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JCECE Engineering JCECE Engineering Solved Paper-2015

  • question_answer
    A weak acid HA after treatment with \[1.2\,\,ml\] of \[0.1\,\,M\] strong base has \[\text{a}\]\[pH\]of\[5\]. At the end point, the volume of same base required is\[26.6\,\,mL\]. The value of \[{{K}_{a}}\] is

    A) \[8.2\times {{10}^{-6}}\]                              

    B) \[6.4\times {{10}^{-6}}\]

    C) \[5.3\times {{10}^{-5}}\]                              

    D)  \[2.4\times {{10}^{-6}}\]

    Correct Answer: A

    Solution :

    For complete neutralisation, Total mill equivalent of acid = mill equivalent of base\[=26.6\times 0.1=2.66\] For partial neutralisation \[HA+BOH\xrightarrow[{}]{{}}BA+{{H}_{2}}O\] 2.66   1.2                0       0 before reaction 1.46   0                   1.2    1.2 after reaction The resultant mixture has HA and BA and thus act as buffer. \[\therefore \]  \[pH=-\log {{K}_{a}}+\log \frac{[Salt]}{[Acid]}\] \[5=-\log {{K}_{a}}+\log \frac{1.2}{1.46}=-\log {{K}_{a}}-0.86\] \[\therefore \]  \[\log {{K}_{a}}=-5.86\] or            \[{{K}_{a}}=\text{anti}\,\,\log \,\,(-5.86)=8.21\times {{10}^{-6}}\]


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