Question

Calculate $\left[F^{-}\right]$ in a solution saturated with respect to both $Mg{F}_2$ and $Sr{F}_2$ assuming no hydrolysis of $F^{-}$. $K_{sp}$ of $Mg{F}_2=6.5\times {10}^{-9}$; $K_{sp}$ of $Sr{F}_2=2.9\times {10}^{-9}$.

• A. $2.35\times {10}^{-3}M$
• B. $2.65\times {10}^{-3}M$
• C. $2.95\times {10}^{-3}M$
• D. None of the above

Answer 1

The correct option is B $2.65\times {10}^{-3}M$

$Mg{F}_2\to \underset{\text{a}}{M{g}^{2+}}+\underset{\text{2a}}{2F^{-}}$
$Sr{F}_2\to \underset{\text{b}}{M{g}^{2+}}+\underset{\text{2b}}{2F^{-}}$
$\left[M{g}^{2+}\right]=a$
$\left[S{r}^{2+}\right]=b$
$\left[F^{-}\right]=2(a+b)$
$K_{sp\left(Mg{F}_2\right)}=\left[M{g}^{2+}\right][F^{-}{]}^2$
$6.5\times {10}^{-9}=a(2a+2b{)}^2=4a(a+b)$......(1)
$K_{sp\left(Sr{F}_2\right)}=\left[S{r}^{2+}\right][F^{-}{]}^2$
$2.9\times {10}^{-9}=b(2a+2b{)}^2=4b(a+b)$......(2)
Divide equation (1) by equation (2)
$\frac{6.5}{2.9}=\frac{a}{b}$
$a=2.24b$ ......(3)
Substitute (3) in (1)
$6.5\times {10}^{-9}=2.24b(2\times 2.24b+2b{)}^2$
$b^3=6.9\times {10}^{-11}$
$b=4.1\times {10}^{-4}$......(4)
Substitute (4) in (3)
$a=2.24b=2.24\times 4.1\times {10}^{-4}=9.19\times {10}^{-4}$
$\left[F^{-}\right]=2(a+b)=2(9.19\times {10}^{-4}+4.1\times {10}^{-4})=2.65\times {10}^{-3}M$