In a solution prepared by adding excess PbI₂(s) [K_sp = 1.4 × 10^-8] to water, [I^-] at equilibrium is A) 2.4 × 10^-3 mol/L. B) 1.5 × 10^-3 mol/L. C) 8.4 × 10^-5 mol/L. D) 3.0 × 10^-3 mol/L. E) 1.2 × 10^-3 mol/L.

Question

Quizplus · Accepted Answer

The dissolution of PbI2 in water can be represented as PbI2(s) ⇌ Pb^2+(aq) + 2I^-(aq). The solubility product expression is Ksp = [Pb^2+][I^-]^2. Let the solubility of PbI2 be s mol/L. Then, [Pb^2+] = s and [I^-] = 2s. Substituting into the Ksp expression: 1.4 × 10^-8 = s(2s)^2 = 4s^3. Solving for s gives s = 1.5 × 10^-3 mol/L, so [I^-] = 2s = 3.0 × 10^-3 mol/L.

In a Solution Prepared by Adding Excess PbI2(s) [Ksp =

In a Solution Prepared by Adding Excess PbI₂(s) [K_sp =