The minimum $pH$ at which $Mg {(OH)}_{2}$ is precipitated from $0.1 M Mg {({NO}_{3})}_{2}$ solution is (Given: $K_{s} (Mg {(OH)}_{2}) =$ $9 \times 10^{- 15} M^{3})$

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Detailed Solution

$K_{s} (Mg (OH)_{2}) = [{Mg}^{2 +}] {[{OH}^{-}]}^{2} .$ hence $[{OH}^{-}] = {(\frac{K_{s}}{[{Mg}^{2 +}]})}^{1 / 2} = {(\frac{9 \times 10^{- 15} M^{3}}{0.1 M})}^{1 / 2} = 3.0 \times 10^{- 7} M$

$pOH = - \log (3.0 \times 10^{- 7}) = 6.52 pH = 14 - 6.52 = 7.48$

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