Question

The solubility of $Ca{F}_2$ in a solution buffered at $pH=3.0$ is........
$K_a$ for $HF=6.3\times {10}^{-4}$ and $K_{sp}$ of $Ca{F}_2=3.45\times {10}^{-11}$.

• A. $4.80\times {10}^{-8}$
• B. $9.60\times {10}^{-9}$
• C. $9.60\times {10}^{-8}$
• D. $3.60\times {10}^{-4}$

Answer 1

The correct option is D $3.60\times {10}^{-4}$

$\left[C{a}^{2+}\right][F^{⊝}{]}^2=3.45\times {10}^{-11}$
The $F^{⊝}$ reacts with $H^{\oplus }(pH=3.0)$ to produce HF
$K_{a_{HF}}=\frac{\left[H^{\oplus }\right]\left[F^{⊝}\right]}{\left[HF\right]}$ or $6.3\times {10}^{-4}=\frac{{10}^{-3}\left[F^{⊝}\right]}{\left[HF\right]}$
$HF=1.58\times \left[F^{⊝}\right]$
Also the solution contains $\left[HF\right]+\left[F^{⊝}\right]=2\times \left[C{a}^{2+}\right]$ or $1.58\times \left[F^{⊝}\right]+\left[F^{⊝}\right]=2\times \left[C{a}^{2+}\right]$
$\therefore \left[F^{⊝}\right]=\frac{2}{2.58}\times \left[C{a}^{2+}\right]=0.775\left[C{a}^{2+}\right]$
Let solubility of $Ca{f}_2$ be S mol $L^{-1}\therefore \left[C{a}^{2+}\right]=S$
$\therefore \left[F^{⊝}\right]=0.7775\times S$
Thus: $S\times (0.775\times S{)}^2=3.45\times {10}^{-11}$
$S=3.86\times {10}^{-4}M$