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 equilibrium** $$CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$$ **Step 2: Express ion concentrations in terms of solubility** If solubility of $CaF_2$ is $s$ moles/litre: - $[Ca^{2+}] = s$ M - $[F^-] = 2s$ M (stoichiometry: 2 moles of $F^-$ per mole of $CaF_2$) **Step 3: Apply the solubility product expression** $$K_{sp} = [Ca^{2+}][F^-]^2$$ **Step 4: Substitute values** $$K_{sp} = (s)(2s)^2 = s \cdot 4s^2 = 4s^3$$ **Answer: $K_{sp} = 4s^3$** This is typically option B in standard chemistry problems. The key is recognizing that fluoride ions have a coefficient of 2 in the equilibrium, which must be reflected in the stoichiometry of the solubility product expression.