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JEE Main 201912 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

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

The molar solubility of Cd (OH) _ 2 is 1.84 × 10 ^ - 5 M in water....

The molar solubility of Cd(OH)2Cd{\left(OH\right)}_{2} is 1.84×10−5M1.84 \times {10}^{- 5}M in water. The expectes solunility of Cd(OH)2Cd{\left(OH\right)}_{2} in a buffer solution of pH=12pH = 12 is:

Options

  1. A2.49×10−10M2.49 \times {10}^{- 10}M
  2. B2.491.84×10−9M\dfrac{2.49}{1.84} \times {10}^{- 9}M
  3. C1.84×10−9M1.84 \times {10}^{- 9}M
  4. D6.23×10−11M6.23 \times {10}^{- 11}M

Correct answer

A. 2.49×10−10M2.49 \times {10}^{- 10}M

Step-by-step solution

Ksp=4(s)3{K}_{sp} = 4{\left(s\right)}^{3}
=4×(1.84×10−5)3= 4 \times {\left(1.84 \times {10}^{- 5}\right)}^{3}
CdS′(OH)2⇌Cd+2S′+2OH−(10−2+S′)≃10−2\mathop{Cd}_{{S}^{'}}{\left(OH\right)}_{2} ⇌ \mathop{C{d}^{+ 2}}_{{S}^{'}} + \mathop{2O{H}^{-}}_{\left({10}^{- 2} + {S}^{'}\right) ≃ {10}^{- 2}}
S′×(10−2)2=4×(1.84×10−5)3{S}^{'} \times {\left({10}^{- 2}\right)}^{2} = 4 \times {\left(1.84 \times {10}^{- 5}\right)}^{3}
S′=4×(1.84)3×10−11{S}^{'} = 4 \times {\left(1.84\right)}^{3} \times {10}^{- 11}
(S′)=2.491×10−10M( {S}^{'} ) = 2.491 \times {10}^{- 10}M
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