Ionic EquilibriumMedium
Question
The solubility product of Mg(OH)2 is 9.0 × 10−12. The pH of an aqueous saturated solution of Mg(OH)2 is (log 1.8 = 0.26, log 3 = 0.48)
Options
A.3.58
B.10.42
C.3.88
D.6.76
Solution
$Mg(OH)_{2}(s) \rightleftharpoons \underset{SM}{Mg^{2 +}} + \underset{2SM}{2OH^{-}}$
$\text{Now, }K_{sp} = 4S^{3} \Rightarrow S = \left( \frac{9 \times 10^{- 12}}{4} \right)^{1/3}M$
$\text{Now, }\left\lbrack OH^{-} \right\rbrack = 2S = 2 \times \left( \frac{9 \times 10^{- 12}}{4} \right)^{1/3} = \left( 18 \times 10^{- 12} \right)^{1/3}M $$$\therefore P^{OH} = - {\log\left( 18 \times 10^{- 12} \right)}^{1/3} = 3.58 \Rightarrow P^{H} = 10.42$$
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