Question

Determine [Zn2 ], [CN–], and [HCN] in a saturated solution of Zn(CN)2 with a fixed pH of 2.890. The Ksp for Zn(CN)2 is 3.0 × 10–16. The Ka for HCN is 6.2 × 10–10.

Answer 1

Answer: According to question here we know that Zn(CN)2 will dissociates into Zn²⁺ and 2CN-

hence , Ksp = S [2S]² = 3.0 * 10^-16

4S³ = 3.0 * 10^-16

solving for S we get = 0.1 * 10^-5

= 1.0 * 10^-6

Hence the concentration of [Zn²⁺] = S = 10^-6 M

[ CN^-] = 2S = 2 * 10^-6 M

Concentration of H+ ion ,

PH = -log [H+] = 2.890

[H+] = 0.001288

Hence the concentration of H+ ion is high then all CN^- will combine with H+ to form HCN hence the concentration of [HCN ] = 2 * 10^-6 M and no CN- is found .

Hence it is all about the given question . Thank you :)