BCECE Engineering BCECE Engineering Solved Paper-2006

  • question_answer
    A monoprotic acid in 1.00 M solution is 0.01% ionized. 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, \[\alpha =0.01%=0.0001=1\times {{10}^{-4}}\] \[V=\frac{1}{C}=\frac{1}{1.0}=1\,L\] \[\therefore \]  \[{{K}_{a}}=\frac{{{\alpha }^{2}}}{V}=\frac{{{(1\times {{10}^{-4}})}^{2}}}{1}=1\times {{10}^{-8}}\]


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