JCECE Medical JCECE Medical Solved Paper-2002

  • question_answer
    0.1M\[C{{H}_{3}}COOH\]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 = ? 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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