The solubility of lead(II) iodide is 0.064 g/100 mL at 20°C. What is the solubility product for lead(II) iodide?

Question

Quizplus · Accepted Answer

The solubility product (Ksp) can be calculated using the solubility of PbI2. First, convert the solubility from g/100 mL to mol/L: (0.064 g/100 mL) * (1 mol/461 g) * (1000 mL/1 L) = 0.00139 mol/L. The dissolution of PbI2 is represented by PbI2 ⇌ Pb²⁺ + 2I⁻, so [Pb²⁺] = 0.00139 M and [I⁻] = 2 * 0.00139 M. Ksp = [Pb²⁺][I⁻]² = (0.00139)(2*0.00139)² = 1.1 × 10⁻⁸.

The Solubility of Lead(II) Iodide Is 0