RI 2025 Tutorial 16 Solubility Equilibria - Answers to Self-check

Suggested Answers to Self–Check Questions 1 (a) Refer to IVY Solubility Equilibria Check Point 1 (Question 1) for worked solution. Ksp = 1.77 x 1014 mol4 dm12. (b) Refer to IVY Solubility Equilibria Check Point 1 (Question 2) for worked solution. Solubility = 0.0729 g dm3. (c) Let the solubility of Ca3(PO4)2 in water be s mol dm-3. Ca 3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43–(aq) At equilibrium in the saturated solution, [Ca 2+] = 3s mol dm-3 [PO43–] = 2s mol dm–3 23 32 3 2 2 6 5 1 5 34 2 4 52 6 126 5 63 of Ca (PO ) [Ca ] [PO ] (3 ) (2 ) 1.0 10 mol dm 108s 1.0 10 1.0 10 2.474 10 mol dm108 spKs s s               Hence [Ca2+] in the saturated solution = (3)(2.474 x 10-6) = 7.42 x 10–6 mol dm–3 2 Refer to IVY Solubility Equilibria Check Point 2 for explanation. Since ionic product < Ksp, precipitation of CaSO4 will not occur. 3 (a) Upon mixing the two solutions and assuming no reaction,   22 123 1000 1000[Ca ] 0.50 0.3333 mol dm     12 1 3 1000 1000[NaOH] [OH ] 2.0 0.6667 mol dm     222 3 9Ionic product Ca OH (0.3333)(0.6667) 0.148 mol dm     Since ionic product > Ksp of Ca(OH)2, precipitation of Ca(OH)2 will occur. A white precipitate will be observed in the test-tube. (b) Upon mixing the two solutions and assuming no reaction,   22 123 1000 1000[Ca ] 0.50 0.3333 mol dm     12 1 3 3 1000 1000[NH ] 2.0 0.6667 mol dm    NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq) At equilibrium, [NH 4+] = [OH–] [NH 3]eqm  [NH3]initial = 0.6667 mol dm–3, since NH3 is a weak base with a small Kb,   2 4 33 53 3 NH OH OH NH NH OH (1.74 10 )(0.6667) 3.406 10 mol dm bK                 223 2 6 3 9Ionic product Ca OH (0.3333)(3.406 10 ) 3.87 10 mol dm        Since ionic product < Ksp of Ca(OH)2, precipitation of Ca(OH)2 will not occur. No white precipitate will be observed in the test-tube.