Answer
Molar solubility $(CaF_2) = $ $2.2 \times 10^{-4}$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ CaF_2(s) \lt -- \gt 1Ca^{2+}(aq) + 2F^{-}(aq)$
$4 \times 10^{-11} = [Ca^{2+}]^ 1[F^{-}]^ 2$
2. Considering a pure solution: $[Ca^{2+}] = 1x$ and $[F^{-}] = 2x$
$4 \times 10^{-11}= ( 1x)^ 1 \times ( 2x)^ 2$
$4 \times 10^{-11} = 4x^ 3$
$1 \times 10^{-11} = x^ 3$
$ \sqrt [ 3] {1 \times 10^{-11}} = x$
$2.154 \times 10^{-4} = x$
- This is the molar solubility value for this salt.
