A) \[Mg{{(OH)}_{2}}\] will be precipitated
B) \[Mg{{(OH)}_{2}}\] is not precipitated
C) \[Mg{{\left( OH \right)}_{3}}\] will be precipitated
D) \[Mg{{\left( OH \right)}_{3}}\] is not precipitated
Correct Answer: B
Solution :
[b] \[pH=9;\left[ {{H}^{+}} \right]={{10}^{-9}};\left[ O{{H}^{-}} \right]={{10}^{-5}};\] \[[M{{g}^{2+}}]=1\times {{10}^{-3}};[M{{g}^{2+}}]{{[O{{H}^{-}}]}^{2}}=1\times {{10}^{-13}}\] given \[{{K}_{sp}}\] of \[Mg{{\left( OH \right)}_{2}}=8.9\times {{10}^{-12}}\] which is more than\[1\times {{10}^{-13}}\]. Hence \[Mg{{(OH)}_{2}}\] will not precipitate
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