Question

What is the solubility of AgCl in 10 M NH3? The Kf of Ag(NH3)2+ from AgCl(s) in aqueous NH3 is 0.0031, which is for the balanced reaction shown below.

AgCl(s) + 2 NH3(aq) → Ag(NH3)2+(aq) + Cl-(aq)

Work:
Ksp = x^2 /(2x+10)^2 = 1.8 x 10^-10
Now I am confused.......
At first I got 0.557 an now.. I don't think this is correct... :(

I used the ICE chart to calculate, but it seems that I am doing this incorrectly...

Answer 1

The first thing wrong is you are using Ksp. You still haven't answered my question about the Kf. Kf is the formation constant for the complex.
Ag^+ + 2NH3 ==> [Ag(NH3)2]^+
The literature shows that as about 1.7E7 and your question gives it as 0.0031. I think that 0.0031 is the equilibrium constant for the reaction
AgCl(s) + 2 NH3(aq) → Ag(NH3)2+(aq) + Cl-(aq) and that is NOT Kf.

In addition, if I calculate the value of Keq for that reaction it is Keq = Kf*Ksp = 0.0031 which adds some credibility to what I've said above. Here is the way I would set up the ICE chart.
.AgCl + 2NH3 =>[Ag(NH3)2]^+ + Cl^-
I..solid..10.....0.............0
C....s....-2x....x.............x
E..s....10-2x....x.............x

K = 0.0031 = (x)(x)/(10-2x)^2
Solve for x.