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192, , CHEMISTRY, , UNIT 7, , EQUILIBRIUM, , After studying this unit you will be, able to, • identify dynamic nature of, equilibrium involved in physical, and chemical processes;, • state the law of equilibrium;, • explain characteristics of, equilibria involved in physical, and chemical processes;, • write, expressions, for, equilibrium constants;, • establish a relationship between, Kp and K c ;, • explain various factors that, affect the equilibrium state of a, reaction;, • classify substances as acids or, bases according to Arrhenius,, Bronsted-Lowry and Lewis, concepts;, • classify acids and bases as weak, or strong in terms of their, ionization constants;, • explain the dependence of, degree of ionization on, concentration of the electrolyte, and that of the common ion;, • describe, pH, scale, for, representing hydrogen ion, concentration;, • explain ionisation of water and, its duel role as acid and base;, • describe ionic product (Kw ) and, pKw for water;, • appreciate use of buffer, solutions;, • calculate solubility product, constant., , Chemical equilibria are important in numerous biological, and environmental processes. For example, equilibria, involving O2 molecules and the protein hemoglobin play a, crucial role in the transport and delivery of O2 from our, lungs to our muscles. Similar equilibria involving CO, molecules and hemoglobin account for the toxicity of CO., When a liquid evaporates in a closed container,, molecules with relatively higher kinetic energy escape the, liquid surface into the vapour phase and number of liquid, molecules from the vapour phase strike the liquid surface, and are retained in the liquid phase. It gives rise to a constant, vapour pressure because of an equilibrium in which the, number of molecules leaving the liquid equals the number, returning to liquid from the vapour. We say that the system, has reached equilibrium state at this stage. However, this, is not static equilibrium and there is a lot of activity at the, boundary between the liquid and the vapour. Thus, at, equilibrium, the rate of evaporation is equal to the rate of, condensation. It may be represented by, H2O (l)
H2O (vap), The double half arrows indicate that the processes in, both the directions are going on simultaneously. The mixture, of reactants and products in the equilibrium state is called, an equilibrium mixture., Equilibrium can be established for both physical, processes and chemical reactions. The reaction may be fast, or slow depending on the experimental conditions and the, nature of the reactants. When the reactants in a closed vessel, at a particular temperature react to give products, the, concentrations of the reactants keep on decreasing, while, those of products keep on increasing for some time after, which there is no change in the concentrations of either of, the reactants or products. This stage of the system is the, dynamic equilibrium and the rates of the forward and, , 2021-22

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EQUILIBRIUM, , 193, , reverse reactions become equal. It is due to, this dynamic equilibrium stage that there is, no change in the concentrations of various, species in the reaction mixture. Based on the, extent to which the reactions proceed to reach, the state of chemical equilibrium, these may, be classified in three groups., (i) The reactions that proceed nearly to, completion and only negligible, concentrations of the reactants are left. In, some cases, it may not be even possible to, detect these experimentally., (ii) The reactions in which only small amounts, of products are formed and most of the, reactants remain unchanged at, equilibrium stage., (iii) The reactions in which the concentrations, of the reactants and products are, comparable, when the system is in, equilibrium., The extent of a reaction in equilibrium, varies with the experimental conditions such, as concentrations of reactants, temperature,, etc. Optimisation of the operational conditions, is very important in industry and laboratory, so that equilibrium is favorable in the, direction of the desired product. Some, important aspects of equilibrium involving, physical and chemical processes are dealt in, this unit along with the equilibrium involving, ions in aqueous solutions which is called as, ionic equilibrium., 7.1 EQUILIBRIUM, IN, PHYSICAL, PROCESSES, The characteristics of system at equilibrium, are better understood if we examine some, physical processes. The most familiar, examples are phase transformation, processes, e.g.,, solid, liquid, solid, , liquid, gas, gas, , 7.1.1 Solid-Liquid Equilibrium, Ice and water kept in a perfectly insulated, thermos flask (no exchange of heat between, its contents and the surroundings) at 273K, , and the atmospheric pressure are in, equilibrium state and the system shows, interesting characteristic features. We observe, that the mass of ice and water do not change, with time and the temperature remains, constant. However, the equilibrium is not, static. The intense activity can be noticed at, the boundary between ice and water., Molecules from the liquid water collide against, ice and adhere to it and some molecules of ice, escape into liquid phase. There is no change, of mass of ice and water, as the rates of transfer, of molecules from ice into water and of reverse, transfer from water into ice are equal at, atmospheric pressure and 273 K., It is obvious that ice and water are in, equilibrium only at particular temperature, and pressure. For any pure substance at, atmospheric pressure, the temperature at, which the solid and liquid phases are at, equilibrium is called the normal melting point, or normal freezing point of the substance., The system here is in dynamic equilibrium and, we can infer the following:, (i) Both the opposing processes occur, simultaneously., (ii) Both the processes occur at the same rate, so that the amount of ice and water, remains constant., 7.1.2 Liquid-Vapour Equilibrium, This equilibrium can be better understood if, we consider the example of a transparent box, carrying a U-tube with mercury (manometer)., Drying agent like anhydrous calcium chloride, (or phosphorus penta-oxide) is placed for a, few hours in the box. After removing the, drying agent by tilting the box on one side, a, watch glass (or petri dish) containing water is, quickly placed inside the box. It will be, observed that the mercury level in the right, limb of the manometer slowly increases and, finally attains a constant value, that is, the, pressure inside the box increases and reaches, a constant value. Also the volume of water in, the watch glass decreases (Fig. 7.1). Initially, there was no water vapour (or very less) inside, the box. As water evaporated the pressure in, the box increased due to addition of water, , 2021-22

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194, , CHEMISTRY, , Fig.7.1 Measuring equilibrium vapour pressure of water at a constant temperature, , molecules into the gaseous phase inside the, box. The rate of evaporation is constant., However, the rate of increase in pressure, decreases with time due to condensation of, vapour into water. Finally it leads to an, equilibrium condition when there is no net, evaporation. This implies that the number of, water molecules from the gaseous state into, the liquid state also increases till the, equilibrium is attained i.e.,, rate of evaporation= rate of condensation, H2O (vap), H2O(l), At equilibrium the pressure exerted by the, water molecules at a given temperature, remains constant and is called the equilibrium, vapour pressure of water (or just vapour, pressure of water); vapour pressure of water, increases with temperature. If the above, experiment is repeated with methyl alcohol,, acetone and ether, it is observed that different, liquids have different equilibrium vapour, pressures at the same temperature, and the, liquid which has a higher vapour pressure is, more volatile and has a lower boiling point., If we expose three watch glasses, containing separately 1mL each of acetone,, ethyl alcohol, and water to atmosphere and, repeat the experiment with different volumes, of the liquids in a warmer room, it is observed, that in all such cases the liquid eventually, disappears and the time taken for complete, evaporation depends on (i) the nature of the, liquid, (ii) the amount of the liquid and (iii) the, temperature. When the watch glass is open to, the atmosphere, the rate of evaporation, remains constant but the molecules are, , dispersed into large volume of the room. As a, consequence the rate of condensation from, vapour to liquid state is much less than the, rate of evaporation. These are open systems, and it is not possible to reach equilibrium in, an open system., Water and water vapour are in equilibrium, position at atmospheric pressure (1.013 bar), and at 100°C in a closed vessel. The boiling, point of water is 100°C at 1.013 bar pressure., For any pure liquid at one atmospheric, pressure (1.013 bar), the temperature at, which the liquid and vapours are at, equilibrium is called normal boiling point of, the liquid. Boiling point of the liquid depends, on the atmospheric pressure. It depends on, the altitude of the place; at high altitude the, boiling point decreases., 7.1.3 Solid – Vapour Equilibrium, Let us now consider the systems where solids, sublime to vapour phase. If we place solid iodine, in a closed vessel, after sometime the vessel gets, filled up with violet vapour and the intensity of, colour increases with time. After certain time the, intensity of colour becomes constant and at this, stage equilibrium is attained. Hence solid iodine, sublimes to give iodine vapour and the iodine, vapour condenses to give solid iodine. The, equilibrium can be represented as,, I2 (vapour), I2(solid), Other examples showing this kind of, equilibrium are,, Camphor (solid)
Camphor (vapour), , 2021-22, , NH4Cl (solid)
NH4Cl (vapour), ,

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EQUILIBRIUM, , 195, , 7.1.4 Equilibrium Involving Dissolution of, Solid or Gases in Liquids, Solids in liquids, We know from our experience that we can, dissolve only a limited amount of salt or sugar, in a given amount of water at room, temperature. If we make a thick sugar syrup, solution by dissolving sugar at a higher, temperature, sugar crystals separate out if we, cool the syrup to the room temperature. We, call it a saturated solution when no more of, solute can be dissolved in it at a given, temperature. The concentration of the solute, in a saturated solution depends upon the, temperature. In a saturated solution, a, dynamic equilibrium exits between the solute, molecules in the solid state and in the solution:, Sugar (solution), , Sugar (solid), and, , the rate of dissolution of sugar = rate of, crystallisation of sugar., Equality of the two rates and dynamic, nature of equilibrium has been confirmed with, the help of radioactive sugar. If we drop some, radioactive sugar into saturated solution of, non-radioactive sugar, then after some time, radioactivity is observed both in the solution, and in the solid sugar. Initially there were no, radioactive sugar molecules in the solution, but due to dynamic nature of equilibrium,, there is exchange between the radioactive and, non-radioactive sugar molecules between the, two phases. The ratio of the radioactive to nonradioactive molecules in the solution increases, till it attains a constant value., Gases in liquids, When a soda water bottle is opened, some of, the carbon dioxide gas dissolved in it fizzes, out rapidly. The phenomenon arises due to, difference in solubility of carbon dioxide at, different pressures. There is equilibrium, between the molecules in the gaseous state, and the molecules dissolved in the liquid, under pressure i.e.,, CO2 (gas), , pressure of the gas above the solvent. This, amount decreases with increase of, temperature. The soda water bottle is sealed, under pressure of gas when its solubility in, water is high. As soon as the bottle is opened,, some of the dissolved carbon dioxide gas, escapes to reach a new equilibrium condition, required for the lower pressure, namely its, partial pressure in the atmosphere. This is how, the soda water in bottle when left open to the, air for some time, turns ‘flat’. It can be, generalised that:, liquid equilibrium, there is, (i) For solid, only one temperature (melting point) at, 1 atm (1.013 bar) at which the two phases, can coexist. If there is no exchange of heat, with the surroundings, the mass of the two, phases remains constant., vapour equilibrium, the, (ii) For liquid, vapour pressure is constant at a given, temperature., (iii) For dissolution of solids in liquids, the, solubility is constant at a given, temperature., (iv) For dissolution of gases in liquids, the, concentration of a gas in liquid is, proportional, to, the, pressure, (concentration) of the gas over the liquid., These observations are summarised in, Table 7.1, Table 7.1, , Some Features, Equilibria, , Process, Liquid, H2O (l), Solid, H2O (s), Solute(s), Sugar(s), , Vapour, H2O (g), Liquid, H2O (l), , Gas(g), , Gas (aq), , CO2(g), , CO2(aq), , 2021-22, , Physical, , Conclusion, pH2Oconstant at given, temperature, Melting point is fixed at, constant pressure, , Solute Concentration of solute, (solution) in solution is constant, Sugar, at a given temperature, (solution), , CO2 (in solution), , This equilibrium is governed by Henry’s, law, which states that the mass of a gas, dissolved in a given mass of a solvent at, any temperature is proportional to the, , of, , [gas(aq)]/[gas(g)] is, constant at a given, temperature, [CO 2 (aq)]/[CO 2 (g)] is, constant at a given, temperature

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196, , CHEMISTRY, , 7.1.5 General Characteristics of Equilibria, Involving Physical Processes, For the physical processes discussed above,, following characteristics are common to the, system at equilibrium:, (i) Equilibrium is possible only in a closed, system at a given temperature., (ii) Both the opposing processes occur at the, same rate and there is a dynamic but, stable condition., (iii) All measurable properties of the system, remain constant., (iv) When equilibrium is attained for a physical, process, it is characterised by constant, value of one of its parameters at a given, temperature. Table 7.1 lists such, quantities., (v) The magnitude of such quantities at any, stage indicates the extent to which the, physical process has proceeded before, reaching equilibrium., 7.2 EQUILIBRIUM IN CHEMICAL, PROCESSES – DYNAMIC, EQUILIBRIUM, Analogous to the physical systems chemical, reactions also attain a state of equilibrium., These reactions can occur both in forward, and backward directions. When the rates of, the forward and reverse reactions become, equal, the concentrations of the reactants, and the products remain constant. This is, the stage of chemical equilibrium. This, equilibrium is dynamic in nature as it, consists of a forward reaction in which the, reactants give product(s) and reverse, reaction in which product(s) gives the, original reactants., For a better comprehension, let us, consider a general case of a reversible reaction,, A+B, , C+D, , With passage of time, there is, accumulation of the products C and D and, depletion of the reactants A and B (Fig. 7.2)., This leads to a decrease in the rate of forward, reaction and an increase in he rate of the, reverse reaction,, , Fig. 7.2 Attainment of chemical equilibrium., , Eventually, the two reactions occur at the, same rate and the system reaches a state of, equilibrium., Similarly, the reaction can reach the state of, equilibrium even if we start with only C and D;, that is, no A and B being present initially, as the, equilibrium can be reached from either direction., The dynamic nature of chemical, equilibrium can be demonstrated in the, synthesis of ammonia by Haber’s process. In, a series of experiments, Haber started with, known amounts of dinitrogen and dihydrogen, maintained at high temperature and pressure, and at regular intervals determined the, amount of ammonia present. He was, successful in determining also the, concentration of unreacted dihydrogen and, dinitrogen. Fig. 7.4 (page 191) shows that after, a certain time the composition of the mixture, remains the same even though some of the, reactants are still present. This constancy in, composition indicates that the reaction has, reached equilibrium. In order to understand, the dynamic nature of the reaction, synthesis, of ammonia is carried out with exactly the, same starting conditions (of partial pressure, and temperature) but using D2 (deuterium), in place of H2. The reaction mixtures starting, either with H2 or D2 reach equilibrium with, the same composition, except that D2 and ND3, are present instead of H2 and NH3. After, equilibrium is attained, these two mixtures, , 2021-22

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EQUILIBRIUM, , 197, , Dynamic Equilibrium – A Student’s Activity, Equilibrium whether in a physical or in a chemical system, is always of dynamic, nature. This can be demonstrated by the use of radioactive isotopes. This is not feasible, in a school laboratory. However this concept can be easily comprehended by performing, the following activity. The activity can be performed in a group of 5 or 6 students., Take two 100mL measuring cylinders (marked as 1 and 2) and two glass tubes, each of 30 cm length. Diameter of the tubes may be same or different in the range of, 3-5mm. Fill nearly half of the measuring cylinder -1 with coloured water (for this, purpose add a crystal of potassium permanganate to water) and keep second cylinder, (number 2) empty., Put one tube in cylinder 1 and second in cylinder 2. Immerse one tube in cylinder, 1, close its upper tip with a finger and transfer the coloured water contained in its, lower portion to cylinder 2. Using second tube, kept in 2nd cylinder, transfer the coloured, water in a similar manner from cylinder 2 to cylinder 1. In this way keep on transferring, coloured water using the two glass tubes from cylinder 1 to 2 and from 2 to 1 till you, notice that the level of coloured water in both the cylinders becomes constant., If you continue intertransferring coloured solution between the cylinders, there will, not be any further change in the levels of coloured water in two cylinders. If we take, analogy of ‘level’ of coloured water with ‘concentration’ of reactants and products in the, two cylinders, we can say the process of transfer, which continues even after the constancy, of level, is indicative of dynamic nature of the process. If we repeat the experiment taking, two tubes of different diameters we find that at equilibrium the level of coloured water in, two cylinders is different. How far diameters are responsible for change in levels in two, cylinders? Empty cylinder (2) is an indicator of no product in it at the beginning., , Fig.7.3, , Demonstrating dynamic nature of equilibrium. (a) initial stage (b) final stage after the, equilibrium is attained., , 2021-22

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198, , CHEMISTRY, , Fig 7.4 Depiction of equilibrium for the reaction, , N 2 ( g ) + 3H2 ( g )
2NH3 ( g ), (H 2, N2, NH3 and D2 , N2, ND3) are mixed, together and left for a while. Later, when this, mixture is analysed, it is found that the, concentration of ammonia is just the same as, before. However, when this mixture is, analysed by a mass spectrometer, it is found, that ammonia and all deuterium containing, forms of ammonia (NH3, NH2D, NHD2 and ND3), and dihydrogen and its deutrated forms, (H2, HD and D2) are present. Thus one can, conclude that scrambling of H and D atoms, in the molecules must result from a, continuation of the forward and reverse, reactions in the mixture. If the reaction had, simply stopped when they reached, equilibrium, then there would have been no, mixing of isotopes in this way., Use of isotope (deuterium) in the formation, of ammonia clearly indicates that chemical, reactions reach a state of dynamic, equilibrium in which the rates of forward, and reverse reactions are equal and there, is no net change in composition., Equilibrium can be attained from both, sides, whether we start reaction by taking,, H2(g) and N2(g) and get NH3(g) or by taking, NH3(g) and decomposing it into N2(g) and, H2(g)., N2(g) + 3H2(g), 2NH3(g), , 2NH3(g), N2(g) + 3H2(g), Similarly let us consider the reaction,, H2(g) + I2(g), 2HI(g). If we start with equal, initial concentration of H2 and I2, the reaction, proceeds in the forward direction and the, concentration of H2 and I2 decreases while that, of HI increases, until all of these become, constant at equilibrium (Fig. 7.5). We can also, start with HI alone and make the reaction to, proceed in the reverse direction; the, concentration of HI will decrease and, concentration of H2 and I2 will increase until, they all become constant when equilibrium is, reached (Fig.7.5). If total number of H and I, atoms are same in a given volume, the same, equilibrium mixture is obtained whether we, start it from pure reactants or pure product., , Fig.7.5 Chemical equilibrium in the reaction, H2(g) + I2(g)
2HI(g) can be attained, from either direction, , 7.3 LAW OF CHEMICAL EQUILIBRIUM, AND EQUILIBRIUM CONSTANT, A mixture of reactants and products in the, equilibrium state is called an equilibrium, mixture. In this section we shall address a, number of important questions about the, composition of equilibrium mixtures: What is, the relationship between the concentrations of, reactants and products in an equilibrium, mixture? How can we determine equilibrium, concentrations from initial concentrations?, , 2021-22

