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, res