Answer
(a) $2.4 \times 10^{-4}M$
(b) 0.018 g/L
Work Step by Step
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)$
$0.0183 = mass(g)$
