Answer
Molar solubility: $(CaF_2)$ = $3.32 \times 10^{-4}M$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ CaF_2(s) \lt -- \gt 1Ca^{2+}(aq) + 2{F}^{-}(aq)$
$1.46 \times 10^{-10} = [Ca^{2+}]^ 1[{F}^{-}]^ 2$
2. Considering a pure solution: $[Ca^{2+}] = 1S$ and $[{F}^{-}] = 2S$
$1.46 \times 10^{-10}= ( 1S)^ 1 \times ( 2S)^ 2$
$1.46 \times 10^{-10} = 4S^ 3$
$3.65 \times 10^{-11} = S^ 3$
$ \sqrt [ 3] {3.65 \times 10^{-11}} = S$
$3.32 \times 10^{-4} = S$
