Question

Determine [Zn2 ], [CN–], and [HCN] in a saturated solution of Zn(CN)2 with a fixed pH of 2.550. The Ksp for Zn(CN)2 is 3.0 × 10–16. The Ka for HCN is 6.2 × 10–10.

1) Zn(CN)2 (s) --> Zn 2+ (aq) + 2 CN- (aq)

2) CN- (aq) + H2O (l) --> HCN (aq) + OH- (aq)

3) H2O (l) --> H+ (aq) + OH- (aq)

Answer 1

From equation 1),

with x being the concentration of Zn2+ in solution, the concentration of CN- becomes 2x,

Ksp = 3.0 x 10^-16 = [Zn2+][CN-]^2

Ksp = 3 x 10^-16 = (x)(2x)^2

1 H+ reacts with 1 CN- to form 1 HCN,

CN- + H+ <==> HCN

So, [H+] = [HCN]

pH = 2.550 = -log[H+]

So, [H+] = 2.82 x 10^-3 M

2x = [CN-] + [HCN]

We can write Ka for HCN,

HCN <==> H+ + CN-

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

Feed [H+] from above,

6.2 x 10^-10 = 2.82 x 10^-3[CN-]/[HCN]

[HCN] = 4.55 x 10^6[CN-]

Feed this in the 2x = [CN-] + [HCN] equation,

2x = [CN-] + 4.55 x 10^6[CN-]

x = 2.28 x 10^6[CN-]

So, [CN-] = x/2.28 x 10^6

Feed this is Ksp equation,

Ksp = 3 x 10^-16 = (x)(2x/2.28x 10^6)^2

x^3 = 3.90 x 10^-4

x = [Zn2+] = 0.073 M

[CN-] = 2 x 0.073 = 0.146 M

[HCN] = 0.146 M