Solubility solubility product |IONICEQUILIBRIUMJEEMain2022

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JEE MAIN 2022

27-07-2022 S1

Question

At 310 K , the solubility of $\mathrm{CaF}_2$ in water is $2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}$. The solubility product of $\mathrm{CaF}_2$ is $\_\_\_\_$ × $10^{-8}(\mathrm{~mol} / \mathrm{L})^3 \cdot\left(\right.$ Given molar mass : $\left.\mathrm{CaF}_2=78 \mathrm{~g} \mathrm{~mol}^{-1}\right)$

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Solution

Solubility of $\mathrm{CaF}_2=S$ mole $/ \mathrm{L}$ $$ \begin{aligned} S=\frac{2.34 \times 10^{-3}}{0.1 \times 78} & =\frac{2.34}{78} \times 10^{-2}=3 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \\ K_{S R}\left(C_2 F_2\right)=4 S^3 & =4\left(3 \times 10^{-4}\right)^3 \\ & =108 \times 10^{-12}=0.0108 \times 10^{-8}(\mathrm{~mol} / \mathrm{L})^3 \end{aligned} $$

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