Page 2 : 2. 1.35 × 10-4, 3. 7.2 × 10-8, 4. 13.5 × 10-8, Solution: The solubility product of a sparingly soluble salt AX is 5.2 × 10-13. Its solubility, in mol dm-3 is 7.2 × 10-7., Exercise | Q 1.5 | Page 61, Choose the most correct answer :, Blood in the human body is highly buffered at a pH of ________., 1. 7.4, 2. 7.0, 3. 6.9, 4. 8.1, Solution: Blood in the human body is highly buffered at a pH of 7.4., Exercise | Q 1.6 | Page 61, Choose the most correct answer :, The conjugate base of [Zn(H2O)4]2⊕ is __________., 1. [Zn(H2O)4]2⊕NH3, 2. [Zn(H2O)3]2⊕, 3. [Zn(H2O)3OH]⊕, 4. [Zn(H2O)H]3⊕, Solution: The conjugate base of [Zn(H2O)4]2⊕ is [Zn(H2O)3OH]⊕., Exercise | Q 1.7 | Page 61, Choose the most correct answer :, For pH > 7 the hydronium ion concentration would be _________., 1. 10-7M, 2. < 10-7M, 3. > 10-7M, 4. 10-7M, Solution: For pH > 7 the hydronium ion concentration would be < 10-7M ., Exercise | Q 2.01 | Page 61, Answer the following in one sentence :, Why cations are Lewis acids?

Page 3 : Solution: Cations are electron-deficient species and can accept an electron pair., Hence, cations are Lewis acids., Exercise | Q 2.02 | Page 61, Answer the following in one sentence :, Why is KCl solution neutral to litmus?, Solution: KCl, being salt of a strong acid (HCl) and a strong base (KOH), does not, undergo hydrolysis. Hence, the KCl solution is neutral to litmus., Exercise | Q 2.03 | Page 61, Answer the following in one sentence :, How are basic buffer solutions prepared?, Solution: Basic buffer solutions are prepared by mixing aqueous solutions of a weak, base and its salt with strong acid., Exercise | Q 2.04 | Page 61, Answer the following in one sentence :, The dissociation constant of acetic acid is 1.8 × 10-5. Calculate percent dissociation of, acetic acid in 0.01 M solution., Solution:, Given :, •, •, , Dissociation constant (Ka) = 1.8×10-5,, Concentration (c) = 0.01 M, To find :, , •, , Percent dissociation, Formulae :, , 1. Ka = a2c, 2. Percent dissociation = α × 100, Calculation :

Page 4 : Exercise | Q 2.05 | Page 61, Answer the following in one sentence :, Write one property of a buffer solution., Solution:, Properties of buffer solution:, i. When a small amount of strong acid (or strong base) is added to a buffer solution,, there is no significant change in the value of pH., ii. The pH of a buffer solution is independent of the volume of the solution. Hence, the, dilution of a buffer solution will not change its pH., iii. The pH of a buffer solution does not change even if it is kept for a long time., Exercise | Q 2.06 | Page 61, Answer the following in one sentence :, The pH of a solution is 6.06. Calculate its H⊕ ion concentration., Solution:, Given:, pH of solution = 6.06, To find:, H+ ion concentration

Page 5 : Formula:, pH = -log10[H3O+], Calculation:, From formula,, pH = -log10[H3O+], ∴ log10[H3O+] = -pH, = -6.06, = -6 - 0.06 + 1 - 1, = (- 6 - 1) + 1 - 0.06, , The H+ ion concentration of the solution is 8.710 × 10-7 M., Exercise | Q 2.07 | Page 61, Answer the following in one sentence :, Calculate the pH of 0.01 M sulphuric acid., Solution:, Given: Concentration of sulphuric acid = 0.01 M, To find: pH

Page 6 : Exercise | Q 2.08 | Page 61, Answer the following in one sentence :, The dissociation of H2S is suppressed in the presence of HCl. Name the phenomenon., Solution:, The phenomenon due to which dissociation of H2S is suppressed in the presence of, HCl is known as the common ion effect., Exercise | Q 2.09 | Page 61, Answer the following in one sentence :, Why is it necessary to add H2SO4 while preparing the solution of CuSO4?, Solution:, i., The aqueous solution of CuSO4 is turbid due to the formation of sparingly soluble, Cu(OH)2 by hydrolysis as shown below:, , ii., If H2SO4, that is H3O+ ions are added, the hydrolytic equilibrium shifts to the left., Turbidity of Cu(OH)2 dissolves to give a clear solution. Hence, it is necessary to add, H2SO4 while preparing the solution of CuSO4.

