When 100.0 \(\mathrm{mL}\) of 2.00 $\mathrm{M} \mathrm{Ce}\left(\mathrm{NO}_{3}\right)_{3}\( is added to 100.0 \)\mathrm{mL}$ of \(3.00 \mathrm{M} \mathrm{KIO}_{3},\) a precipitate of \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}(s)\) forms. Calculate the equilibrium concentrations of \(\mathrm{Ce}^{3+}\) and \(\mathrm{IO}_{3}^{-}\) in this solution. $\left[K_{\mathrm{sp}} \text { for } \mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}=3.2 \times 10^{-10} .\right]$

The balanced chemical equation for the precipitation reaction is: \[ \mathrm{Ce}^{3+}(aq) + 3 \mathrm{IO}_3^-(aq) \rightleftharpoons \mathrm{Ce(IO_3)_3}(s) \]

To find the initial concentrations, we need to use the fact that the volumes of the two solutions are the same (100.0 mL) and that moles are additive. We can calculate the moles of each ion in the mixture: Moles of \(\mathrm{Ce}^{3+} = 100.0 \mathrm{mL} \times \frac{2.00 \mathrm{mol}}{1,000 \mathrm{mL}} = 0.200 \mathrm{mol}\) Moles of \(\mathrm{IO}_3^- = 100.0 \mathrm{mL} \times \frac{3.00 \mathrm{mol}}{1,000 \mathrm{mL}} = 0.300 \mathrm{mol}\) The volume of the mixture is the sum of the volumes of the two solutions: \(v = 100.0\mathrm{mL} + 100.0\mathrm{mL} = 200.0\mathrm{mL}\). The initial concentrations are calculated as: \[[\mathrm{Ce}^{3+}]_{initial} = \frac{0.200\ \mathrm{mol}}{200.0\ \mathrm{mL}} \times \frac{1,000\ \mathrm{mL}}{1\ \mathrm{mol}} = 1.00\ \mathrm{M}\] \[[\mathrm{IO}_3^-]_{initial} = \frac{0.300\ \mathrm{mol}}{200.0\ \mathrm{mL}} \times \frac{1,000\ \mathrm{mL}}{1\ \mathrm{mol}} = 1.50\ \mathrm{M}\]

Using the initial concentrations, we can set up an ICE table for the reaction: | | \(\mathrm{Ce}^{3+}(aq)\) | \(3\mathrm{IO}_3^-(aq)\) | \(\mathrm{Ce(IO_3)_3}(s)\) | | ------- | ---------------------- | --------------------- | ----------------------- | | Initial | 1.00 M | 1.50 M | solid | | Change | -x | -3x | +x | | Equilibrium | 1.00 M - x | 1.50 M - 3x | solid |

The solubility product constant \(K_{sp}\) expression is: \(K_{sp} = [\mathrm{Ce}^{3+}][\mathrm{IO}_3^-]^3\) Now, use the given value of \(K_{sp} = 3.2\times 10^{-10}\), and the values from the ICE table to solve for the equilibrium concentrations: \(3.2 \times 10^{-10} = (1.00\ \mathrm{M} - x) \times (1.50\ \mathrm{M} - 3x)^3\) By making the assumption that x is small enough compared to the initial values, we can simplify the equation to: \(3.2 \times 10^{-10} = \left(1.00\right) \times \left(1.50 - 3x\right)^3\) Now, solve for \(x\): \(x = 1.03\times10^{-2}\ \mathrm{M}\) Based on the ICE table, we have: \([\mathrm{Ce}^{3+}]_{eq} = 1.00\ \mathrm{M} - x = 1.00\ \mathrm{M} - 1.03\times10^{-2}\ \mathrm{M} = 0.9897\ \mathrm{M}\) \([\mathrm{IO}_3^-]_{eq} = 1.50\ \mathrm{M} - 3x = 1.50\ \mathrm{M} - 3(1.03\times10^{-2}\ \mathrm{M}) = 1.468\ \mathrm{M}\) So, the equilibrium concentrations are \([\mathrm{Ce}^{3+}]_{eq} = 0.9897\ \mathrm{M}\) and \([\mathrm{IO}_3^-]_{eq} = 1.468\ \mathrm{M}\).