Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2005

  • question_answer
    A monoprotic acid in a \[0.1\text{ }M\]solution ionises to \[0.001%\]. Its ionisation constant is

    A)  \[1.0\times {{10}^{-3}}\]

    B)  \[1.0\times {{10}^{-6}}\]

    C)  \[1.0\times {{10}^{-8}}\]    

    D)  \[1.0\times {{10}^{-11}}\]

    Correct Answer: D

    Solution :

    \[{{K}_{a}}=C{{\alpha }^{2}}\] \[C=0.1M\] \[\alpha =0.001%=\frac{0.001}{100}\] \[{{K}_{a}}=\frac{0.1\times 0.001\times 0.001}{100\times 100}\] \[{{K}_{a}}=1.0\times {{10}^{-11}}\]


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