Question

The solubility products of PbSO4 and SrSO4 are 6.3×10-7 and 3.2×10-7, respectively. What are the values of [SO42−], [Pb2+], and [Sr2+] in a solution at equilibrium with both substances? HINT: Set up the ICE table; [SO42−] is a combination of both compounds. The solution will require solving simultaneous equations.

Answer 1

PbSO4 <-----------------> Pb+2   + SO42-

                                        S              S

Ksp = [Pb+2][SO42-]

6.3×10^-7 = S^2

S = 7.94 x 10^-4 M

[Pb+2] = [SO42-] = 7.94 x 10^-4 M

SrSO4 <-----------------> Sr+2 + SO42-

Ksp = [Sr+2][SO42-]

3.2×10^-7 = [Sr+2] x 7.94 x 10^-4

[Sr+2] = 4.03 x 10^-4 M