AFMC AFMC Solved Paper-2005

  • question_answer
    If 0.1M of a weak acid is taken and its percentage of degree of ionization is 1.34%, then its ionization constant will be:

    A) \[0.8\times {{10}^{-5}}\]                              

    B) \[1.79\times {{10}^{-5}}\]

    C) \[0.182\times {{10}^{-5}}\]                         

    D) None of these

    Correct Answer: B

    Solution :

     According to Ostwald's dilution law\[{{K}_{a}}=\frac{C{{\alpha }^{2}}}{1-\alpha }\]where, \[{{K}_{a}}=\]equilibrium constant for weak acid C = concentration \[\alpha =\]degree of ionisation of weak acid (HA) For very weak acid \[\alpha <<<<1\]or \[1-\alpha \simeq 1\] \[{{K}_{a}}=C{{\alpha }^{2}}\] or                               \[\alpha =\sqrt{\frac{{{K}_{a}}}{C}}\] \[C=0.1\,M,\,\alpha =1.34%=0.0134\] \[{{K}_{a}}=0.1\times {{(1.34\times {{10}^{-2}})}^{2}}\] \[=1.79\times {{10}^{-5}}\]


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