Question

Ignoring activities, determine the molar solubility (S) of Zn(CN)2 in a solution with a pH =...

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.

Homework Answers

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

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