At 25,^oC , solubility product of Mg(OH)₂ is 1.0×10^{-11} . At which pH , will Mg^{2+} ions start

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6-2.Equilibrium-II (Ionic Equilibrium)

hard

At $25\,^oC$, the solubility product of $Mg(OH)_2$ is $1.0\times10^{-11}$. At which $pH$, will $Mg^{2+}$ ions start precipitating in the form of $Mg(OH)_2$ from a solution of $0.001\, M\, Mg^{2+}$ ions?

A

$9$

B

$10$

C

$11$

D

$8$

(AIIMS-2017)

Solution

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

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

$1.0 \times {10^{ – 11}} = {10^{ – 3}} \times {[O{H^ – }]^2}$

$[O{H^ – }] = \sqrt {\frac{{{{10}^{ – 11}}}}{{{{10}^{ – 3}}}}} = {10^{ – 4}}$

$\therefore \,pOH = 4$

$\therefore pH + pOH = 14$

$\therefore \,pH = 10$

Standard 11

Chemistry

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