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BCECE Engineering BCECE Engineering Solved Paper-2013

  • question_answer
    Calculate pH change when 0.01 mol \[\text{C}{{\text{H}}_{\text{3}}}\text{KOO}\,\text{Na}\]solution is added to 1L of \[\text{0}\text{.01}\,\text{M}\,\text{C}{{\text{H}}_{\text{3}}}\text{COOH}\]solution. \[{{K}_{a}}(C{{H}_{3}}COOH)=1.8\times {{10}^{-5}},p{{K}_{a}}=4.74\]

    A) 3.37                 

    B)        1.37                       

    C) 4.74                

    D) 8.01

    Correct Answer: B

    Solution :

    pH of \[\text{0}\text{.01}\,\text{M}\,\text{C}{{\text{H}}_{\text{3}}}\text{COOH}\] \[pH=\frac{1}{2}[p{{K}_{a}}-log\,C]\] \[=\frac{1}{2}(4.74-log\,0.01)\]                 \[=\frac{1}{2}(4.74\,+\,2)\] \[=3.37\] When \[\text{0}\text{.01}\,\text{mol}\,\text{C}{{\text{H}}_{\text{3}}}\text{COONa}\]is added to it, it is now a buffer and \[[[C{{H}_{3}}COONa]=0.01]\,M.\] Now from                 \[pH+p{{K}_{a}}+\log \frac{[C{{H}_{3}}COO]}{[C{{H}_{3}}COOH]}\]                                 \[=4.74+\log \frac{0.01}{0.01}=4.74\] \[\therefore \] Change in \[pH=4.74-3.37=1.37\]


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