Question

The solubility of $Ca{F}_2(K=3.4\times {10}^{-11})$ in $0.1M$ solution of $NaF$ would be:

• A. $3.4\times {10}^{-12}M$
• B. $3.4\times {10}^{-10}M$
• C. $3.4\times {10}^{-9}M$
• D. $3.4\times {10}^{-13}M$

Answer 1

The correct option is C $3.4\times {10}^{-9}M$

$NaF\to N{a}^{+}+F^{-}$
$F^{-}$ is the common ion in $NaF$ and $Ca{F}_2$.
Initially $0.1M$ of $F^{-}$ is present in the solution due to $NaF$.
Now, let $x$ be the solubility of $Ca{F}_2$.
$Ca{F}_2\rightleftharpoons C{a}^{2+}+2F^{-}$
Initial: $0$ $0.1$
Final: $x$ $0.1+2x$
$K=\left[C{a}^{2+}\right][F^{-}{]}^2=x.(0.1+2x{)}^2$

Since $x$ is small, $(0.1+2x)$$\to$ $0.1$
$\Rightarrow x=\frac{K}{0.01}=3.4\times {10}^{-9}$