Question

The solubility of AgCN in water is  1.10×10^-8 M, but increases in the presence of thiosulfate ion. Calculate the solubility of AgCN when the solid is added to a solution that is 0.785 M in thiosulfate ion. For AgCN, K_sp = 1.20×10^-16 and for Ag(S₂O₃)₂^3-, K_f = 2.00×10¹³.

Solubility =  M

Answer 1

AgCN(s) <-------->Ag+(aq) + CN-(aq)

Ksp = [ CN- ] [ Ag+ ]/[ AgCN]

Ag+ ( aq) + 2S2O32- (aq) <--------> [Ag(S2O3)2]3-(aq)

Kf = [[Ag(S2O3)2]3-] /[ Ag+] [ S2O32-]^2

Adding two equation

AgCN(s) + Ag+(aq) + 2S2O32-(aq) <---------> Ag+(aq) + CN-(aq) + [Ag(S2O3)2]3-(aq)

K = [ CN- ] [ Ag(S2O3)2]3-/ [S2O32-]^2

= Ksp × Kf

= 1.20×10^-16 × 2×10^13

= 2.4×10^-3

So,

2.4×10^-3 = [ CN- ] [ Ag(S2O3)2]3-]/[S2O32-]^2

Initial concentration

[ CN- ] = 0

[[ Ag(S2O32)2]3-] = 0

[ S2O32- ] = 0.785M

concentration at equilibrium

[ S2O32-] = 0.785 - 2x

[ CN-] = x

[ [Ag(S2O3)2]3-] = x

Therefore,

2.4×10^-3 = x^2/(0.785 - 2x)^2

x/(0.785 - 2x) = 0.04899

0.03846 - 0.09798x =x

1.09798x = 0.03846

x = 0.0350

So,

[ CN- ] = 0.0350M

[ CN- ] = Solubility of AgCN

Therefore,

Solubility of AgCN in 0.785M S2O32- solution = 0.0350mol/L