Question

Solid copper(II) carbonate and solid copper(II) sulfide are in equilibrium with a solution containing 1.12×10-2 M sodium carbonate. Calculate the concentration of sulfide ion present in this solution. [sulfide] = M

Answer 1

Ksp of CuCO3 = 2.5 x 10^-10

Ksp of CuS = 7.9 x 10^-37

CuCO3 (s) <-------------> Cu2+ (aq) + CO32- (aq)

Ksp = [Cu2+][CO32-]

2.5 x 10^-10 = [Cu2+] (1.12 x 10^-2)

[Cu2+] = 2.23 x 10^-8 M

CuS (s)   ------------> Cu2+ (aq) + S2- (aq)

Ksp = [Cu2+][S2-]

7.9 x 10^-37 = (2.23 x 10^-8 ) [S2]

[S2-] = 3.54 x 10^-29

[sulfide] = 3.54 x 10^-29 M