Question

1. Part (A)

Solid copper(II) sulfide and solid lead sulfide are in equilibrium with a solution containing 1.48×10^-2 M lead acetate.

Calculate the concentration of copper(II) ion present in this solution.
[copper(II)] = ? M

1. Part (B)

Solid nickel(II) sulfide and solid calcium sulfide are in equilibrium with a solution containing 1.05×10^-2 M calcium acetate.

Calculate the concentration of nickel ion present in this solution.
[nickel] = ? M

Answer 1

1)

part A)

PbS   -------------> Pb2+   +   S2-

Ksp = [Pb2+][S2-]

3.2 x 10^-28 = 1.48 x 10^-2 x [S2-]

[S2-] = 2.16 x 10^-26 M

CuS -------------> Cu2+ +   S2-

Ksp =[Cu2+][S2-]

7.9 x 10^-37 = [Cu2+] [ 2.16 x 10^-26]

[Cu2+] = 3.65 x 10^-11 M

part B)

CaS ------------> Ca2+ + S2-

Ksp = [Ca2+] [S2-]

8.0 x 10^-6 = 1.05 x 10^-2 x [S2-]

[S2-] = 7.62 x 10^-4 M

NiS ------------> Ni2+ +   S2-

Ksp = [Ni2+][S2-]

3 x 10^-21 = [Ni2+] (7.62 x 10^-4)

[Ni2+] = 3.94 x 10^-18 M