Ionic EquilibriumHard
Question
An amount of 2.0 M solution of Na2CO3 is boiled in a closed container with excess of CaF2. Very little amount of CaCO3 and NaF are formed. If the solubility product of CaCO3 is ‘x’ and the molar solubility of CaF2 is ‘y’, the molar concentration of F– in the resulting solution after equilibrium is attained is
Options
A.$\sqrt{\frac{2y}{x}}$
B.$\frac{8y^{3}}{x}$
C.$\sqrt{\frac{8y^{3}}{x}}$
D.$\sqrt{\frac{4y^{3}}{x}}$
Solution
$\left\lbrack CO_{3}^{2 -} \right\rbrack = 2.0\text{ M}$
Now, $\frac{K_{sp}\left( CaCO_{3} \right)}{K_{sp}\left( CaF_{2} \right)} = \frac{\left\lbrack CO_{3}^{2 -} \right\rbrack}{\left\lbrack F^{-} \right\rbrack^{2}} \Rightarrow \frac{x}{4y^{3}} = \frac{2}{\left\lbrack F^{-} \right\rbrack} \Rightarrow \left( F^{-} \right) = \sqrt{\frac{8y^{3}}{x}}$
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