Use the following data to calculate the \(K_{\mathrm{sp}}\) value for each solid. a. The solubility of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) is $4.8 \times 10^{-5} \mathrm{mol} / \mathrm{L}$ . b. The solubility of \(\mathrm{BiI}_{3}\) is $1.32 \times 10^{-5} \mathrm{mol} / \mathrm{L}$ .

1. From the given data, the solubility of \(\mathrm{CaC}_{2}\mathrm{O}_{4}\) is \(4.8 \times 10^{-5}\) \(\mathrm{mol} / \mathrm{L}\) which means that for every mole of \(\mathrm{CaC}_{2}\mathrm{O}_{4}\) that dissolves, we will get one mole of \(\mathrm{Ca}^{2+}\) and two moles of \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\). So, the concentrations of the ions in the saturated solution are: \[ [\mathrm{Ca}^{2+}] = 4.8 \times 10^{-5} \mathrm{M} \] \[ [\mathrm{C}_{2}\mathrm{O}_{4}^{2-}] = 2(4.8 \times 10^{-5})\mathrm{M} \] 2. Using these concentrations, calculate the solubility product constant, \(K_{\mathrm{sp}}\), for \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) as follows: \[ K_{\mathrm{sp}} = [\mathrm{Ca}^{2+}][(\mathrm{C}_{2}\mathrm{O}_{4}^{2-})]^2 = (4.8 \times 10^{-5}) (2( 4.8 \times 10^{-5}))^2 \] 3. Solve for \(K_{\mathrm{sp}}\): \[ K_{\mathrm{sp}} = 2.95 \times 10^{-12} \]

1. From the given data, the solubility of \(\mathrm{BiI}_{3}\) is \(1.32 \times 10^{-5}\) \(\mathrm{mol} / \mathrm{L}\) which means that for every mole of \(\mathrm{BiI}_{3}\) that dissolves, we will get one mole of \(\mathrm{Bi}^{3+}\) and three moles of \(\mathrm{I}^{-}\). So, the concentrations of the ions in the saturated solution are: \[ [\mathrm{Bi}^{3+}] = 1.32 \times 10^{-5} \mathrm{M} \] \[ [\mathrm{I}^{-}] = 3(1.32 \times 10^{-5})\mathrm{M} \] 2. Using these concentrations, calculate the solubility product constant, \(K_{\mathrm{sp}}\), for \(\mathrm{BiI}_{3}\) as follows: \[ K_{\mathrm{sp}} = [\mathrm{Bi}^{3+}][(\mathrm{I}^{-})]^3 = (1.32 \times 10^{-5}) (3( 1.32 \times 10^{-5}))^3 \] 3. Solve for \(K_{\mathrm{sp}}\): \[ K_{\mathrm{sp}} = 1.04 \times 10^{-32} \]