Question

At $25{}^{\circ }\text{C},$ the solubility product $\left(K_{sp}\right)$ of $Ca{F}_2$ in water is $3.2\times {10}^{-11}{\text{mol}}^3{\text{L}}^{-3}.$ The solubility (in mole per litre) of the salt at the same temperature (ignore ion pairing) is:

• A. $4.0\times {10}^{-6}$
• B. $3.2\times {10}^{-4}$
• C. $2.5\times {10}^{-4}$
• D. $2.0\times {10}^{-4}$

Answer 1

The correct option is D $2.0\times {10}^{-4}$

$Ca{F}_2\left(s\right)\rightleftharpoons \underset{S}{C{a}^{2+}}+\underset{2S}{2F^{-}}$

$\text{Considering the solubility of}Ca{F}_2{\text{to be S mol L}}^{-1}$,
$K_{sp}=\left[C{a}^{2+}\right][F^{-}{]}^2=S\times (2S{)}^2=4S^3\Rightarrow 3.2\times {10}^{-11}=4S^3\Rightarrow S^3=\frac{3.2\times {10}^{-11}}{4}=8\times {10}^{-12}\therefore S=(8\times {10}^{-12}{)}^{\left(\frac{1}{3}\right)}=2.0\times {10}^{-4}{\text{mol L}}^{-1}$