The solubility of is ‘s’ moles/litre. Then its solubility product is........ — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

# Solubility Product of $CaF_2$ **Step 1: Write the dissolution equation** $$CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$$ **Step 2: Express ion concentrations in terms of solubility 's'** If solubility of $CaF_2$ is 's' moles/litre: - $[Ca^{2+}] = s$ M - $[F^-] = 2s$ M (since 1 mole of $CaF_2$ produces 2 moles of $F^-$) **Step 3: Write the solubility product expression** $$K_{sp} = [Ca^{2+}][F^-]^2$$ **Step 4: Substitute the concentrations** $$K_{sp} = (s)(2s)^2 = s \cdot 4s^2 = 4s^3$$ **Answer: $K_{sp} = 4s^3$** The solubility product depends on the stoichiometry of the salt. Since $CaF_2$ produces one $Ca^{2+}$ ion and two $F^-$ ions, the $[F^-]$ term is squared in the expression, leading to the coefficient of 4 in the final answer.