The {K_{sp}} of Mg(OH)_2 is 1 times 10^{-12}. | vedclass

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Question

(B) The dissociation equilibrium is: $Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-$The solubility product expression is: ${K_{sp}} = [Mg^{2+}][OH^-]^2$

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) The dissociation equilibrium is: $Mg(OH)_2 ightleftharpoons Mg^{2+} + 2OH^-$The solubility product expression is: ${K_{sp}} = [Mg^{2+}][OH^-]^2$Given ${K_{sp}} = 1 	imes 10^{-12}$ and $[Mg^{2+}] = 0.01 \ M = 10^{-2} \ M$.Substituting the values: $1 	imes 10^{-12} = (10^{-2}) 	imes [OH^-]^2$$[OH^-]^2 = 10^{-10} \Rightarrow [OH^-] = 10^{-5} \ M$Using the ionic product of water: $[H^+][OH^-] = 10^{-14}$$[H^+] = 10^{-14} / 10^{-5} = 10^{-9} \ M$$pH = -\log[H^+] = -\log(10^{-9}) = 9$

Formula	Solubility product $(K_{sp})$
$AgBrO_3$	$2.5 \times 10^{-10}$
$MgSO_4$	$2.7 \times 10^{-27}$
$CaCO_3$	$2.7 \times 10^{-18}$
$FeC_2O_4$	$2.4 \times 10^{-15}$

The ${K_{sp}}$ of $Mg(OH)_2$ is $1 \times 10^{-12}$. At what limiting $pH$ will $0.01 \ M$ $Mg(OH)_2$ precipitate?

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