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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2025 · 24 January · Morning Shift · Chemistry · Ionic Equilibrium

K_ sp for Cr(OH)_3 is 1.6 × 10^ -30 . What is the molar solubility...

Ksp \mathrm{K}_{\text{sp }} for Cr(OH)3\mathrm{Cr}(\mathrm{OH})_3 is 1.6×10−301.6 \times 10^{-30}. What is the molar solubility of this salt in water?

Options

  1. A1.8×10−3027\frac{1.8 \times 10^{-30}}{27}
  2. B1.8×10−305\sqrt[5]{1.8 \times 10^{-30}}
  3. C1.6×10−30274\sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}
  4. D1.6×10−302\sqrt[2]{1.6 \times 10^{-30}}

Correct answer

C. 1.6×10−30274\sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}

Step-by-step solution

Cr(OH)3⇌Cr3+s(aq)+3OH−(aq)3 sKsp=s(3 s)31.6×10−30=27 s41.6×10−30274=s\begin{aligned} & \mathrm{Cr}(\mathrm{OH})_3 \rightleftharpoons \underset{ \mathrm{s}}{\mathrm{Cr}^{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}
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