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JEE Main 20209 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
9 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2020 · 9 January · Evening Shift · Chemistry · Ionic Equilibrium

The solubility product of Cr ( OH ) _ 3 at 298K is 6.0 × 10 ^ - 31...

The solubility product of Cr(OH)3Cr{( OH )}_{3} at 298K298K is 6.0×10−31.6.0 \times {10}^{- 31} . The concentration of hydroxide ions in a saturated solution of Cr(OH)3Cr{\left(OH\right)}_{3} will be

Options

  1. A{\left(2.22 \times {10}^{- 31}\right)}^{\dfrac{1}{4}}
  2. B{\left(18 \times {10}^{- 31}\right)}^{\dfrac{1}{4}}
  3. C{\left(18 \times {10}^{- 31}\right)}^{\dfrac{1}{2}}
  4. D{\left(4.86 \times {10}^{- 29}\right)}^{\dfrac{1}{4}}

Correct answer

B. {\left(18 \times {10}^{- 31}\right)}^{\dfrac{1}{4}}

Step-by-step solution

Cr(OH)3→Cr+3+3O−Cr{\left(OH\right)}_{3} \to C{r}^{+ 3} + 3O{}^{-}
Ksp=s.(3S)3{K}_{sp} = s . {\left(3S\right)}^{3}
⇒6×10−31=27.(3s)4⇒ 6 \times {10}^{- 31} = 27 . {\left(3s\right)}^{4}
⇒s=(627×10−31)1/4⇒ s = {\left(\dfrac{6}{27} \times {10}^{- 31}\right)}^{1 / 4}
\left[O{H}^{-}\right] = 3s<br/><br/>= 3 \times {\left(\dfrac{6}{27} \times {10}^{- 31}\right)}^{14} = {\left(18 \times {10}^{- 31}\right)}^{1 / 4}M
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