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: Set up an ICE table with solubility 's' mol/L** | | $Ca^{2+}$ | $F^-$ | |---|---|---| | At equilibrium | $s$ | $2s$ | When 's' moles of $CaF_2$ dissolve per litre, it produces 's' moles of $Ca^{2+}$ and $2s$ moles of $F^-$. **Step 3: Write the $K_{sp}$ expression** $$K_{sp} = [Ca^{2+}][F^-]^2$$ **Step 4: Substitute equilibrium concentrations** $$K_{sp} = (s)(2s)^2 = s \cdot 4s^2 = 4s^3$$ **Answer: $K_{sp} = 4s^3$** The key point is recognizing that $CaF_2$ produces **two** fluoride ions per formula unit, making the $F^-$ concentration $2s$, which becomes $(2s)^2$ in the expression.