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WB JEE Medical WB JEE Medical Solved Paper-2015

  • question_answer
    The ratio of volumes of \[\text{C}{{\text{H}}_{\text{3}}}\text{COOH}\,\text{0}\text{.1}\,\text{(N)}\]to \[\text{C}{{\text{H}}_{\text{3}}}\text{COONa}\,\,\text{0}\text{.1(N)}\]required to prepare a buffer solution of pH 5.74 is (Given \[p{{K}_{a}}\]of \[C{{H}_{3}}COOH\]is 4.74)

    A)  10: 1  

    B)  5:1   

    C)  1 :5   

    D)  1 : 10

    Correct Answer: D

    Solution :

     Given, that \[pH=5.74,\,p{{k}_{a}}=4.74\] Suppose that volume of acid solution = x L Volume of salt solution = y L From Henderson equation, \[pH=p{{K}_{a}}+\log \frac{[Salt]}{[Acid]}\] or, \[pH-pK=\log \frac{[C{{H}_{3}}COONa]}{[C{{H}_{3}}COOH]}\] or, \[5.74-4.74=1=\log \frac{[C{{H}_{3}}COONa]}{[C{{H}_{3}}COOH]}\] or \[\frac{[C{{H}_{3}}COONa]}{[C{{H}_{3}}COOH]}=10\] or \[\frac{[C{{H}_{3}}COOH]}{[C{{H}_{3}}COONa]}=\frac{1}{10}\frac{\frac{0.1x}{x+y}}{\frac{0.1y}{x+y}}\] Thus, \[\frac{x}{y}=\frac{1}{10}\]


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