The solubility of lead(II) chloride is 0.45 g/100 mL of solution. What is the K_sp of PbCl₂?

Question

Quizplus · Accepted Answer

The solubility product (Ksp) of PbCl2 can be calculated using its solubility and the dissociation equation PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq). Given the solubility is 0.45 g/100 mL, first convert this to moles/L (molarity, M) using the molar mass of PbCl2 (207.2 + 35.45*2 = 278.1 g/mol). Solubility in moles/L = 0.45 g / 278.1 g/mol / 0.1 L = 1.62 × 10^-2 M. For each mole of PbCl2 that dissolves, 1 mole of Pb2+ and 2 moles of Cl- are produced. Therefore, [Pb2+] = 1.62 × 10^-2 M and [Cl-] = 2 * 1.62 × 10^-2 M = 3.24 × 10^-2 M. Ksp = [Pb2+][Cl-]^2 = (1.62 × 10^-2)(3.24 × 10^-2)^2 = 1.7 × 10^-5.

The Solubility of Lead(II) Chloride Is 0