RI 2025 Tutorial 16 Solubility Equilibria Ans

1 Tutorial 16: Solubility Equilibria - Answers to Practice Questions 4 (a) (i) Let the solubility of PbI2 in pure water be s mol dm–3. PbI2(s) ⇌ Pb2+(aq) + 2I(aq) At equilibrium in the saturated solution, [Pb 2+] = s mol dm–3 [ I−] = 2s mol dm–3 I 22 92 39 19 3 33 Pb 7.1 10 ( )(2 ) 47 . 1 1 0 7.1 10 1.21 10 mol dm4 spK ss s s                  Hence solubility of PbI2 in pure water = 1.21 x 10−3 mol dm–3 (ii) Let the solubility of PbI2 in 0.10 mol dm–3 BaI2 solution be y mol dm–3. BaI2(aq)  Ba2+(aq) + 2I− (aq) PbI2(s) ⇌ Pb2+(aq) + 2I−(aq) At equilibrium in the saturated solution, [Pb 2+] = y mol dm-3 [ I−] = 2y + (2)(0.10) = (2y + 0.20) mol dm -3 I 222 9 3 9Pb ( )(2 0.20) 7.1 10 mol dmspKy y           Since PbI2 is sparingly soluble in water and the presence of I− ions from Ba I2 further suppresses its solubility, 2y << 0.20. Thus, (2y + 0.20)  0.20. 29 9 73 2 () ( 0 . 2 0 ) 7 . 11 0 7.1 10 1.78 10 mol dm(0.20) y y         Hence solubility of PbI2 in 0.10 mol dm–3 BaI2 solution = 1.78 x 10−7 mol dm–3 (iii) Let the solubility of PbI2 in 0.20 mol dm–3 Pb(NO3)2 solution be w mol dm–3. Pb(NO3)2(aq)  Pb2+(aq) + 2NO3(aq) PbI2(s) ⇌ Pb2+(aq) + 2I(aq) At equilibrium in the saturated solution, [Pb 2+] = (w + 0.20) mol dm-3 [ I] = 2w mol dm-3 I 222 9 3 9Pb ( 0.20)(2 ) 7.1 10 mol dmspKw w          

2 Since Pb I2 is sparingly soluble in water and the presence of Pb 2+ ions from Pb(NO3)2 further suppresses its solubility, w << 0.20. Thus, (w + 0.20)  0.20. 29 9 53 2 (0.20)(2 ) 7.1 10 7.1 10 9.42 10 mol dm(0.20)(2) w w        solubility of PbI2 in 0.20 mol dm–3 Pb(NO3)2 solution = 9.42 x 10−5 mol dm–3 (b) (i) Let [KI] be x mol dm-3. KI(aq)  K+(aq) + I(aq) PbI2(s) ⇌ Pb2+(aq) + 2I(aq) At equilibrium in the saturated solution, [Pb 2+] = 1.0 x 10–4 mol dm–3 [ I] = (2.0 x 10–4 + x) mol dm–3 I                   22 94 4 2 9 43 3 4 Pb 7.1 10 (1.0 10 )(2.0 10 ) 7.1 10 2 . 01 0 8 . 2 31 0 m o l d m1.0 10 spK x x Hence [KI] = 8.23 x 10–3 mol dm–3 (ii) When PbI2 is shaken with water, the following equilibrium is established: PbI2(s) ⇌ Pb2+(aq) + 2I(aq) -----(1) When a large excess of KI is added and the mixture shaken, the Pb2+ ions react with I to form the soluble complex, [PbI4]2, as shown below. Pb2+(aq) + 4I(aq) ⇌ [PbI4]2(aq) -----(2) The formation of [PbI4]2 decreases the uncomplexed [Pb2+] in the solution. To counteract the decrease in [Pb 2+], the equilibrium position of reaction (1) shifts to the right, resulting in more Pb I2 dissolving and the solubility of Pb I2 is increased. When sufficient K I is added, the ionic product, [Pb 2+][I]2, will be less than Ksp and hence all the PbI2 dissolves.

3 5 (a) When ZnF2 is shaken with water and the undissolved ZnF 2 filtered off, a saturated solution of ZnF2 is produced. Let solubility of ZnF2 be s mol dm3. [Zn2+] = s mol dm3 [F] = 2s mol dm3 Ksp = [Zn2+][F]2 = (s)(2s)2 = 4s3 = 3.2 x 102 mol3 dm–9 s = 0.2 mol dm3 [F] = 2s = 0.400 mol dm3 (b) When BaF2 just precipitates, [F] = 0.4 mol dm3. Ksp of BaF2 = Ionic product of BaF2 1.6 x 107 = [Ba2+][F]2 = [Ba 2+](0.4)2 [Ba2+] = 1.00 x 106 mol dm3 6 (a) In the resultant solution before adding solid KF, [Ca2+] = (½)(0.100) = 0.0500 mol dm-3 [Ba2+] = (½)(0.100) = 0.0500 mol dm-3 To ppt out max. amount of CaF2 with no BaF2 ppt, ionic product of BaF2 cannot be greater than Ksp of BaF2, i.e. [F-] in the solution is just high enough to make the ionic product of BaF2 = Ksp of BaF2. 22 7 33 2 Ba F 1.84 10F1 . 9 2 1 0 m o l d m0.0500Ba sp sp K K                 (b) Ksp of CaF2 = [Ca2+] [F-]2 When [F-] = 1.918 x 10-3 mol dm-3, 22 11 26 3 23 2 Ca F 3.45 10Ca 9.38 10 mol dm(1.918 10 )F sp sp K K                 (c) Percentage of Ca2+ remaining in solution = 69.38 10 100% 0.0188%0.05   The separation was very effective as almost 100% of the Ca 2+ has been removed from solution. *concentration is halved as volume is doubled when the two solution are mixed to gether.

