Solid is added to a solution containing 0.1 mole of solution (1 L) until equilibrium is reached. If — Ionic Equilibrium Chemistry Question
Question
Solid $\text{BaF}_2$ is added to a solution containing 0.1 mole of $\text{Na}_2\text{C}_2\text{O}_4$ solution (1 L) until equilibrium is reached. If the $K_{sp}$ of $\text{BaF}_2$ and $\text{BaC}_2\text{O}_4$ is $10^{-6} \text{ mol}^3 \text{ L}^{-3}$ and $10^{-10} \text{ mol}^2 \text{ L}^{-2}$, respectively, find the equilibrium concentration of $\text{Ba}^{2+}$ in the solution. Assume addition of $\text{BaF}_2$ does not cause any change in volume.
💡 Solution & Explanation
The reaction is $\text{BaF}_2(s) + \text{C}_2\text{O}_4^{2-}(aq) \rightleftharpoons \text{BaC}_2\text{O}_4(s) + 2\text{F}^-(aq)$. The equilibrium constant $K = \frac{K_{sp}(\text{BaF}_2)}{K_{sp}(\text{BaC}_2\text{O}_4)} = \frac{10^{-6}}{10^{-10}} = 10^4$. Since $K$ is very large, the reaction essentially goes to completion. The initial $0.1 \text{ M}$ of $\text{C}_2\text{O}_4^{2-}$ is entirely converted into $2 \times 0.1 = 0.2 \text{ M}$ of $\text{F}^-$. At equilibrium, $[\text{F}^-] \approx 0.2 \text{ M}$. Using the $K_{sp}$ for $\text{BaF}_2$: $[\text{Ba}^{2+}][\text{F}^-]^2 = 10^{-6} \implies [\text{Ba}^{2+}](0.2)^2 = 10^{-6} \implies [\text{Ba}^{2+}](0.04) = 10^{-6} \implies [\text{Ba}^{2+}] = \frac{10^{-6}}{0.04} = 2.5 \times 10^{-5} \text{ M}$. Therefore, correct answer is C.