# Chapter 17 - Questions and Problems - Page 826: 17.54

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.

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