In a solution prepared by adding excess PbI₂(s) [K_sp = 1.4 *10^-⁸] to water, [I^-] at equilibrium is

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Quizplus · Accepted Answer

The solubility product expression for PbI2 is Ksp = [Pb^2+][I^-]^2. Let the solubility of PbI2 be s mol/L. Then, [Pb^2+] = s and [I^-] = 2s. Substituting these into the Ksp expression gives 1.4 x 10^-8 = s(2s)^2 = 4s^3. Solving for s gives s = 1.5 x 10^-3 mol/L, so [I^-] = 2s = 3.0 x 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 =