When pure water is saturated with ${CaCO}_{3}$ and $8 x 10^{- 5} M$ ${CaC}_{2} O_{4}$ , the concentration of calcium ion in the solution under equilibrium is 8 x 10- 5 M. If the ratio of the solubility product of ${CaCO}_{3}$ to that of ${CaC}_{2} O_{4}$ is 3, what is the solubility of ${CaCO}_{3}$ in pure water?

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$9.60 \times 10^{- 9}$

$4.80 \times 10^{- 8}$

$9.62 \times 10^{- 8}$

$6.93 \times 10^{- 5}$

answer is D.

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Detailed Solution

Let solubility of ${CaCO}_{3}$ be s1 and ${CaC}_{2} O_{4}$ be $s_{2}$

$K_{s p_{1}} = s_{1} (s_{1} + s_{2}); K_{s p_{2}} = s_{2} (s_{1} + s_{2})$

$\begin{array}{l} \frac{s_{1}}{s_{2}} = 3 and s_{1} + s_{2} = 8 \times 10^{- 5} \\ \Rightarrow s_{2} = 2 \times 10^{- 5}, s_{1} = 6 \times 10^{- 5} \\ \Rightarrow K_{s p_{1}} = 48 \times 10^{- 10} \end{array}$

Solubility in pure water

$\begin{aligned} = \sqrt{48} \times 10^{- 5} M \\ = 6.93 \times 10^{- 5} \end{aligned}$

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