Chemistry

The key here is recognizing that $\text{PbI}_2$ dissolves into one $\text{Pb}^{2+}$ and two $\text{I}^-$ ions, but you're not starting from pure water—there's already iodide present from KI. Set up the solubility equilibrium with $s$ as the molar solubility of $\text{PbI}_2$, account for the common ion effect (the initial $0.0010\,\text{M}$ $\text{I}^-$ suppresses dissolution), and use $K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2$ to solve for $s$.

The key here is recognizing that $\text{PbI}_2$ dissolves into one $\text{Pb}^{2+}$ and two $\text{I}^-$ ions, but you're not starting from pure water—there's already iodide present from KI. Set up the solubility equilibrium with $s$ as the molar solubility of $\text{PbI}_2$, account for the common ion effect (the initial $0.0010\,\text{M}$ $\text{I}^-$ suppresses dissolution), and use $K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2$ to solve for $s$.