The solubility of lead(II)sulfate is 4.0 × 10^-2 g/L.What is the solubility product constant for lead(II)sulfate?

Question

Quizplus · Accepted Answer

The solubility product constant (Ksp) for lead(II) sulfate, PbSO4, can be calculated from its solubility. Given the solubility is 4.0 × 10^-2 g/L, we first convert this to moles per liter (M) using the molar mass of PbSO4 (303.26 g/mol). Solubility in mol/L = 4.0 × 10^-2 g/L / 303.26 g/mol ≈ 1.32 × 10^-4 M. Since PbSO4 dissociates into Pb^2+ and SO4^2- in a 1:1 ratio, the concentrations of Pb^2+ and SO4^2- are each 1.32 × 10^-4 M. The Ksp = [Pb^2+][SO4^2-] = (1.32 × 10^-4)^2 ≈ 1.7 × 10^-8.

The Solubility of Lead(II)sulfate Is 4