BCECE Medical BCECE Medical Solved Papers-2006

  • question_answer
    A monoprotic acid in 1.00 M solution is 0.01% ionised. The dissociation constant of this acid is :

    A)  \[1\times {{10}^{-8}}\]       

    B)  \[1\times {{10}^{-4}}\]

    C)  \[1\times {{10}^{-6}}\]

    D)  \[1\times {{10}^{-5}}\]

    Correct Answer: A

    Solution :

    For weak electrolytes, according to Ostwalds dilution law                 \[\alpha =\sqrt{KV}\] Here, \[\alpha =0.01%=0.0001=1\times {{10}^{-4}}\]                 \[V=\frac{1}{C}=\frac{1}{1.0}=1\,L\] \[\therefore \] \[{{K}_{a}}=\frac{{{\alpha }^{2}}}{V}=\frac{{{(i\times {{10}^{-4}})}^{2}}}{1}=1\times {{10}^{-8}}\]


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