Question

$2M$ solution of $N{a}_{2}C{O}_3$ is boiled in a closed container with excess of $Ca{F}_2.$ Very little amount of $CaC{O}_3$ and $NaF$ are formed. If the solubility product $(K_{sp}$ of $CaC{O}_3$ is x and molar solubility of $Ca{F}_2$ is y, find the molar concentration of $F^{-}$ in resulting solution after equilibrium is attained.

• A. $\frac{4y^3}{x}$
• B. $\sqrt{\frac{4y^3}{x}}$
• C. $\frac{8y^3}{x}$
• D. $\sqrt{\frac{8y^3}{x}}$

Answer 1

The correct option is D $\sqrt{\frac{8y^3}{x}}$

Given, $K_{sp}\left(CaC{O}_3\right)=x$ and $\text{molar solubility of}Ca{F}_2=y,\left[N{a}_{2}C{O}_3\right]=2M$

$N{a}_{2}C{O}_3\left(aq\right)+Ca{F}_2\left(s\right)\rightleftharpoons 2NaF\left(aq\right)+CaC{O}_3\left(s\right)t=02---t=eq2-a-2a-$
where a is very small
For $CaC{O}_3,$
$CaC{O}_3\left(s\right)\rightleftharpoons C{a}^{2+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)$
$K_{sp}=x=\left[C{a}^{2+}\right]\left[C{O}_3^{2-}\right]=\left[C{a}^{2+}\right]\times 2(\because C{O}_3^{2-}$ mainly coming from $N{a}_{2}C{O}_3$
$\therefore \left[C{a}^{2+}\right]=\frac{x}{2}.......\left(1\right)$
For $Ca{F}_2,$
$Ca{F}_2\left(s\right)\rightleftharpoons C{a}^{2+}\left(aq\right)+2F^{-}\left(aq\right)1001-yy2y$
$K_{sp}=\left[C{a}^{2+}\right][F^{-}{]}^2$
$K_{sp}=y\times (2y{)}^2=4y^3$
$\left[C{a}^{2+}\right][F^{-}{]}^2=4y^3$
$\left(\frac{x}{2}\right)[F^{-}{]}^2=4y^3..........(\text{from equation 1})$
$\Rightarrow \left[F^{-}\right]=\sqrt{\frac{8y^3}{x}}$