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Ionic Equilibrium question

2025 · 24 Jan · Shift 1 · Q9
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  5. /2025 · 24 Jan · Shift 1 · Q9

Ionic Equilibrium question

2025 · 24 Jan · Shift 1 · Q9

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
Ksp\mathrm{K}_{\mathrm{sp}}Ksp​ for Cr(OH)3\mathrm{Cr}(\mathrm{OH})_3Cr(OH)3​ is 1.6×10−301.6 \times 10^{-30}1.6×10−30. What is the molar solubility of this salt in water?
  1. A
    1.8×10−305\sqrt[5]{1.8 \times 10^{-30}}51.8×10−30​
  2. B
    1.8×10−3027\frac{1.8 \times 10^{-30}}{27}271.8×10−30​
  3. C
    1.6×10−30274\sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}4271.6×10−30​​
  4. D
    1.6×10−302\sqrt[2]{1.6 \times 10^{-30}}21.6×10−30​
View written solutionFree

Correct answer: C

  1. Write the dissolution equilibrium
Cr(OH)3(s)⇌Cr3+(aq)+3OH−(aq)\mathrm{Cr(OH)_3(s)} \rightleftharpoons \mathrm{Cr^{3+}(aq)} + 3\mathrm{OH^- (aq)}Cr(OH)3​(s)⇌Cr3+(aq)+3OH−(aq)
  1. Assume molar solubility = sss mol L−1^{-1}−1

Then at equilibrium:

[Cr3+]=s,[OH−]=3s[\mathrm{Cr^{3+}}] = s, \qquad [\mathrm{OH^-}] = 3s[Cr3+]=s,[OH−]=3s
  1. Write the solubility product expression
Ksp=[Cr3+][OH−]3K_{sp} = [\mathrm{Cr^{3+}}][\mathrm{OH^-}]^3Ksp​=[Cr3+][OH−]3

Substitute the concentrations:

Ksp=s(3s)3=s⋅27s3=27s4K_{sp} = s(3s)^3 = s \cdot 27s^3 = 27s^4Ksp​=s(3s)3=s⋅27s3=27s4
  1. Substitute the given value of KspK_{sp}Ksp​
1.6×10−30=27s41.6 \times 10^{-30} = 27s^41.6×10−30=27s4

So,

s4=1.6×10−3027s^4 = \frac{1.6 \times 10^{-30}}{27}s4=271.6×10−30​

Therefore,

s=(1.6×10−3027)1/4s = \left(\frac{1.6 \times 10^{-30}}{27}\right)^{1/4}s=(271.6×10−30​)1/4
  1. Match with the options

This corresponds to:

1.6×10−30274\sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}4271.6×10−30​​

So the correct option is C.

  1. Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

They agree.

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