The solubility product constant K_{sp} of Mg( | vedclass

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Question

(D) The dissolution equilibrium for $Mg(OH)_2$ is given by: $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$.The solubility product expression

Vedclass · Accepted Answer

$3.0 	imes 10^{-5} \ M$

Vedclass · Answer

(D) The dissolution equilibrium for $Mg(OH)_2$ is given by: $Mg(OH)_2(s) ightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$.The solubility product expression is $K_{sp} = [Mg^{2+}][OH^-]^2$.Given $K_{sp} = 9.0 	imes 10^{-12}$ and $[Mg^{2+}] = 0.010 \ M$.Substituting these values into the expression: $9.0 	imes 10^{-12} = (0.010)[OH^-]^2$.Solving for $[OH^-]^2$: $[OH^-]^2 = \frac{9.0 	imes 10^{-12}}{0.010} = 9.0 	imes 10^{-10}$.Taking the square root: $[OH^-] = \sqrt{9.0 	imes 10^{-10}} = 3.0 	imes 10^{-5} \ M$.Therefore,the maximum hydroxide ion concentration is $3.0 	imes 10^{-5} \ M$.

The solubility product constant $K_{sp}$ of $Mg(OH)_2$ is $9.0 \times 10^{-12}.$ If a solution is $0.010 \ M$ with respect to $Mg^{2+}$ ion,what is the maximum hydroxide ion concentration which could be present without causing the precipitation of $Mg(OH)_2$?

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