4 7 (a) Ksp of Ca(OH)2 = [Ca2+][OH–]2 (b) (i) OH–(aq) + HCl(aq)  H2O(l) + Cl–(aq)      3 3 Amount of OH in 25.0 cm filtrate 20.0 Amount of HCl used 0.050 1.00 10 mol1000     3 31.00 10[OH ] in the filtrate 1000 0.0400 mol dm25.0 (ii) Let the solubility of Ca(OH)2 in 0.010 mol dm-3 NaOH be y mol dm-3. Concentration/mol dm-3 Ca(OH)2(s) ⇌ Ca2+(aq) + 2OH–(aq) Initial - 0.010 Change +y +2y Equilibrium y (0.010 + 2y) Ksp of Ca(OH)2 = [Ca2+][OH–]2 = (y)(0.010 + 2y)2    0.010 2y 0.0400 y0 . 0 1 5 0 Ksp of Ca(OH)2 = (y)(0.010 + 2y)2 = (0.0150)(0.0400)2 = 2.40 x 10–5 mol3 dm–9 (c) Let the [NaOH] be y mol dm-3. Upon mixing the two solutions and assuming no reaction,  3[NaOH] mol dm2 y  3 2 0.010[Ca(OH) ] 0.0050 mol dm2 Hence in the resultant solution,  23[Ca ] 0.0050 mol dm           3[OH ] 2 0.0050 0.010 mol dm22 yy Ca(OH)2(s) ⇌ Ca2+(aq) + 2OH-(aq) For precipitation to takes place, solution must first be saturated with respect to Ca(OH)2, i.e. ionic product of Ca(OH)2 = Ksp of Ca(OH)2 *concentration is halved as volume is doubled when the two solution are mixed together.

5         22 5 2 5 [Ca ][OH ] 2.40 10 (0.0050) 0.010 2.40 102 0.119 y y Hence [NaOH(aq)] = 0.119 mol dm–3. (d) No effect on Ksp as temperature is kept constant at 25 oC (Ksp only changes with temperature). Ca(OH) 2(s) ⇌ Ca2+(aq) + 2OH–(aq) ----- (1) When solid Ca(NO3)2 is added, the Ca 2+ ions from Ca(NO 3)2 exerts a common ion effect and causes the equilibrium position of (1) to shift left, reducing the solubility of Ca(OH)2 in water. (e) (i) At 25 oC, pOH = 14 – 12.3 = 1.7 [OH] at 25 oC = 101.7 = 2.00 x 102 mol dm3 At 32 oC, Kw = 1.70 x 10–14 mol2 dm–6  pKw = –lg Kw = 13.77 pOH at 32 oC = 13.77 – 11.7 = 2.07 [OH] at 32 oC = 102.07 = 8.51 x 103 mol dm3 Alternative workings to calculate [OH–] at 32C (simpler): [OH] at 32 oC = KW [H+] = 1.70×10ష14 10ష11.7 = 8.52 x 103 mol dm3 At a higher temperature, [OH ] decreases. Hence, there must be lesser extent of dissolution of Ca(OH)2(s) at a higher temperature. According to Le Chatelier’s Principle, an increase in temperature will favour the endothermic reaction. Therefore, the dissolution of Ca(OH)2 is exothermic. (ii) Ksp decreases (from (e)(i), when temperature increases, there is lesser extent of dissolution of Ca(OH)2(s), i.e. solubility decreases). (iii) At infinite dilution, the pH of the solution approaches that of pure water at the same temperature such that [H+] ≈ [OH–].

6 Kw = [H+][OH–] = 1.70 x 10–14 mol2 dm–6 (at 32 oC) [H+] = 141.70 x 10 = 1.304 x 10–7 mol dm–3 pH = –lg (1.304 x 10–7) = 6.88 (f) Add to soil to reduce acidity in soil. (https://www.chemguideforcie.co.uk/2016section10/learning10p2a.html) 8 (a) When BaCO 3(s) and BaSO 4(s) are added separately to water to form saturated solutions, the following equilibria are established. BaCO 3(s) ⇌ Ba2+(aq) + CO32–(aq) ………………….(1) BaSO 4(s) ⇌ Ba2+(aq) + SO42–(aq) ………………….(2) When HCl(aq) is added, an acid-base reaction takes place, causing the [CO