(C) The solubility product expression for $Mg(OH)_2$ is given by: $K_{sp} = [Mg^{2+}][OH^{-}]^2$ Given $K_{sp} = 5.6 \times 10^{-12}$ and $[Mg^{2+}] =

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(C) The solubility product expression for $Mg(OH)_2$ is given by: $K_{sp} = [Mg^{2+}][OH^{-}]^2$ Given $K_{sp} = 5.6 \times 10^{-12}$ and $[Mg^{2+}] = 10^{-10} \ M$. Substituting the values: $5.6 \times 10^{-12} = (10^{-10}) \times [OH^{-}]^2$ $[OH^{-}]^2 = \frac{5.6 \times 10^{-12}}{10^{-10}} = 5.6 \times 10^{-2}$ $[OH^{-}] = \sqrt{5.6 \times 10^{-2}} \approx 0.2366 \ M \approx 0.24 \ M$ Thus,the concentration of $OH^{-}$ is $0.24 \ M$.

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

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$Mg(OH)_2$ is precipitated when $NaOH$ is added to a solution of $Mg^{2+}$. If the final concentration of $Mg^{2+}$ is $10^{-10} \ M$,the concentration of $OH^{-} \ (M)$ in the solution is (Given: Solubility product for $Mg(OH)_2 = 5.6 \times 10^{-12}$)

KVPY 2017 All KVPY

A
$0.056$
B
$0.12$
C
$0.24$
D
$0.025$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Solubility product

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$Mg(OH)_2$ is precipitated when $NaOH$ is added to a solution of $Mg^{2+}$. If the final concentration of $Mg^{2+}$ is $10^{-10} \ M$,the concentration of $OH^{-} \ (M)$ in the solution is (Given: Solubility product for $Mg(OH)_2 = 5.6 \times 10^{-12}$)

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