Question

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

• A. $3.56\times {10}^{-4}M$
• B. $3.86\times {10}^{-4}M$
• C. $4.16\times {10}^{-4}M$
• D. none of these

Answer 1

The correct option is B $3.86\times {10}^{-4}M$

$\begin{array}{ccccccccc}Ca{F}_2& & \rightleftharpoons & & C{a}^{2+}& & +& & 2F^{-}\\ & & & & x& & & & 2x\\ & & & & x& & & & 2x-y\end{array}$

Since this is a buffer solution with $pH=3$, the conc. of $H^{+}$ remains constant at ${10}^{-3}$

$\begin{array}{ccccccc}{H}^{+}& & +& & F^{-}\rightleftharpoons & & HF\\ {10}^{-3}& & & & 2x& & \\ {10}^{-3}& & & & 2x-y& & y\end{array}$

$K_a=\frac{\left[H^{+}\right]\left[F^{-}\right]}{\left[HF\right]}$

$\Rightarrow \frac{\left({10}^{-3}\right)(2x-y)}{y}=6.3\times {10}^{-3}$

On solving, we get $y=\frac{2x}{1.63}$

$K_{sp}=\left[C{a}^{2+}\right][F^{-}{]}^2$

$\Rightarrow x\times (2x-y{)}^2=3.45\times {10}^{-11}$

Substituting $y=\frac{2x}{1.63}$,

${(2x-\frac{2x}{1.63})}^2\times x=3.45\times {10}^{-11}$

On solving, we get $x=3.86\times {10}^{-4}$