Question

Calculate the solubility of silver chloride in a solution that is 0.150 M in NH3.

Answer 1

First, we must note that there are 2 types of reactions...

Precipitation:

AgCl(s) <-> Ag+ + Cl- Ksp = 1.8*10^-10

Complex formation reaction

Ag+ + 2NH3(aq) → Ag(NH3)2+ Kf = 1.7*10^7

therefore, overall reaction:

AgCl(s) + 2NH3(aq) ↔ Ag(NH3)2+ + Cl-

the equilibrium:

K = [Ag(NH3)+2][Cl-] / [NH3]2

calcualte K overall as follows:

K = Ksp × Kf = 1.8*10^-10 * (1.7*10^7) =0.00306

you get..

K = 3.06*10^-3

substitute, in equilibrium

[Ag(NH3)+2] = [Cl-] = x

[NH3] = M-x

K = = S^2/(M-2s)^2

M = 0.15 so

K = = x^2/(0.15 -2x)^2

solve for x

sqrt(3.1*10^-3) = S/(0.15-2s)

0.0556(0.15-2s) = s

0.00834 - 0.111s = s

1.11s = 0.00834

s = 0.00834 /1.1 = 0.00758181818 M

overall solubility = 0.0075818 M