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EQUILIBRIUM, , 199, , What factors can be exploited to alter the, composition of an equilibrium mixture? The, last question in particular is important when, choosing conditions for synthesis of industrial, chemicals such as H2, NH3, CaO etc., To answer these questions, let us consider, a general reversible reaction:, A+B, C+D, where A and B are the reactants, C and D are, the products in the balanced chemical, equation. On the basis of experimental studies, of many reversible reactions, the Norwegian, chemists Cato Maximillian Guldberg and Peter, Waage proposed in 1864 that the, concentrations in an equilibrium mixture are, related by the following equilibrium, equation,, , Kc =, , [C ][D], [ A ][B ], , (7.1) where Kc is the equilibrium constant, and the expression on the right side is called, the equilibrium constant expression., The equilibrium equation is also known as, the law of mass action because in the early, days of chemistry, concentration was called, “active mass”. In order to appreciate their work, better, let us consider reaction between, gaseous H2 and I2 carried out in a sealed vessel, at 731K., 2HI(g), H2(g) + I2(g), 1 mol 1 mol, 2 mol, , Six sets of experiments with varying initial, conditions were performed, starting with only, gaseous H2 and I2 in a sealed reaction vessel, in first four experiments (1, 2, 3 and 4) and, only HI in other two experiments (5 and 6)., Experiment 1, 2, 3 and 4 were performed, taking different concentrations of H2 and / or, I2, and with time it was observed that intensity, of the purple colour remained constant and, equilibrium was attained. Similarly, for, experiments 5 and 6, the equilibrium was, attained from the opposite direction., Data obtained from all six sets of, experiments are given in Table 7.2., It is evident from the experiments 1, 2, 3, and 4 that number of moles of dihydrogen, reacted = number of moles of iodine reacted =, ½ (number of moles of HI formed). Also,, experiments 5 and 6 indicate that,, [H2(g)]eq = [I2(g)]eq, Knowing the above facts, in order to, establish, a, relationship, between, concentrations of the reactants and products,, several combinations can be tried. Let us, consider the simple expression,, [HI(g)]eq / [H2(g)]eq [I2(g)]eq, It can be seen from Table 7.3 that if we, put the equilibrium concentrations of the, reactants and products, the above expression, , Table 7.2 Initial and Equilibrium Concentrations of H2, I2 and HI, , 2021-22

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200, , CHEMISTRY, , Table 7.3, , Expression, Involving, the, Equilibrium Concentration of, Reactants, H2(g) + I2(g), 2HI(g), , The equilibrium constant for a general, reaction,, cC + dD, aA + bB, is expressed as,, c, , d, , a, , b, , Kc = [C] [D] / [A] [B], (7.4), where [A], [B], [C] and [D] are the equilibrium, concentrations of the reactants and products., Equilibrium constant for the reaction,, 4NH3(g) + 5O2(g), as, , 4NO(g) + 6H2O(g) is written, , 4, , 6, , 4, , 5, , Kc = [NO] [H2O] / [NH3] [O2], is far from constant. However, if we consider, the expression,, [HI(g)]2eq / [H2(g)]eq [I2(g)]eq, we find that this expression gives constant, value (as shown in Table 7.3) in all the six, cases. It can be seen that in this expression, the power of the concentration for reactants, and products are actually the stoichiometric, coefficients in the equation for the chemical, reaction. Thus, for the reaction H2(g) + I2(g), 2HI(g), following equation 7.1, the equilibrium, constant Kc is written as,, 2, , Kc = [HI(g)]eq / [H2(g)]eq [I2(g)]eq, , (7.2), , Generally the subscript ‘eq’ (used for, equilibrium) is omitted from the concentration, terms. It is taken for granted that the, concentrations in the expression for Kc are, equilibrium values. We, therefore, write,, Kc = [HI(g)]2 / [H2(g)] [I2(g)], , (7.3), , The subscript ‘c’ indicates that K c is, expressed in concentrations of mol L–1., At a given temperature, the product of, concentrations of the reaction products, raised to the respective stoichiometric, coefficient in the balanced chemical, equation divided by the product of, concentrations of the reactants raised to, their individual stoichiometric coefficients, has a constant value. This is known as, the Equilibrium Law or Law of Chemical, Equilibrium., , Molar concentration of different species is, indicated by enclosing these in square bracket, and, as mentioned above, it is implied that these, are equilibrium concentrations. While writing, expression for equilibrium constant, symbol for, phases (s, l, g) are generally ignored., Let us write equilibrium constant for the, reaction, H2(g) + I2(g), 2HI(g), (7.5), as, Kc = [HI]2 / [H2] [I2] = x, (7.6), The equilibrium constant for the reverse, H2(g) + I2(g), at the same, reaction, 2HI(g), temperature is,, K′c = [H2] [I2] / [HI]2 = 1/ x = 1 / Kc, Thus, K′c = 1 / Kc, , (7.7), (7.8), , Equilibrium constant for the reverse, reaction is the inverse of the equilibrium, constant for the reaction in the forward, direction., If we change the stoichiometric coefficients, in a chemical equation by multiplying, throughout by a factor then we must make, sure that the expression for equilibrium, constant also reflects that change. For, example, if the reaction (7.5) is written as,, ½ H2(g) + ½ I2(g), HI(g), (7.9), the equilibrium constant for the above reaction, is given by, K″c = [HI] / [H2], , 1/2, , 1/2, , [I2], , 2, , 1/2, , = {[HI] / [H2][I2]}, 1/2, , 1/2, , = x = Kc, (7.10), On multiplying the equation (7.5) by n, we get, , 2021-22

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EQUILIBRIUM, , 201, , nH2(g) + nI2(g) D, 2nHI(g), (7.11), Therefore, equilibrium constant for the, n, reaction is equal to Kc . These findings are, summarised in Table 7.4. It should be noted, that because the equilibrium constants Kc and, K ′c have different numerical values, it is, important to specify the form of the balanced, chemical equation when quoting the value of, an equilibrium constant., , 800K. What will be Kc for the reaction, N2(g) + O2(g), 2NO(g), Solution, For the reaction equilibrium constant,, Kc can be written as,, , Kc =, , Table 7.4 Relations between Equilibrium, Constants for a General Reaction, and its Multiples., Chemical equation, c C + dD, , Kc, , cC+dD, , aA+bB, , K′c =(1/Kc ), , na A + nb B, , ncC + ndD, , K′″c = (Kc ), n, , Problem 7.1, The following concentrations were, obtained for the formation of NH3 from N2, and H 2 at equilibrium at 500K., [N2] = 1.5 × 10–2M. [H2] = 3.0 ×10–2 M and, [NH3] = 1.2 ×10–2M. Calculate equilibrium, constant., Solution, The equilibrium constant for the reaction,, 2NH3(g) can be written, N2(g) + 3H2(g), as,, ,  NH3 ( g ), 3,  N 2 (g )  H2 ( g ), 2, , Kc =, , (1.2 × 10 ), (1.5 × 10 ) (3.0 × 10 ), −2 2, , =, , −2, , −2 3, , = 0.106 × 104 = 1.06 × 103, Problem 7.2, At equilibrium, the concentrations of, N2=3.0 × 10 –3M, O2 = 4.2 × 10–3M and, NO= 2.8 × 10–3M in a sealed vessel at, , (2.8 × 10 M ), (3.0 × 10 M) (4.2 × 10, -3, , =, , Equilibrium, constant, , aA+bB, , [NO]2, [ N 2 ][O2 ], −3, , 2, , −3, , M, , ), , = 0.622, 7.4 HOMOGENEOUS EQUILIBRIA, In a homogeneous system, all the reactants, and products are in the same phase. For, example, in the gaseous reaction,, 2NH3(g), reactants and, N 2(g) + 3H 2(g), products are in the homogeneous phase., Similarly, for the reactions,, CH3COOC2H5 (aq) + H2O (l), CH3COOH (aq), + C2H5OH (aq), –, , and, Fe3+ (aq) + SCN (aq), , Fe(SCN)2+ (aq), , all the reactants and products are in, homogeneous solution phase. We shall now, consider equilibrium constant for some, homogeneous reactions., 7.4.1 Equilibrium Constant in Gaseous, Systems, So far we have expressed equilibrium constant, of the reactions in terms of molar, concentration of the reactants and products,, and used symbol, Kc for it. For reactions, involving gases, however, it is usually more, convenient to express the equilibrium, constant in terms of partial pressure., The ideal gas equation is written as,, pV = n RT, , n, RT, V, Here, p is the pressure in Pa, n is the number, of moles of the gas, V is the volume in m3 and, T is the temperature in Kelvin, ⇒ p=, , 2021-22

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202, , CHEMISTRY, , Therefore,, n/V is concentration expressed in mol/m3, If concentration c, is in mol/L or mol/dm3,, and p is in bar then, p = cRT,, We can also write p = [gas]RT., Here, R= 0.0831 bar litre/mol K, At constant temperature, the pressure of, the gas is proportional to its concentration i.e.,, p ∝ [gas], For reaction in equilibrium, H2(g) + I2(g), 2HI(g), We can write either, ,  NH3 ( g ) [RT ], −2, = , = K c ( RT ), 3,  N 2 ( g )  H 2 ( g ), 2, , or K p = K c ( RT ), , −2, , aA + bB, , Kp, , ( p )( p ), H2, , (7.12), , Further, since p HI =  HI (g ) RT, pI2 =  I2 ( g ) RT, pH2 =  H2 ( g ) RT, Therefore,, , Kp =, ,  HI ( g ) [RT ], =,  H2 ( g ) RT . I2 ( g ) RT, 2, , ( p )( p ), H2, , I2, ,  HI ( g ), = Kc,  H2 (g ) I2 ( g ), , 2, , c, , d, , c +d ), , a, , b, , a +b ), , A, , B, , [C]c [D]d (RT )∆n, [ A ]a [B ]b, , ∆n, , = K c (RT ), , (7.15), , Table 7.5 Equilibrium Constants, Kp for a, Few Selected Reactions, , 2, , =, , d, , D, b, , where ∆n = (number of moles of gaseous, products) – (number of moles of gaseous, reactants) in the balanced chemical equation., It is necessary that while calculating the value, of Kp, pressure should be expressed in bar, because standard state for pressure is 1 bar., We know from Unit 1 that :, –2, 1pascal, Pa=1Nm , and 1bar = 105 Pa, Kp values for a few selected reactions at, different temperatures are given in Table 7.5, , I2, , ( pHI )2, , c, , C, a, , [C ]c [D]d (RT )(c +d )−(a +b ), [ A ]a [B ]b, , =, , 2, , cC + dD, , p )( p ) [C] [D] ( RT )(, (, =, =, ( p )( p ) [ A ] [B] (RT )(, =, , 2, , or K c =, , (7.14), , Similarly, for a general reaction, ,  HI ( g ), Kc =, H2 ( g ) I2 ( g ), , ( p HI ), , −2, , (7.13), , In this example, K p = K c i.e., both, equilibrium constants are equal. However, this, is not always the case. For example in reaction, N2(g) + 3H2(g), 2NH3(g), , (p ), =, ( p )( p ), 2, , Kp, , NH 3, , 3, , N2, , H2, , Problem 7.3, ,  NH3 ( g ) [RT ], =, 3, 3,  N 2 ( g ) RT .  H 2 ( g ) ( RT ), 2, , 2, , PCl5, PCl3 and Cl2 are at equilibrium at, 500 K and having concentration 1.59M, PCl3, 1.59M Cl2 and 1.41 M PCl5., , 2021-22

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EQUILIBRIUM, , 203, , Calculate Kc for the reaction,, PCl5, , the value 0.194 should be neglected, because it will give concentration of the, reactant which is more than initial, concentration., Hence the equilibrium concentrations are,, [CO2] = [H2-] = x = 0.067 M, [CO] = [H2O] = 0.1 – 0.067 = 0.033 M, , PCl3 + Cl2, , Solution, The equilibrium constant Kc for the above, reaction can be written as,, , [PCl ] [Cl ] = (1.59), (1.41), [PCl ], , 2, , Kc =, , 3, , 2, , = 1.79, , 5, , Problem 7.5, For the equilibrium,, , Problem 7.4, The value of Kc = 4.24 at 800K for the reaction,, CO (g) + H2O (g), CO2 (g) + H2 (g), Calculate equilibrium concentrations of, CO2, H2, CO and H2O at 800 K, if only CO, and H 2 O are present initially at, concentrations of 0.10M each., Solution, For the reaction,, CO2 (g) + H2 (g), CO (g) + H2O (g), Initial concentration:, 0.1M, 0.1M, 0, 0, Let x mole per litre of each of the product, be formed., At equilibrium:, (0.1-x) M (0.1-x) M, xM, xM, where x is the amount of CO2 and H2 at, equilibrium., Hence, equilibrium constant can be, written as,, Kc = x2/(0.1-x)2 = 4.24, x2 = 4.24(0.01 + x2-0.2x), x2 = 0.0424 + 4.24x2-0.848x, 3.24x2 – 0.848x + 0.0424 = 0, a = 3.24, b = – 0.848, c = 0.0424, (for quadratic equation ax2 + bx + c = 0,, , (− b ±, x=, , b2 − 4ac, , ), , 2a, x = 0.848±√(0.848)2– 4(3.24)(0.0424)/, (3.24×2), x = (0.848 ± 0.4118)/ 6.48, x1 = (0.848 – 0.4118)/6.48 = 0.067, x2 = (0.848 + 0.4118)/6.48 = 0.194, , 2NO(g) + Cl2(g), 2NOCl(g), the value of the equilibrium constant, Kc, is 3.75 × 10–6 at 1069 K. Calculate the Kp, for the reaction at this temperature?, Solution, We know that,, ∆n, Kp = Kc(RT), For the above reaction,, ∆n = (2+1) – 2 = 1, Kp = 3.75 ×10–6 (0.0831 × 1069), Kp = 0.033, 7.5 HETEROGENEOUS EQUILIBRIA, Equilibrium in a system having more than one, phase is called heterogeneous equilibrium., The equilibrium between water vapour and, liquid water in a closed container is an, example of heterogeneous equilibrium., H2O(g), H2O(l), In this example, there is a gas phase and a, liquid phase. In the same way, equilibrium, between a solid and its saturated solution,, Ca(OH)2 (s) + (aq), Ca2+ (aq) + 2OH–(aq), is a heterogeneous equilibrium., Heterogeneous equilibria often involve pure, solids or liquids. We can simplify equilibrium, expressions for the heterogeneous equilibria, involving a pure liquid or a pure solid, as the, molar concentration of a pure solid or liquid, is constant (i.e., independent of the amount, present). In other words if a substance ‘X’ is, involved, then [X(s)] and [X(l)] are constant,, whatever the amount of ‘X’ is taken. Contrary, , 2021-22

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204, , CHEMISTRY, , to this, [X(g)] and [X(aq)] will vary as the, amount of X in a given volume varies. Let us, take thermal dissociation of calcium carbonate, which is an interesting and important example, of heterogeneous chemical equilibrium., CaCO3 (s), , CaO (s) + CO2 (g), , (7.16), , On the basis of the stoichiometric equation,, we can write,, , CaO ( s) CO2 ( g ), Kc =, CaCO 3 ( s), , 2, , Kp = PCO2 = 2 × 105 Pa/105 Pa = 2.00, Similarly, in the equilibrium between, nickel, carbon monoxide and nickel carbonyl, (used in the purification of nickel),, , Since [CaCO3(s)] and [CaO(s)] are both, constant, therefore modified equilibrium, constant for the thermal decomposition of, calcium carbonate will be, K´c = [CO2(g)], or Kp = pCO, , This shows that at a particular, temperature, there is a constant concentration, or pressure of CO2 in equilibrium with CaO(s), and CaCO3(s). Experimentally it has been, found that at 1100 K, the pressure of CO2 in, equilibrium with CaCO3(s) and CaO(s), is, 2.0 ×105 Pa. Therefore, equilibrium constant, at 1100K for the above reaction is:, , (7.17), (7.18), , Units of Equilibrium Constant, The value of equilibrium constant Kc can, be calculated by substituting the, concentration terms in mol/L and for K p, partial pressure is substituted in Pa, kPa,, bar or atm. This results in units of, equilibrium constant based on molarity or, pressure, unless the exponents of both the, numerator and denominator are same., For the reactions,, H2(g) + I2(g), 2HI, Kc and Kp have no unit., , Ni(CO)4 (g),, Ni (s) + 4 CO (g), the equilibrium constant is written as, ,  Ni (CO)4 , Kc = , [CO]4, It must be remembered that for the, existence of heterogeneous equilibrium pure, solids or liquids must also be present, (however small the amount may be) at, equilibrium, but their concentrations or, partial pressures do not appear in the, expression of the equilibrium constant. In the, reaction,, 2AgNO3(aq) +H2O(l), Ag2O(s) + 2HNO3(aq), , Kc =, , [ AgNO ], [HNO ], , 2, , 3, , 2, , 3, , N2O4(g), 2NO2 (g), Kc has unit mol/L and, Kp has unit bar, , Problem 7.6, , Equilibrium constants can also be, expressed as dimensionless quantities if, the standard state of reactants and, products are specified. For a pure gas, the, standard state is 1bar. Therefore a pressure, of 4 bar in standard state can be expressed, as 4 bar/1 bar = 4, which is a, dimensionless number. Standard state (c0), for a solute is 1 molar solution and all, concentrations can be measured with, respect to it. The numerical value of, equilibrium constant depends on the, standard state chosen. Thus, in this, system both K p and Kc are dimensionless, quantities but have different numerical, values due to different standard states., , The value of Kp for the reaction,, CO2 (g) + C (s), 2CO (g), is 3.0 at 1000 K. If initially PCO = 0.48 bar, 2, and PCO = 0 bar and pure graphite is, present, calculate the equilibrium partial, pressures of CO and CO2., Solution, For the reaction,, let ‘x’ be the decrease in pressure of CO2,, then, CO2(g) + C(s), 2CO(g), Initial, pressure: 0.48 bar, 0, , 2021-22