Page 7 : Exercise | Q 2.1 | Page 61, Answer the following in one sentence :, Classify the following buffers into different types :, a. CH3COOH + CH3COONa, b. NH4OH + NH4Cl, c. Sodium benzoate + benzoic acid, d. Cu(OH)2 + CuCl2, Solution:, Buffer, , Type, , i., , CH3COOH + CH3COONa, , Acidic buffer, , ii., , NH4OH + NH4Cl, , Basic buffer, , iii., , Sodium benzoate + benzoic acid, , iv., , Cu(OH)2 + CuCl2, , Acidic buffer, Basic buffer, , Exercise | Q 3.01 | Page 62, Answer the following in brief :, What are acids and bases according to Arrhenius theory?, Solution:, According to Arrhenius theory, acids and bases are defined as follows:, i., Acid: An acid is a substance that contains hydrogen and gives H+ ions in an aqueous, solution., e.g., , ii.

Page 8 : Exercise | Q 3.02 | Page 62, Answer the following in brief :, What is meant by conjugate acid-base pair?, Solution:, i. The base produced by accepting the proton from acid is the conjugate base of that, acid., ii. Similarly, the acid produced when a base accepts a proton is called the conjugate, acid of that base., iii. A pair of an acid and a base differing by a proton is said to be a conjugate acid-base, pair., , Exercise | Q 3.03 | Page 62, Answer the following in brief :, Label the conjugate acid-base pair in the following reaction:, , Solution:, , Exercise | Q 3.03 | Page 62, Answer the following in brief :, Label the conjugate acid-base pair in the following reaction:, , Solution:, , Exercise | Q 3.04 | Page 62

Page 9 : Answer the following in brief :, Write a reaction in which water acts as a base., Solution:, , Exercise | Q 3.05 | Page 62, Answer the following in brief :, Ammonia serves as a Lewis base whereas AlCl3 is Lewis acid. Explain., Solution:, i., According to Lewis's theory, an acid is a substance that can accept a share in an, electron pair. In the AlCl3 molecule, the octet of Al is incomplete. Therefore, it can, accept an electron pair to complete its octet. Hence, AlCl3 acts as a Lewis acid., ii., According to Lewis's theory, a base is a substance that can donate an electron pair. In, the ammonia (NH3) molecule, the nitrogen atom has one lone pair of electrons to, donate. Hence, NH3 acts as a Lewis base., Exercise | Q 3.06 | Page 62, Answer the following in brief :, Acetic acid is 5% ionised in its decimolar solution. Calculate the dissociation constant of, acid., Solution:, Given: Percent dissociation = 5%, Concentration (c) = 1 decimolar, To find: Dissociation constant of acid (Ka), Formulae:, i. Percent dissociation = α × 100, ii. Ka = α2 c, Calculation:, Using formula (i),

Page 10 : Exercise | Q 3.07 | Page 62, Answer the following in brief :, Derive the relation pH + pOH = 14., Solution:, Relationship between pH and pOH:, The ionic product of water is given as:, Kw = [H3O+][OH–], , Exercise | Q 3.08 | Page 62, Answer the following in brief :, The aqueous solution of sodium carbonate is alkaline whereas the aqueous solution of, ammonium chloride is acidic. Explain.

Page 11 : Solution:, i., Sodium carbonate (Na2CO3) is a salt of weak acid H2CO3 and strong base NaOH. When, dissolved in water, it dissociates completely., , ii., The Na+ ions of salt have no tendency to react with OH− ions of water since the possible, product of the reaction is NaOH, a strong electrolyte., iii., , iv., As a result of excess OH− ions produced, the resulting solution of Na2CO3 is alkaline., v., Similarly, ammonium chloride (NH4Cl) is salt of strong acid HCl and weak base NH4OH., When NH4Cl is dissolved in water, it dissociates completely as,

Page 12 : Exercise | Q 3.09 | Page 62, Answer the following in brief :, The pH of a weak monobasic acid is 3.2 in its 0.02 M solution. Calculate its dissociation, constant., Solution:, Given:, pH of weak monobasic acid = 3.2,, Concentration of solution (c) = 0.02 M, To find: Dissociation constant (Ka), Formula:, i. pH = -log10[H3O+], ii. Ka = α2c, Calculation:, From the formula (i),, pH = -log10[H3O+], ∴ log10[H3O+] = -3.2

