Question

$3M$ 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 y and molar solubility of $Ca{F}_2$ in water is x, find the molar concentration of $F^{-}$ in resulting solution after equilibrium is attained.

• A. $\sqrt{\frac{8y^3}{x}}$
• B. $\sqrt{\frac{8x^3}{y}}$
• C. $\sqrt{\frac{3y^3}{x}}$
• D. $\sqrt{\frac{12x^3}{y}}$

Answer 1

The correct option is D $\sqrt{\frac{12x^3}{y}}$

Given, $K_{sp}\left(CaC{O}_3\right)=y$ and $\text{Molar solubility of}Ca{F}_2=x,\left[N{a}_{2}C{O}_3\right]=3M$

$N{a}_{2}C{O}_3\left(aq\right)+Ca{F}_2\left(s\right)\rightleftharpoons 2NaF\left(aq\right)+CaC{O}_3\left(s\right)t=03---t=eq3-a-2aa$
where $a$ is very small
For $CaC{O}_3,$
$CaC{O}_3\left(s\right)\rightleftharpoons C{a}^{2+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)$
$K_{sp}=y=\left[C{a}^{2+}\right]\left[C{O}_3^{2-}\right]=\left[C{a}^{2+}\right]\times 3(\because C{O}_3^{2-}$ mainly coming from $N{a}_{2}C{O}_3$
$\therefore \left[C{a}^{2+}\right]=\frac{y}{3}.......\left(1\right)$
For $Ca{F}_2,$
$Ca{F}_2\left(s\right)\rightleftharpoons C{a}^{2+}\left(aq\right)+2F^{-}\left(aq\right)C00C-xx2x$
$K_{sp}=\left[C{a}^{2+}\right][F^{-}{]}^2$
$K_{sp}=x\times (2x{)}^2=4x^3$
$\left[C{a}^{2+}\right][F^{-}{]}^2=4x^3$
$\left(\frac{y}{3}\right)[F^{-}{]}^2=4x^3..........(\text{from equation 1})$
$\Rightarrow \left[F^{-}\right]=\sqrt{\frac{12x^3}{y}}$