SATHEE: Equilibrium Question 543

Equilibrium Question 543

Question: 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 pH [DPMT 2005]

Options:

A) 3

B) 9

C) 5

D) 8

Show Answer

Answer:

Correct Answer: B

Solution:

$M g {(O H)}_{2}$ ⇌ $M g^{2 +} + 2 O H^{-}$

$K_{s p} = [M g^{2 +}] {[O H^{-}]}^{2}$

$1 \times 10^{- 12} = 0.01, {[O H^{-}]}^{2}$

${[O H^{-}]}^{2} = 1 \times 10^{- 10}$

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

$[H^{+}] = 10^{- 14} / 10^{- 5} = 10^{9}$

$p H = - \log [H^{+}] = - \log [10^{- 9}] = 9$

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Equilibrium Question 543

Question: The Ksp of Mg(OH)2 is 1×10−12,,0.01M,Mg(OH)2 will precipitate at the limiting pH [DPMT 2005]

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Question: 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 pH [DPMT 2005]