The \(K_{\mathrm{sp}}\) for lead iodide \(\left(\mathrm{PbI}_{2}\right)\) is \(1.4 \times 10^{-8} .\) Calculate the solubility of lead iodide in each of the following. \(\mathbf{a}\). water b. \(0.10 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\)

We will begin by writing the balanced chemical equation for the dissolution of lead iodide in water and the corresponding Ksp expression. For the dissolution of \(\mathrm{PbI_{2}}\) in water, we have \(\mathrm{PbI}_{2}\left(\mathrm{s}\right) \rightleftharpoons \mathrm{Pb}^{2+}\left(\mathrm{aq}\right)+2 \mathrm{I}^{-}\left(\mathrm{aq}\right)\) The solubility product constant (Ksp) expression is \(K_\mathrm{sp} = [\mathrm{Pb}^{2+}][\mathrm{I}^{-}]^2\) We are given that \(K_\mathrm{sp} = 1.4 \times 10^{-8}\).

In water, let the solubility of \(\mathrm{PbI}_{2}\) be S molar. From the stoichiometry of the balanced chemical equation, if S moles of \(\mathrm{PbI_{2}}\) dissolve, S moles of \(\mathrm{Pb}^{2+}\) ions and 2S moles of \(\mathrm{I}^{-}\) ions would be formed. Hence, using the Ksp expression and substituting the given value: \(1.4 \times 10^{-8} = [S][2S]^2\) Now, solve for S: \(1.4 \times 10^{-8} = 4S^3\) \(S^3 = \frac{1.4 \times 10^{-8}}{4}\) \(S^3 = 3.5 \times 10^{-9}\) \(S = \sqrt[3]{3.5 \times 10^{-9}}\) \(S = 1.52 \times 10^{-3}\) Thus, the solubility of \(\mathrm{PbI_{2}}\) in water is approximately \(1.52 \times 10^{-3}\) M.

In the presence of \(0.10\,\mathrm{M}\,\mathrm{Pb}(\mathrm{NO}_3)_2\), the added Pb²⁺ ions will affect the solubility of \(\mathrm{PbI_{2}}\). The \(\mathrm{Pb}^{2+}\) ions in solution will come from both the \(\mathrm{Pb}(\mathrm{NO}_3)_{2}\) and the dissolved \(\mathrm{PbI}_{2}\). Let the solubility of \(\mathrm{PbI_{2}}\) in the presence of \(0.10\,\mathrm{M}\) Pb²⁺ be S' molar. Then, the concentration of \(\mathrm{Pb}^{2+}\) ions from the dissolved \(\mathrm{PbI}_{2}}\) would be S' and from \(\mathrm{Pb}(\mathrm{NO}_3)_2\) would be 0.10 M. Hence, using the Ksp expression in this case: \(1.4 \times 10^{-8} = [S' + 0.10][2S']^2\) Now, solve for S': \(1.4 \times 10^{-8} = (0.10 + S')[4S'^2]\) Assuming \(S' \ll 0.10\), we can approximate and simplify the equation as: \(1.4 \times 10^{-8} \approx 0.10[4S'^2]\) \(S'^2 \approx 3.5 \times 10^{-8}\) \(S' \approx \sqrt{3.5 \times 10^{-8}}\) \(S' \approx 5.92 \times 10^{-5}\) Thus, the solubility of \(\mathrm{PbI_{2}}\) in the presence of \(0.10\,\mathrm{M}\) Pb(NO₃)₂ is approximately \(5.92 \times 10^{-5}\) M.