JEE MAIN - Chemistry (2025 - 24th January Morning Shift - No. 9)
$\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{Cr}(\mathrm{OH})_3$ is $1.6 \times 10^{-30}$. What is the molar solubility of this salt in water?
$\sqrt[5]{1.8 \times 10^{-30}}$
$\frac{1.8 \times 10^{-30}}{27}$
$\sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}$
$\sqrt[2]{1.6 \times 10^{-30}}$
Explanation
$\mathrm{Cr}(\mathrm{OH})_{3(\mathrm{~s})} \rightleftharpoons \mathop {\mathrm{Cr_{(aq)}^{ + 3}}}\limits_s + \mathop {\mathrm{3OH_{(aq)}^ - }}\limits_{3s}$
At eq :
$$\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=(\mathrm{s}) \cdot(3 \mathrm{~s})^3=27 \mathrm{~s}^4 \\ & 27 \mathrm{~s}^4=1.6 \times 10^{-30} \\ & \mathrm{~s}=\left(\frac{1.6}{27} \times 10^{-30}\right)^{1 / 4} \end{aligned}$$
Option (1)
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