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Chemistry · Ionic Equilibrium

JEE Main 2025 — 24 January, Morning Shift — Question 31

Ksp 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?

  1. Option A:

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

    Correct
  2. Option B:

    1.8×10−3027\frac{1.8 \times 10^{-30}}{27}

  3. Option C:

    1.8×10−305\sqrt[5]{1.8 \times 10^{-30}}

  4. Option D:

    1.6×10−302\sqrt[2]{1.6 \times 10^{-30}}

Answer: A

Step-by-step solution

Cr(OH)3( s)⇌Cr(aq)+3+3OH(aq)−\mathrm{Cr}(\mathrm{OH})_{3(\mathrm{~s})} \rightleftharpoons \mathrm{Cr}_{(\mathrm{aq})}^{+3}+3 \mathrm{OH}_{(\mathrm{aq})}^{-}

At eq: s3 s\mathrm{s} \quad \quad 3 \mathrm{~s}

Ksp=(s).(3 s)3=27 s4\mathrm{K}_{\mathrm{sp}}=(\mathrm{s}) .(3 \mathrm{~s})^{3}=27 \mathrm{~s}^{4}

27 s4=1.6×10−3027 \mathrm{~s}^{4}=1.6 \times 10^{-30}

s=(1.627×10−30)1/4\mathrm{s}=\left(\frac{1.6}{27} \times 10^{-30}\right)^{1 / 4}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
24 January, Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium
Topic
Sparingly Soluble Salts, Solubility Product & Precipitation Conditions
Ksp for Cr ( OH ) 3 is 1.6 × 10 -30 . What is the molar solubility of… | JEE Main 2025 PYQ with Solution · DhiX AI