The solubility product of CaF is . What mass of CaF will dissolve in 500 ml water in order to make a — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

For $\text{CaF}_2$, $K_{sp} = 4S^3 = 1.08 \times 10^{-10} = 108 \times 10^{-12}$. Solving gives $S^3 = 27 \times 10^{-12} \implies S = 3.0 \times 10^{-4} \text{ M}$. In 500 ml (0.5 L), the moles dissolved $= 0.5 \times 3.0 \times 10^{-4} = 1.5 \times 10^{-4} \text{ moles}$. The molar mass of $\text{CaF}_2 = 40 + 2(19) = 78 \text{ g/mol}$. The mass dissolved $= 1.5 \times 10^{-4} \times 78 = 117 \times 10^{-4} \text{ g} = 1.17 \times 10^{-2} \text{ g}$.