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WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    At \[25{{\,}^{o}}C,\] the molar conductance of 0.007 M hydrofluoric acid is 150 mho \[\text{c}{{\text{m}}^{\text{2}}}\,\text{mo}{{\text{l}}^{-1}}\]and its \[A_{m}^{o}=500\,\,\text{mho}\,\text{c}{{\text{m}}^{2}}\text{mo}{{\text{l}}^{-1}}.\]The value of the dissociation constant of the acid at the given concentration at \[25{{\,}^{o}}C\]is

    A)  \[7\times {{10}^{-4}}M\]

    B)  \[7\times {{10}^{-5}}M\]

    C)  \[9\times {{10}^{-3}}M\]

    D)  \[9\times {{10}^{-4}}M\]

    Correct Answer: D

    Solution :

    Degree of dissociation, \[\alpha =\frac{{{\lambda }^{o}}_{c}}{{{\lambda }^{o}}_{m}}=\frac{150}{500}=0.3\] Given,      \[C=0.007M\] Hydrofluoric acid dissociates in the following manner \[HF\xrightarrow{{}}{{H}^{+}}+{{F}^{-}}\] \[\underset{C(1-\alpha )}{\mathop{\underset{C-C\alpha }{\mathop{C}}\,\,\,\,}}\,\,\,\,\,\,\underset{C\alpha }{\mathop{0}}\,\,\,\,\,\,\,\,\,\underset{C\alpha }{\mathop{0}}\,\,\,\,\underset{\text{At}\,\text{time}\,\text{t}}{\mathop{\text{Initially}}}\,\] Dissociation constant, \[{{K}_{a}}=\frac{[{{H}^{+}}][{{F}^{-}}]}{[HF]}=\frac{C\alpha .C\alpha }{C(1-\alpha )}=\frac{C{{\alpha }^{2}}}{(1-\alpha )}\] On substituting values, we get \[{{K}_{a}}=\frac{0.007\times {{(0.3)}^{2}}}{(1-0.3)}=\frac{63\times {{10}^{-3}}\times {{10}^{-2}}}{0.7}\] \[=9\times {{10}^{-4}}M\]


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