Question

What is the molar solubility of AgCl in 0.30 M NH3? Ksp for AgCl is 1.8 x 10-10 and Kf for Ag(NH3)2 + is 1.7 x 107 .

Answer 1

AgCl--->Ag+ Cl- Ksp= 1.8x10^-10

Ag + 2NH3 ---> Ag(NH3)2 Kf= 1.6x10^7

net reaction: AgCl + 2NH3 ----> Cl- + Ag(NH3)2

so Ksp * Kf = .00288 = net equilbrium constant

so, for the formation of the complex, you would use the net equilibrium constant for the values of your ICE chart of the formation of the Ag(NH3)2 and solve for x

Kf = [Ag(NH3)2] / [Ag][NH3]^2

so Knet = [Ag(NH3)2+][Cl-] / [NH3]^2

Knet = 0.00288 = x^2 / 0.1 - 2x

x^2 + 0.00576x - 0.000288 = 0

x =0.0143M = molar solubility of AgCl in NH3

Best Of luck