(C) The solubility product expression for $Mg(OH)_2$ is: $K_{sp} = [Mg^{2+}][OH^-]^2$. Given $K_{sp} = 1.0 \times 10^{-11}$ and $[Mg^{2+}] = 0.001 \,

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(C) The solubility product expression for $Mg(OH)_2$ is: $K_{sp} = [Mg^{2+}][OH^-]^2$. Given $K_{sp} = 1.0 \times 10^{-11}$ and $[Mg^{2+}] = 0.001 \, M = 10^{-3} \, M$. Substituting the values: $1.0 \times 10^{-11} = (10^{-3})[OH^-]^2$. $[OH^-]^2 = \frac{1.0 \times 10^{-11}}{10^{-3}} = 10^{-8}$. $[OH^-] = \sqrt{10^{-8}} = 10^{-4} \, M$. Now,$pOH = -\log[OH^-] = -\log(10^{-4}) = 4$. Since $pH + pOH = 14$,we have $pH = 14 - 4 = 10$.

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

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At $25 \, ^\circ C$,the solubility product $(K_{sp})$ of $Mg(OH)_2$ is $1.0 \times 10^{-11}$. What is the minimum $pH$ required to precipitate $Mg(OH)_2$ from a $0.001 \, M \, Mg^{2+}$ solution?

A
$8$
B
$9$
C
$10$
D
$11$

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At $25 \, ^\circ C$,the solubility product $(K_{sp})$ of $Mg(OH)_2$ is $1.0 \times 10^{-11}$. What is the minimum $pH$ required to precipitate $Mg(OH)_2$ from a $0.001 \, M \, Mg^{2+}$ solution?

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Silver ions are added to a solution with $[Br^{-}] = [Cl^{-}] = [CO_3^{2-}] = [AsO_4^{3-}] = 0.1 \ M$. Which compound will precipitate with the lowest $[Ag^{+}]$ concentration?

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