Page 13 : = - 3 - 0.2 + 1 - 1, = (-3 - 1) + 1 - 0.2, , Exercise | Q 3.1 | Page 62, Answer the following in brief :, In NaOH solution [OH-] is 2.87 × 10-4. Calculate the pH of the solution., Solution:, Given: [OH-] = 2.87×10-4 M, To find: pH of the solution, Formulae:, i. pOH = -log10[OH-], ii. pH + pOH = 14, Calculation:, From formula (i),, pOH = -log10[OH-], ∴ pOH = -log10[2.87 × 10-4] = -log102.87 - log1010-4

Page 14 : = -log102.87 + 4 = 4 - 0.4579, pOH = 3.5421, From formula (ii),, pH + pOH = 14, pH = 14 - pOH = 14 - 3.5421 = 10.4579,, pH of the solution is 10.4579, Exercise | Q 4.01 | Page 62, Answer the following :, Define the degree of dissociation, Solution:, The degree of dissociation (α) of an electrolyte is defined as a fraction of the total, number of moles of the electrolyte that dissociates into its ions when the equilibrium is, attained., Exercise | Q 4.01 | Page 62, Answer the following :, Derive Ostwald's dilution law for the CH3COOH., Solution:, i., Consider an equilibrium of weak acid CH3COOH that exists in solution partly as the, undissociated species CH3COOH and partly H+ and CH3COO– ions. Then, , iii., Suppose 1 mol of acid CH3COOH is initially present in volume V dm3 of the solution. At, equilibrium, the fraction dissociated would be α, where α is the degree of dissociation of, the acid. The fraction of an acid that remains undissociated would be (1 - α).

Page 15 : vi., If c is the initial concentration of CH3COOH in mol dm–3 and V is the volume in dm3 mol–, 1 then c = 1/V. Replacing 1/V in equation (2) by c,, we get

Page 16 : Equation (5) implies that the degree of dissociation of a weak acid (CH3COOH) is, inversely proportional to the square root of its concentration or directly proportional to, the square root of the volume of the solution containing 1 mol of the weak acid., Exercise | Q 4.02 | Page 62, Answer the following :, Define pH and pOH., Solution:, i., The pH of a solution is defined as the negative logarithm to the base 10, of the, concentration of H+ ions in solution in mol dm–3., pH is expressed mathematically as, pH = -log10[H+] or pH = -log10[H3O+], ii., Similarly, pOH of a solution can be defined as the negative logarithm to the base 10, of, the molar concentration of OH− ions in solution., pOH = -log10[OH-], Exercise | Q 4.02 | Page 62, Answer the following :

Page 17 : Derive the relationship between pH and pOH., Solution:, Relationship between pH and pOH:, The ionic product of water is given as:, Kw = [H3O+][OH-], Now, Kw = 1×10-14 at 298 K, Thus, [H3O+][OH-] = 1.0 × 10-14, Taking the logarithm of both the sides, we write, log10[H3O+] + log10[OH-] = -14, -log10[H3O+] + {-log10[OH-]} = 14, Now, pH = -log10[H3O+] and pOH = -log10[OH-], ∴ pH + pOH = 14, Exercise | Q 4.03 | Page 62, Answer the following :, What is meant by hydrolysis?, Solution:, Hydrolysis of salt is defined as the reaction in which cations or anions or both ions of a, salt react with ions of water to produce acidity or alkalinity (or sometimes even, neutrality)., Exercise | Q 4.03 | Page 62, Answer the following :, A solution of CH3COONH4 is neutral. why?, Solution:, i., CH3COONH4 is a salt of a weak acid, CH3COOH (Ka = 1.8 × 10–5), and weak base,, NH4OH (Kb = 1.8 × 10–5). When the salt CH3COONH4 is dissolved in water, it undergoes, hydrolysis:

Page 18 : Exercise | Q 4.04 | Page 62, Answer the following :, The dissociation of HCN is suppressed by the addition of HCl. Explain., Solution:, i. HCN and HCl both dissociate to produce H+ ions which are common to both., ii. HCN is a weak electrolyte. It dissociates to a little extent.

Page 19 : Exercise | Q 4.05 | Page 62, Answer the following :, Derive the relationship between the degree of dissociation and dissociation constant in, weak electrolytes., Solution1:, i. Consider 1 mol of weak base BOH dissolved in V dm3 of solution. The base, dissociates partially as, , iii. Let the fraction dissociated at equilibrium is α and that remains undissociated is (1 −, α).

