The K s p of M g O H 2 is 1 × 10 – 12 . 0.01 M M g O H 2 will precipitate at the limiting p H -

[ M g 2 + ] [ O H − ] 2 = 10 − 12 0.01 × [ O H − ] 2 = 10 − 12 O H – = 10 – 5 M ∴ H + = 10 – 9 M and p H = – log ⁡ 10 – 9 = 9

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Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductEasy2 minPYQ_2020

ChemistryEasysingle choice

The $K_{s p}$ of $M g {(O H)}_{2}$ is $1 \times 10^{- 12}$ . $0.01 M M g {(O H)}_{2}$ will precipitate at the limiting $p H$ -

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Answer:

Solution:

$[M g^{2 +}] {[O H^{-}]}^{2} = 10^{- 12}$

$0.01 \times {[O H^{-}]}^{2} = 10^{- 12}$

$[{O H}^{-}] = 10^{- 5} M$

$∴ [H^{+}] = 10^{- 9} M$ and $p H = - log 10^{- 9} = 9$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: PYQ_2020

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