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MGIMS WARDHA MGIMS WARDHA Solved Paper-2015

  • question_answer
    Equivalent conductance of 0.0128 N acetic acid solution, at\[25{}^\circ C\]is 14 \[oh{{m}^{-1}}e{{q}^{-1}}\]and at infinite dilution is 391\[oh{{m}^{-1}}e{{q}^{-1}}\]. The degree of dissociation of acetic acid in the solution and its dissociation constant, respectively are

    A)  0.036 and \[1.66\times {{10}^{-5}}\]

    B)  0.121 and \[1.78\times {{10}^{-4}}\]

    C)  0.213 and \[1.84\times {{10}^{-5}}\]

    D)  0.054 and \[1.94\times {{10}^{-4}}\]

    Correct Answer: A

    Solution :

                    Given,                   \[{{\lambda }_{V}}=14oh{{m}^{-1}}e{{q}^{-1}}\] \[{{\lambda }_{\infty }}=391oh{{m}^{-1}}e{{q}^{-1}}\] \[C=0.0128N\] \[\therefore \]  \[\alpha =\frac{{{\lambda }_{v}}}{{{\lambda }_{\infty }}}=\frac{14}{391}=0.036\] Now,     \[{{K}_{a}}=C{{\alpha }^{2}}\]                 \[=0.0128\times {{(0.036)}^{2}}\]                 \[=1.66\times {{10}^{-5}}\]


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