A solution of 0.01 M MgCl_2 will form a prec | vedclass

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(A) The concentration of $Mg^{2+}$ ions is $[Mg^{2+}] = 0.01 \ M = 10^{-2} \ M$.The solubility product expression for $Mg(OH)_2$ is $K_{sp} = [Mg^{2+}

Vedclass · Accepted Answer

$9$

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(A) The concentration of $Mg^{2+}$ ions is $[Mg^{2+}] = 0.01 \ M = 10^{-2} \ M$.The solubility product expression for $Mg(OH)_2$ is $K_{sp} = [Mg^{2+}][OH^-]^2$.Substituting the given values: $1 	imes 10^{-12} = (10^{-2}) 	imes [OH^-]^2$.Solving for $[OH^-]$: $[OH^-]^2 = 10^{-10}$,so $[OH^-] = 10^{-5} \ M$.Calculating $pOH$: $pOH = -\log[OH^-] = -\log(10^{-5}) = 5$.Finally,calculating $pH$: $pH = 14 - pOH = 14 - 5 = 9$.

$A$ solution of $0.01 \ M$ $MgCl_2$ will form a precipitate of $Mg(OH)_2$ at a limiting $pH$ of: (Given: $K_{sp}$ of $Mg(OH)_2 = 1 \times 10^{-12}$)

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