(B) The solubility product expression for $Mg(OH)_2$ is: $K_{sp} = [Mg^{+2}][OH^{-}]^2$ Given: $K_{sp} = 9 \times 10^{-12}$ $[Mg^{+2}] = 0.01 \ M = 10

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(B) The solubility product expression for $Mg(OH)_2$ is: $K_{sp} = [Mg^{+2}][OH^{-}]^2$ Given: $K_{sp} = 9 \times 10^{-12}$ $[Mg^{+2}] = 0.01 \ M = 10^{-2} \ M$ Substituting the values: $9 \times 10^{-12} = (10^{-2}) \times [OH^{-}]^2$ $[OH^{-}]^2 = \frac{9 \times 10^{-12}}{10^{-2}} = 9 \times 10^{-10}$ Taking the square root on both sides: $[OH^{-}] = \sqrt{9 \times 10^{-10}} = 3 \times 10^{-5} \ M$

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Solubility product

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The solubility product of $Mg(OH)_2$ is $9 \times 10^{-12}$. If the concentration of $Mg^{+2}$ is $0.01 \ M$,then the concentration of $OH^{-}$ is = .....

A
$9 \times 10^{-10}$
B
$3 \times 10^{-5}$
C
$9 \times 10^{-12}$
D
$3 \times 10^{-7}$

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The solubility product of $Mg(OH)_2$ is $9 \times 10^{-12}$. If the concentration of $Mg^{+2}$ is $0.01 \ M$,then the concentration of $OH^{-}$ is = .....

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Assertion : Addition of silver ions to a mixture of aqueous sodium chloride and sodium bromide solution will first precipitate $AgBr$ rather than $AgCl$.
Reason : $K_{sp}$ of $AgCl < K_{sp}$ of $AgBr$.