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EQUILIBRIUM, , 205, , At equilibrium:, (0.48 – x)bar, , Kp =, , 5. The equilibrium constant K for a reaction, is related to the equilibrium constant of the, corresponding reaction, whose equation is, obtained by multiplying or dividing the, equation for the original reaction by a small, integer., Let us consider applications of equilibrium, constant to:, • predict the extent of a reaction on the basis, of its magnitude,, • predict the direction of the reaction, and, • calculate equilibrium concentrations., , 2x bar, , 2, CO, , p, pC O2, , Kp = (2x)2/(0.48 – x) = 3, 4x2 = 3(0.48 – x), 4x2 = 1.44 – x, 4x2 + 3x – 1.44 = 0, a = 4, b = 3, c = –1.44, , (−b ±, x=, , b2 − 4 ac, , ), , 2a, = [–3 ± √(3)2– 4(4)(–1.44)]/2 × 4, = (–3 ± 5.66)/8, = (–3 + 5.66)/ 8 (as value of x cannot be, negative hence we neglect that value), x = 2.66/8 = 0.33, The equilibrium partial pressures are,, pCO = 2x = 2 × 0.33 = 0.66 bar, 2, , pCO = 0.48 – x = 0.48 – 0.33 = 0.15 bar, 2, , 7.6 APPLICATIONS OF EQUILIBRIUM, CONSTANTS, Before considering the applications of, equilibrium constants, let us summarise the, important features of equilibrium constants as, follows:, 1. Expression for equilibrium constant is, applicable only when concentrations of the, reactants and products have attained, constant value at equilibrium state., 2. The value of equilibrium constant is, independent of initial concentrations of the, reactants and products., 3. Equilibrium constant is temperature, dependent having one unique value for a, particular reaction represented by a, balanced equation at a given temperature., 4. The equilibrium constant for the reverse, reaction is equal to the inverse of the, equilibrium constant for the forward, reaction., , 7.6.1 Predicting the Extent of a Reaction, The numerical value of the equilibrium, constant for a reaction indicates the extent of, the reaction. But it is important to note that, an equilibrium constant does not give any, information about the rate at which the, equilibrium is reached. The magnitude of Kc, or K p is directly proportional to the, concentrations of products (as these appear, in the numerator of equilibrium constant, expression) and inversely proportional to the, concentrations of the reactants (these appear, in the denominator). This implies that a high, value of K is suggestive of a high concentration, of products and vice-versa., We can make the following generalisations, concerning the composition of, equilibrium mixtures:, • If Kc > 103, products predominate over, reactants, i.e., if Kc is very large, the reaction, proceeds nearly to completion. Consider, the following examples:, (a) The reaction of H2 with O2 at 500 K has a, very large equilibrium c o n s t a n t ,, Kc = 2.4 × 1047., (b) H2(g) + Cl2(g), Kc = 4.0 × 1031., , 2HCl(g) at 300K has, , (c) H 2(g) + Br 2(g), Kc = 5.4 × 1018, , 2HBr (g) at 300 K,, , •, , 2021-22, , If Kc < 10–3, reactants predominate over, products, i.e., if Kc is very small, the reaction, proceeds rarely. Consider the following, examples:

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206, , CHEMISTRY, , (a) The decomposition of H2O into H2 and O2, at 500 K has a very small equilibrium, constant, Kc = 4.1 × 10 –48, (b) N2(g) + O2(g), 2NO(g),, at 298 K has Kc = 4.8 ×10 – 31., • If K c is in the range of 10 – 3 to 10 3 ,, appreciable concentrations of both, reactants and products are present., Consider the following examples:, (a) For reaction of H2 with I2 to give HI,, Kc = 57.0 at 700K., (b) Also, gas phase decomposition of N2O4 to, NO2 is another reaction with a value, of Kc = 4.64 × 10 –3 at 25°C which is neither, too small nor too large. Hence,, equilibrium mixtures contain appreciable, concentrations of both N2O4 and NO2., These generarlisations are illustrated in, Fig. 7.6, , Fig.7.6 Dependence of extent of reaction on Kc, , 7.6.2 Predicting the Direction of the, Reaction, The equilibrium constant helps in predicting, the direction in which a given reaction will, proceed at any stage. For this purpose, we, calculate the reaction quotient Q. The, reaction quotient, Q (Q c with molar, concentrations and QP with partial pressures), is defined in the same way as the equilibrium, constant Kc except that the concentrations in, Qc are not necessarily equilibrium values., For a general reaction:, cC+dD, (7.19), aA+bB, c, d, a, b, Qc = [C] [D] / [A] [B], (7.20), , If Qc = Kc, the reaction mixture is already, at equilibrium., Consider the gaseous reaction of H 2, with I2,, H2(g) + I2(g), 2HI(g); Kc = 57.0 at 700 K., Suppose we have molar concentrations, [H2]t=0.10M, [I2]t = 0.20 M and [HI]t = 0.40 M., (the subscript t on the concentration symbols, means that the concentrations were measured, at some arbitrary time t, not necessarily at, equilibrium)., Thus, the reaction quotient, Qc at this stage, of the reaction is given by,, Qc = [HI]t2 / [H2]t [I2]t = (0.40)2/ (0.10)×(0.20), = 8.0, Now, in this case, Qc (8.0) does not equal, Kc (57.0), so the mixture of H2(g), I2(g) and HI(g), is not at equilibrium; that is, more H2(g) and, I2(g) will react to form more HI(g) and their, concentrations will decrease till Qc = Kc., The reaction quotient, Q c is useful in, predicting the direction of reaction by, comparing the values of Qc and Kc., Thus, we can make the following, generalisations concerning the direction of the, reaction (Fig. 7.7) :, , Fig. 7.7 Predicting the direction of the reaction, , •, •, •, , Then,, If Qc > Kc, the reaction will proceed in the, direction of reactants (reverse reaction)., If Qc < Kc, the reaction will proceed in the, direction of the products (forward reaction)., , 2021-22, , If Qc < Kc, net reaction goes from left to right, If Qc > Kc, net reaction goes from right to, left., If Qc = Kc, no net reaction occurs., Problem 7.7, The value of Kc for the reaction, 2A, B + C is 2 × 10–3. At a given time,, the composition of reaction mixture is, [A] = [B] = [C] = 3 × 10–4 M. In which, direction the reaction will proceed?

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EQUILIBRIUM, , 207, , Solution, For the reaction the reaction quotient Qc, is given by,, Qc = [B][C]/ [A]2, as [A] = [B] = [C] = 3 × 10 –4M, Qc = (3 ×10 –4)(3 × 10 –4) / (3 ×10 –4)2 = 1, as Qc > Kc so the reaction will proceed in, the reverse direction., , The total pressure at equilbrium was, found to be 9.15 bar. Calculate Kc, Kp and, partial pressure at equilibrium., , 7.6.3 Calculating Equilibrium, Concentrations, In case of a problem in which we know the, initial concentrations but do not know any of, the equilibrium concentrations, the following, three steps shall be followed:, Step 1. Write the balanced equation for the, reaction., Step 2. Under the balanced equation, make a, table that lists for each substance involved in, the reaction:, (a) the initial concentration,, (b) the change in concentration on going to, equilibrium, and, (c) the equilibrium concentration., In constructing the table, define x as the, concentration (mol/L) of one of the substances, that reacts on going to equilibrium, then use, the stoichiometry of the reaction to determine, the concentrations of the other substances in, terms of x., Step 3. Substitute the equilibrium, concentrations into the equilibrium equation, for the reaction and solve for x. If you are to, solve a quadratic equation choose the, mathematical solution that makes chemical, sense., Step 4. Calculate the equilibrium, concentrations from the calculated value of x., Step 5. Check your results by substituting, them into the equilibrium equation., Problem 7.8, 13.8g of N2O4 was placed in a 1L reaction, vessel at 400K and allowed to attain, equilibrium, N 2O4 (g), , 2NO2 (g), , 2021-22, , Solution, We know pV = nRT, Total volume (V ) = 1 L, Molecular mass of N2O4 = 92 g, Number of moles = 13.8g/92 g = 0.15, of the gas (n), Gas constant (R) = 0.083 bar L mol–1K–1, Temperature (T ) = 400 K, pV = nRT, p × 1L = 0.15 mol × 0.083 bar L mol–1K–1, × 400 K, p = 4.98 bar, N2O4, 2NO2, Initial pressure: 4.98 bar, 0, At equilibrium: (4.98 – x) bar 2x bar, Hence,, ptotal at equilibrium = pN O + pNO, 2 4, 2, 9.15 = (4.98 – x) + 2x, 9.15 = 4.98 + x, x = 9.15 – 4.98 = 4.17 bar, Partial pressures at equilibrium are,, pN O = 4.98 – 4.17 = 0.81bar, 2 4, , pNO = 2x = 2 × 4.17 = 8.34 bar, 2, , (, , K p = p NO2, Kp, , ), , 2, , / p N 2O4, , = (8.34)2/0.81 = 85.87, ∆n, = Kc(RT), , 85.87 = Kc(0.083 × 400)1, Kc = 2.586 = 2.6, Problem 7.9, 3.00 mol of PCl5 kept in 1L closed reaction, vessel was allowed to attain equilibrium, at 380K. Calculate composition of the, mixture at equilibrium. Kc= 1.80, Solution, PCl5, PCl3 + Cl2, Initial, concentration: 3.0, 0, 0

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208, , CHEMISTRY, 0, , K = e–∆G /RT, (7.23), Hence, using the equation (7.23), the, reaction spontaneity can be interpreted in, terms of the value of ∆G0., 0, 0, • If ∆G < 0, then –∆G /RT is positive, and, >1, making K >1, which implies, a spontaneous reaction or the reaction, which proceeds in the forward direction to, such an extent that the products are, present predominantly., 0, 0, • If ∆G > 0, then –∆G /RT is negative, and, , Let x mol per litre of PCl5 be dissociated,, At equilibrium:, (3-x), x, x, Kc = [PCl3][Cl2]/[PCl5], 1.8 = x2/ (3 – x), x2 + 1.8x – 5.4 = 0, x = [–1.8 ± √(1.8)2 – 4(–5.4)]/2, x = [–1.8 ± √3.24 + 21.6]/2, x = [–1.8 ± 4.98]/2, x = [–1.8 + 4.98]/2 = 1.59, [PCl5] = 3.0 – x = 3 –1.59 = 1.41 M, [PCl3] = [Cl2] = x = 1.59 M, , < 1, that is , K < 1, which implies, , 7.7 RELATIONSHIP BETWEEN, EQUILIBRIUM CONSTANT K,, REACTION QUOTIENT Q AND, GIBBS ENERGY G, The value of Kc for a reaction does not depend, on the rate of the reaction. However, as you, have studied in Unit 6, it is directly related, to the thermodynamics of the reaction and, in particular, to the change in Gibbs energy,, ∆G. If,, • ∆G is negative, then the reaction is, spontaneous and proceeds in the forward, direction., • ∆G is positive, then reaction is considered, non-spontaneous. Instead, as reverse, reaction would have a negative ∆G, the, products of the forward reaction shall be, converted to the reactants., • ∆G is 0, reaction has achieved equilibrium;, at this point, there is no longer any free, energy left to drive the reaction., A mathematical expression of this, thermodynamic view of equilibrium can be, described by the following equation:, 0, ∆G = ∆G + RT lnQ, (7.21), 0, where, G is standard Gibbs energy., At equilibrium, when ∆G = 0 and Q = Kc,, the equation (7.21) becomes,, 0, ∆G = ∆G + RT ln K = 0, 0, ∆G = – RT lnK, (7.22), 0, lnK = – ∆G / RT, Taking antilog of both sides, we get,, , a non-spontaneous reaction or a reaction, which proceeds in the forward direction to, such a small degree that only a very minute, quantity of product is formed., Problem 7.10, 0, The value of ∆G for the phosphorylation, of glucose in glycolysis is 13.8 kJ/mol., Find the value of Kc at 298 K., Solution, 0, ∆G = 13.8 kJ/mol = 13.8 × 103J/mol, 0, Also, ∆G = – RT lnKc, Hence, ln Kc = –13.8 × 103J/mol, (8.314 J mol –1K –1 × 298 K), ln Kc = – 5.569, Kc = e–5.569, Kc = 3.81 × 10 –3, Problem 7.11, Hydrolysis of sucrose gives,, Glucose + Fructose, Sucrose + H2O, Equilibrium constant Kc for the reaction, 0, is 2 ×1013 at 300K. Calculate ∆G at, 300K., Solution, 0, ∆G = – RT lnKc, 0, ∆G = – 8.314J mol–1K–1×, 300K × ln(2×1013), 0, 4, ∆G = – 7.64 ×10 J mol–1, 7.8 FACTORS AFFECTING EQUILIBRIA, One of the principal goals of chemical synthesis, is to maximise the conversion of the reactants, , 2021-22

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EQUILIBRIUM, , 209, , to products while minimizing the expenditure, of energy. This implies maximum yield of, products at mild temperature and pressure, conditions. If it does not happen, then the, experimental conditions need to be adjusted., For example, in the Haber process for the, synthesis of ammonia from N2 and H2, the, choice of experimental conditions is of real, economic importance. Annual world, production of ammonia is about hundred, million tones, primarily for use as fertilizers., Equilibrium constant, Kc is independent of, initial concentrations. But if a system at, equilibrium is subjected to a change in the, concentration of one or more of the reacting, substances, then the system is no longer at, equilibrium; and net reaction takes place in, some direction until the system returns to, equilibrium once again. Similarly, a change in, temperature or pressure of the system may, also alter the equilibrium. In order to decide, what course the reaction adopts and make a, qualitative prediction about the effect of a, change in conditions on equilibrium we use, Le Chatelier’s principle. It states that a, change in any of the factors that, determine the equilibrium conditions of a, system will cause the system to change, in such a manner so as to reduce or to, counteract the effect of the change. This, is applicable to all physical and chemical, equilibria., , “When the concentration of any of the, reactants or products in a reaction at, equilibrium is changed, the composition, of the equilibrium mixture changes so as, to minimize the effect of concentration, changes”., Let us take the reaction,, H2(g) + I2(g), 2HI(g), If H2 is added to the reaction mixture at, equilibrium, then the equilibrium of the, reaction is disturbed. In order to restore it, the, reaction proceeds in a direction wherein H2 is, consumed, i.e., more of H2 and I2 react to form, HI and finally the equilibrium shifts in right, (forward) direction (Fig.7.8). This is in, accordance with the Le Chatelier’s principle, which implies that in case of addition of a, reactant/product, a new equilibrium will be, set up in which the concentration of the, reactant/product should be less than what it, was after the addition but more than what it, was in the original mixture., , We shall now be discussing factors which, can influence the equilibrium., 7.8.1 Effect of Concentration Change, In general, when equilibrium is disturbed by, the addition/removal of any reactant/, products, Le Chatelier’s principle predicts that:, • The concentration stress of an added, reactant/product is relieved by net reaction, in the direction that consumes the added, substance., • The concentration stress of a removed, reactant/product is relieved by net reaction, in the direction that replenishes the, removed substance., or in other words,, , Fig. 7.8, , Effect of addition of H2 on change of, concentration for the reactants and, products, in, the, reaction,, H2(g) + I2 (g), 2HI(g), , The same point can be explained in terms, of the reaction quotient, Qc,, 2, Qc = [HI] / [H2][I2], , 2021-22

