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J & K CET Engineering J and K - CET Engineering Solved Paper-2015

  • question_answer
    HA is a weak acid. At\[\text{25}{{\,}^{\text{o}}}\text{C,}\]the molar conductivity   of   0.02   M   HA   is \[150\,{{\Omega }^{-1}}c{{m}^{2}}mo{{l}^{-1}}.\]If its\[\text{ }\!\!\Lambda\!\!\text{ }_{\text{m}}^{\text{o}}\] is \[300\,{{\Omega }^{-1}}\,c{{m}^{2}}mo{{l}^{-1}},\]then equilibrium constant of  HA dissociation is

    A)  0.001  

    B)  0.005  

    C)  0.01   

    D)  0.02

    Correct Answer: C

    Solution :

     Given, molar conductivity \[({{\Lambda }_{m}})=300\,{{\Omega }^{-1}}\,c{{m}^{2}}mo{{l}^{-1}}\] Limiting molar conductivity \[({{\Lambda }^{o}}_{m})=150\,{{\Omega }^{-1}}\,c{{m}^{2}}\,mo{{l}^{-1}}\] Concentration \[(C)=0.02\,M\] \[\therefore \]Degree of dissociation \[(\alpha )=\frac{{{\Lambda }_{m}}}{{{\Lambda }^{o}}_{m}}\] \[\alpha =\frac{150}{300}=0.5\] Now, by using following reaction, we can determine the value of its dissociation constant \[{{K}_{a}}=\frac{C{{\alpha }^{2}}}{(1-\alpha )}\] \[{{K}_{a}}=\frac{0.02\times {{(0.5)}^{2}}}{(1-0.5)}\] \[=\frac{0.02\times 0.5\times .5}{0.5}=0.01\]


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