JEE Main & Advanced Sample Paper JEE Main Sample Paper-47

  • question_answer
    A solution of acetic acid (0.1 M) is titrated against a solution of NaOH (0.5M). Calculate the change in pH between one fourth and three fourth stages of neutralization of the acid

    A)  0.9542                       

    B)  0.5924           

    C)  0.9245                       

    D)  0.5294

    Correct Answer: A

    Solution :

    Once NaOH is added, the solution becomes a buffer. \[C{{H}_{3}}COOH+NaOH\xrightarrow{\,}\,C{{H}_{3}}COONa\] \[+{{H}_{2}}O\] 0.1 M               0.5 M               0 When \[\frac{1}{4}\]  \[0.1\frac{-0.1\times 1}{4}\]   \[\frac{0.1\times 1}{4}\] is neutralized When \[\frac{3}{4}\]  \[0.1-\frac{0.1\times 3}{4}\]   \[\frac{0.1\times 3}{4}\] Is neutralized \[{{(pH)}_{1/4}}=p{{K}_{a}}+\log \,\frac{[C{{H}_{3}}COONa]}{[C{{H}_{3}}COOH]}\] \[=p{{K}_{a}}+\log \,\frac{\frac{0.1}{4}}{\frac{0.4-0.1}{4}}\] \[=pKa+\log \,\frac{1}{3}\] Similarly, \[{{(pH)}_{3/4}}=p{{K}_{a}}+\log \,\frac{\frac{0.3}{4}}{\frac{0.4-0.3}{4}}\] \[=p{{K}_{a}}+\log \,\frac{3}{1}\] \[\therefore \]    \[\Delta pH\] (by Henderson?s equation) \[=\,\log \,\left( \frac{3}{1} \right)\,-\log \,\left( \frac{1}{3} \right)=2\,\log \,3\] \[=2\times 0.4771=0.9542\]

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