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BCECE Engineering BCECE Engineering Solved Paper-2015

  • question_answer
    When 0.1 M \[\text{NaOH}\] 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 :

    \[\underset{2}{\mathop{NaOH}}\,+\underset{2}{\mathop{HA}}\,\to \underset{-}{\mathop{NaA}}\,+\underset{-}{\mathop{{{H}_{2}}O}}\,\] At end point \[\equiv \] 0.1 x 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. \[C=[NaA]=\frac{2}{20+20}=0.05M\]                 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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