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210, , CHEMISTRY, , Addition of hydrogen at equilibrium results, in value of Qc being less than Kc . Thus, in order, to attain equilibrium again reaction moves in, the forward direction. Similarly, we can say, that removal of a product also boosts the, forward reaction and increases the, concentration of the products and this has, great commercial application in cases of, reactions, where the product is a gas or a, volatile substance. In case of manufacture of, ammonia, ammonia is liquified and removed, from the reaction mixture so that reaction, keeps moving in forward direction. Similarly,, in the large scale production of CaO (used as, important building material) from CaCO3,, constant removal of CO2 from the kiln drives, the reaction to completion. It should be, remembered that continuous removal of a, product maintains Qc at a value less than Kc, and reaction continues to move in the forward, direction., Effect of Concentration – An experiment, This can be demonstrated by the following, reaction:, –, , 2+, , Fe3+(aq)+ SCN (aq), [Fe(SCN)] (aq), yellow, colourless, deep red, , (7.24), , (7.25), A reddish colour appears on adding two, drops of 0.002 M potassium thiocynate, solution to 1 mL of 0.2 M iron(III) nitrate, solution due to the formation of [Fe(SCN)]2+., The intensity of the red colour becomes, constant on attaining equilibrium. This, equilibrium can be shifted in either forward, or reverse directions depending on our choice, of adding a reactant or a product. The, equilibrium can be shifted in the opposite, direction by adding reagents that remove Fe3+, –, or SCN ions. For example, oxalic acid, (H2C2O4), reacts with Fe3+ ions to form the, stable complex ion [Fe(C 2 O 4) 3 ] 3 – , thus, decreasing the concentration of free Fe3+(aq)., In accordance with the Le Chatelier’s principle,, the concentration stress of removed Fe3+ is, relieved by dissociation of [Fe(SCN)] 2+ to, , replenish the Fe 3+ ions. Because the, concentration of [Fe(SCN)]2+ decreases, the, intensity of red colour decreases., Addition of aq. HgCl2 also decreases red, –, colour because Hg2+ reacts with SCN ions to, 2–, form stable complex ion [Hg(SCN)4] . Removal, –, of free SCN (aq) shifts the equilibrium in, equation (7.24) from right to left to replenish, –, SCN ions. Addition of potassium thiocyanate, on the other hand increases the colour, intensity of the solution as it shift the, equilibrium to right., 7.8.2 Effect of Pressure Change, A pressure change obtained by changing the, volume can affect the yield of products in case, of a gaseous reaction where the total number, of moles of gaseous reactants and total, number of moles of gaseous products are, different. In applying Le Chatelier’s principle, to a heterogeneous equilibrium the effect of, pressure changes on solids and liquids can, be ignored because the volume (and, concentration) of a solution/liquid is nearly, independent of pressure., Consider the reaction,, CH4(g) + H2O(g), CO(g) + 3H2(g), Here, 4 mol of gaseous reactants (CO + 3H2), become 2 mol of gaseous products (CH4 +, H2O). Suppose equilibrium mixture (for above, reaction) kept in a cylinder fitted with a piston, at constant temperature is compressed to one, half of its original volume. Then, total pressure, will, be, doubled, (according, to, pV = constant). The partial pressure and, therefore, concentration of reactants and, products have changed and the mixture is no, longer at equilibrium. The direction in which, the reaction goes to re-establish equilibrium, can be predicted by applying the Le Chatelier’s, principle. Since pressure has doubled, the, equilibrium now shifts in the forward, direction, a direction in which the number of, moles of the gas or pressure decreases (we, know pressure is proportional to moles of the, gas). This can also be understood by using, reaction quotient, Qc. Let [CO], [H2], [CH4] and, [H 2 O] be the molar concentrations at, equilibrium for methanation reaction. When, , 2021-22

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EQUILIBRIUM, , 211, , volume of the reaction mixture is halved, the, partial pressure and the concentration are, doubled. We obtain the reaction quotient by, replacing each equilibrium concentration by, double its value., , CH4 ( g )  H2 O ( g ), Qc = , 3, CO ( g )  H2 (g ), As Qc < Kc , the reaction proceeds in the, forward direction., In reaction C(s) + CO2(g), 2CO(g), when, pressure is increased, the reaction goes in the, reverse direction because the number of moles, of gas increases in the forward direction., 7.8.3 Effect of Inert Gas Addition, ,
, , If the volume is kept constant and an inert gas, such as argon is added which does not take, part in the reaction, the equilibrium remains, undisturbed. It is because the addition of an, inert gas at constant volume does not change, the partial pressures or the molar, concentrations of the substance involved in the, reaction. The reaction quotient changes only, if the added gas is a reactant or product, involved in the reaction., 7.8.4 Effect of Temperature Change, Whenever an equilibrium is disturbed by a, change in the concentration, pressure or, volume, the composition of the equilibrium, mixture changes because the reaction, quotient, Qc no longer equals the equilibrium, constant, Kc. However, when a change in, temperature occurs, the value of equilibrium, constant, Kc is changed., , Production of ammonia according to the, reaction,, N2(g) + 3H2(g), 2NH3(g) ;, ∆H= – 92.38 kJ mol–1, is an exothermic process. According to, Le Chatelier’s principle, raising the, temperature shifts the equilibrium to left and, decreases the equilibrium concentration of, ammonia. In other words, low temperature is, favourable for high yield of ammonia, but, practically very low temperatures slow down, the reaction and thus a catalyst is used., Effect of Temperature – An experiment, Effect of temperature on equilibrium can be, demonstrated by taking NO2 gas (brown in, colour) which dimerises into N 2 O 4 gas, (colourless)., N2O4(g); ∆H = –57.2 kJ mol–1, 2NO2(g), NO 2 gas prepared by addition of Cu, turnings to conc. HNO3 is collected in two, 5 mL test tubes (ensuring same intensity of, colour of gas in each tube) and stopper sealed, with araldite. Three 250 mL beakers 1, 2 and, 3 containing freezing mixture, water at room, temperature and hot water (363 K ),, respectively, are taken (Fig. 7.9). Both the test, tubes are placed in beaker 2 for 8-10 minutes., After this one is placed in beaker 1 and the, other in beaker 3. The effect of temperature, on direction of reaction is depicted very well, in this experiment. At low temperatures in, beaker 1, the forward reaction of formation of, N2O4 is preferred, as reaction is exothermic, and, thus, intensity of brown colour due to NO2, decreases. While in beaker 3, high, temperature favours the reverse reaction of, , In general, the temperature dependence of, the equilibrium constant depends on the sign, of ∆H for the reaction., •, , The equilibrium constant for an exothermic, reaction (negative ∆H) decreases as the, temperature increases., , •, , The equilibrium constant for an, endothermic reaction (positive ∆H), increases as the temperature increases., , Temperature changes affect the, equilibrium constant and rates of reactions., , Fig. 7.9 Effect of temperature on equilibrium for, the reaction, 2NO2 (g), N2O4 (g), , 2021-22

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212, , CHEMISTRY, , formation of NO2 and thus, the brown colour, intensifies., Effect of temperature can also be seen in, an endothermic reaction,, 3+, , –, , [Co(H2O) 6] (aq) + 4Cl (aq), pink, , colourless, , 2–, , [CoCl4] (aq) +, 6H2O(l), blue, , At room temperature, the equilibrium, mixture is blue due to [CoCl4]2–. When cooled, in a freezing mixture, the colour of the mixture, turns pink due to [Co(H2O)6]3+., 7.8.5 Effect of a Catalyst, A catalyst increases the rate of the chemical, reaction by making available a new low energy, pathway for the conversion of reactants to, products. It increases the rate of forward and, reverse reactions that pass through the same, transition state and does not affect, equilibrium. Catalyst lowers the activation, energy for the forward and reverse reactions, by exactly the same amount. Catalyst does not, affect the equilibrium composition of a, reaction mixture. It does not appear in the, balanced chemical equation or in the, equilibrium constant expression., Let us consider the formation of NH3 from, dinitrogen and dihydrogen which is highly, exothermic reaction and proceeds with, decrease in total number of moles formed as, compared to the reactants. Equilibrium, constant decreases with increase in, temperature. At low temperature rate, decreases and it takes long time to reach at, equilibrium, whereas high temperatures give, satisfactory rates but poor yields., German chemist, Fritz Haber discovered, that a catalyst consisting of iron catalyse the, reaction to occur at a satisfactory rate at, temperatures, where the equilibrium, concentration of NH3 is reasonably favourable., Since the number of moles formed in the, reaction is less than those of reactants, the, yield of NH3 can be improved by increasing, the pressure., Optimum conditions of temperature and, pressure for the synthesis of NH 3 using, catalyst are around 500 °C and 200 atm., , Similarly, in manufacture of sulphuric, acid by contact process,, 2SO2(g) + O2(g), , 2SO3(g); Kc = 1.7 × 1026, , though the value of K is suggestive of reaction, going to completion, but practically the oxidation, of SO2 to SO3 is very slow. Thus, platinum or, divanadium penta-oxide (V2O5) is used as, catalyst to increase the rate of the reaction., Note: If a reaction has an exceedingly small, K, a catalyst would be of little help., 7.9 IONIC EQUILIBRIUM IN SOLUTION, Under the effect of change of concentration on, the direction of equilibrium, you have, incidently come across with the following, equilibrium which involves ions:, Fe3+(aq) + SCN–(aq), , [Fe(SCN)]2+(aq), , There are numerous equilibria that involve, ions only. In the following sections we will, study the equilibria involving ions. It is well, known that the aqueous solution of sugar, does not conduct electricity. However, when, common salt (sodium chloride) is added to, water it conducts electricity. Also, the, conductance of electricity increases with an, increase in concentration of common salt., Michael Faraday classified the substances into, two categories based on their ability to conduct, electricity. One category of substances, conduct electricity in their aqueous solutions, and are called electrolytes while the other do, not and are thus, referred to as nonelectrolytes. Faraday further classified, electrolytes into strong and weak electrolytes., Strong electrolytes on dissolution in water are, ionized almost completely, while the weak, electrolytes are only partially dissociated., For example, an aqueous solution of, sodium chloride is comprised entirely of, sodium ions and chloride ions, while that, of acetic acid mainly contains unionized, acetic acid molecules and only some acetate, ions and hydronium ions. This is because, there is almost 100% ionization in case of, sodium chloride as compared to less, than 5% ionization of acetic acid which is, a weak electrolyte. It should be noted, that in weak electrolytes, equilibrium is, , 2021-22

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EQUILIBRIUM, , 213, , established between ions and the unionized, molecules. This type of equilibrium involving, ions in aqueous solution is called ionic, equilibrium. Acids, bases and salts come, under the category of electrolytes and may act, as either strong or weak electrolytes., 7.10, , ACIDS, BASES AND SALTS, , Acids, bases and salts find widespread, occurrence in nature. Hydrochloric acid, present in the gastric juice is secreted by the, lining of our stomach in a significant amount, of 1.2-1.5 L/day and is essential for digestive, processes. Acetic acid is known to be the main, constituent of vinegar. Lemon and orange, juices contain citric and ascorbic acids, and, tartaric acid is found in tamarind paste. As, most of the acids taste sour, the word “acid”, has been derived from a latin word “acidus”, meaning sour. Acids are known to turn blue, litmus paper into red and liberate dihydrogen, on reacting with some metals. Similarly, bases, are known to turn red litmus paper blue, taste, bitter and feel soapy. A common example of a, base is washing soda used for washing, purposes. When acids and bases are mixed in, the right proportion they react with each other, to give salts. Some commonly known, examples of salts are sodium chloride, barium, sulphate, sodium nitrate. Sodium chloride, (common salt ) is an important component of, our diet and is formed by reaction between, hydrochloric acid and sodium hydroxide. It, , exists in solid state as a cluster of positively, charged sodium ions and negatively charged, chloride ions which are held together due to, electrostatic interactions between oppositely, charged species (Fig.7.10). The electrostatic, forces between two charges are inversely, proportional to dielectric constant of the, medium. Water, a universal solvent, possesses, a very high dielectric constant of 80. Thus,, when sodium chloride is dissolved in water,, the electrostatic interactions are reduced by a, factor of 80 and this facilitates the ions to move, freely in the solution. Also, they are wellseparated due to hydration with water, molecules., , Fig.7.10 Dissolution of sodium chloride in water., Na+ and Cl – ions are stablised by their, hydration with polar water molecules., , Comparing, the ionization of hydrochloric, acid with that of acetic acid in water we find, that though both of them are polar covalent, , Faraday was born near London into a family of very limited means. At the age of 14 he, was an apprentice to a kind bookbinder who allowed Faraday to read the books he, was binding. Through a fortunate chance he became laboratory assistant to Davy, and, during 1813-4, Faraday accompanied him to the Continent. During this trip he gained, much from the experience of coming into contact with many of the leading scientists of, the time. In 1825, he succeeded Davy as Director of the Royal Institution laboratories,, and in 1833 he also became the first Fullerian Professor of Chemistry. Faraday’s first Michael Faraday, (1791–1867), important work was on analytical chemistry. After 1821 much of his work was on, electricity and magnetism and different electromagnetic phenomena. His ideas have led to the establishment, of modern field theory. He discovered his two laws of electrolysis in 1834. Faraday was a very modest, and kind hearted person. He declined all honours and avoided scientific controversies. He preferred to, work alone and never had any assistant. He disseminated science in a variety of ways including his, Friday evening discourses, which he founded at the Royal Institution. He has been very famous for his, Christmas lecture on the ‘Chemical History of a Candle’. He published nearly 450 scientific papers., , 2021-22

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214, , CHEMISTRY, , molecules, former is completely ionized into, its constituent ions, while the latter is only, partially ionized (< 5%). The extent to which, ionization occurs depends upon the strength, of the bond and the extent of solvation of ions, produced. The terms dissociation and, ionization have earlier been used with different, meaning. Dissociation refers to the process of, separation of ions in water already existing as, such in the solid state of the solute, as in, sodium chloride. On the other hand, ionization, corresponds to a process in which a neutral, molecule splits into charged ions in the, solution. Here, we shall not distinguish, between the two and use the two terms, interchangeably., , Hydronium and Hydroxyl Ions, Hydrogen ion by itself is a bare proton with, very small size (~10 –15 m radius) and, intense electric field, binds itself with the, water molecule at one of the two available, +, lone pairs on it giving H3O . This species, has been detected in many compounds, +, –, (e.g., H3O Cl ) in the solid state. In aqueous, solution the hydronium ion is further, +, +, hydrated to give species like H5O2 , H7O3 and, +, H9O4 . Similarly the hydroxyl ion is hydrated, –, –, to give several ionic species like H3O2 , H5O3, –, and H7O4 etc., , 7.10.1 Arrhenius Concept of Acids and, Bases, According to Arrhenius theory, acids are, substances that dissociates in water to give, +, hydrogen ions H (aq) and bases are, substances that produce hydroxyl ions, –, OH (aq). The ionization of an acid HX (aq) can, be represented by the following equations:, HX (aq) → H (aq) + X (aq), or, +, –, HX(aq) + H2O(l) → H3O (aq) + X (aq), +, , –, , A bare proton, H+ is very reactive and, cannot exist freely in aqueous solutions. Thus,, it bonds to the oxygen atom of a solvent water, molecule to give trigonal pyramidal, +, +, hydronium ion, H3O {[H (H2O)] } (see box)., +, +, In this chapter we shall use H (aq) and H3O (aq), interchangeably to mean the same i.e., a, hydrated proton., , +, , H9O4, , 7.10.2 The Brönsted-Lowry Acids and, Bases, The Danish chemist, Johannes Brönsted and, the English chemist, Thomas M. Lowry gave a, more general definition of acids and bases., According to Brönsted-Lowry theory, acid is, a substance that is capable of donating a, hydrogen ion H+ and bases are substances, capable of accepting a hydrogen ion, H+. In, short, acids are proton donors and bases are, proton acceptors., Consider the example of dissolution of NH3, in H2O represented by the following equation:, , Similarly, a base molecule like MOH, ionizes in aqueous solution according to the, equation:, MOH(aq) → M (aq) + OH (aq), The hydroxyl ion also exists in the hydrated, form in the aqueous solution. Arrhenius, concept of acid and base, however, suffers, from the limitation of being applicable only to, aqueous solutions and also, does not account, for the basicity of substances like, ammonia, which do not possess a hydroxyl group., +, , –, , The basic solution is formed due to the, presence of hydroxyl ions. In this reaction,, water molecule acts as proton donor and, ammonia molecule acts as proton acceptor, and are thus, called Lowry-Brönsted acid and, , 2021-22

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EQUILIBRIUM, , 215, , Arrhenius was born near Uppsala, Sweden. He presented his thesis, on the conductivities, of electrolyte solutions, to the University of Uppsala in 1884. For the next five years he, travelled extensively and visited a number of research centers in Europe. In 1895 he was, appointed professor of physics at the newly formed University of Stockholm, serving its, rector from 1897 to 1902. From 1905 until his death he was Director of physical chemistry, at the Nobel Institute in Stockholm. He continued to work for many years on electrolytic, solutions. In 1899 he discussed the temperature dependence of reaction rates on the, basis of an equation, now usually known as Arrhenius equation., He worked in a variety of fields, and made important contributions to, immunochemistry, cosmology, the origin of life, and the causes of ice age. He was the, Svante Arrhenius, first to discuss the ‘green house effect’ calling by that name. He received Nobel Prize in, (1859-1927), Chemistry in 1903 for his theory of electrolytic dissociation and its use in the development, of chemistry., , base, respectively. In the reverse reaction, H+, is transferred from NH4+ to OH – . In this case,, –, NH4+ acts as a Bronsted acid while OH acted, as a Brönsted base. The acid-base pair that, differs only by one proton is called a conjugate, –, acid-base pair. Therefore, OH is called the, +, conjugate base of an acid H2O and NH4 is, called conjugate acid of the base NH 3. If, Brönsted acid is a strong acid then its, conjugate base is a weak base and viceversa. It may be noted that conjugate acid, has one extra proton and each conjugate base, has one less proton., Consider the example of ionization of, hydrochloric acid in water. HCl(aq) acts as an, acid by donating a proton to H2O molecule, which acts as a base., , ammonia it acts as an acid by donating a, proton., , It can be seen in the above equation, that, water acts as a base because it accepts the, proton. The species H3O+ is produced when, water accepts a proton from HCl. Therefore,, –, Cl is a conjugate base of HCl and HCl is the, –, conjugate acid of base Cl . Similarly, H2O is a, +, +, conjugate base of an acid H3O and H3O is a, conjugate acid of base H2O., It is interesting to observe the dual role of, water as an acid and a base. In case of reaction, with HCl water acts as a base while in case of, , 2021-22, , Problem 7.12, What will be the conjugate bases for the, following Brönsted acids: HF, H2SO4 and, –, HCO3 ?, Solution, The conjugate bases should have one, proton less in each case and therefore the, –, corresponding conjugate bases are: F ,, –, 2–, HSO4 and CO3 respectively., Problem 7.13, Write the conjugate acids for the following, –, –, Brönsted bases: NH2 , NH3 and HCOO ., Solution, The conjugate acid should have one extra, proton in each case and therefore the, corresponding conjugate acids are: NH3,, +, NH4 and HCOOH respectively., Problem 7.14, –, –, The species: H2O, HCO3 , HSO4 and NH3, can act both as Bronsted acids and bases., For each case give the corresponding, conjugate acid and conjugate base., Solution, The answer is given in the following Table:, Species Conjugate, Conjugate, acid, base, –, +, H2O, H3O, OH, –, HCO3, H2CO3, CO32–, –, HSO4, H2SO4, SO42–, –, NH3, NH4+, NH2

