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 \[Mg{{(OH)}_{2}}=[M{{g}^{2+}}]{{[O{{H}^{-}}]}^{2}}\] \[1\times {{10}^{-12}}=0.01{{[O{{H}^{-}}]}^{2}}\] \[{{[O{{H}^{-}}]}^{2}}=1\times {{10}^{-10}}\] \[{{[O{{H}^{-}}]}^{2}}={{10}^{-5}}\] We know \[[{{H}^{+}}][O{{H}^{-}}]={{10}^{-14}}\] \[[{{H}^{+}}][{{10}^{-5}}]={{10}^{-14}}\] \[[{{H}^{+}}]={{10}^{-14}}/{{10}^{-5}}={{10}^{-9}}\] \[pH=-\log [{{H}^{+}}]=-\log {{10}^{-9}}=9\]You need to login to perform this action.
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