JCECE Engineering JCECE Engineering Solved Paper-2002

  • question_answer
    \[0.1\,\,M\,\,C{{H}_{3}}COOH\] is \[1.3%\] ionised. The dissociation constant of it will be:

    A) \[1.69\times {{10}^{-5}}\]                            

    B) \[1.69\times {{10}^{-6}}\]

    C) \[1.69\times {{10}^{-4}}\]                            

    D)  None of these

    Correct Answer: A

    Solution :

    Key Idea:\[K=C{{\alpha }^{2}}\] where\[K=\]dissociation constant of acid\[K==?\] \[C=\]concentration of acid\[~=0.1\,\,M\]                 \[\alpha =1.3%\]                 \[K=0.1\times {{\left( \frac{1.3}{100} \right)}^{2}}\]                    \[=0.169\times {{10}^{-4}}\]                    \[=1.69\times {{10}^{-5}}\]


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