Question

Find the solubility of agi in 2.5 m nh3 [ksp of agi = 8.3 × 10−17; kf of ag(nh3)2+ = 1.7 × 107].

Answer 1

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