Answer
$K_{sp}=[Mg^{2+}][F^{-}]^{2}=7.0\times10^{-11}$
Work Step by Step
Equilibrium equation of the reaction is:
$MgF_{2}\rightleftharpoons Mg^{2+}+2F^{-}$
Therefore, $K_{sp}=[Mg^{2+}][F^{-}]^{2}$
If S= solubility of $MgF_{2}$, then
$[Mg^{2+}]=2.6\times10^{-4}\,mol/L=S$;
$[F^{-}]=2S$ and
$K_{sp}=(S)(2S)^{2}=4S^{3}=4(2.6\times10^{-4})=7.0\times10^{-11}$