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216, , CHEMISTRY, , 7.10.3 Lewis Acids and Bases, G.N. Lewis in 1923 defined an acid as a, species which accepts electron pair and base, which donates an electron pair. As far as, bases are concerned, there is not much, difference between Brönsted-Lowry and Lewis, concepts, as the base provides a lone pair in, both the cases. However, in Lewis concept, many acids do not have proton. A typical, example is reaction of electron deficient species, BF3 with NH3., BF3 does not have a proton but still acts, as an acid and reacts with NH3 by accepting, its lone pair of electrons. The reaction can be, represented by,, BF3 + :NH3 → BF3:NH3, Electron deficient species like AlCl3, Co3+,, Mg , etc. can act as Lewis acids while species, –, like H2O, NH3, OH etc. which can donate a pair, of electrons, can act as Lewis bases., 2+, , Problem 7.15, Classify the following species into Lewis, acids and Lewis bases and show how, these act as such:, –, –, +, (a) HO (b)F, (c) H, (d) BCl3, Solution, (a) Hydroxyl ion is a Lewis base as it can, –, donate an electron lone pair (:OH )., (b) Flouride ion acts as a Lewis base as, it can donate any one of its four, electron lone pairs., (c) A proton is a Lewis acid as it can, accept a lone pair of electrons from, bases like hydroxyl ion and fluoride, ion., (d) BCl3 acts as a Lewis acid as it can, accept a lone pair of electrons from, species like ammonia or amine, molecules., 7.11 IONIZATION OF ACIDS AND BASES, Arrhenius concept of acids and bases becomes, useful in case of ionization of acids and bases, as mostly ionizations in chemical and, biological systems occur in aqueous medium., Strong acids like perchloric acid (HClO4),, , hydrochloric acid (HCl), hydrobromic acid, (HBr), hyrdoiodic acid (HI), nitric acid (HNO3), and sulphuric acid (H2SO4) are termed strong, because they are almost completely, dissociated into their constituent ions in an, aqueous medium, thereby acting as proton, (H +) donors. Similarly, strong bases like, lithium hydroxide (LiOH), sodium hydroxide, (NaOH), potassium hydroxide (KOH), caesium, hydroxide (CsOH) and barium hydroxide, Ba(OH)2 are almost completely dissociated into, ions in an aqueous medium giving hydroxyl, ions, OH – . According to Arrhenius concept, they are strong acids and bases as they are, +, able to completely dissociate and produce H3O, –, and OH ions respectively in the medium., Alternatively, the strength of an acid or base, may also be gauged in terms of BrönstedLowry concept of acids and bases, wherein a, strong acid means a good proton donor and a, strong base implies a good proton acceptor., Consider, the acid-base dissociation, equilibrium of a weak acid HA,, HA(aq) + H2O(l), H3O+(aq) + A–(aq), conjugate conjugate, acid, base, acid, base, In section 7.10.2 we saw that acid (or base), dissociation equilibrium is dynamic involving, a transfer of proton in forward and reverse, directions. Now, the question arises that if the, equilibrium is dynamic then with passage of, time which direction is favoured? What is the, driving force behind it? In order to answer, these questions we shall deal into the issue of, comparing the strengths of the two acids (or, bases) involved in the dissociation equilibrium., Consider the two acids HA and H3O+ present, in the above mentioned acid-dissociation, equilibrium. We have to see which amongst, them is a stronger proton donor. Whichever, exceeds in its tendency of donating a proton, over the other shall be termed as the stronger, acid and the equilibrium will shift in the, direction of weaker acid. Say, if HA is a, stronger acid than H3O+, then HA will donate, protons and not H3O+, and the solution will, mainly contain A – and H 3 O + ions. The, equilibrium moves in the direction of, formation of weaker acid and weaker base, , 2021-22

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EQUILIBRIUM, , 217, , because the stronger acid donates a proton, to the stronger base., It follows that as a strong acid dissociates, completely in water, the resulting base formed, would be very weak i.e., strong acids have, very weak conjugate bases. Strong acids like, perchloric acid (HClO4), hydrochloric acid, (HCl), hydrobromic acid (HBr), hydroiodic acid, (HI), nitric acid (HNO3) and sulphuric acid, –, (H2SO4) will give conjugate base ions ClO4 , Cl,, – –, –, –, Br , I , NO3 and HSO4 , which are much weaker, bases than H2O. Similarly a very strong base, would give a very weak conjugate acid. On the, other hand, a weak acid say HA is only partially, dissociated in aqueous medium and thus, the, solution mainly contains undissociated HA, molecules. Typical weak acids are nitrous acid, (HNO2), hydrofluoric acid (HF) and acetic acid, (CH3COOH). It should be noted that the weak, acids have very strong conjugate bases. For, 2–, –, –, example, NH2 , O and H are very good proton, acceptors and thus, much stronger bases than, H2O., Certain water soluble organic compounds, like phenolphthalein and bromothymol blue, behave as weak acids and exhibit different, colours in their acid (HIn) and conjugate base, –, (In ) forms., +, , –, , HIn(aq) + H2O(l), H3O (aq) + In (aq), acid, conjugate conjugate, indicator, acid, base, colour A, colourB, Such compounds are useful as indicators, +, in acid-base titrations, and finding out H ion, concentration., 7.11.1 The Ionization Constant of Water, and its Ionic Product, Some substances like water are unique in their, ability of acting both as an acid and a base., We have seen this in case of water in section, 7.10.2. In presence of an acid, HA it accepts a, proton and acts as the base while in the, –, presence of a base, B it acts as an acid by, donating a proton. In pure water, one H2O, molecule donates proton and acts as an acid, and another water molecules accepts a proton, and acts as a base at the same time. The, following equilibrium exists:, , H3O+(aq) + OH–(aq), conjugate, conjugate, acid, base, , H2O(l) + H2O(l), acid, base, , The dissociation constant is represented by,, –, , K = [H3O+] [OH ] / [H2O], (7.26), The concentration of water is omitted from, the denominator as water is a pure liquid and, its concentration remains constant. [H2O] is, incorporated within the equilibrium constant, to give a new constant, Kw, which is called the, ionic product of water., +, , –, , Kw = [H ][OH ], , (7.27), +, , The concentration of H has been found, out experimentally as 1.0 × 10–7 M at 298 K., And, as dissociation of water produces equal, –, number of H+ and OH ions, the concentration, –, +, of hydroxyl ions, [OH ] = [H ] = 1.0 × 10 –7 M., Thus, the value of Kw at 298K,, +, , –, , Kw = [H3O ][OH ] = (1 × 10–7)2 = 1 × 10–14 M 2, (7.28), The value of Kw is temperature dependent, as it is an equilibrium constant., The density of pure water is 1000 g / L, and its molar mass is 18.0 g /mol. From this, the molarity of pure water can be given as,, [H2O] = (1000 g /L)(1 mol/18.0 g) = 55.55 M., Therefore, the ratio of dissociated water to that, of undissociated water can be given as:, –9, , –9, , 10 –7 / (55.55) = 1.8 × 10 or ~ 2 in 10 (thus,, equilibrium lies mainly towards undissociated, water), We can distinguish acidic, neutral and, basic aqueous solutions by the relative values, of the H3O+ and OH– concentrations:, –, Acidic: [H3O+] > [OH ], –, , Neutral: [H3O+] = [OH ], –, , Basic : [H3O+] < [OH ], 7.11.2 The pH Scale, Hydronium ion concentration in molarity is, more conveniently expressed on a logarithmic, scale known as the pH scale. The pH of a, solution is defined as the negative logarithm, , ( ), , to base 10 of the activity a H+, , 2021-22, , of hydrogen

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218, , CHEMISTRY, , ion. In dilute solutions (< 0.01 M), activity of, +, hydrogen ion (H ) is equal in magnitude to, +, molarity represented by [H ]. It should, be noted that activity has no units and is, defined as:, +, , = [H ] / mol L–1, From the definition of pH, the following, can be written,, +, , pH = – log aH+ = – log {[H ] / mol L–1}, Thus, an acidic solution of HCl (10–2 M), will have a pH = 2. Similarly, a basic solution, –, –4, +, of NaOH having [OH ] =10 M and [H3O ] =, –10, 10, M will have a pH = 10. At 25 °C, pure, water has a concentration of hydrogen ions,, +, –7, [H ] = 10 M. Hence, the pH of pure water is, given as:, , change in pH by just one unit also means, change in [H+] by a factor of 10. Similarly, when, the hydrogen ion concentration, [H+] changes, by a factor of 100, the value of pH changes by, 2 units. Now you can realise why the change, in pH with temperature is often ignored., Measurement of pH of a solution is very, essential as its value should be known when, dealing with biological and cosmetic, applications. The pH of a solution can be found, roughly with the help of pH paper that has, different colour in solutions of different pH., Now-a-days pH paper is available with four, strips on it. The different strips have different, colours (Fig. 7.11) at the same pH. The pH in, the range of 1-14 can be determined with an, accuracy of ~0.5 using pH paper., , pH = –log(10–7) = 7, Acidic solutions possess a concentration, +, of hydrogen ions, [H ] > 10–7 M, while basic, solutions possess a concentration of hydrogen, +, ions, [H ] < 10–7 M. thus, we can summarise, that, Acidic solution has pH < 7, Basic solution has pH > 7, Neutral solution has pH = 7, Now again, consider the equation (7.28) at, 298 K, –, +, Kw = [H3O ] [OH ] = 10–14, Taking negative logarithm on both sides, of equation, we obtain, , Fig.7.11 pH-paper with four strips that may, have different colours at the same pH, , For greater accuracy pH meters are used., pH meter is a device that measures the, pH-dependent electrical potential of the test, solution within 0.001 precision. pH meters of, the size of a writing pen are now available in, the market. The pH of some very common, substances are given in Table 7.5 (page 212)., , –, , –log Kw = – log {[H3O+] [OH ]}, , The concentration of hydrogen ion in a, sample of soft drink is 3.8 × 10–3M. what, is its pH ?, , –, , = – log [H3O+] – log [OH ], = – log 10 –14, pKw = pH +, , pOH = 14, , Problem 7.16, , (7.29), , Note that although Kw may change with, temperature the variations in pH with, temperature are so small that we often, ignore it., pK w is a very important quantity for, aqueous solutions and controls the relative, concentrations of hydrogen and hydroxyl ions, as their product is a constant. It should be, noted that as the pH scale is logarithmic, a, , 2021-22, , Solution, pH = – log[3.8 × 10 –3], = – {log[3.8] + log[10 –3]}, = – {(0.58) + (– 3.0)} = – { – 2.42} = 2.42, Therefore, the pH of the soft drink is 2.42, and it can be inferred that it is acidic., Problem 7.17, Calculate pH of a 1.0 × 10, of HCl., , –8, , M solution, , a

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EQUILIBRIUM, , 219, , Table 7.5 The pH of Some Common Substances, Name of the Fluid, Saturated solution of NaOH, 0.1 M NaOH solution, Lime water, Milk of magnesia, Egg white, sea water, Human blood, Milk, Human Saliva, , pH, , Name of the Fluid, , ~15, 13, 10.5, 10, 7.8, 7.4, 6.8, 6.4, , Black Coffee, Tomato juice, Soft drinks and vinegar, Lemon juice, Gastric juice, 1M HCl solution, Concentrated HCl, , –, , 2H2O (l), H3O (aq) + OH (aq), –, Kw = [OH ][H3O+], = 10–14, –, +, Let, x = [OH ] = [H3O ] from H2O. The H3O+, concentration is generated (i) from, the ionization of HCl dissolved i.e.,, +, , –, , HCl(aq) + H2O(l), H3O (aq) + Cl (aq),, and (ii) from ionization of H2O. In these, very dilute solutions, both sources of, H3O+ must be considered:, –8, [H3O+] = 10 + x, –8, –14, Kw = (10 + x)(x) = 10, –8, or x2 + 10 x – 10–14 = 0, –, –8, [OH ] = x = 9.5 × 10, So, pOH = 7.02 and pH = 6.98, 7.11.3 Ionization Constants of Weak Acids, Consider a weak acid HX that is partially, ionized in the aqueous solution. The, equilibrium can be expressed by:, +, , 5.0, ~4.2, ~3.0, ~2.2, ~1.2, ~0, ~–1.0, , constant for the above discussed aciddissociation equilibrium:, Ka = c2α2 / c(1-α) = cα2 / 1-α, Ka is called the dissociation or ionization, constant of acid HX. It can be represented, alternatively in terms of molar concentration, as follows,, +, –, Ka = [H ][X ] / [HX], (7.30), At a given temperature T, K a is a, measure of the strength of the acid HX, i.e., larger the value of Ka, the stronger is, the acid. Ka is a dimensionless quantity, with the understanding that the standard, state concentration of all species is 1M., The values of the ionization constants of, some selected weak acids are given in, Table 7.6., , Solution, +, , pH, , Table 7.6 The Ionization Constants of Some, Selected Weak Acids (at 298K), Acid, , –, , H3O (aq) + X (aq), HX(aq) + H2O(l), Initial, concentration (M), c, 0, 0, Let α be the extent of ionization, Change (M), -cα, +cα, +cα, Equilibrium concentration (M), c-cα, cα, cα, Here, c = initial concentration of the, undissociated acid, HX at time, t = 0. α = extent, up to which HX is ionized into ions. Using, these notations, we can derive the equilibrium, , Ionization Constant,, Ka, –4, , Hydrofluoric Acid (HF), , 3.5 × 10, , Nitrous Acid (HNO2), , 4.5 × 10 –4, , Formic Acid (HCOOH), , 1.8 × 10, , –4, , Niacin (C5H4NCOOH), , 1.5 × 10, , –5, , Acetic Acid (CH3COOH), , 1.74 × 10, , Benzoic Acid (C6H5COOH), , 6.5 × 10 –5, , Hypochlorous Acid (HCIO), , 3.0 × 10, , –8, , Hydrocyanic Acid (HCN), , 4.9 × 10, , –10, , Phenol (C6H5OH), , 1.3 × 10, , –10, , –5, , The pH scale for the hydrogen ion, concentration has been so useful that besides, pKw, it has been extended to other species and, , 2021-22

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220, , CHEMISTRY, , quantities. Thus, we have:, pKa = –log (Ka), (7.31), Knowing the ionization constant, Ka of an, acid and its initial concentration, c, it is, possible to calculate the equilibrium, concentration of all species and also the degree, of ionization of the acid and the pH of the, solution., A general step-wise approach can be, adopted to evaluate the pH of the weak, electrolyte as follows:, Step 1. The species present before dissociation, are, identified, as, Brönsted-Lowry, acids / bases., Step 2. Balanced equations for all possible, reactions i.e., with a species acting both as, acid as well as base are written., Step 3. The reaction with the higher Ka is, identified as the primary reaction whilst the, other is a subsidiary reaction., Step 4. Enlist in a tabular form the following, values for each of the species in the primary, reaction, (a) Initial concentration, c., (b) Change in concentration on proceeding to, equilibrium in terms of α, degree of, ionization., (c) Equilibrium concentration., Step 5. Substitute equilibrium concentrations, into equilibrium constant equation for, principal reaction and solve for α., Step 6. Calculate the concentration of species, in principal reaction., Step 7. Calculate pH = – log[H3O+], The above mentioned methodology has, been elucidated in the following examples., Problem 7.18, The ionization constant of HF is, 3.2 × 10 –4 . Calculate the degree of, dissociation of HF in its 0.02 M solution., Calculate the concentration of all species, +, –, present (H3O , F and HF) in the solution, and its pH., , Solution, The following proton transfer reactions are, possible:, 1) HF + H2O, , H3O+ + F, , –, , Ka = 3.2 × 10–4, –, 2) H2O + H2O, H3O + OH, Kw = 1.0 × 10–14, As Ka >> Kw, [1] is the principle reaction., –, H3O+ + F, HF + H2O, Initial, concentration (M), 0.02, 0, 0, Change (M), –0.02α, +0.02α +0.02α, Equilibrium, concentration (M), 0.02 – 0.02 α, 0.02 α 0.02α, Substituting equilibrium concentrations, in the equilibrium reaction for principal, reaction gives:, Ka = (0.02α)2 / (0.02 – 0.02α), = 0.02 α2 / (1 –α) = 3.2 × 10–4, We obtain the following quadratic, equation:, α2 + 1.6 × 10–2α – 1.6 × 10–2 = 0, The quadratic equation in α can be solved, and the two values of the roots are:, α = + 0.12 and – 0.12, The negative root is not acceptable and, hence,, α = 0.12, This means that the degree of ionization,, α = 0.12, then equilibrium concentrations, –, +, of other species viz., HF, F and H3O are, given by:, –, [H3O+] = [F ] = cα = 0.02 × 0.12, = 2.4 × 10–3 M, [HF] = c(1 – α) = 0.02 (1 – 0.12), = 17.6 × 10-3 M, pH = – log[H+] = –log(2.4 × 10–3) = 2.62, +, , Problem 7.19, The pH of 0.1M monobasic acid is 4.50., +, Calculate the concentration of species H ,, , 2021-22

