Q23 / 180 · KCET 2024 · Chemistry
Solubility product of $\mathrm{CaC}_2 \mathrm{O}_4$ at a given temperature in pure water is $4 \times 10^{-9}\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^2$. Solubility of $\mathrm{CaC}_2 \mathrm{O}_4$ at the same temperature is
- Correct answerA$6.3 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}$
- B$2 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}$
- C$2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$
- D$6.3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$
Solution
Given,
$K_{\text {sp }}=4 \times 10^{-9}\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^2$
$S=$ ?
$\mathrm{CaC}_2 \mathrm{O}_4 \longrightarrow \mathrm{Ca}^{2+}+\mathrm{C}_2 \mathrm{O}_4^{2-}$
$K_{\mathrm{sp}}=(S)(S)$
$4 \times 10^{-9}=S^2$
$\Rightarrow \quad S=\sqrt{4 \times 10^{-9}}$
$=6.3 \times 10^{-5} \mathrm{~mol} / \mathrm{L}$
Thus, the solubility of $\mathrm{CaC}_2 \mathrm{O}_4$ is $6.3 \times 10^{-5} \mathrm{~mol} / \mathrm{L}$.
Hence, the correct answer is option: 'A'.