Question

Calculate the molar solubility of AgCl in a 0.15 M solution of NH3(aq).
(Ksp(AgCl) = 1.6 x 10^-10; Kf(Ag(NH3)2+) = 1.5 x 10^7)

Answer 1

.......AgCl(s) ==> Ag^+ + Cl^-....................Ksp = (Ag^+)(Cl^-)
.......Ag^+ + 2NH3 ==> [Ag(NH3)2]^+....Kf = [Ag(NH3)2]^+/(Ag^+)(NH3)^2
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.......AgCl(s) + 2NH3 ==> Cl^- + [Ag(NH3)2]^+.... Keq = Ksp*Kf
I........solid........0.15...........0............0
C......solid..........-2x............x.............x
E......solid.........0.15-x........x..............x
Plug the E line into the Keq expression and solve for x = solubility of AgCl in mols/L
Post your work if you get stuck.