##### J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 16.26

Calculate Ksp values of the following, for which a reference book lists the indicated solubilities: (a) Ce(IO3)4, solubility 1.8×10–4 mol/L; (b) Hg2SO4, 8.9×10–4 mol/L; and (c) barium chromate, 0.0010 g BaCrO4/100 mL H2O.

Strategy: write the chemical reaction, write the mass action expression, set up a table of concentrations, convert the given solubility to molarity if necessary, and then use this value to find Ksp.

(a)

 Ce(IO3)4(s) Ce4+(aq) + 4 IO3–(aq) Ksp = [Ce4+]e[IO3–]e4 Initial 0 0 Change + x + 4x Equilibrium x 4x

Molar solubility of Ce(IO3)4 = x = 1.8×10–4 mol/L

[Ce4+]e = x = 1.8×10–4 mol/L

[IO3]e = 4x = 4(1.8×10–4 mol/L) = 7.2×10–4 mol/L

Ksp = [Ce4+]e[IO3]e4 = [1.8×10–4 mol/L][7.2×10–4 mol/L]4 = 4.8×10–17

(b)

 Hg2SO4(s) Hg22+(aq) + SO42–(aq) Ksp = [Hg22+]e[SO42–]e Initial 0 0 Change + x + x Equilibrium x x

Molar solubility of Hg2SO4 = x = 8.9×10–4 mol/L

[Hg22+]e = x = 8.9×10–4 mol/L

[SO42–]e = x = 8.9×10–4 mol/L

Ksp = [Hg22+]e[SO42–]e = [8.9×10–4 mol/L][8.9×10–4 mol/L] = 7.9×10–7

(c)

 BaCrO4(s) Ba2+(aq) + CrO42–(aq) Ksp = [Ba2+]e[CrO42–]e Initial 0 0 Change + x + x Equilbrium x x

The variable x must be found from the solubility given, after converting to molarity.

BaCrO4 = 137.3 + 52.0 + 4(16.0) = 253.3 g/mol

0.0010 g/100 mL = 0.010 g/L = 0.010 g/(253.3 g/mol)/L = 3.9×10–5 M

Since the stoichiometry between barium chromate and the ions is 1:1, x = 3.9×10–5 M.

Now this value can be used in the mass action expression for Ksp to give:

Ksp = [x][x] = [3.9×10–5] [3.9×10–5] = 1.6×10–9