Aim # 14: How can we determine the solubility of slightly soluble salts? H.W. # 14 Study pp. 759 – 765 (up to the common ion effect) Ans. ques. p. 781 # 19,21,22,27 Do Now: Zumdahl (8th ed.) p. 741 # 90,87

I For the system PbCl2(s) Pb2+ (aq) + 2Cl-(aq) Ksp = [Pb2+][Cl-]2 The equilibrium constant for the equilibrium expression between a solid salt and its ions in solution is called the solubility product constant (or solubility constant). It is represented by the symbol Ksp. Problem: Write the expression for the solubility product constant for a) calcium chloride b) barium carbonate

Ans: a) Ca3(PO4)2 (s) 3Ca2+(aq) + 2PO43-(aq) at 250C, Ksp = [Ca2+]3[PO43-]2 = 1.3 x 10-32 b) BaCO3(s) = Ba2+(aq) + CO32-(aq) at 250C, Ksp = [Ba2+][CO32-] = 1.6 x 10-9

Selected Solubility Product Constants, Ksp, at Room Temperature Compound Formula Ksp aluminum hydroxide Al(OH)3 5 x 10-33 barium carbonate BaCO3 1.6 x 10-9 barium chromate BaCrO4 8.5 x 10-11 barium sulfate BaSO4 1.5 x 10-9 cadmium sulfide CdS 1.0 x 10-28 calcium carbonate CaCO3 4.7 x 10-9 calcium sulfate CaSO4 2.4 x 10-5 copper(II) iodide CuI 1.1 x 10-12 iron(II) sulfide FeS 4 x 10-19 lead(II) choride PbCl2 1.6 x 10-5 lead(II) chromate PbCrO4 2 x 10-16 lead(II) sulfate PbSO4 1.3 x 10-8 lead(II) sulfide PbS 7 x 10-29 magnesium hydroxide Mg(OH)2 8.8 x 10-12 silver bromide AgBr 5.0 x 10-13 silver chloride AgCl 1.7 x 10-10 silver chromate Ag2CrO4 1.1 x 10-12 silver iodide AgI 8.5 x 10-17

The values of Ksp change with temperature The values of Ksp change with temperature. The values given in reference tables are usually for 250C The solubility of a salt (the concentrations of its ions) changes with changes in concentrations of other solutes- salts with common ions, as well as with pH. However, for a given temperature, Ksp DOESN’T CHANGE. II Solubility and Ksp Problem: Find the solubility product constant, Ksp, for lead (II) bromide at 250C. Its solubility is 1.0 x 10-2 M.

Ans: PbBr2(s) Pb2+(aq) + 2Br-(aq) Ksp = [Pb2+][Br-]2 [Br-] = 2[Pb2+] = 2(1.0 x 10-2) Ksp = (1.0 x 10-2)(2.0 x 10-2)2 Ksp = 4.0 x 10-6 Note: Units are omitted for Ksp values Problem: Calculate the Ksp value for bismuth sulfide, Bi2S3, which has a solubility of 1.0 x 10-15 M at 250C.

Ans: Bi2S3(s) 2Bi3+(aq) + 3s2-(aq) Ksp = [Bi3+][S2-] Ksp = [2(1.0 x 10-15)]2[3(1.0 x 10-15)]3 Ksp = 1.1 x 10-73 B. Calculating solubility from Ksp Problem: The Ksp value for copper(II) iodate, Cu(IO3)2, is 1.4 x 10-7 at 250c. Calculate its solubility at 250C. Ans: Cu(IO3)2(s) Cu2+(aq) + 2IO3-(aq) Ksp = [Cu2+][IO3-]2 = 1.4 x 10-7 Let [Cu2+] = x then [IO3-] = 2x

1.4 x 10-7 = (x)(2x)2 = 4x3 x = 3√ 3.5 x 10-3 = 3.3 x 10-3 mol/L Problem: The value of Ksp for cerium hydroxide, Ce(OH)3, is 1.5 x 10-20. What is the molar solubility of Ce(OH)3 in a solution that contains 0.10 M NaOH? Ans: Ce(OH)3(s) Ce3+(aq) + 3OH-(aq) Ksp = [Ce3+][OH-]3 = 1.5 x 10-20 NaOH is a strong base, and dissociates completely. from NaoH: [OH-] = 0.10 M from Ce(OH)3: [Ce3+] = x [OH-] = 3x

1.5 x 10-20 = x(.10 + 3x)3 since x is small, [OH-] ≈ 0.10 1.5 x 10-20 = x(.10)3 = .001x x = 1.5 x 10-17 = [Ce3+] = [Ce(OH)3] What would the solubility be if Ce(OH)3 was the only species in solution? Ans: 1.5 x 10-20 = (x)(3x)3 = 27x4 x4 = 5.6 x 10-22 x = [Ce3+] = [Ce(OH)3] = 4.9 x 10-6

Practice Problems Zumdahl (8th ed.) p. 766 # 24,26,28 p. 769 # 89 Note: Since Ksp is a constant, adding OH- ion to the solution (from NaOH) decreases the amount of Ce3+ ion which can remain in solution (and the solubility of Ce(OH)3). Practice Problems Zumdahl (8th ed.) p. 766 # 24,26,28 p. 769 # 89