Question

Find the molar solubility of ZnS solid in a 0.500 M NaCN(aq)solution. The Ksp of ZnS...

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

Homework Answers

Answer #1

ZnS (s) ↔ Zn2+ + S2-

Ksp = [Zn2+][S2-]

Zn2+ + 4CN- ↔ Zn(CN)42-

Kf = [Zn(CN)42-]/[Zn2+][CN-]4

ZnS (s) + 4CN- ↔ Zn(CN)42- + S2-

Koverall = Kf xKsp = [Zn(CN)42-][S2-]/[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)42-] = s ; [S2-] = s and [CN-] = (0.5 - 4s)

[Zn(CN)42-][S2-]/[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

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