Question

A solution is saturated with respect to $SrC{O}_3$ and $Sr{F}_2$. The $\left[C{O}_3^{2-}\right]$ was found to be $1.2\times {10}^{-3}M$. The values of the solubility products of $SrC{O}_3$ and $Sr{F}_2$ are $7.0\times {10}^{-10}$ and $7.9\times {10}^{-10}$ respectively. The concentration of $F^{-}$ in the solution would be:

• A. $1.3\times {10}^{-3}M$
• B. $3.7\times {10}^{-2}M$
• C. $5.8\times {10}^{-7}M$
• D. $2.6\times {10}^{-2}M$

Answer 1

The correct option is B $3.7\times {10}^{-2}M$

For strontium carbonate, $K_{sp}=\left[S{r}^{2+}\right]\left[C{O}_3^{2-}\right]$.

Substitute values in the above expression.

$7.0\times {10}^{-10}=\left[S{r}^{2+}\right]\times 1.2\times {10}^{-3}$

Hence, $\left[S{r}^{2+}\right]=5.8\times {10}^{-7}$.

For strontium fluoride, $K_{sp}=\left[S{r}^{2+}\right][F^{-}{]}^2$.

Substitute values in the above expression.

$7.9\times {10}^{-10}=5.8\times {10}^{-7}[F^{-}{]}^2$

$\left[F^{-}\right]=3.7\times {10}^{-2}M$.