Question

Determine [Zn²⁺ ], [CN²], and [HCN] in a saturated solution of Zn(CN)₂ with a fixed pH of 2.730. The Ksp for Zn(CN)₂ is 3.0 × 10–16. The Ka for HCN is 6.2 × 10–10.

What are the concentrations of [Zn²⁺] [CN^-] and [HCN]

Answer 1

pH = 2.73 so [H+] = 10^-pH = 10^-2.73 = 0.00186

The equations involved is

Zn(CN)₂ = Zn²⁺(aq) + 2CN^-(aq) Ksp = 3 x 10^-16 ----(1)

HCN = H+(aq) + CN-(aq) Ka = 6.2 x 10^-10

This needs to be reversed as the amount of CN- formed will combine to form HCN

CN-(aq) + H+(aq) = HCN K = 1/Ka = 1/6.2 x 10-10 = 1.6 x 10⁹

2CN^-(aq) + 2H+(aq) = 2HCN K = 2 x 1.6 x 10⁹ = 3.2 x 10⁹ ---(2)

So overall reaction is adding (1) and (2)

Zn(CN)₂ + H⁺(aq) = Zn²⁺(aq) + 2HCN K = 3.2 x 10⁹ x 3 x 10^-16 = 9.68 x 10^-7

As per the equation HCN is 2x when Zn2+ is x

K = [Zn2+][HCN]²/[H+]

9.68 x 10^-7 = x* (2x)²/0.00186

9.68 x 10^-7 =4x³/0.00186

x = 7.66 x 10^-4 M is Zn²⁺ and HCN is 0.00153 M

Since HCN is 0.00153 and H+ is 0.00186 CN- will be

HCN = CN- + H+

Ka = [CN-][H+]/[HCN]

6.2 x 10^-10 = [CN-] * 0.00186/0.00153

[CN-] = 5.1 x 10^-10 M