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VMMC VMMC Medical Solved Paper-2015

  • question_answer
    The molar conductivity of 0.009 M aqueous solution of a weak acid (HA) is \[\text{0}\text{.005}\,\text{S}\,{{\text{m}}^{\text{2}}}\,\text{mo}{{\text{l}}^{\text{-1}}}\]and the limiting molar conductivity of HA is \[\text{0}\text{.005}\,\text{S}\,{{\text{m}}^{\text{2}}}\,\text{mo}{{\text{l}}^{\text{-1}}}\]at 298 K. Assuming activity coefficients to be unity, the acid dissociation constant \[({{K}_{a}})\] of HA at this temperature is

    A) \[1\times {{10}^{-4}}\]                  

    B) \[0.1\]

    C) \[9\times {{10}^{-4}}\]                  

    D) \[1.1\times {{10}^{-5}}\]

    Correct Answer: A

    Solution :

    \[{{\lambda }_{{{m}^{o}}}}=0.05S{{m}^{2}}mo{{l}^{-1}}\] C= 0.009 m \[{{\lambda }_{m}}=0.05S\,{{m}^{2}}mo{{l}^{-1}}\] \[\alpha =\frac{{{\lambda }_{m}}}{{{\lambda }_{{{m}^{o}}}}}=0.1\] For weak acid, \[{{K}_{a}}=\frac{{{\alpha }^{2}}c}{1-\alpha }=\frac{{{(0.1)}^{2}}\times 0.009}{(1-0.1)}={{10}^{-4}}\]


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