 seconds, m, 40, =, Z=, n × 96500 2 × 96500, , (Option (b)), 9. m=ZIt, , 3Mn2+  Mn  2Mn3+ E cell  ?, o, Eocell = ( Eoox ) + ( Ered, ), , t=, =, , and non spontaneous, , Since E is –ve DG is +ve and the given, o, , m, ZI, , (mass of 1 mole of Cl 2 gas = 71), ( ∴ mass of 0.1mole of Cl 2 gas = 7.1 g mol −1 ), , 7. 1, (2 Cl - → Cl 2 +2e- ), 71, ×3, 2 × 96500, , 2 × 96500 × 7.1, 71 × 3, = 6433.33sec, =107.2 min, , =, , forward cell reaction is non – spontaneous., (Option (b)), 3. Anodic oxidation: (Reverse the given reaction), , ( E ) = 0.76V cathodic reduction, ∴ E = (E ) + (E ), o, ox, , o, ox, , = ( Λ  )HCl + ( Λ  ) NaOAC  - ( Λ  ) NaCl, = (426.2 + 91) −(126.5), = 390.7, , ch arg eof 6.022 × 1023 e−, , 6.22 × 10, × 9650 = 6.022 × 1022, 96500, , , cell, , ∞ HoAC, , 6. 1F = 96500 C = charge of 1 mole of e- =, , 23, , = −1.51 −1.18, = −2.69V, , (Λ ), , (Option (c)), , Unit 9 Electro Chemistry, , 2., , κ, × 10−3 mol −1 m 3, M, , 5.76 × 10−3 S c m −1 × 10−3, =, mol −1 m 3, 0.5, −3, 5.76 × 10 × 10−3 × 106, =, S cm −1mol −1 cm 3 ., 0. 5, = 11.52 S cm 2 mol −1, , = antilog off (-11 + 0.74), , ∴ 9650 C =, , Λ=, , (Option (b)), , o, red, , = 0.76 + 0.34 = 1.1V, (Option (c)), , 306, , Answers.indd 306, , 2/19/2020 5:08:29 PM

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www.tntextbooks.in, , 10., , Q =It, = 1A × 60S, 96500 C charge ≡ 6.022 × 1023 electrons, 6.022 × 1023, 60 C charge ≡, × 60, 96500, = 3.744 × 1020 electrons, (Option (C)), , 11. In general, specific conductance of an electrolyte decreases with dilution. So, 0.002N solution has least, specific conductance., (Option (b)), 12. Charging : anode : PbSO 4 (s)+ 2e → Pb (s) + SO 4 (aq), −, , -2, , Cathode : PbSO 4 (s)+ 2H2O (l) → PbO2 (s) + SO 4 (aq)+2e, -2, , −, , (Option (C)), 13. Option (a) I and IV, 14. E Zn2+ Zn = −0.76V and EFe2+ Fe = −0.44 V Zinc has higher negative electrode potential than iron, iron, o, , o, , cannot be coated on zinc., Option (d), 15. Both are false, i) Dry air has no reaction with iron, ii) Rust has the composition Fe2O3 . x H2O, (Option (d)), 16. (Option (a)), 17., , α=, , Λ, 6, =, Λ o 400, , K a =α2C, 6, 6, 1, ×, ×, =, 400 400 36, = 6.25 × 10−6, Option (b), 18., , l, A, R, cell constant =, ρ, R=ρ., , 1, = κ.R  =, ρ, , , κ, , , = 1.25 × 10−3 Ω−1cm −1 × 800 Ω, = 1 cm−1, Option (c), 19. Option (d), , 307, , Answers.indd 307, , 2/19/2020 5:08:30 PM

