Calculate the molar solubility of () in a 0.1 M aqueous solution of NaF. — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

In 0.1 M NaF, $[F^-]$ is heavily dominated by the salt, so $[F^-] \approx 0.1 \text{ M}$. The equilibrium is $CaF_2 \rightleftharpoons Ca^{2+} + 2F^-$. $K_{sp} = [Ca^{2+}][F^-]^2 = S \times (0.1)^2 = 5.3 \times 10^{-11}$. Thus, $S = 5.3 \times 10^{-9} \text{ M}$.