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EQUILIBRIUM, , 221, , –, , A and HA at equilibrium. Also, determine, the value of Ka and pKa of the monobasic, acid., Solution, , Percent dissociation, = {[HOCl]dissociated / [HOCl]initial }× 100, = 1.41 × 10–3 × 102/ 0.08 = 1.76 %., pH = –log(1.41 × 10–3) = 2.85., , +, , pH = – log [H ], Therefore, [H+] = 10 –pH = 10 –4.50, = 3.16 × 10, +, , –, , [H ] = [A ] = 3.16 × 10, Thus,, , 7.11.4 Ionization of Weak Bases, –5, , –5, , Ka = [H+][A-] / [HA], , [HA]eqlbm = 0.1 – (3.16 × 10-5) ∫ 0.1, Ka = (3.16 × 10–5)2 / 0.1 = 1.0 × 10–8, pKa = – log(10–8) = 8, Alternatively, “Percent dissociation” is, another useful method for measure of, strength of a weak acid and is given as:, Percent dissociation, = [HA]dissociated/[HA]initial × 100%, , (7.32), , Problem 7.20, Calculate the pH of 0.08M solution of, hypochlorous acid, HOCl. The ionization, constant of the acid is 2.5 × 10 –5 ., Determine the percent dissociation of, HOCl., Solution, HOCl(aq) + H2O (l), H3O+(aq) + ClO–(aq), Initial concentration (M), 0.08, 0, 0, Change to reach, equilibrium concentration, (M), –x, +x, +x, equilibrium concentartion (M), 0.08 – x, x, x, Ka = {[H3O+][ClO–] / [HOCl]}, = x2 / (0.08 –x), As x << 0.08, therefore 0.08 – x ∫ 0.08, x2 / 0.08 = 2.5 × 10–5, x2 = 2.0 × 10–6, thus, x = 1.41 × 10–3, [H+] = 1.41 × 10–3 M., Therefore,, , The ionization of base MOH can be represented, by equation:, –, M+(aq) + OH (aq), MOH(aq), In a weak base there is partial ionization, –, of MOH into M+ and OH , the case is similar to, that of acid-dissociation equilibrium. The, equilibrium constant for base ionization is, called base ionization constant and is, represented by Kb. It can be expressed in terms, of concentration in molarity of various species, in equilibrium by the following equation:, –, Kb = [M+][OH ] / [MOH], (7.33), Alternatively, if c = initial concentration of, base and α = degree of ionization of base i.e., the extent to which the base ionizes. When, equilibrium is reached, the equilibrium, constant can be written as:, Kb = (cα)2 / c (1-α) = cα2 / (1-α), The values of the ionization constants of, some selected weak bases, Kb are given in, Table 7.7., Table 7.7, , The Values of the Ionization, Constant of Some Weak Bases at, 298 K, , Base, , Kb, –4, , Dimethylamine, (CH3)2NH, , 5.4 × 10, , Triethylamine, (C2H5)3N, , 6.45 × 10, , Ammonia, NH3 or NH4OH, , 1.77 × 10, , Quinine, (A plant product), , 1.10 × 10, , Pyridine, C5H5N, , 1.77 × 10, , Aniline, C6H5NH2, , 4.27 × 10, , Urea, CO (NH2)2, , –5, –5, –6, –9, , –10, , –14, , 1.3 × 10, , Many organic compounds like amines are, weak bases. Amines are derivatives of, ammonia in which one or more hydrogen, atoms are replaced by another group. For, example, methylamine, codeine, quinine and, , 2021-22

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222, , CHEMISTRY, , nicotine all behave as very weak bases due to, their very small Kb. Ammonia produces OH–, in aqueous solution:, +, –, NH3(aq) + H2O(l), NH4 (aq) + OH (aq), The pH scale for the hydrogen ion, concentration has been extended to get:, pKb = –log (Kb), (7.34), , Kb = 10–4.75 = 1.77 × 10–5 M, , [H+] = antilog (–pH), = antilog (–9.7) = 1.67 ×10–10, [OH–] = Kw / [H+] = 1 × 10–14 / 1.67 × 10–10, = 5.98 × 10–5, The concentration of the corresponding, hydrazinium ion is also the same as that, of hydroxyl ion. The concentration of both, these ions is very small so the, concentration of the undissociated base, can be taken equal to 0.004M., Thus,, +, , –, , Kb = [NH2NH3 ][OH ] / [NH2NH2], = (5.98 × 10–5)2 / 0.004 = 8.96 × 10–7, , Calculate the pH of the solution in which, 0.2M NH4Cl and 0.1M NH3 are present. The, pKb of ammonia solution is 4.75., Solution, +, , The ionization constant of NH3,, Kb = antilog (–pKb) i.e., , +, , OH, , –, , 0.10, , 0.20, , 0, , +x, , +x, , 0.20 + x, , x, , Change to reach, equilibrium (M), –x, At equilibrium (M), 0.10 – x, +, , –, , Kb = [NH4 ][OH ] / [NH3], = (0.20 + x)(x) / (0.1 – x) = 1.77 × 10–5, As K b is small, we can neglect x in, comparison to 0.1M and 0.2M. Thus,, –, , [OH ] = x = 0.88 × 10–5, Therefore, [H+] = 1.12 × 10–9, pH = – log[H+] = 8.95., 7.11.5 Relation between K a and K b, As seen earlier in this chapter, K a and K b, represent the strength of an acid and a base,, respectively. In case of a conjugate acid-base, pair, they are related in a simple manner so, that if one is known, the other can be deduced., +, Considering the example of NH4 and NH3, we see,, H3O+(aq) + NH3(aq), NH4+(aq) + H2O(l), Ka = [H3O+][ NH3] / [NH4+] = 5.6 × 10–10, –, NH4+(aq) + OH (aq), NH3(aq) + H2O(l), –, Kb =[ NH4+][ OH ] / NH3 = 1.8 × 10–5, –, H3O+(aq) + OH (aq), Net: 2 H2O(l), –, –14, Kw = [H3O+][ OH ] = 1.0 × 10 M, +, , Problem 7.22, , NH4, , OH, , Where, K a represents the strength of NH4 as, an acid and K b represents the strength of NH3, as a base., , pKb = –logKb = –log(8.96 × 10–7) = 6.04., , NH3 + H2O, , +, , Initial concentration (M), , Problem 7.21, The pH of 0.004M hydrazine solution is, 9.7. Calculate its ionization constant Kb, and pKb., Solution, –, NH2NH3+ + OH, NH2NH2 + H2O, From the pH we can calculate the, hydrogen ion concentration. Knowing, hydrogen ion concentration and the ionic, product of water we can calculate the, concentration of hydroxyl ions. Thus we, have:, , +, , NH4, , NH3 + H2O, , –, , It can be seen from the net reaction that, the equilibrium constant is equal to the, product of equilibrium constants K a and K b, for the reactions added. Thus,, +, , +, , K a × K b = {[H3O+][ NH3] / [NH4 ]} × {[NH4 ], –, [ OH ] / [NH3]}, –, , = [H3O+][ OH ] = Kw, = (5.6x10–10) × (1.8 × 10–5) = 1.0 × 10–14 M, , 2021-22

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EQUILIBRIUM, , 223, , This can be extended to make a, generalisation. The equilibrium constant for, a net reaction obtained after adding two, (or more) reactions equals the product of, the equilibrium constants for individual, reactions:, K NET = K1 × K2 × ……, , The value of α is small, therefore the, quadratic equation can be simplified by, neglecting α in comparison to 1 in the, denominator on right hand side of the, equation,, , (7.36), , Thus,, Kb = c α2 or α = √ (1.77 × 10–5 / 0.05), = 0.018., –, , [OH ] = c α = 0.05 × 0.018 = 9.4 × 10–4M., –, , [H+] = Kw / [OH ] = 10–14 / (9.4 × 10–4), , –, , BH+(aq) + OH (aq), , = 1.06 × 10–11, , –, , K b = [BH ][OH ] / [B], , pH = –log(1.06 × 10–11) = 10.97., , As the concentration of water remains, constant it has been omitted from the, denominator and incorporated within the, dissociation constant. Then multiplying and, dividing the above expression by [H+], we get:, –, , +, , Now, using the relation for conjugate, acid-base pair,, Ka × Kb = Kw, using the value of K b of NH 3 from, Table 7.7., , +, , Kb = [BH ][OH ][H ] / [B][H ], –, , +, , +, , –, , Kb = 0.05 α2 / (1 – α), , Alternatively, the above expression, K w = K a × K b , can also be obtained by, considering the base-dissociation equilibrium, reaction:, , +, , OH, , –, , Knowing one, the other can be obtained. It, should be noted that a strong acid will have, a weak conjugate base and vice-versa., , +, , +, , [OH ] = c α = 0.05 α, , Similarly, in case of a conjugate acid-base, pair,, , B(aq) + H2O(l), , NH4, , We use equation (7.33) to calculate, hydroxyl ion concentration,, , (3.35), , K a × Kb = K w, , +, , NH3 + H2O, , +, , ={[ OH ][H ]}{[BH ] / [B][H ]}, = K w / Ka, or K a × K b = K w, It may be noted that if we take negative, logarithm of both sides of the equation, then, pK values of the conjugate acid and base are, related to each other by the equation:, pK a + pK b = pK w = 14 (at 298K), Problem 7.23, Determine the degree of ionization and pH, of a 0.05M of ammonia solution. The, ionization constant of ammonia can be, taken from Table 7.7. Also, calculate the, ionization constant of the conjugate acid, of ammonia., Solution, The ionization of NH 3 in water is, represented by equation:, , We can determine the concentration of, +, conjugate acid NH4, Ka = Kw / Kb = 10–14 / 1.77 × 10–5, = 5.64 × 10–10., 7.11.6 Di- and Polybasic Acids and Di- and, Polyacidic Bases, Some of the acids like oxalic acid, sulphuric, acid and phosphoric acids have more than one, ionizable proton per molecule of the acid., Such acids are known as polybasic or, polyprotic acids., The ionization reactions for example for a, dibasic acid H 2X are represented by the, equations:, +, –, H2X(aq), H (aq) + HX (aq), –, +, 2–, HX (aq), H (aq) + X (aq), And the corresponding equilibrium, constants are given below:, +, –, Ka = {[H ][HX ]} / [H2X] and, 1, , 2021-22

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224, , CHEMISTRY, , Ka2 = {[H+][X2-]} / [HX-], Here, Ka1and Ka2are called the first and second, ionization constants respectively of the acid H2, X. Similarly, for tribasic acids like H3PO4 we, have three ionization constants. The values of, the ionization constants for some common, polyprotic acids are given in Table 7.8., Table 7.8 The Ionization Constants of Some, Common Polyprotic Acids (298K), , It can be seen that higher order ionization, constants (Ka2, Ka3) are smaller than the lower, order ionization constant (Ka1) of a polyprotic, acid. The reason for this is that it is more, difficult to remove a positively charged proton, from a negative ion due to electrostatic forces., This can be seen in the case of removing a, proton from the uncharged H 2 CO 3 as, –, compared from a negatively charged HCO3 ., Similarly, it is more difficult to remove a proton, 2–, from a doubly charged HPO 4 anion as, –, compared to H2PO4 ., Polyprotic acid solutions contain a mixture, 2–, –, of acids like H2A, HA and A in case of a, diprotic acid. H2A being a strong acid, the, primary reaction involves the dissociation of, H2 A, and H3O+ in the solution comes mainly, from the first dissociation step., 7.11.7 Factors Affecting Acid Strength, Having discussed quantitatively the strengths, of acids and bases, we come to a stage where, we can calculate the pH of a given acid, solution. But, the curiosity rises about why, should some acids be stronger than others?, What factors are responsible for making them, stronger? The answer lies in its being a, complex phenomenon. But, broadly speaking, we can say that the extent of dissociation of, an acid depends on the strength and polarity, of the H-A bond., , In general, when strength of H-A bond, decreases, that is, the energy required to break, the bond decreases, HA becomes a stronger, acid. Also, when the H-A bond becomes more, polar i.e., the electronegativity difference, between the atoms H and A increases and, there is marked charge separation, cleavage, of the bond becomes easier thereby increasing, the acidity., But it should be noted that while, comparing elements in the same group of the, periodic table, H-A bond strength is a more, important factor in determining acidity than, its polar nature. As the size of A increases, down the group, H-A bond strength decreases, and so the acid strength increases. For, example,, Size increases, HF << HCl << HBr << HI, Acid strength increases, Similarly, H2S is stronger acid than H2O., But, when we discuss elements in the same, row of the periodic table, H-A bond polarity, becomes the deciding factor for determining, the acid strength. As the electronegativity of A, increases, the strength of the acid also, increases. For example,, Electronegativity of A increases, CH4 < NH3 < H2O < HF, Acid strength increases, 7.11.8 Common Ion Effect in the, Ionization of Acids and Bases, Consider an example of acetic acid dissociation, equilibrium represented as:, –, CH3COOH(aq), H+(aq) + CH3COO (aq), –, or HAc(aq), H+ (aq) + Ac (aq), –, Ka = [H+][Ac ] / [HAc], Addition of acetate ions to an acetic acid, solution results in decreasing the, concentration of hydrogen ions, [H+]. Also, if, H+ ions are added from an external source then, the equilibrium moves in the direction of, undissociated acetic acid i.e., in a direction of, reducing the concentration of hydrogen ions,, +, [H ]. This phenomenon is an example of, , 2021-22

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EQUILIBRIUM, , 225, , common ion effect. It can be defined as a, shift in equilibrium on adding a substance, that provides more of an ionic species already, present in the dissociation equilibrium. Thus,, we can say that common ion effect is a, phenomenon based on the Le Chatelier’s, principle discussed in section 7.8., In order to evaluate the pH of the solution, resulting on addition of 0.05M acetate ion to, 0.05M acetic acid solution, we shall consider, the acetic acid dissociation equilibrium once, again,, –, HAc(aq), H+(aq) + Ac (aq), Initial concentration (M), 0.05, 0, 0.05, Let x be the extent of ionization of acetic, acid., Change in concentration (M), –x, +x, +x, Equilibrium concentration (M), 0.05-x, x, 0.05+x, Therefore,, –, K a= [H+][Ac ]/[H Ac] = {(0.05+x)(x)}/(0.05-x), As Ka is small for a very weak acid, x<<0.05., , Thus, x = 1.33 × 10–3 = [OH–], Therefore,[H+] = Kw / [OH–] = 10–14 /, (1.33 × 10–3) = 7.51 × 10–12, pH = –log(7.5 × 10–12) = 11.12, On addition of 25 mL of 0.1M HCl, solution (i.e., 2.5 mmol of HCl) to 50 mL, of 0.1M ammonia solution (i.e., 5 mmol, of NH3), 2.5 mmol of ammonia molecules, are neutralized. The resulting 75 mL, solution contains the remaining, unneutralized 2.5 mmol of NH3 molecules, +, and 2.5 mmol of NH4 ., NH3, , Problem 7.24, Calculate the pH of a 0.10M ammonia, solution. Calculate the pH after 50.0 mL, of this solution is treated with 25.0 mL of, 0.10M HCl. The dissociation constant of, ammonia, K b = 1.77 × 10–5, Solution, H2O, , →, , NH4+ + OH–, , Kb = [NH4+][OH–] / [NH3] = 1.77 × 10–5, Before neutralization,, –, [NH4+] = [OH ] = x, [NH3] = 0.10 – x ∫ 0.10, x2 / 0.10 = 1.77 × 10–5, , → NH4+ + Cl–, , The final 75 mL solution after, neutralisation already contains, 2.5 m mol NH4+ ions (i.e. 0.033M), thus, +, total concentration of NH4 ions is given as:, +, [NH4 ] = 0.033 + y, As y is small, [NH4OH] ∫ 0.033 M and, [NH4+] ∫ 0.033M., We know,, Kb = [NH4+][OH–] / [NH4OH], = y(0.033)/(0.033) = 1.77 × 10–5 M, Thus, y = 1.77 × 10–5 = [OH–], [H+] = 10–14 / 1.77 × 10–5 = 0.56 × 10–9, Hence, pH = 9.24, , Thus,, 1.8 × 10–5 = (x) (0.05 + x) / (0.05 – x), –5, = x(0.05) / (0.05) = x = [H+] = 1.8 × 10 M, pH = – log(1.8 × 10–5) = 4.74, , +, , HCl, , 2.5, 2.5, 0, 0, At equilibrium, 0, 0, 2.5, 2.5, The resulting 75 mL of solution contains, +, 2.5 mmol of NH4 ions (i.e., 0.033 M) and, 2.5 mmol (i.e., 0.033 M ) of uneutralised, NH3 molecules. This NH3 exists in the, following equilibrium:, +, –, NH4 + OH, NH4OH, 0.033M – y, y, y, –, +, where, y = [OH ] = [NH4 ], , Hence, (0.05 + x) ≈ (0.05 – x) ≈ 0.05, , NH3, , +, , 7.11.9 Hydrolysis of Salts and the pH of, their Solutions, Salts formed by the reactions between acids, and bases in definite proportions, undergo, ionization in water. The cations/anions formed, , 2021-22

