Calculate the solublity for PbCrO_4 given a Ksp = 0.0000109 in a 0.176 M Na_2CrO_4 solution

Question

Calculate the solublity for PbCrO_4 given a Ksp = 0.0000109 in a 0.176 M Na_2CrO_4 solution

Answer 1

Similar to your previous question about PbF₂ just write the Ksp expression, and fill in the values. Then solve.

PbCrO₄(s) <==> Pb²⁺(aq) + CrO₄^2-(aq)

Ksp = 1.09x10^-5 = [Pb²⁺][CrO₄^2-]

Common ion is CrO₄^2- from the Na₂CrO₄ so [CrO₄^2-] = 0.176 M

1.09x10^-5 = [Pb²⁺][0.176]

Solve for [Pb²⁺] and that is the molar solubility (6.19x10^-5 M)