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JEE Main 201912 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2019 · 12 April · Morning Shift · Chemistry · Ionic Equilibrium

What is the molar solubility of AI (OH) _ 3 in 0.2 MNaOH solution?...

What is the molar solubility of AI(OH)3AI{\left(OH\right)}_{3} in 0.20.2 MNaOHMNaOH solution? Given that, solubility product of Al(OH)3=2.4×10−24Al{\left(OH\right)}_{3} = 2.4 \times {10}^{- 24} :

Options

  1. A3×10−193 \times {10}^{- 19}
  2. B12×10−2112 \times {10}^{- 21}
  3. C12×10−2312 \times {10}^{- 23}
  4. D3×10−223 \times {10}^{- 22}

Correct answer

D. 3×10−223 \times {10}^{- 22}

Step-by-step solution

Concept of common −ion- ion will be applicable for OH−O{H}^{-} group.

Al(OH)3⇌Al3+S+3(OH−)(3s+0.2)\text{Al}{\left(\text{OH}\right)}_{3} ⇌ \mathop{\text{A}{\text{l}}^{\text{3+}}}_{\text{S}} + \mathop{3\left(\text{O}{\text{H}}^{-}\right)}_{\left(3s + 0.2\right)}

KspKsp for Al(OH)3=2.4×10−24=s(3s+0.2)3Al{\left(OH\right)}_{3} = 2.4 \times {10}^{- 24} = s{\left(3s + 0.2\right)}^{3}

Looking at Ksp,Ksp, it is clear that ss will be much lesser than 0.2.0.2 .

Or, s≪0.2s ≪ 0.2

s×(0.2)3=2.4×10−24⇒s=2.4×10−24/0.008s \times {\left(0.2\right)}^{3} = 2.4 \times {10}^{- 24} ⇒ s = 2.4 \times {10}^{- 24} / 0.008
or, s=3×10−22s = 3 \times {10}^{- 22}

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