An amount of 2.0 M solution of is boiled in a closed container with excess of . Very little amount o — Ionic Equilibrium Chemistry Question
Question
An amount of 2.0 M solution of $\text{Na}_2\text{CO}_3$ is boiled in a closed container with excess of $\text{CaF}_2$. Very little amount of $\text{CaCO}_3$ and NaF are formed. If the solubility product of $\text{CaCO}_3$ is '$x$' and molar solubility of $\text{CaF}_2$ is '$y$', the molar concentration of $\text{F}^-$ in the resulting solution after equilibrium is attained is
💡 Solution & Explanation
The equilibrium reaction is $\text{CaF}_2(s) + \text{CO}_3^{2-}(aq) \rightleftharpoons \text{CaCO}_3(s) + 2\text{F}^-(aq)$. The equilibrium constant is $K = \frac{[\text{F}^-]^2}{[\text{CO}_3^{2-}]} = \frac{K_{sp}(\text{CaF}_2)}{K_{sp}(\text{CaCO}_3)}$. Given $K_{sp}(\text{CaCO}_3) = x$. For $\text{CaF}_2$, with molar solubility $y$, $K_{sp}(\text{CaF}_2) = 4y^3$. Thus, $K = \frac{4y^3}{x}$. Because very little product is formed, $[\text{CO}_3^{2-}]$ remains approximately $2.0 \text{ M}$. We have $\frac{[\text{F}^-]^2}{2.0} = \frac{4y^3}{x} \implies [\text{F}^-]^2 = \frac{8y^3}{x}$. Taking the square root yields $[\text{F}^-] = \sqrt{\frac{8y^3}{x}}$. Therefore, correct answer is C.