Answer
$ K_{sp} (CaF_2) = (6.3 \times 10^{-12})$
Work Step by Step
1. Calculate the molar mass:
40.08* 1 + 19* 2 = 78.08g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 9.1\times 10^{- 3}}{ 78.08}$
$n(moles) = 1.165\times 10^{- 4}$
Therefore:
$C(mol/L) = 1.165\times 10^{- 4}$
4. Write the $K_{sp}$ expression:
$ CaF_2(s) \lt -- \gt 1Ca^{2+}(aq) + 2F^{-}(aq)$
$ K_{sp} = [Ca^{2+}]^ 1[F^{-}]^ 2$
5. Determine the ions concentrations:
$[Ca^{2+}] = [CaF_2] * 1 = [1.165 \times 10^{-4}] * 1 = 1.165 \times 10^{-4}$
$[F^{-}] = [CaF_2] * 2 = 2.331 \times 10^{-4}$
6. Calculate the $K_{sp}$:
$ K_{sp} = (1.165 \times 10^{-4})^ 1 \times (2.331 \times 10^{-4})^ 2$
$ K_{sp} = (1.165 \times 10^{-4}) \times (5.433 \times 10^{-8})$
$ K_{sp} = (6.332 \times 10^{-12})$
