Answer
The $K_{sp}$ for $MgF_2$ is equal to $7.0 \times 10^{-11}$
Work Step by Step
1. Write the $K_{sp}$ expression, and find its value:
$ MgF_2(s) \lt -- \gt 1Mg^{2+}(aq) + 2F^{-}(aq)$
$K_{sp} = [Mg^{2+}]^ 1[F^{-}]^ 2$
$K_{sp} = 2.6 \times 10^{-4} \times (5.2 \times 10^{-4})^2$
$K_{sp} = 2.6 \times 10^{-4} \times 2.7 \times 10^{-7} = 7.0 \times 10^{-11}$