{K_{sp}} of Mg{(OH)₂} is 1 × {10^{ - 12}},,0.01,M,Mg{(OH)₂} will precipitate at limiting

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6-2.Equilibrium-II (Ionic Equilibrium)

hard

The ${K_{sp}}$ of $Mg{(OH)_2}$ is $1 \times {10^{ - 12}},\,0.01\,M\,Mg{(OH)_2}$ will precipitate at the limiting $pH$

A

$3$

B

$9$

C

$5$

D

$8$

Solution

(b) $Mg{(OH)_2}$ $ \rightleftharpoons $ $M{g^{2 + }} + 2O{H^ – }$

${K_{sp}} = [M{g^{2 + }}]{[O{H^ – }]^2}$

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

${[O{H^ – }]^2} = 1 \times {10^{ – 10}}$ $ \Rightarrow [O{H^ – }] = {10^{ – 5}}$

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

$pH = – \log [{H^ + }] = – \log [{10^{ – 9}}] = 9$

Standard 11

Chemistry

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