J & K CET Medical J & K - CET Medical Solved Paper-2012

  • question_answer
    pH of 0.0002 M formic acid\[[{{K}_{a}}=2\times {{10}^{-4}}]\] approximately is

    A)  1.35             

    B)  0.5

    C)  3.7               

    D)  1.85

    Correct Answer: C

    Solution :

     \[{{[{{H}^{+}}]}_{HCOOH}}=\sqrt{{{K}_{a}}.C}\] \[=\sqrt{2\times {{10}^{-4}}\times 0.0002}\] \[pH=-log[{{H}^{+}}]=-\log (2\times {{10}^{-4}})\] \[=3.69-3.7\]


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