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www.tntextbooks.in, , 1.5 × 10−3 S m 2 mol −1, 300 × 10−4 S m 2 mol −1, α = 0.05, α2 c, Ka =, 1-α, (0.05)2 (0.01), =, 1-0.05, 25 × 10−4 × 10−2, =, 95 × 10−2, = 0.26 × 10−4, α=, , = 2.6 × 10−5., 13. Given, , V=250 mL, , I = 1.608A; t = 50 min = 50 × 60 ,   = 3000S, , C = 0.5 M, , η = 100%, , Calculate the number of faradays of electricity passed through the CuSO 4 solution, , ⇒ Q=It, Q = 1.608 × 3000, Q = 4824C, ∴ number of Faradays of electicity =, Electrolysis of CuSO4, , 4824 C, = 0.05 F, 96500 C, , Cu 2+ (aq)+2e− → Cu(s)., The above equation shows that 2F electricity will deposit 1 mole of Cu to Cu., 2+, , ∴ 0.05F electricity will, deposit, , 1mol, × 0.05 F = 0.025 mol, 2F, , 0.55, × 250mL, 1000 mL, = 0.125 mol, 2+, ∴ number of moles of Cu after electrolysis = 0.125 - 0.025, = 0.1 mol, 0.1 mol, ∴ Concentration of Cu2+ =, × 1000 mL, 250 mL, = 0.4 M, Initial number of molar of Cu 2+ in 250 ml of solution =, , 14. Required half cell reaction, , 2 Br - → Br2 + 2e −, , 2 Fe3+ + 2e − → 2Fe 2+, 2Fe3+ + 2Br − → 2Fe2+ +Br2, , ( E ) = −1.09V, ( E ) = + 0.771V, (E ) = ?, o, ox, , , red, , o, cell, , 309, , Answers.indd 309, , 2/19/2020 5:08:34 PM

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www.tntextbooks.in, , E ocell = ( E oox ) + ( E ored ), = −1.09 + 0.771, = −0.319V, , Ecell is – ve; DG is +ve and the cell reaction is non spontaneous. Hence Fe3+ cannot oxidises Br - to Br2, ., 15., , (E ), , o, ox Fe Fe2+, , o, = 0.44 V and ( Ered, )Cu, , 2+, , Cu, , = 0.34 V ., , These +ve emf values shows that iron will oxidise and copper will get reduced i.e., the vessel will dissolve., Hence it is not possible to store copper sulphate in an iron vessel., 16. Metals having higher oxidation potential will liberate H2 from H2SO 4 . Hence, the metal, M1 having + xV, oxidation potential will liberate H2 from H2SO4 ., 17. oxidation potential of M1 is more +ve than the oxidation potential of Fe which indicates that it will, prevent iron from rusting, 18. Cell reactions:, Oxidation at anode: Cd (s) → Cd (aq) + 2e, 2+, , −, , o, ox Cd Cd 2+, , Reduction at cathode: Cu (aq) + 2e → Cu (s), 2+, , (E ), (E ), , -, , = 0. 4 V, , o, red Cu2+ Cu, , = 0..34V, , Cd ( s ) + 2e- → Cd 2+ (aq) + Cu(s), o, Ecell = ( Eoox ) + ( Ered, )cathode, , = 0.4 + 0.34, = 0.74 V., , emf is +ve, so DG is (-)ve, the reaction is feasible., 19. Oxidation at anode:, , 2H2 (g) + 4OH- (aq) → 4H2O (l) + 4e−, , 1 mole of hydrogen gas produces 2 moles of electrons at 250 C and 1 atm pressure, 1 mole of hydrogen, gas occupies = 22.4 litres, , 1 mole, × 44.8 litres, 22.4 litres, = 2 moles of hydrogen, , ∴ no. of moles of hydrogen gas produced =, , ∴ 2 of moles of hydrogen produces 4 moles of electron i.e., 4F charge., , We know that Q= It, , I=, , Q, t, , 4F, 10 mins, 4 × 96500 C, =, 10 × 60 s, I=643.33 A, =, , Electro deposition of copper, , Cu 2+ (aq)+2e- → Cu(s), , 2F charge is required to deposit, , 310, , Answers.indd 310, , 2/19/2020 5:08:41 PM