Page 20 : vi. If c is the initial concentration of base in mol dm–3 and V is the volume in dm3 mol–, 1 then c = 1/V. Replacing 1/V in equation (2) by c,, we get, , The degree of dissociation of a weak base is inversely proportional to square root of its, concentration and is directly proportional to square root of volume of the solution, containing 1 mol of weak base.

Page 21 : Solution2:, i. Consider an equilibrium of weak acid HA that exists in solution partly as the, undissociated species HA and partly H+ and A– ions. Then, , iii. Suppose 1 mol of acid HA is initially present in volume V dm3 of the solution. At, equilibrium, the fraction dissociated would be α, where α is the degree of dissociation of, the acid. The fraction of an acid that remains undissociated would be (1 − α).

Page 22 : The equation (5) implies that the degree of dissociation of a weak acid is inversely, proportional to the square root of its concentration or directly proportional to the square, root of volume of the solution containing 1 mol of the weak acid., Exercise | Q 4.06 | Page 62, Answer the following :, Sulfides of the cation of group II are precipitated in acidic solution (H2S + HCl) whereas, sulfides of cations of group IIIB are precipitated in the ammoniacal solution of H 2S., Comment on the relative values of the solubility product of sulfides of these., Solution:, Group II and group IIIB cations are precipitated as their sulphides. However, the, solubility product of sulphides of group II cations is lower than group IIIB cations., Therefore, for the precipitation of cations of group II, only a small concentration of, sulphide ion is required. This is achieved by passing H2S gas in the presence of strong, electrolyte HCl, which has a common ion (H+) with H2S. Due to the common ion effect,, the dissociation of H2S is suppressed and thus, the concentration of S2- ions decreases.

Page 23 : This results only in the precipitation of sulphides of group II while sulphides of the higher, group remain in solution as they require a higher concentration of S2- ions for, precipitation., Exercise | Q 4.07 | Page 62, Answer the following :, The solubility of a sparingly soluble salt gets affected in the presence of a soluble salt, having one common ion. Explain., Solution:, i. The presence of a common ion affects the solubility of a sparingly soluble salt., ii. Consider, the solubility equilibrium of AgCl,, , iii. Suppose AgNO3 is added to the saturated solution of AgCl. The salt AgNO3 being a, strong electrolyte dissociates completely in the solution., , iv. The dissociation of AgCl and AgNO3 produce a common Ag+ ion. The concentration, of Ag+ ion in the solution increases owing to complete dissociation of AgNO3., v. According to Le-Chatelier's principle, the addition of Ag+ ions from AgNO3 to the, solution of AgCl shifts the solubility equilibrium of AgCl from right to left. The reverse, reaction in which AgCl precipitates is favoured until the solubility equilibrium is reestablished., vi. However, the value of Ksp remains the same since it is an equilibrium constant. Thus,, the solubility of a sparingly soluble compound decreases with the presence of a, common ion in solution., Exercise | Q 4.08 | Page 62, Answer the following :, The pH of rainwater collected in a certain region of Maharashtra on a particular day was, 5.1. Calculate the H+ ion concentration of the rainwater and its percent dissociation., Solution:, Given: pH of rainwater = 5.1

Page 24 : To find:, i. H+ ion concentration, ii. Percent dissociation, Formula:, i. pH = -log10[H3O+], ii. Percent dissociation = α × 100, Calculation: From the formula (i),, pH = -log10[H3O+], ∴ log10[H3O+] = -5.1, = -5 - 0.1 + 1 - 1, = (-5 - 1) + 1 - 0.1, , Considering that the pH of rainwater is due to the dissociation of a monobasic strong, acid (HA), we have, , Exercise | Q 4.09 | Page 62, Answer the following :, Explain the relation between ionic product and solubility product to predict whether a, precipitate will form when two solutions are mixed?

Page 25 : Solution:, Condition of precipitation:, The ionic product (IP) of an electrolyte is defined in the same way as solubility product, (Ksp). The only difference is that the ionic product expression contains a concentration, of ions under any condition whereas the expression of Ksp contains only equilibrium, concentrations. If,, i. IP = Ksp; the solution is saturated and solubility equilibrium exists., ii. IP > Ksp; the solution is supersaturated and hence precipitation of the compound will, occur., iii. If IP < Ksp, the solution is unsaturated and precipitation will not occur.