The solubility product constant of PbI₂(s)is 7.1 × 10^-9.How many moles of PbI₂ will precipitate if 250 mL of a 0.200 M solution of NaI(aq)are added to 150 mL of 0.100 M solution of Pb(NO₃)₂(aq)? You may neglect hydrolysis.

Question

Quizplus · Accepted Answer

When the solutions are mixed, the concentration of I⁻ is 0.125 M and Pb²⁺ is 0.0375 M. The reaction quotient Q = [Pb²⁺][I⁻]² = (0.0375)(0.125)² = 5.86 × 10⁻⁴, which is greater than Ksp, so PbI₂ will precipitate. The moles of PbI₂ precipitated is calculated by the difference in initial and equilibrium concentrations, resulting in 0.015 mol.

The Solubility Product Constant of PbI2(s)is 7

The Solubility Product Constant of PbI₂(s)is 7