Question

Calculate $F^{-}$ in a solution saturated with respect of both $Mg{F}_2$ and $Sr{F}_2$.
$K_{sp}\left(Mg{F}_2\right)=9.5\times {10}^{-9},K_{sp}\left(Sr{F}_2\right)=4\times {10}^{-9}$

Answer 1

$Mg{F}_2\leftrightharpoons M{g}^{2+}+2F^{-}$

$x$ $2x$

$Sr{F}_2\leftrightharpoons S{r}^{2+}+2F^{-}$

$y$ $2y$

Total, $\left[M{g}^{2+}\right]=x;\left[S{r}^{2+}\right]=y\left[F^{-}\right]=2(x+y)Ksp\left(Mg{F}_2\right)=\left[M{g}^{2+}\right](2{\left[F^{-}\right])}^{2}9.5\times {10}^{-9}=(x\left)\right(4\left){(x+y)}^{2}Ksp\right(Sr{F}_2)=[S{r}^{2+}\left]\right({\left[F^{-}\right])}^{2}4\times {10}^{-9}=\left(y\right)4{(x+y)}^2\left(x\right){(x+y)}^2=\frac{95.}{4}\times {10}^{-9}..........\left(1\right)\left(y\right){(x+y)}^2={10}^{-9}.....................\left(2\right)$

$\frac{\left(1\right)}{\left(2\right)}⟹\frac{x}{y}=\frac{9.5}{4}$

$x=2.375y...............\left(3\right)$

$\left(3\right)$ in $\left(2\right)$

$y(3.375y{)}^2={10}^{-9}$

$y^3=\frac{{10}^{-9}}{(3.375{)}^2}$

$y=\frac{1}{2250}$ $x=\frac{19}{18000}$

$\left[F^{-}\right]=2(x+y)=2\left(3.375y\right)=0.003M$

$⟹\left[F^{-}\right]=0.03M$