## Chemistry and Chemical Reactivity (9th Edition)

(a) $2.4 \times 10^{-4}M$ (b) 0.018 g/L
1. Write the $K_{sp}$ expression: $CaF_2(s) \lt -- \gt 1Ca^{2+}(aq) + 2F^-(aq)$ $5.3 \times 10^{-11} = [Ca^{2+}]^ 1[F^-]^ 2$ 2. Considering a pure solution: $[Ca^{2+}] = 1S$ and $[F^-] = 2S$ $5.3 \times 10^{-11}= ( 1S)^ 1 \times ( 2S)^ 2$ $5.3 \times 10^{-11} = 4S^ 3$ $1.3 \times 10^{-11} = S^ 3$ $\sqrt [ 3] {1.3 \times 10^{-11}} = S$ $2.35 \times 10^{-4} = S$ - This is the molar solubility value for this salt. 4. Determine the molar mass of this compound (CaF_2): 40.08* 1 + 19* 2 = 78.08g/mol 5. Calculate the mass $mm(g/mol) = \frac{mass(g)}{n(mol)}$ $mm(g/mol) * n(mol) = mass(g)$ $78.08 * 2.35 \times 10^{-4} = mass(g)$ $0.0183 = mass(g)$