A) 3
B) 9
C) 5
D) 8
Correct Answer: B
Solution :
Key Idea: First find \[[O{{H}^{-}}]\] from \[{{K}_{sp}}\]. Then find \[[{{H}^{+}}]\] by using formula\[[{{H}^{+}}][O{{H}^{-}}]={{10}^{-14}}\]. Finally calculate pH by using following formula. \[pH=-\log [{{H}^{+}}]\] Given, \[{{K}_{sp}}\] \[Mg{{(OH)}_{2}}=1\times {{10}^{-12}}\] Concentration \[Mg{{(OH)}_{2}}=0.01\,M\] \[Mg{{(OH)}_{2}}M{{g}^{2+}}+2O{{H}^{-}}\] \[\therefore \] \[{{K}_{sp}}=[M{{g}^{2+}}+2O{{H}^{-}}\] or \[1\times {{10}^{-12}}=0.01\times {{[O{{H}^{-}}]}^{2}}\] or \[{{[O{{H}^{-}}]}^{2}}=1\times {{10}^{-10}}\] or \[[O{{H}^{-}}]=1\times {{10}^{-5}}\] \[[{{H}^{+}}]\,[O{{H}^{-}}]={{10}^{-14}}\] or \[[{{H}^{+}}]\,[1\times {{10}^{-5}}]={{10}^{-14}}\] or \[[{{H}^{+}}]=\frac{{{10}^{-14}}}{{{10}^{-5}}}={{10}^{-9}}\] \[pH=-\log \,[{{H}^{+}}]\] \[=-\log {{10}^{-9}}\] = 9 \[\therefore \] \[Mg{{(OH)}_{2}}\] will precipitate at limited pH of 9.
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