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NEET 2019
Exam
NEET
Year
2019
Subject
Chemistry
Chapter
Ionic Equilibrium

NEET 2019 · Chemistry · Ionic Equilibrium

pH of a saturated solution of Ca (OH) _ 2 is 9 . The solubility...

pH of a saturated solution of Ca(OH)2Ca{\left(OH\right)}_{2} is 99 . The solubility product \left({K}_{sp}\right) of Ca{\left(OH\right)}_{2} is:

Options

  1. A0.5×10−150.5 \times {10}^{- 15}
  2. B0.25×10−100.25 \times {10}^{- 10}
  3. C0.125×10−150.125 \times {10}^{- 15}
  4. D0.5×10−100.5 \times {10}^{- 10}

Correct answer

A. 0.5×10−150.5 \times {10}^{- 15}

Step-by-step solution

Ca{\left(OH\right)}_{2} ⇌ \mathop{C{a}^{+ 2}}_{\left(S\right)} + \mathop{2O{H}^{-}}_{\left(2S\right)}<br/><br />pHofsaturatedsolutionof saturated solution= 9<br/><br />pH = 9, \left[{H}^{+}\right] = {10}^{- 9}<br/>Hence[OH−]=10−1410−9=10−5<br />Hence \left[O{H}^{-}\right] = \dfrac{{10}^{- 14}}{{10}^{- 9}} = {10}^{- 5}
2s=10−5,s=10−522s = {10}^{- 5}, s = \dfrac{{10}^{- 5}}{2}
ksp{k}_{sp} for Ca(OH)2⇒4s3Ca{\left(OH\right)}_{2} ⇒ 4{s}^{3}
⇒4×(10−52)3⇒ 4 \times {\left(\dfrac{{10}^{-5}}{2}\right)}^{3}
=10−152=0.5×10−15= \dfrac{{10}^{-1 5}}{2} = 0.5 \times {10}^{- 15}
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