A) 3
B) 9
C) 5
D) 8
Correct Answer: B
Solution :
\[Mg{{(OH)}_{2}}\] ⇌ \[M{{g}^{2+}}+2O{{H}^{-}}\] \[{{K}_{sp}}=[M{{g}^{2+}}]{{[O{{H}^{-}}]}^{2}}\] \[1\times {{10}^{-12}}=0.01\,{{[O{{H}^{-}}]}^{2}}\] \[{{[O{{H}^{-}}]}^{2}}=1\times {{10}^{-10}}\] \[\Rightarrow [O{{H}^{-}}]={{10}^{-5}}\] \[[{{H}^{+}}]={{10}^{-14}}/{{10}^{-5}}={{10}^{9}}\] \[pH=-\log [{{H}^{+}}]=-\log [{{10}^{-9}}]=9\]You need to login to perform this action.
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