$\mathrm{K}_{\text {sp }}$ for $\mathrm{Cr}(\mathrm{OH})_3$ is $1.6 \times 10^{-30}$. What is the molar…

$\mathrm{K}_{\text {sp }}$ for $\mathrm{Cr}(\mathrm{OH})_3$ is $1.6 \times 10^{-30}$. What is the molar solubility of this salt in water?
  1. $\frac{1.8 \times 10^{-30}}{27}$
  2. $\sqrt[5]{1.8 \times 10^{-30}}$
  3. $\sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}$
  4. $\sqrt[2]{1.6 \times 10^{-30}}$

Solution

$\begin{aligned} & \mathrm{Cr}(\mathrm{OH})_3 \rightleftharpoons \mathrm{Cr}_5^{3+}(\mathrm{aq})+\underset{3 \mathrm{~s}}{3 \mathrm{OH}^{-}(\mathrm{aq})} \\ & \mathrm{K}_{\mathrm{sp}}=\mathrm{s}(3 \mathrm{~s})^3 \\ & 1.6 \times 10^{-30}=27 \mathrm{~s}^4 \\ & \sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}=\mathrm{s}\end{aligned}$

Asked in: JEE Main 2025 (24 Jan Shift 1)

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