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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 $M{g^{2 + }}$ ion, what is the maximum hydroxide ion concentration which could be present without causing the precipitation of $Mg\,{(OH)_2}$
$1.5 \times {10^{ - 7}}\,M$
$3.0 \times {10^{ - 7}}\,M$
$1.5 \times {10^{ - 5}}\,M$
$3.0 \times {10^{ - 5}}\,M$
Solution
(d) $\mathop {Mg{{(OH)}_2}}\limits_{{K_{sp}}} \mathop {}\limits_ = $ $ \rightleftharpoons $ $\mathop {M{g^{ + 2 }}}\limits_{S\,\,\,\,} + \mathop {2O{H^ – }}\limits_{{{(2S)}^2}} $
${K_{sp}} = S \times 4{S^2}$
$\frac{{{K_{sp}}}}{{S \times 4}} = {S^2} = \frac{{9 \times {{10}^{ – 12}}}}{{.010 \times 4}} = 2.25 \times {10^{ – 10}}$
$S = \sqrt {2.25 \times {{10}^{ – 10}}} = 1.5 \times {10^{ – 5}}$ $m/l$
