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BCECE Medical BCECE Medical Solved Papers-2015

  • question_answer
    When 0.1 M NaOH is titrated with 0.1 M, 20 mL HA till the end point, \[{{K}_{a}}(HA)=6\times {{10}^{-6}}\] and degree of dissociation of HA is small as compared to unity. What is the pH of the resulting solution at the end point?

    A)  6.23              

    B)  9.22

    C)  7.21              

    D)  8.95

    Correct Answer: D

    Solution :

    \[\underset{2}{\mathop{NaOH}}\,+\underset{2}{\mathop{HA}}\,\xrightarrow{{}}\underset{-}{\mathop{NaA}}\,+H\underset{-}{\mathop{_{2}}}\,O\] At end point \[=0.1\times 20=2\] \[\because \,\,20\,\,mL\] of \[NaOH\] is required for the complete neutralisation of HA. NaA is a salt of strong base and weak acid. Thus, will undergo hydrolysis and solution will become basic.                 \[C=[NaA]=\frac{2}{20+20}=0.05\,M\] and        \[p{{K}_{a}}=-\log \,(6\times {{10}^{-6}})=5.2\] pH at the end point \[=7+\frac{1}{2}\,(p{{K}_{a}}+\log \,C)\] \[7+\frac{1}{2}(5.2+\log \,0.05)=8.95\]


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