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www.tntextbooks.in, , 1 mole of copper i.e., 63.5 g, If the entire current produced in the fuel cell ie., 4 F is utilised for electrolysis, then, 2× 63.5 i.e., 127.0 g copper will be deposited at cathode., 20. Ni (aq) + 2e → Ni (s), 2+, , -, , Cr 2+ (aq)+3e− → Cr (s), , The above reaction indicates that 2F charge is required to deposit 58.7g of Nickel form nickel nitrate and, 3F charge is required to deposit 52g of chromium., Given that 2.935 gram of Nickel is deposited, ∴The amount of charge passed through the cell =, , 2F, × 2.935 g, 58.7 g, , = 0.1F, ∴ if 0.1F charge is passed through chromium nitrate the amount of chromium deposited, =, , = 1.733g, , 52 g, × 0.1F, 3F, , 21. Given that, , Cu 2+  = 0.1M, , = 0.34, ECu, Cu, 2+, , Ecell = ?, Cell reaction is, , Cu 2+ (aq) + 2e− → Cu (s), , 0.0591, [Cu ], log, n, Cu 2+ , 0.0591, 1, log, = 0.34 −, 2, 0. 1, = 0.34 −0.0296, , Ecell = Eo -, , = 0.31V, , 22. oxidation at anode, , Mg → Mg2+ +2e- ..............(1), Reduction at cathode, , Ag + + e- → Ag ........... (2), , ( E ) =2.37V, o, ox, , 0.80V, (E ) =, o, red, , ∴ E ocell = ( E oox )anode + ( E ored )cathode, = 2.37+0.80, = 3.17V, Overall reaction, , Equation (1) + 2 × equation ( 2 ) ⇒, Mg+ 2Ag 2+ → Mg 2+ + 2Ag, , 311, , Answers.indd 311, , 2/19/2020 5:08:45 PM

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www.tntextbooks.in, , ∆G = -nfE, = −2 × 96500 × 3.17, = −611810 J, ∆G = −6.12 × 105 J, o, , W = 6.12 × 105 J, ∆G o =−2.303 RT log K C, 6.12 × 105, 2.303 × 8.314 × 298, K c = Antilog of (107.2), , ⇒ log K c =, , 23. Electrolysis of water, At anode:, , 2H 2 O → 4H+ + O 2 + 4e−.........(1), , At cathode:, , 2H2O+2e− → H2 + 2OH+, Overall reaction 6H2O → 4H + 4OH + 2H2 + O2, (or), , Equation (1) +(2) ×2 ⇒ 2H 2 O → 2H 2 + O 2, ∴ According to faradays Law of electrolysis, to electrolyse two mole of Water (36g  36 mL of H2O), 4F, charge is required alternatively, when 36 mL of water is electrolysed, the charge generated = 4 × 96500 C., ∴ When the whole water which is available on the lake is completely electrolysed the amount of charge, , 4 × 96500 C, × 9 × 1012 L, 36 mL, 12, 4 × 96500 × 9 × 10, =, C, 36 × 10−3, = 96500 × 1015 C, ∴Given that in 1 second, 2 × 106 C is generated therefore, the time required to generate, 1S, 96500 × 1015 C is =, × 96500 × 1015 C, 6, 2 × 10 C, generated is equal to, , , , = 48250 × 109 S, , 48250 × 109, 365 × 24 × 60 × 60, = 1.5299 × 106 years, , ∴ Number of years =, , 1 year = 365 days, = 365 × 24 hours, = 365 × 24 × 60 min, = 365 × 24 × 60 × 60 sec., , 312, , Answers.indd 312, , 2/19/2020 5:08:50 PM

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www.tntextbooks.in, , Unit 10 Surface Chemistry, S.No., , Answers, , 1., , (c), , 1, x, =k.p n, m, x, 1, ⇒ log( )=logk+ logp, m, n, , y=c+mx, , m=, , 1, and c= logk, n, , 2., , The incorrect statement is option (b), Physisorption is an exothermic process. Hence increase in temperature decreases the, physisorption., , 3., , (d), Adsorption leads to decrease in randomness (entropy).i.e. DS< 0 for the adsorption to occur, DG, should be -ve. We know that DG=DH-TDS if DS is -ve, TDS is +ve. It means that ∆G will, become negative only when ∆H is -ve and DH>TDS, , 4., , (c) dispersion medium-gas, dispersed phase-liquid, , 13., , pyroxylin(nitro cellulose), , 5., , (a) (Hardy-Schulze rule), , 14., , (d) Both reactant and catalyst are in same, phase. i.e(l), , 6., , (b), , 15., , (a), , 7., , (b) Emulsion dispersed phase, Dispersion medium -liquid, , 16., , (a), , 8., , (b) Gel-butter, , 17., , (d) DS is -ve, , 9., , (d) As 2S3 is a -vely charged colloid. It will, be most effectively coagulated by the cation, 3+, with greater valency. i.e., Al ., , 18., , (d), , 10., , (b), , 19., , (a), , 11., , (d) Tyndall effect-scattering of light, , 20., , (d), , 12., , (b), , coagulating power α, , 1, coagulation value, , 313, , Answers.indd 313, , 2/19/2020 5:08:53 PM

