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JEE Main & Advanced Sample Paper JEE Main Sample Paper-27

  • question_answer
    \[0.1\text{ }M\] acetic acid solution is titrated against \[0.1\text{ }M\] \[NaOH\] solution What would be the difference in \[\mathbf{pH}\] between \[1/4\] and \[3/4\] stages of neutralisation of acid?

    A)  \[2\text{ }log\text{ }3/4\]          

    B)  \[2\text{ }log\text{ }1/4\]

    C)  \[log\text{ }1/3\]                       

    D)  \[2\text{ }log\text{ }3\]

    Correct Answer: D

    Solution :

    \[\underset{V=1}{\mathop{C{{H}_{3}}}}\,COOH+\underset{0.1}{\mathop{Na}}\,OH\,\rightleftharpoons \underset{0.1}{\mathop{C{{H}_{3}}}}\,COONa+{{H}_{2}}O\] \[\frac{1}{4}\] th stage      0.75       0.75              0.25 \[pH\,=pKa\,+\log \,\frac{(C{{H}_{3}}COONa)}{(C{{H}_{3}}\,COOH)}\,=pKa\,+\log \left( \frac{1}{3} \right)\] \[\underset{V=1}{\mathop{C{{H}_{3}}}}\,COOH+\underset{0.1}{\mathop{NaOH}}\,\,\underset{0.1}{\mathop{\rightleftharpoons }}\,C{{H}_{3}}COONa\,+{{H}_{2}}O\] \[\frac{3}{4}\] th stage         0.75  0.75                0.25 \[p'H\,=pKa+\log \,\left( \frac{3}{1} \right)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\Delta pH\,=\log \,3-\log \left( \frac{1}{3} \right)\,=2\log \,3\].


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