JEE MAIN - Chemistry (2022 - 27th July Morning Shift - No. 19)
At $$310 \mathrm{~K}$$, 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 ____________ $$\times 10^{-8}(\mathrm{~mol} / \mathrm{L})^{3}$$. (Give molar mass : $$\mathrm{CaF}_{2}=78 \mathrm{~g} \mathrm{~mol}^{-1}$$)
Answer
0
Explanation
$\mathrm{CaF}_{2} \stackrel{\mathrm{s}}{\rightleftharpoons} \underset{ \mathrm{s}}{\mathrm{Ca}^{2+}}+\underset{2 \mathrm{~s}}{2 \mathrm{~F}^{-}}$
$$ \begin{aligned} \mathrm{K}_{\mathrm{sp}} &=\mathrm{s}(2 \mathrm{~s})^{2} \\\\ &=4 \mathrm{~s}^{3} \end{aligned} $$
Solubility $(\mathrm{s})=2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}$
$=\frac{2 \cdot 34 \times 10^{-3} \times 10}{78}$ mole $/$ lit
$=3 \times 10^{-4} \mathrm{~mole} / \mathrm{lit}$
$\therefore \mathrm{K}_{\mathrm{sp}}=4 \times\left(3 \times 10^{-4}\right)^{3}$
$$ \begin{aligned} &=108 \times 10^{-12} \\\\ &=0.0108 \times 10^{-8}(\mathrm{~mole} / \mathrm{lit})^{3} \end{aligned} $$
$$ \begin{aligned} & \therefore x \approx 0 \end{aligned} $$
$$ \begin{aligned} \mathrm{K}_{\mathrm{sp}} &=\mathrm{s}(2 \mathrm{~s})^{2} \\\\ &=4 \mathrm{~s}^{3} \end{aligned} $$
Solubility $(\mathrm{s})=2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}$
$=\frac{2 \cdot 34 \times 10^{-3} \times 10}{78}$ mole $/$ lit
$=3 \times 10^{-4} \mathrm{~mole} / \mathrm{lit}$
$\therefore \mathrm{K}_{\mathrm{sp}}=4 \times\left(3 \times 10^{-4}\right)^{3}$
$$ \begin{aligned} &=108 \times 10^{-12} \\\\ &=0.0108 \times 10^{-8}(\mathrm{~mole} / \mathrm{lit})^{3} \end{aligned} $$
$$ \begin{aligned} & \therefore x \approx 0 \end{aligned} $$
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