Question

Find the molar solubility of ZnS solid in a 0.500 M NaCN(aq)solution. The Ksp of ZnS is 2.0×10^-25 and the Kf for the Zn(CN)4 2- complex ion is 2.1×10^19

Answer 1

ZnS (s) ↔ Zn²⁺ + S^2-

K_sp = [Zn²⁺][S^2-]

Zn²⁺ + 4CN^- ↔ Zn(CN)₄^2-

K_f = [Zn(CN)₄^2-]/[Zn²⁺][CN^-]⁴

ZnS (s) + 4CN^- ↔ Zn(CN)₄^2- + S^2-

K_overall = K_f xK_sp = [Zn(CN)₄^2-][S^2-]/[CN^-]⁴ = 2.0 x 10^-25 x 2.1 x 10¹⁹ = 4.2 x 10^-6

If 's' is the solubility then, [Zn(CN)₄^2-] = s ; [S^2-] = s and [CN^-] = (0.5 - 4s)

[Zn(CN)₄^2-][S^2-]/[CN^-]⁴ = s x s/(0.5 - 4s)⁴ = 4.2 x 10^-6

s/(0.5 - 4s)² = sqrt(4.2 x 10^-6) = 2.05 x 10^-3

s = 2.05 x 10^-3(0.25 + 16s² - 4s)

2.05 x 10^-3 x 16s² - (1+4 x 2.05 x 10^-3)s + 2.05 x 10^-3 x 0.25 = 0

s = 0.000508 M = 5.08 x 10^-4 moles/lit