Question

The solubility of AgCN (s) in 0.200M NH3 (aq) is 8.8×10−6mol/L.

Calculate Ksp for AgCN.

ANSWER Ksp = 1.2×10⁻¹⁶

Answer 1

When we dissolve AgCn in NH₃, it forms a complex as [Ag(NH₃)₂]⁺

Given that solubility of AgCN in NH₃ is 8.8 *10^-6 mol/L

and reaction is:

AgCN(s) + 2NH₃(aq) <=> [Ag(NH₃)₂]⁺(aq) + CN^-(aq)

Solubility =  [Ag(NH₃)₂]⁺ = [CN^-] = 8.8 *10^-6

We have

AgCN(s) <=> Ag⁺(aq) + CN^-(aq) .... Ksp =?

Ag⁺(aq) +2NH₃(aq)---> [Ag(NH₃)₂]⁺(aq)...K_f = 1.6 *10⁷ ... From Literature

Overall reaction

AgCN(s) + 2NH₃(aq) <=> [Ag(NH₃)₂]⁺(aq) + CN^-(aq) .....K = K_sp*K_f

K =  [Ag(NH₃)₂]⁺ [CN^-] / [NH₃]²

Solids have activity =1. So, not included in reaction

K = (8.8 *10^-6)² / (0.200) = 1.9 *10^-9

Ksp = K/Kf = (1.9 *10^-9) / ( 1.6 *10⁷) = 1.2 *10^-16