Question

Calculate the solubility (in g·L–1) of CuBr(s) (Ksp = 6.3× 10–9) in 0.57 M NH3(aq).

Answer 1

Cu⁺(aq) + 2NH₃(aq) <-------> Cu(NH₃)₂(aq)

Kf = 6.3 x 10¹⁰

CuBr(aq)     <------->    Cu⁺(aq) + Br^-(aq)

Ksp of CuBr = 6.3 x 10^-9

Adding both equations,

CuBr + 2NH₃  <-------> Cu(NH₃)₂ + Br^-

K = (6.3 x 10¹⁰) x (6.3 x 10^-9) = 396.9

                      CuBr   +   2NH₃     <------->    Cu(NH₃)₂ +    Br-
I                                     0.57 0                   0
C                                   -2x +x                  +x
E                                   (0.57-2x) x                   x

K = x²/(0.57 - 2x)² = 396.9

x/(0.57 - 2x) =19.9

x = 11.34 - 39.8 x

40.8 x = 11.34

x = 0.277 M

Solubility (in g/L) of CuBr(s) = 0.277 x (143.45 g/mol) = 39.87 g/L