## Chemistry (7th Edition)

$K_{sp}=[Mg^{2+}][F^{-}]^{2}=7.0\times10^{-11}$
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 $K_{sp}=(S)(2S)^{2}=4S^{3}=4(2.6\times10^{-4})=7.0\times10^{-11}$