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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2000

  • question_answer
    The dissociation constants of two acid AB and \[A{{B}_{2}}\]are \[3.4\times {{10}^{-4}}\]and \[1.8\times {{10}^{5}}\] respectively. The relative strengths of the acids will approximately be:

    A) \[16:1\]               

    B)        \[1:4\]                  

    C) \[1:16\]               

    D)        \[4:1\]

    Correct Answer: D

    Solution :

    \[\frac{{{x}_{1}}}{{{x}^{2}}}=\sqrt{\frac{K{{a}_{1}}}{K{{a}_{2}}}}\] \[=\sqrt{\frac{3.4\times {{10}^{-4}}}{1.8\times {{10}^{-5}}}}\] \[=\frac{4}{1}=4:1\]


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