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JEE Main 202125 FebruaryEvening Shift
Exam
JEE Main
Year
2021
Date
25 February
Shift
Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2021 · 25 February · Evening Shift · Chemistry · Ionic Equilibrium

The solubility of Ca (OH) _ 2 in water is : [Given : The solubility...

The solubility of Ca(OH)2Ca{\left(OH\right)}_{2} in water is :
[Given : The solubility product of Ca(OH)2Ca{\left(OH\right)}_{2} in water =5.5×10−6= 5 . 5 \times {10}^{- 6}]

Options

  1. A1.77×10−21 . 77 \times {10}^{- 2}
  2. B1.77×10−61 . 77 \times {10}^{- 6}
  3. C1.11×10−21 . 11 \times {10}^{- 2}
  4. D1.11×10−61 . 11 \times {10}^{- 6}

Correct answer

C. 1.11×10−21 . 11 \times {10}^{- 2}

Step-by-step solution

Given that,

The solubility product of Ca(OH)2Ca{\left(OH\right)}_{2} in water =5.5×10−6= 5 . 5 \times {10}^{- 6}

Ca(OH)2⇌Ca2+(aq)s+2OH−(aq)2sCa{\left(OH\right)}_{2} ⇌ \mathop{{Ca}^{2 +}\left(aq\right)}_{s} + \mathop{2{OH}^{-}\left(aq\right)}_{2s}

ksp=s(2s)2⇒5.5×10−6=4s3{k}_{sp} = s{\left(2s\right)}^{2} ⇒ 5 . 5 \times {10}^{- 6} = 4{s}^{3}

4s3=5.5×10−64{s}^{3} = 5 . 5 \times {10}^{- 6}

s=(5.54)1/3×10−2s = {\left(\dfrac{5 . 5}{4}\right)}^{1 / 3} \times {10}^{- 2}
s=1.11×10−2s = 1 . 11 \times {10}^{- 2}

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