Question

The sparingly soluble salt $Sr{F}_2$ has a $K_{sp}$ value of $2.45\times {10}^{–9}$ at room temperature. Calculate its solubility in $0.1M$ highly soluble $NaF$ aqueous solution.

• A. $2.45\times {10}^{-7}mol{L}^{-1}$
• B. $9\times {10}^{-9}mol{L}^{-1}$
• C. $4\times {10}^{-10}mol{L}^{-1}$
• D. $2.9\times {10}^{-8}mol{L}^{-1}$

Answer 1

The correct option is A $2.45\times {10}^{-7}mol{L}^{-1}$

Lets consider solid $Sr{F}_2$ is in contact with its saturated aqueous solution.The equilibrium is established between undissolved $Sr{F}_2$ and its ions. The equilibrium reaction is given by :
$Sr{F}_2\rightleftharpoons S{r}^{2+}+2F^{-}t=t_{eq}c-ss2s$

When $NaF$ is added , as it is a strong electrolyte , So it completely ionized. It shall provide $\left[F^{-}\right]$ ion concentration i.e. $\left[F^{-}\right]=0.1M$

$\left[S{r}^{2+}\right]=s\left[F^{-}\right]=2s+0.1$

Solubility product $K_{sp}=\left[S{r}^{2+}\right][F^{-}{]}^2$
Since NaF is a strong electrolyte , so it completely dissociates into ions. $2s+0.1\approx 0.1$

$K_{sp}=\left[S{r}^{2+}\right][F^{-}{]}^2$
$2.45\times {10}^{-9}=s(0.1{)}^2$
$s=\frac{2.45\times {10}^{-9}}{0.01}$
$s=2.45\times {10}^{-7}mol{L}^{-1}$