Question

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

• A. $\left[F^{-}\right]=1.5\times {10}^{-3}M$
• B. $\left[F^{-}\right]=3\times {10}^{-3}M$
• C. $\left[F^{-}\right]=6\times {10}^{-3}M$
• D. $\left[F^{-}\right]=9\times {10}^{-3}M$

Answer 1

The correct option is B $\left[F^{-}\right]=3\times {10}^{-3}M$

$\underset{a}{Mg{F}_2}\to \underset{a}{M{g}^{2+}}+\underset{2a+2b}{2F^{-}}\underset{b}{Sr{F}_2}\to \underset{b}{S{r}^{2+}}+\underset{2a+2b}{2F^{-}}$
$K_{SP,Mg{F}_2}=\left[M{g}^{2+}\right][F^{-}{]}^{2}9.5\times {10}^{-9}=a(2a+2b{)}^2$......(1)
$K_{SP,Sr{F}_2}=\left[S{r}^{2+}\right][F^{-}{]}^{2}4\times {10}^{-9}=b(2a+2b{)}^2$.....(2)
Divide equation (1) by equation (2)
$\frac{9.5\times {10}^{-9}}{4\times {10}^{-9}}=\frac{a(2a+2b{)}^2}{b(2a+2b{)}^2}$
$2.375=\frac{a}{b}$
$a=2.375b$......(3)
Substitute (3) in (2)
$4\times {10}^{-9}=b\left(2\right(2.375b)+2b{)}^2$
$4\times {10}^{-9}=45.5625b^3$
$b=0.0004444$
Substitute this in (3)
$a=2.375\times 0.0004444=0.001056$
$\left[F^{-}\right]=(2a+2b)=(2\times 0.001056+2\times 0.0004444)=3\times {10}^{-3}$