Question

Calculate the solubility of silver chloride in a solution that is 0.120 M in NH3 (initial concentration).

Answer 1

Ksp = 1.8 x 10^-10

Kf = 1.6 x 10^7

AgCl (s) ------------------> Ag+(aq) + Cl-(aq)

Ag+(aq) + 2NH3 ----------> Ag(NH3)2+

------------------------------------------------------------------------------------

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

                    0.120                                 0                        0

                     -2x                                    +x                     +x

                  0.120-2x                                  x                      x

K = Kf x Ksp = (1.6 x 10^7 ) x (1.8 x 10^-10 )

K = 2.88 x 10^-3 = x^2 / (0.120-2x)^2

0.0537 = x / 0.120 -2x

x = 6.44 x 10^-3 - 0.1074 x

1.1074 x = 6.44 x 10^-3

x = 5.82 x 10^-3

solubility = 5.82 x 10^-3 M