Question

$K_{sp}$ of $Sr{F}_2\left(s\right)$ in water is $3.2\times {10}^{-11}$. The solubility of $Sr{F}_2\left(s\right)$ in $0.1M$ $NaCl$ solution is:

• A. $3.2\times {10}^{-9}M$
• B. $2\times {10}^{-4}M$
• C. $4\times {10}^{-4}M$
• D. Slightly higher than $2\times {10}^{-4}M$

Answer 1

The correct option is D Slightly higher than $2\times {10}^{-4}M$

$Sr{F}_2\left(s\right)\rightleftharpoons S{r}^{2+}+2F^{-}$
$s$ $2s$
(where $s$ is the solubility)

$\therefore 4s^3=32\times {10}^{-12}$
or $s=2\times {10}^{-4}M$

But practically the solubility of $Sr{F}_2\left(s\right)$ in $NaCl$ solution is slightly greater than $2\times {10}^{-4}$ because $NaCl$ increases ionic strength of the solution.