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NEET AIPMT SOLVED PAPER SCREENING 2010

  • question_answer
    What is \[\text{ }\!\![\!\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\!\!]\!\!\text{ }\] in mol/L of a solution that is 0.20 M in \[\text{C}{{\text{H}}_{\text{3}}}\text{COONa}\] and 0.10 M in \[\text{C}{{\text{H}}_{\text{3}}}\text{COOH}\] ? (Ka for \[\text{C}{{\text{H}}_{\text{3}}}\text{COOH=1}\text{.8}\times \text{1}{{\text{0}}^{\text{-5}}}\])

    A) \[3.5\times {{10}^{-4}}\]                              

    B) \[1.1\times {{10}^{-5}}\]                              

    C) \[1.8\times {{10}^{-5}}\]              

    D)        \[9.0\times {{10}^{-6}}\]

    Correct Answer: D

    Solution :

    Key Idea \[C{{H}_{3}}COOH\](weak acid) and \[C{{H}_{3}}COONa\](conjugated salt) form acidic buffer and for acidic buffer, \[pH=p{{K}_{a}}+\log \frac{[salt]}{acid}\] and        \[[{{H}^{+}}]=-anti\log pH\]                 \[pH=-\log {{K}_{a}}+\log \frac{[salt]}{acid}\] \[[\because p{{K}_{a}}=-\log {{K}_{a}}]\] \[=-\log (1.8\times {{10}^{-5}})+\log \frac{(0.20)}{(0.10)}\] \[=4.74+\log 2\] \[=4.74+0.3010=5.041\] Now,     \[[{{H}^{+}}]=anti\log (-5.045)\]                 \[=9.2\times {{10}^{-6}}mol/L\]


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