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JCECE Engineering JCECE Engineering Solved Paper-2003

  • question_answer
    The degree of hydrolysis of \[0.01\,\,M\,\,N{{H}_{4}}Cl\] 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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