A) 3
B) 9
C) 5
D) 8
Correct Answer: B
Solution :
\[Mg{{(OH)}_{2}}M{{g}^{2+}}+2O{{H}^{-}}\] The solubility product, \[{{K}_{sp}}\]of\[Mg{{(OH)}_{2}}={{[Mg]}^{2+}}{{[O{{H}^{-}}]}^{2}}\] \[1\times {{10}^{-12}}=[0.01]{{[O{{H}^{-}}]}^{2}}\] \[{{[O{{H}^{-}}]}^{2}}=\frac{1\times {{10}^{-12}}}{0.01}=1\times {{10}^{-10}}\] \[[O{{H}^{-}}]={{10}^{-5}}\] \[\because \] \[[{{H}^{+}}][O{{H}^{-}}]={{10}^{-14}}\] \[[{{H}^{+}}][{{10}^{-5}}]={{10}^{-14}}\] \[[{{H}^{+}}]=\frac{{{10}^{-14}}}{{{10}^{-5}}}={{10}^{-9}}\] \[pH=-\log [{{H}^{+}}]\] \[=-\log {{10}^{-9}}=9\]
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