Question

Solubility product of $Sr{F}_2$ in water is $8\times {10}^{-10}$. Its solubility in 0.1 M NaF aqueous solution is

• A. $8\times {10}^{-11}M$
• B. $8\times {10}^{-12}M$
• C. $8\times {10}^{-8}M$
• D. $8\times {10}^{-9}M$

Answer 1

The correct option is C $8\times {10}^{-8}M$

As NaF is a strong electrolyte, it ionizes completely

$\left[F^{-}\right]=0.1M+x\left(fromSr{F}_2\right)$

$=0.1M(\because x<<0.1)$

$K_{sp}\left(Sr{F}_2\right)=\left[S{r}^{2+}\right][F^{-}{]}^2$

$8\times {10}^{-10}=\left[S{r}^{2+}\right][0.1{]}^2$

$\left[S{r}^{2+}\right]=\frac{8\times {10}^{-10}}{(0.1{)}^2}=8\times {10}^{-8}M$