How many times solubility of CaF is decreased in M – KF(aq) solution as compared to pure water at 25 — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Let $S_0$ be the solubility in pure water: $4S_0^3 = 3.2 \times 10^{-11} \implies S_0^3 = 8 \times 10^{-12} \implies S_0 = 2 \times 10^{-4}\text{ M}$. Let $S$ be the solubility in $4 \times 10^{-3}\text{ M}$ KF. Due to the common ion effect, $[\text{F}^-] \approx 4 \times 10^{-3}\text{ M}$. Thus, $S(4 \times 10^{-3})^2 = 3.2 \times 10^{-11} \implies 16 \times 10^{-6} S = 3.2 \times 10^{-11} \implies S = 0.2 \times 10^{-5} = 2 \times 10^{-6}\text{ M}$. The ratio is $S_0 / S = (2 \times 10^{-4}) / (2 \times 10^{-6}) = 100$.