Question

Ignoring activities, determine the molar solubility (S) of Zn(CN)2 in a solution with a pH = 4.94. The Ksp for Zn(CN)2 is 3.0 × 10-16. The Ka for HCN is 6.2 × 10-10.

Answer 1

pH = 4.94

-log [H⁺] = 4.94

[H⁺] = 10^-4.94 = 1.15 x 10^-5

Zn(CN)₂ (s) Zn²⁺(aq) + 2CN^-(aq)

Ksp = [Zn²⁺] [CN^-]² = 3 x 10^-16

CN^-(aq) + H⁺(aq) HCN(aq)

Ka = [CN^-] [H⁺] / [HCN] = 6.2 x 10^-10

[CN^-]_Total = [CN^-] + [HCN] = 2[Zn²⁺]

Zn(CN)₂(s) + 2H⁺ ==⇒ Zn⁺² + 2HCN

Need Keq for the above reaction:

[Zn⁺²] [HCN]² / [H⁺]²

Now,

Ka = [CN^-] [H⁺] / [HCN] = 6.2 x 10^-10

Square the above:

[CN^-]² [H⁺]² / [HCN]² = (6.2 x 10^-10)² = 3.84 x 10^-19

[HCN]² / [CN^-]²[ H⁺]² = 1 / (3.84 x 10^-19) = 2.6 x 10¹⁸

Multiply the above equation by the Ksp

([Zn⁺²] [CN^-]²) x ( [HCN]²/ [CN^-]² [H⁺]² ) = (2.6 x 10¹⁸) x (3.0 x 10^-16)

[Zn⁺²] [HCN]² / [H⁺]² = 780

Let x = molar solubility of Zn(CN)2
Then [Zn⁺²] = x ; [HCN] = 2x and [H⁺] = 1.15 x 10^-5
x(2x)² / (1.15 x 10^-5)² = 780

4x³ = 780 x (1.15 x 10^-5)²

4x³ = 1.03 x 10^-7

x³ = 2.58 x 10^-8

x = 3.0 x 10^-3