Question

Calculate the solubility of silver chloride in a solution that is 0.200 M in NH3.

Answer 1

The equilibrium for your equation:
AgCl_(s) + 2 NH_3(aq) <=> [Ag (NH₃)₂] + Cl^-
where
K = [Ag (NH₃)₂] [Cl^-] / [NH₃]²
is a combination of 2 equilibriums:

A) AgCl_(s) <=> Ag⁺ + Cl^-
Ksp = [Ag+] [Cl-]
Ksp = 1.8 X 10^-10 (from books)

B) Ag+ 2NH₃ <=> [Ag(NH₃)₂]
Kf = [Ag(NH₃)₂] / [Ag⁺] [NH3]²
Kf = 1.6 X 10⁷ (Again from books)

if multiplying
the ratio (A) time (B) gives your ratio:
( [Ag⁺] [Cl^-] ) times ( [Ag(NH₃)₂] / [Ag⁺] [NH3]² ) = [Ag (NH₃)₂] [Cl^-] / [NH₃]²

then multiplying the Ksp times the Kf gives your K
(1.8 X 10^-10) (1.6 X 10⁷) = 2.88 X 10^-3
(with a constant that large, we need to use the quadratic formula)

so your set up for (remember that solides doesn´t matter directly)
            AgCl_(s) + 2 NH_3(aq) <=> [Ag (NH₃)₂]   + Cl^-
IS              -            0.200 -2x            x                    x             (2x because there are 2 NH3)
K = [Ag (NH3)2] [Cl-] / [NH3]^2

2.88 x 10^-3 = [Ag (NH₃)₂] [Cl^-] / [NH₃]²

2.88 x 10^-3 = [X] [X] / [0.200 - 2X]²

2.88 x 10^-3 [0.200 - 2X]² = [X] [X]

0.00288 [0.04 - 0.8X + 4X²] = X²

1.152 x 10^-4 - 2.304 x 10^-3 X + 0.01152 X² = X²

0.98848 X² + 2.304 x 10^-3 X - 1.152 x 10^-4 = 0

using the quadratic formula
X = 9.693 x 10^-3 M

X=Cl^-=Ag⁺=AgCl= 9.693 x 10^-3 M