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JCECE Medical JCECE Medical Solved Paper-2003

  • question_answer
    The degree of hydrolysis of \[\text{0}\text{.01 M N}{{\text{H}}_{\text{4}}}\text{C1}\]is:\[({{K}_{h}}=2.5\times {{10}^{-9}})\]

    A)  \[5\times {{10}^{-5}}\]

    B) \[5\times {{10}^{-4}}\]

    C)  \[5\times {{10}^{-3}}\]

    D) \[5\times {{10}^{-7}}\]

    Correct Answer: B

    Solution :

     Key Idea: Degree of hydrolysis \[(h)=\sqrt{\frac{{{K}_{h}}}{C}}\] where\[{{K}_{h}}=\]hydrolysis constant\[=2.5\times {{10}^{-9}}\] C = concentration = 0.01 M \[\therefore \] \[h=\sqrt{\frac{2.5\times {{10}^{-9}}}{0.01}}\] \[=5\times {{10}^{-4}}\]


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