Answer
$ K_{sp} (MgF_2) = (7.44 \times 10^{-11})$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ MgF_2(s) \lt -- \gt 1Mg^{2+}(aq) + 2F^{-}(aq)$
$ K_{sp} = [Mg^{2+}]^ 1[F^{-}]^ 2$
2. Determine the ions concentrations:
$[Mg^{2+}] = [MgF_2] * 1 = [2.65 \times 10^{-4}] * 1 = 2.65 \times 10^{-4}$
$[F^{-}] = [MgF_2] * 2 = 5.3 \times 10^{-4}$
3. Calculate the $K_{sp}$:
$ K_{sp} = (2.65 \times 10^{-4})^ 1 \times (5.3 \times 10^{-4})^ 2$
$ K_{sp} = (2.65 \times 10^{-4}) \times (2.81 \times 10^{-7})$
$ K_{sp} = (7.44 \times 10^{-11})$
