When the first reaction equation is:
AgI(S) ↔ Ag+(Aq) + I-(Aq)
So, the Ksp expression = [Ag+][I-]
∴Ksp = [Ag+][I-] = 8.3 x 10^-17
Then the second reaction equation is:
Ag+(aq) + 2NH3(aq) ↔ Ag(NH3)2+
So, Kf expression = [Ag(NH3)2+] / [Ag+] [NH3]^2
∴Kf = [Ag(NH3)2+] /[Ag+] [NH3]^2 = 1.7 x 10^7
by combining the two equations and solve for Ag+:
and by using ICE table:
AgI(aq) + 2NH3 ↔ Ag(NH3)2+ + I-
initial 2.5 0 0
change -2X +X +X
Equ (2.5-2X) X X
so K = [Ag(NH3)2+] [I-] / [NH3]^2
Kf * Ksp = X^2 / (2.5-2X)
8.3 x 10^-17 * 1.7 x10^7 = X^2 / (2.5-2X) by solving for X
∴ X = 5.9 x 10^-5
∴ the solubility of AgI = X = 5.9 x 10^-5 M
