JEE Main · 2025 · Shift-IeasyIEQ-086

Ksp for Cr(OH)3 is 1.6 10-30. What is the molar solubility of this salt in water? (a) [4]1.6 10-3027 (b) 1.8 10-3027…

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

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

(a) 1.6×10−30274\sqrt[4]{\dfrac{1.6 \times 10^{-30}}{27}} \quad (b) 1.8×10−3027\dfrac{1.8 \times 10^{-30}}{27} \quad (c) 1.8×10−305\sqrt[5]{1.8 \times 10^{-30}} \quad (d) 1.6×10−304\sqrt[4]{1.6 \times 10^{-30}}

Options
  1. a

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

    ✓
  2. b

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

  3. c

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

  4. d

    1.6×10−304\sqrt[4]{1.6 \times 10^{-30}}

Correct Answera

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

Detailed Solution

Dissolution equilibrium:

Cr(OH)3⇌Cr3++3OH−\text{Cr(OH)}_3 \rightleftharpoons \text{Cr}^{3+} + 3\text{OH}^-

Let molar solubility = ss:

  • [Cr3+]=s[\text{Cr}^{3+}] = s
  • [OH−]=3s[\text{OH}^-] = 3s

Expression for KspK_{sp}:

Ksp=[Cr3+][OH−]3=s⋅(3s)3=27s4K_{sp} = [\text{Cr}^{3+}][\text{OH}^-]^3 = s \cdot (3s)^3 = 27s^4

Solving for ss:

s4=Ksp27=1.6×10−3027s^4 = \frac{K_{sp}}{27} = \frac{1.6 \times 10^{-30}}{27}

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

Answer: (a)

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