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J & K CET Medical J & K - CET Medical Solved Paper-2009

  • question_answer
    The solubility product of a sparingly soluble metal hydroxide\[M{{(OH)}^{2}}\]at 298 K is\[5\times {{10}^{-16}}\] \[mo{{l}^{3}}d{{m}^{-9}}\]. The pH value of its aqueous and saturated solution is

    A)  5                                            

    B)  9

    C)  11.5                                      

    D)  2.5

    Correct Answer: B

    Solution :

                    \[M{{(OH)}_{2}}\underset{s}{\mathop{{{M}^{+}}}}\,+\underset{2s}{\mathop{2O{{H}^{-}}}}\,\] \[{{K}_{sp}}=(s){{(2s)}^{2}}=4{{s}^{2}}=5\times {{10}^{-16}}\] \[\therefore \]  \[s=\sqrt[3]{\frac{5\times {{10}^{-16}}}{4}}=5\times {{10}^{-6}}\] Cone. of \[O{{H}^{-}}=2\times 5\times {{10}^{-6}}={{10}^{-5}}mol\text{ }d{{m}^{-3}}\] \[pOH=-\log \,[O{{H}^{-}}]\] \[=-\text{ }\log \text{ }{{10}^{-5}}=5\] \[pH=14-pOH=14-5=9.\]


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