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226, , CHEMISTRY, , on ionization of salts either exist as hydrated, ions in aqueous solutions or interact with, water to reform corresponding acids/bases, depending upon the nature of salts. The later, process of interaction between water and, cations/anions or both of salts is called, hydrolysis. The pH of the solution gets affected, +, +, by this interaction. The cations (e.g., Na , K ,, 2+, 2+, Ca , Ba , etc.) of strong bases and anions, –, –, –, –, (e.g., Cl , Br , NO3, ClO4 etc.) of strong acids, simply get hydrated but do not hydrolyse, and, therefore the solutions of salts formed from, strong acids and bases are neutral i.e., their, pH is 7. However, the other category of salts, do undergo hydrolysis., We now consider the hydrolysis of the salts, of the following types :, (i) salts of weak acid and strong base e.g.,, CH3COONa., (ii) salts of strong acid and weak base e.g.,, NH4Cl, and, (iii) salts of weak acid and weak base, e.g.,, CH3COONH4., , increased of H+ ion concentration in solution, making the solution acidic. Thus, the pH of, NH4Cl solution in water is less than 7., Consider the hydrolysis of CH3COONH4 salt, formed from weak acid and weak base. The, ions formed undergo hydrolysis as follow:, –, +, CH3COOH +, CH3COO + NH4 + H2O, NH4OH, CH3COOH and NH4OH, also remain into, partially dissociated form:, –, CH3COO + H+, CH3COOH, +, –, NH4OH, NH4 + OH, –, H2O, H+ + OH, Without going into detailed calculation, it, can be said that degree of hydrolysis is, independent of concentration of solution, and, pH of such solutions is determined by their pK, values:, pH = 7 + ½ (pKa – pKb), (7.38), The pH of solution can be greater than 7,, if the difference is positive and it will be less, than 7, if the difference is negative., , In the first case, CH3COONa being a salt of, weak acid, CH3COOH and strong base, NaOH, gets completely ionised in aqueous solution., , Problem 7.25, The pK a of acetic acid and pK b of, ammonium hydroxide are 4.76 and 4.75, respectively. Calculate the pH of, ammonium acetate solution., , CH3COONa(aq) → CH3COO (aq)+ Na (aq), –, , +, , Acetate ion thus formed undergoes, hydrolysis in water to give acetic acid and OH–, ions, –, , CH3COO (aq)+H2O(l), , pH = 7 + ½ [pKa – pKb], , –, , CH3COOH(aq)+OH (aq), , Acetic acid being a weak acid, (Ka = 1.8 × 10–5) remains mainly unionised in, –, solution. This results in increase of OH ion, concentration in solution making it alkaline., The pH of such a solution is more than 7., Similarly, NH4Cl formed from weak base,, NH 4OH and strong acid, HCl, in water, dissociates completely., +, , Solution, , –, , NH4Cl(aq) → NH 4(aq) +Cl (aq), Ammonium ions undergo hydrolysis with, water to form NH4OH and H+ ions, +, NH 4 (aq) + H2O (1), NH4OH(aq) + H+(aq), Ammonium hydroxide is a weak base, (Kb = 1.77 × 10–5) and therefore remains almost, unionised in solution. This results in, , = 7 + ½ [4.76 – 4.75], = 7 + ½ [0.01] = 7 + 0.005 = 7.005, 7.12 BUFFER SOLUTIONS, Many body fluids e.g., blood or urine have, definite pH and any deviation in their pH, indicates malfunctioning of the body. The, control of pH is also very important in many, chemical and biochemical processes. Many, medical and cosmetic formulations require, that these be kept and administered at a, particular pH. The solutions which resist, change in pH on dilution or with the, addition of small amounts of acid or alkali, are called Buffer Solutions. Buffer solutions, , 2021-22

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EQUILIBRIUM, , 227, , of known pH can be prepared from the, knowledge of pK a of the acid or pK of base, b, and by controlling the ratio of the salt and acid, or salt and base. A mixture of acetic acid and, sodium acetate acts as buffer solution around, pH 4.75 and a mixture of ammonium chloride, and ammonium hydroxide acts as a buffer, around pH 9.25. You will learn more about, buffer solutions in higher classes., 7.12.1 Designing Buffer Solution, Knowledge of pK a, pK b and equilibrium, constant help us to prepare the buffer solution, of known pH. Let us see how we can do this., Preparation of Acidic Buffer, To prepare a buffer of acidic pH we use weak, acid and its salt formed with strong base. We, develop the equation relating the pH, the, equilibrium constant, Ka of weak acid and ratio, of concentration of weak acid and its conjugate, base. For the general case where the weak acid, HA ionises in water,, HA + H2O Ç H3O+ + A–, For which we can write the expression, , conjugate base (anion) of the acid and the acid, present in the mixture. Since acid is a weak, acid, it ionises to a very little extent and, concentration of [HA] is negligibly different from, concentration of acid taken to form buffer. Also,, most of the conjugate base, [A—], comes from, the ionisation of salt of the acid. Therefore, the, concentration of conjugate base will be, negligibly different from the concentration of, salt. Thus, equation (7.40) takes the form:, , [Salt], [Acid], In the equation (7.39), if the concentration, of [A—] is equal to the concentration of [HA],, then pH = pKa because value of log 1 is zero., Thus if we take molar concentration of acid and, salt (conjugate base) same, the pH of the buffer, solution will be equal to the pKa of the acid. So, for preparing the buffer solution of the required, pH we select that acid whose pKa is close to the, required pH. For acetic acid pKa value is 4.76,, therefore pH of the buffer solution formed by, acetic acid and sodium acetate taken in equal, molar concentration will be around 4.76., A similar analysis of a buffer made with a, weak base and its conjugate acid leads to the, result,, pH=pK a + log, , Rearranging the expression we have,, , pOH= pK b +log, , Taking logarithm on both the sides and, rearranging the terms we get —, , Or, , (7.41), pH of the buffer solution can be calculated, by using the equation pH + pOH =14., We know that pH + pOH = pK w and, pKa + pKb = pKw. On putting these values in, equation (7.41) it takes the form as follows:, pK w - pH = pK w − pK a + log, , (7.39), , (7.40), The expression (7.40) is known as, Henderson–Hasselbalch equation. The, is the ratio of concentration of, , [Conjugate acid,BH + ], [Base,B], , or, pH = pK a + log, , quantity, , [Conjugate acid,BH+ ], [Base,B], , [Conjugate acid,BH+ ], [Base,B], , (7.42), , If molar concentration of base and its, conjugate acid (cation) is same then pH of the, buffer solution will be same as pKa for the base., pKa value for ammonia is 9.25; therefore a, buffer of pH close to 9.25 can be obtained by, taking ammonia solution and ammonium, , 2021-22

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228, , CHEMISTRY, , chloride solution of same molar concentration., For a buffer solution formed by ammonium, chloride and ammonium hydroxide, equation, (7.42) becomes:, , [Conjugate acid,BH + ], pH = 9.25 + log, [Base,B], pH of the buffer solution is not affected by, dilution because ratio under the logarithmic, term remains unchanged., 7.13 SOLUBILITY, EQUILIBRIA, OF, SPARINGLY SOLUBLE SALTS, We have already known that the solubility of, ionic solids in water varies a great deal. Some, of these (like calcium chloride) are so soluble, that they are hygroscopic in nature and even, absorb water vapour from atmosphere. Others, (such as lithium fluoride) have so little solubility, that they are commonly termed as insoluble., The solubility depends on a number of factors, important amongst which are the lattice, enthalpy of the salt and the solvation enthalpy, of the ions in a solution. For a salt to dissolve, in a solvent the strong forces of attraction, between its ions (lattice enthalpy) must be, overcome by the ion-solvent interactions. The, solvation enthalpy of ions is referred to in terms, of solvation which is always negative i.e. energy, is released in the process of solvation. The, amount of solvation enthalpy depends on the, nature of the solvent. In case of a non-polar, (covalent) solvent, solvation enthalpy is small, and hence, not sufficient to overcome lattice, enthalpy of the salt. Consequently, the salt does, not dissolve in non-polar solvent. As a general, rule , for a salt to be able to dissolve in a, particular solvent its solvation enthalpy must, be greater than its lattice enthalpy so that the, latter may be overcome by former. Each salt has, its characteristic solubility which depends on, temperature. We classify salts on the basis of, their solubility in the following three categories., Category I, , Soluble, , Solubility > 0.1M, , Category II, , Slightly, Soluble, , 0.01M<Solubility< 0.1M, , Category III, , Sparingly, Soluble, , Solubility < 0.01M, , We shall now consider the equilibrium, between the sparingly soluble ionic salt and, its saturated aqueous solution., 7.13.1 Solubility Product Constant, Let us now have a solid like barium sulphate, in contact with its saturated aqueous solution., The equilibrium between the undisolved solid, and the ions in a saturated solution can be, represented by the equation:, BaSO4(s), , Ba2+(aq) + SO42–(aq),, , The equilibrium constant is given by the, equation:, K = {[Ba2+][SO42–]} / [BaSO4], For a pure solid substance the, concentration remains constant and we can, write, Ksp = K[BaSO4] = [Ba2+][SO42–], (7.43), We call Ksp the solubility product constant, or simply solubility product. The experimental, value of Ksp in above equation at 298K is, –10, 1.1 × 10 . This means that for solid barium, sulphate in equilibrium with its saturated, solution, the product of the concentrations of, barium and sulphate ions is equal to its, solubility, product, constant., The, concentrations of the two ions will be equal to, the molar solubility of the barium sulphate. If, molar solubility is S, then, 1.1 × 10–10 = (S)(S) = S2, or, S = 1.05 × 10–5., Thus, molar solubility of barium sulphate, will be equal to 1.05 × 10–5 mol L–1., A salt may give on dissociation two or more, than two anions and cations carrying different, charges. For example, consider a salt like, zirconium phosphate of molecular formula, 4+, 3–, (Zr )3(PO4 )4. It dissociates into 3 zirconium, cations of charge +4 and 4 phosphate anions, of charge –3. If the molar solubility of, zirconium phosphate is S, then it can be seen, from the stoichiometry of the compound that, [Zr4+] = 3S and [PO43–] = 4S, and Ksp = (3S)3 (4S)4 = 6912 (S)7, or S = {Ksp / (33 × 44)}1/7 = (Ksp / 6912)1/7, , 2021-22

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EQUILIBRIUM, , 229, , A solid salt of the general formula, q−, M X y with molar solubility S in equilibrium, with its saturated solution may be represented, by the equation:, p+, q–, M x X y (s), xM (aq) + yX (aq), (where x × p+ = y × q–), And its solubility product constant is given, by:, p+ x, q– y, x, y, Ksp = [M ] [X ] = (xS) (yS), (7.44), = xx . yy . S(x + y), S(x + y) = Ksp / xx . yy, S = (Ksp / xx . yy)1 / x + y, (7.45), The term Ksp in equation is given by Qsp, (section 7.6.2) when the concentration of one, or more species is not the concentration under, equilibrium. Obviously under equilibrium, conditions Ksp = Qsp but otherwise it gives the, direction of the processes of precipitation or, dissolution. The solubility product constants, of a number of common salts at 298K are given, in Table 7.9., p+, x, , Table 7.9 The Solubility Product Constants,, Ksp of Some Common Ionic Salts at, 298K., , Problem 7.26, Calculate the solubility of A2 X3 in pure, water, assuming that neither kind of ion, reacts with water. The solubility product, of A 2X 3 , Ksp = 1.1 × 10–23., Solution, A2X3 → 2A3+ + 3X2–, Ksp = [A3+]2 [X2–]3 = 1.1 × 10–23, If S = solubility of A 2X 3 , then, [A3+] = 2S; [X2–] = 3S, therefore, Ksp = (2S)2(3S)3 = 108S5, = 1.1 × 10–23, thus, S5 = 1 × 10–25, S = 1.0 × 10–5 mol/L., Problem 7.27, The values of Ksp of two sparingly soluble, salts Ni(OH)2 and AgCN are 2.0 × 10–15, and 6 × 0–17 respectively. Which salt is, more soluble? Explain., Solution, AgCN, , –, , Ag+ + CN, , 2021-22

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230, , CHEMISTRY, , Ksp = [Ag+][CN–] = 6 × 10–17, Ni(OH)2, Ni2+ + 2OH–, Ksp = [Ni2+][OH–]2 = 2 × 10–15, Let [Ag+] = S1, then [CN-] = S1, Let [Ni2+] = S2, then [OH–] = 2S2, S12 = 6 × 10–17 , S1 = 7.8 × 10–9, (S2)(2S2)2 = 2 × 10–15, S2 = 0.58 × 10–4, Ni(OH)2 is more soluble than AgCN., , Dissolution of S mol/L of Ni(OH)2 provides, –, S mol/L of Ni2+ and 2S mol/L of OH , but, –, the total concentration of OH = (0.10 +, 2S) mol/L because the solution already, contains 0.10 mol/L of OH– from NaOH., –, , 7.13.2 Common Ion Effect on Solubility, of Ionic Salts, It is expected from Le Chatelier’s principle that, if we increase the concentration of any one of, the ions, it should combine with the ion of its, opposite charge and some of the salt will be, precipitated till once again Ksp = Qsp. Similarly,, if the concentration of one of the ions is, decreased, more salt will dissolve to increase, the concentration of both the ions till once, again Ksp = Qsp. This is applicable even to, soluble salts like sodium chloride except that, due to higher concentrations of the ions, we, use their activities instead of their molarities, in the expression for Qsp. Thus if we take a, saturated solution of sodium chloride and, pass HCl gas through it, then sodium chloride, is precipitated due to increased concentration, (activity) of chloride ion available from the, dissociation of HCl. Sodium chloride thus, obtained is of very high purity and we can get, rid of impurities like sodium and magnesium, sulphates. The common ion effect is also used, for almost complete precipitation of a particular, ion as its sparingly soluble salt, with very low, value of solubility product for gravimetric, estimation. Thus we can precipitate silver ion, as silver chloride, ferric ion as its hydroxide, (or hydrated ferric oxide) and barium ion as, its sulphate for quantitative estimations., Problem 7.28, Calculate the molar solubility of Ni(OH)2, in 0.10 M NaOH. The ionic product of, Ni(OH)2 is 2.0 × 10–15., Solution, Let the solubility of Ni(OH)2 be equal to S., , Ksp = 2.0 × 10–15 = [Ni2+] [OH ]2, = (S) (0.10 + 2S)2, As Ksp is small, 2S << 0.10,, thus, (0.10 + 2S) ≈ 0.10, Hence,, 2.0 × 10–15 = S (0.10)2, S = 2.0 × 10–13 M = [Ni2+], The solubility of salts of weak acids like, phosphates increases at lower pH. This is, because at lower pH the concentration of the, anion decreases due to its protonation. This, in turn increase the solubility of the salt so, that Ksp = Qsp. We have to satisfy two equilibria, simultaneously i.e.,, Ksp = [M+] [X–],, , –, , [X ] / [HX] = Ka / [H+], Taking inverse of both side and adding 1, we get, , [HX ] + 1 =,  X − , ,  H+ , +1, Ka, , [HX ] + H − ,  X − , ,  H+  + K a, =, Ka, , Now, again taking inverse, we get, –, –, [X ] / {[X ] + [HX]} = f = Ka / (Ka + [H+]) and it, can be seen that ‘f’ decreases as pH decreases. If, S is the solubility of the salt at a given pH then, Ksp = [S] [f S] = S2 {Ka / (Ka + [H+])} and, S = {Ksp ([H+] + Ka ) / Ka }1/2, (7.46), Thus solubility S increases with increase, in [H+] or decrease in pH., , 2021-22

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EQUILIBRIUM, , 231, , SUMMARY, When the number of molecules leaving the liquid to vapour equals the number of, molecules returning to the liquid from vapour, equilibrium is said to be attained and is, dynamic in nature. Equilibrium can be established for both physical and chemical, processes and at this stage rate of forward and reverse reactions are equal. Equilibrium, constant, Kc is expressed as the concentration of products divided by reactants, each, term raised to the stoichiometric coefficient., For reaction, a A + b B, c C +d D, c, d, a, b, Kc = [C] [D] /[A] [B], Equilibrium constant has constant value at a fixed temperature and at this stage, all the macroscopic properties such as concentration, pressure, etc. become constant., For a gaseous reaction equilibrium constant is expressed as Kp and is written by replacing, concentration terms by partial pressures in Kc expression. The direction of reaction can, be predicted by reaction quotient Qc which is equal to Kc at equilibrium. Le Chatelier’s, principle states that the change in any factor such as temperature, pressure,, concentration, etc. will cause the equilibrium to shift in such a direction so as to reduce, or counteract the effect of the change. It can be used to study the effect of various, factors such as temperature, concentration, pressure, catalyst and inert gases on the, direction of equilibrium and to control the yield of products by controlling these factors., Catalyst does not effect the equilibrium composition of a reaction mixture but increases, the rate of chemical reaction by making available a new lower energy pathway for, conversion of reactants to products and vice-versa., All substances that conduct electricity in aqueous solutions are called electrolytes., Acids, bases and salts are electrolytes and the conduction of electricity by their aqueous, solutions is due to anions and cations produced by the dissociation or ionization of, electrolytes in aqueous solution. The strong electrolytes are completely dissociated. In, weak electrolytes there is equilibrium between the ions and the unionized electrolyte, molecules. According to Arrhenius, acids give hydrogen ions while bases produce, hydroxyl ions in their aqueous solutions. Brönsted-Lowry on the other hand, defined, an acid as a proton donor and a base as a proton acceptor. When a Brönsted-Lowry, acid reacts with a base, it produces its conjugate base and a conjugate acid corresponding, to the base with which it reacts. Thus a conjugate pair of acid-base differs only by one, proton. Lewis further generalised the definition of an acid as an electron pair acceptor, and a base as an electron pair donor. The expressions for ionization (equilibrium), constants of weak acids (Ka) and weak bases (Kb) are developed using Arrhenius definition., The degree of ionization and its dependence on concentration and common ion are, discussed. The pH scale (pH = -log[H+]) for the hydrogen ion concentration (activity) has, –, been introduced and extended to other quantities (pOH = – log[OH ]) ; pKa = –log[Ka] ;, pKb = –log[Kb]; and pKw = –log[Kw] etc.). The ionization of water has been considered and, we note that the equation: pH + pOH = pKw is always satisfied. The salts of strong acid, and weak base, weak acid and strong base, and weak acid and weak base undergo, hydrolysis in aqueous solution.The definition of buffer solutions, and their importance, are discussed briefly. The solubility equilibrium of sparingly soluble salts is discussed, and the equilibrium constant is introduced as solubility product constant (Ksp). Its, relationship with solubility of the salt is established. The conditions of precipitation of, the salt from their solutions or their dissolution in water are worked out. The role of, common ion and the solubility of sparingly soluble salts is also discussed., , 2021-22

