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)

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At $25\,^oC,$ the solubility product of $Mg(OH)_2$ is $1.0 \times 10^{-11}.$ 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

$8$

(AIEEE-2010)

Solution

$M g(O H)_{2} \rightleftharpoons M g^{++}+2 O H$

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

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

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

$\therefore p O H=4$

$\therefore p H+p O H=14$

$\therefore p H=10$

Standard 11

Chemistry

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