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MGIMS WARDHA MGIMS WARDHA Solved Paper-2006

  • question_answer
    The\[{{K}_{sp}}\]of\[Mg{{(OH)}_{2}}\]is\[1\times {{10}^{-12}}.\text{ }0.01\text{ }M\] \[Mg{{(OH)}_{2}}\]will precipitate at the limited pH:

    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 \[[{{H}^{+}}][O{{H}^{-}}]={{10}^{-14}}\] \[[{{H}^{+}}][{{10}^{-5}}]={{10}^{-14}}\]                 \[[{{H}^{+}}]={{10}^{-14}}/{{10}^{-5}}={{10}^{-9}}\]                 \[pH=-\log [{{H}^{+}}]=-\log {{10}^{-9}}=9\]                


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