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232, , CHEMISTRY, , SUGGESTED ACTIVITIES FOR STUDENTS REGARDING THIS UNIT, (a) The student may use pH paper in determining the pH of fresh juices of various, vegetables and fruits, soft drinks, body fluids and also that of water samples available., (b) The pH paper may also be used to determine the pH of different salt solutions and, from that he/she may determine if these are formed from strong/weak acids and, bases., (c) They may prepare some buffer solutions by mixing the solutions of sodium acetate, and acetic acid and determine their pH using pH paper., (d) They may be provided with different indicators to observe their colours in solutions, of varying pH., (e) They may perform some acid-base titrations using indicators., (f) They may observe common ion effect on the solubility of sparingly soluble salts., (g) If pH meter is available in their school, they may measure the pH with it and compare, the results obtained with that of the pH paper., , EXERCISES, 7.1, , A liquid is in equilibrium with its vapour in a sealed container at a fixed, temperature. The volume of the container is suddenly increased., , a), , What is the initial effect of the change on vapour pressure?, , b), , How do rates of evaporation and condensation change initially?, , c), , What happens when equilibrium is restored finally and what will be the final, vapour pressure?, , 7.2, , What is Kc for the following equilibrium when the equilibrium concentration of, each substance is: [SO2]= 0.60M, [O2] = 0.82M and [SO3] = 1.90M ?, , 7.3, , At a certain temperature and total pressure of 105Pa, iodine vapour contains 40%, by volume of I atoms, I2 (g), 2I (g), , 2SO2(g) + O2(g), , 2SO3(g), , Calculate Kp for the equilibrium., 7.4, , Write the expression for the equilibrium constant, Kc for each of the following, reactions:, (i), , 7.5, , 7.6, , 2NOCl (g), , 2NO (g) + Cl2 (g), , (ii), , 2Cu(NO3)2 (s), , (iii), , CH3COOC2H5(aq) + H2O(l), 3+, , (iv), , Fe, , (v), , I2 (s) + 5F2, , 2CuO (s) + 4NO2 (g) + O2 (g), –, , (aq) + 3OH (aq), , CH3COOH (aq) + C2H5OH (aq), Fe(OH)3 (s), , 2IF5, , Find out the value of Kc for each of the following equilibria from the value of Kp:, (i), , 2NOCl (g), , 2NO (g) + Cl2 (g); Kp= 1.8 × 10–2 at 500 K, , (ii), , CaCO3 (s), , CaO(s) + CO2(g); Kp= 167 at 1073 K, , For the following equilibrium, Kc= 6.3 × 1014 at 1000 K, , NO (g) + O3 (g), , NO2 (g) + O2 (g), , Both the forward and reverse reactions in the equilibrium are elementary, bimolecular reactions. What is Kc, for the reverse reaction?, , 2021-22

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EQUILIBRIUM, , 233, , 7.7, , Explain why pure liquids and solids can be ignored while writing the equilibrium, constant expression?, , 7.8, , Reaction between N2 and O2– takes place as follows:, 2N2 (g) + O2 (g), , 2N2O (g), , If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction, vessel and allowed to form N2O at a temperature for which Kc= 2.0 × 10 –37,, determine the composition of equilibrium mixture., 7.9, , Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given, below:, 2NOBr (g), , 2NO (g) + Br2 (g), , When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at, constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate, equilibrium amount of NO and Br2 ., 7.10, , At 450K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium., 2SO2(g) + O2(g), , 2SO3 (g), , What is Kc at this temperature ?, 7.11, , A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the, partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium ?, 2HI (g), , H2 (g) + I2 (g), , 7.12, , A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into, a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant,, Kc for the reaction N2 (g) + 3H2 (g), 2NH3 (g) is 1.7 × 102. Is the reaction mixture, at equilibrium? If not, what is the direction of the net reaction?, , 7.13, , The equilibrium constant expression for a gas reaction is,, , Kc =, , 4, 5, [NH3 ] [O2 ], 6, 4, [ NO] [H2O], , Write the balanced chemical equation corresponding to this expression., 7.14, , One mole of H2O and one mole of CO are taken in 10 L vessel and heated to, 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the, equation,, H2O (g) + CO (g), , H2 (g) + CO2 (g), , Calculate the equilibrium constant for the reaction., 7.15, , At 700 K, equilibrium constant for the reaction:, H2 (g) + I2 (g), , 2HI (g), , is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the, concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and, allowed it to reach equilibrium at 700 K?, 7.16, , What is the equilibrium concentration of each of the substances in the, equilibrium when the initial concentration of ICl was 0.78 M ?, 2ICl (g), , 7.17, , I2 (g) + Cl2 (g);, , Kc = 0.14, , Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium, concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed, to come to equilibrium?, C2H6 (g), , C2H4 (g) + H2 (g), , 2021-22

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234, , CHEMISTRY, , 7.18, , Ethyl acetate is formed by the reaction between ethanol and acetic acid and the, equilibrium is represented as:, CH3COOH (l) + C2H5OH (l), , CH3COOC2H5 (l) + H2O (l), , (i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note:, water is not in excess and is not a solvent in this reaction), (ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol,, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate, the equilibrium constant., (iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining, it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium, been reached?, 7.19, , A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After, equilibrium was attained, concentration of PCl 5 was found to be, 0.5 × 10–1 mol L–1. If value of Kc is 8.3 × 10–3, what are the concentrations of PCl3, and Cl2 at equilibrium?, PCl5 (g), PCl3 (g) + Cl2(g), , 7.20, , One of the reaction that takes place in producing steel from iron ore is the, reduction of iron(II) oxide by carbon monoxide to give iron metal and CO2., FeO (s) + CO (g), , Fe (s) + CO2 (g);, , Kp = 0.265 atm at 1050K, , What are the equilibrium partial pressures of CO and CO2 at 1050 K if the, initial partial pressures are: pCO = 1.4 atm and, 7.21, , = 0.80 atm?, , Equilibrium constant, Kc for the reaction, N2 (g) + 3H2 (g), , 2NH3 (g) at 500 K is 0.061, , At a particular time, the analysis shows that composition of the reaction mixture, is 3.0 mol L–1 N2, 2.0 mol L–1 H2 and 0.5 mol L–1 NH3. Is the reaction at equilibrium?, If not in which direction does the reaction tend to proceed to reach equilibrium?, 7.22, , Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches, the equilibrium:, 2BrCl (g), Br2 (g) + Cl2 (g), for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of, 3.3 × 10–3 mol L–1, what is its molar concentration in the mixture at equilibrium?, , 7.23, , At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium, with soild carbon has 90.55% CO by mass, C (s) + CO2 (g), , 2CO (g), , Calculate Kc for this reaction at the above temperature., 7.24, , 0, , Calculate a) ∆G and b) the equilibrium constant for the formation of NO2 from, NO and O2 at 298K, NO (g) + ½ O2 (g), , NO2 (g), , where, 0, , ∆fG (NO2) = 52.0 kJ/mol, 0, , ∆fG (NO) = 87.0 kJ/mol, 0, , ∆fG (O2) = 0 kJ/mol, 7.25, , Does the number of moles of reaction products increase, decrease or remain, same when each of the following equilibria is subjected to a decrease in pressure, by increasing the volume?, , (a), , PCl5 (g), , PCl3 (g) + Cl2 (g), , 2021-22

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EQUILIBRIUM, , 235, , (b), , CaO (s) + CO2 (g), , CaCO3 (s), , (c), , 3Fe (s) + 4H2O (g), , Fe3O4 (s) + 4H2 (g), , 7.26, , Which of the following reactions will get affected by increasing the pressure?, Also, mention whether change will cause the reaction to go into forward or, backward direction., , (i), , COCl2 (g), , (ii), , CH4 (g) + 2S2 (g), , (iii), , CO2 (g) + C (s), , (iv), , 2H2 (g) + CO (g), , (v), , CaCO3 (s), , (vi), , 4 NH3 (g) + 5O2 (g), , 7.27, , The equilibrium constant for the following reaction is 1.6 ×105 at 1024K, , CO (g) + Cl2 (g), , H2(g) + Br2(g), , CS2 (g) + 2H2S (g), 2CO (g), CH3OH (g), CaO (s) + CO2 (g), 4NO (g) + 6H2O(g), 2HBr(g), , Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a, sealed container at 1024K., 7.28, , Dihydrogen gas is obtained from natural gas by partial oxidation with steam as, per following endothermic reaction:, CH4 (g) + H2O (g), , CO (g) + 3H2 (g), , (a) Write as expression for Kp for the above reaction., (b) How will the values of Kp and composition of equilibrium mixture be affected, by, (i), , increasing the pressure, , (ii) increasing the temperature, (iii) using a catalyst ?, 7.29, , Describe the effect of :, a), b), c), d), on, , addition of H2, addition of CH3OH, removal of CO, removal of CH3OH, the equilibrium of the reaction:, 2H2(g) + CO (g), , 7.30, , CH3OH (g), , At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride,, PCl5 is 8.3 ×10-3. If decomposition is depicted as,, PCl5 (g), , 0, , ∆rH = 124.0 kJ mol–1, , PCl3 (g) + Cl2 (g), , a), , write an expression for Kc for the reaction., , b), , what is the value of Kc for the reverse reaction at the same temperature ?, , c), , what would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased, (iii) the temperature is increased ?, , 7.31, , Dihydrogen gas used in Haber’s process is produced by reacting methane from, natural gas with high temperature steam. The first stage of two stage reaction, involves the formation of CO and H2. In second stage, CO formed in first stage is, reacted with more steam in water gas shift reaction,, CO (g) + H2O (g), , CO2 (g) + H2 (g), , 2021-22

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236, , CHEMISTRY, , If a reaction vessel at 400 °C is charged with an equimolar mixture of CO and, steam such that pco = pH O = 4.0 bar, what will be the partial pressure of H2 at, 2, equilibrium? Kp= 10.1 at 400°C, 7.32, , Predict which of the following reaction will have appreciable concentration of, reactants and products:, 2Cl (g) Kc = 5 ×10–39, , a) Cl2 (g), , b) Cl2 (g) + 2NO (g), , 2NOCl (g), , Kc = 3.7 × 108, , c) Cl2 (g) + 2NO2 (g), , 2NO2Cl (g), , Kc = 1.8, , 7.33, , The value of Kc for the reaction 3O2 (g), 2O3 (g) is 2.0 ×10–50 at 25°C. If the, equilibrium concentration of O 2 in air at 25°C is 1.6 ×10 –2, what is the, concentration of O3?, , 7.34, , The reaction, CO(g) + 3H2(g), , CH4(g) + H2O(g), , is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol, of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine, the concentration of CH4 in the mixture. The equilibrium constant, Kc for the, reaction at the given temperature is 3.90., 7.35, , What is meant by the conjugate acid-base pair? Find the conjugate acid/base, for the following species:, –, , –, , –, , 2–, , 2–, , HNO2, CN , HClO4, F , OH , CO3 , and S, , +, , +, , 7.36, , Which of the followings are Lewis acids? H2O, BF3, H , and NH4, , 7.37, , What will be the conjugate bases for the Brönsted acids: HF, H2SO4 and HCO 3?, , 7.38, , –, , –, , –, , Write the conjugate acids for the following Brönsted bases: NH2 , NH3 and HCOO ., –, , –, , 7.39, , The species: H2O, HCO3, HSO4 and NH3 can act both as Brönsted acids and, bases. For each case give the corresponding conjugate acid and base., , 7.40, , Classify the following species into Lewis acids and Lewis bases and show how, –, –, these act as Lewis acid/base: (a) OH (b) F (c) H+ (d) BCl3 ., , 7.41, , The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3 M. what, is its pH?, , 7.42, , The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen, ion in it., , 7.43, , The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10 –4,, 1.8 × 10–4 and 4.8 × 10–9 respectively. Calculate the ionization constants of the, corresponding conjugate base., , 7.44, , The ionization constant of phenol is 1.0 × 10–10. What is the concentration of, phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization, if the solution is also 0.01M in sodium phenolate?, , 7.45, , The first ionization constant of H2S is 9.1 × 10–8. Calculate the concentration of, –, HS ion in its 0.1M solution. How will this concentration be affected if the solution, is 0.1M in HC l also ? If the second dissociation constant of H 2 S is, 1.2 × 10–13, calculate the concentration of S2– under both conditions., , 7.46, , The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of, dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of, acetate ion in the solution and its pH., , 7.47, , It has been found that the pH of a 0.01M solution of an organic acid is 4.15., Calculate the concentration of the anion, the ionization constant of the acid, and its pKa., , 2021-22

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EQUILIBRIUM, , 7.48, , 237, , Assuming complete dissociation, calculate the pH of the following solutions:, (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH, , 7.49, , Calculate the pH of the following solutions:, a), , 2 g of TlOH dissolved in water to give 2 litre of solution., , b), , 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution., , c), , 0.3 g of NaOH dissolved in water to give 200 mL of solution., , d), , 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution., , 7.50, , The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate, the pH of the solution and the pKa of bromoacetic acid., , 7.51, , The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization, constant and pKb., , 7.52, , What is the pH of 0.001M aniline solution ? The ionization constant of aniline, can be taken from Table 7.7. Calculate the degree of ionization of aniline in the, solution. Also calculate the ionization constant of the conjugate acid of aniline., , 7.53, , Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74., How is the degree of dissociation affected when its solution also contains, (a) 0.01M (b) 0.1M in HCl ?, , 7.54, , The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of, ionization in its 0.02M solution. What percentage of dimethylamine is ionized if, the solution is also 0.1M in NaOH?, , 7.55, , Calculate the hydrogen ion concentration in the following biological fluids whose, pH are given below:, (a) Human muscle-fluid, 6.83, , (b), , Human stomach fluid, 1.2, , (c) Human blood, 7.38, , (d), , Human saliva, 6.4., , 7.56, , The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8,, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion, concentration in each., , 7.57, , If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K., Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What, is its pH?, , 7.58, , The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the, concentrations of strontium and hydroxyl ions and the pH of the solution., , 7.59, , The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of, ionization of the acid in its 0.05M solution and also its pH. What will be its, degree of ionization if the solution is 0.01M in HCl also?, , 7.60, , The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization, constant of the acid and its degree of ionization in the solution., , 7.61, , The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M, sodium nitrite solution and also its degree of hydrolysis., , 7.62, , A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the, ionization constant of pyridine., , 7.63, , Predict if the solutions of the following salts are neutral, acidic or basic:, NaCl, KBr, NaCN, NH4NO3, NaNO2 and KF, , 7.64, , The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of, 0.1M acid and its 0.1M sodium salt solution?, , 2021-22

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238, , CHEMISTRY, , 7.65, , Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at, this temperature?, , 7.66, , Calculate the pH of the resultant mixtures:, a), , 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl, , b), , 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2, , c), , 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH, , 7.67, , Determine the solubilities of silver chromate, barium chromate, ferric hydroxide,, lead chloride and mercurous iodide at 298K from their solubility product, constants given in Table 7.9. Determine also the molarities of individual ions., , 7.68, , The solubility product constant of Ag 2CrO 4 and AgBr are 1.1 × 10 –12 and, 5.0 × 10–13 respectively. Calculate the ratio of the molarities of their saturated, solutions., , 7.69, , Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are, mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate, Ksp = 7.4 × 10–8 )., , 7.70, , The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate, is 2.5 × 10–13. How many times is silver benzoate more soluble in a buffer of pH, 3.19 compared to its solubility in pure water?, , 7.71, , What is the maximum concentration of equimolar solutions of ferrous sulphate, and sodium sulphide so that when mixed in equal volumes, there is no, precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 × 10–18)., , 7.72, , What is the minimum volume of water required to dissolve 1g of calcium sulphate, at 298 K? (For calcium sulphate, Ksp is 9.1 × 10–6)., , 7.73, , The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen, sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of, the following: FeSO4, MnCl 2, ZnCl2 and CdCl2. in which of these solutions, precipitation will take place?, , 2021-22