The solubility of \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) in a $0.20-M \mathrm{KIO}_{3}\( solution is \)4.4 \times 10^{-8} \mathrm{mol} / \mathrm{L}$ . Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}.\)

First, write the balanced chemical equation for the dissolution of \(\ce{Ce(IO3)3}\) in water: \[\ce{Ce(IO3)3(s) <=> Ce^{3+}(aq) + 3 IO3^-(aq)}\]

Next, write the solubility product constant (\(K_{sp}\)) expression for the chemical equation: \[K_{sp} = [\ce{Ce^{3+}}][\ce{IO3^-}]^{3}\]

Since the solubility of \(\ce{Ce(IO3)3}\) is given in a \(0.20\,\mathrm{M}\) \(\ce{KIO3}\) solution, there will be a common ion effect. For every mole of \(\ce{Ce(IO3)3}\) that dissolves, there will be \(3\) moles of \(\ce{IO3^-}\) ions produced. Therefore, the concentration of \(\ce{IO3^-}\) ions will be the sum of the ions produced from the dissolution of \(\ce{Ce(IO3)3}\) and the concentration of \(\ce{IO3^-}\) ions provided by the \(\ce{KIO3}\) solution. Let \(x\) be the solubility of \(\ce{Ce(IO3)3}\) in moles per liter. Then: \[[\ce{Ce^{3+}}]=x\] \[[\ce{IO3^-}]=(0.20\,\text{M}) + 3x\] The given solubility of \(\ce{Ce(IO3)3}\) in the \(0.20\, \text{M}\) \(\ce{KIO3}\) solution is \(4.4 \times 10^{-8}\, \mathrm{mol/L}\). Thus, \[x = 4.4 \times 10^{-8}\, \mathrm{mol/L}\]

Substitute the solubility values into the \(K_{sp}\) expression: \[K_{sp} = (4.4 \times 10^{-8})(0.20 + 3(4.4 \times 10^{-8}))^3\] Now, simplify and solve for \(K_{sp}\): \[K_{sp} = (4.4 \times 10^{-8})(0.20 + 1.32 \times 10^{-7})^3\] \[K_{sp} = (4.4 \times 10^{-8})(2.052 \times 10^{-7})^3\] \[K_{sp} = 2.195 \times 10^{-26}\] Solution: The \(K_{sp}\) for \(\ce{Ce(IO3)3}\) is approximately \(2.195 \times 10^{-26}\).