Question

The precipitate of $Ca{F}_2$ with $K_{sp}$ equal to $1.7\times {10}^{-10}$ is obtained when equal volumes of the following are mixed?

• A. ${10}^{-4}MC{a}^{2+}+{10}^{-4}M{F}^{-}$
• B. ${10}^{-2}MC{a}^{2+}+{10}^{-3}M{F}^{-}$
• C. ${10}^{-5}MC{a}^{2+}+{10}^{-3}M{F}^{-}$
• D. ${10}^{-3}MC{a}^{2+}+{10}^{-5}M{F}^{-}$

Answer 1

The correct option is B ${10}^{-2}MC{a}^{2+}+{10}^{-3}M{F}^{-}$

For precipitation, $Q_{sp}$ (product of molar concentration) > $K_{sp}$ (solubility product).
In water, $Ca{F}_2$ is in equilibrium with its ions as follows:
$Ca{F}_2\left(s\right)\rightleftharpoons C{a}^{2+}\left(aq\right)+2F^{-}\left(aq\right)$
Solubility product of $Ca{F}_2$
$K_{sp}$ = $\left[C{a}^{2+}\right]\left[{F^{-}}^2\right]$ = $1.7\times {10}^{-10}$
In a), $Q_{sp}=\left[\frac{{10}^{-4}}{2}\right][\frac{{10}^{-4}}{2}{]}^2=1.25\times {10}^{-13}$
In b), $Q_{sp}=\left[\frac{{10}^{-2}}{2}\right][\frac{{10}^{-3}}{2}{]}^2=1.25\times {10}^{-9}$

In c), $Q_{sp}=\left[\frac{{10}^{-5}}{2}\right][\frac{{10}^{-3}}{2}{]}^2=1.25\times {10}^{-12}$

In d), $Q_{sp}=\left[\frac{{10}^{-3}}{2}\right][\frac{{10}^{-5}}{2}{]}^2=1.25\times {10}^{-14}$
The option b) only has the ionic product greater than the solubility product.