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www.tntextbooks.in, , 17. (d) triethyl amine ( 3 amine), , , 18. Option (b ) CH3 is a+I group, all other – I group. +I group increase the electron density on NH2 and, hence increases the basic nature., 19. Option (a) Ethanol, ammonium hydroxide, 20. Option (d), 21. b) O, , C, , CH3, , OCH3, PO, , 2 5, 4, 2, 22. (b) C 6 H5COONH 4 Heat, , → C 6 H5 -C ≡ N  , → C 6 H5CH2 NH2  , → C 6 H5CH2OH, 23. Option (a), , LiAlH, , HNO, , Unit 9 Electro chemistry, 1. (c), 2. (b) 3. (c) 4. (b) 5. (c) 6. (b) 7. (a) 8. (b) 9. (b) 10. (c) 11. (b) 12. (c), 13. (a), 14. (d) 15.(d) 16. (a) 17. (b) 18. (c) 19.(d) 20. (b) 21.(d) 22. (a) 23. (b) 24. (a), 25. (a), Unit 10 – Surface Chemistry, 1. (c), 2.(b) 3.(d) 4. (c) 5. (a) 6. (b) 7. (b) 8.(b) 9.(d) 10.(b) 11. (d) 12. (b), 13. (d), , 14. (d) 15. (a) 16. (a) 17. (d) 18. (d) 19. (a) 20. (d), , Unit – 11 – Alcohols and Ethers, 1. (a), , 2. (c) 3. (a) 4. (c) 5. (c) 6.(b), , 7.(a), , 8.(c), , 9.(b), , 10.(a) 11.(a) 12.(d), , 13. (c), , 14. (d) 15. (a) 16.(c) 17.(d) 18.(c) 19.(d) 20.(a) 21.(b) 22. (b), , Unit – 12 Carbonyl Compounds and Carboxylic Acids, 1. (b) 2. (d) 3. (c) 4. (b) 5. (c) 6. (a) 7.(a), , 8.(a), , 9. (c) 10. (c) 11. (a) 12.(a), , 13. (b) 14. (a) 15. (d) 16. (a) 17.(b) 18.(b) 19. (a) 20.(b) 21. (c) 22. (d) 23.(c) 24. (d), Unit – 13 Organic Nitrogen compounds, 1. (a) 2. (b) 3. (a) 4. (d) 5. (c) 6. (c) 7.(c), 8.(c), 9. (b) 10. (d) 11. (d) 12.(a), 13. (a) 14. (d) 15. (b) 16. (b) 17.(d) 18.(b) 19. (a) 20.(d) 21. (b) 22.(b) 23.(a) 24. (b), 25. (b), Unit – 14 Bio molecules, , 1. (c) 2. (d) 3. (b) 4. (a) 5. (a) 6. (c) 7. (a) 8. (c) 9. (d) 10. (d) 11. (d) 12. (d), 13. (a) 14. (c) 15. (c) 16. (d) 17. (d) 18. (d) 19. (c) 20. (b) 21. (a) 22. (c) 23. (b), 24. (d) 25. (d), , Unit – 15 Chemistry in Action, 1. (c) 2. (a) 3. (a) 4. (a) 5.(d), 13.(d) 14.(b) 15. (d), , 6.(c), , 7. (a) 8. (c) 9. (a) 10.(d) 11. (d) 12. (c), , 319, , Answers.indd 319, , 2/19/2020 5:09:01 PM