RAJASTHAN ­ PET Rajasthan PET Solved Paper-2001

  • question_answer
    If the dissociation constant of two acids are \[2.7\times {{10}^{-4}}\]and\[3\times {{10}^{-5}}\]then the ratio of their relative strength will be

    A)  \[9:1\]              

    B)  \[3:1\]

    C)  \[3:2\]              

    D)  \[9:2\]

    Correct Answer: B

    Solution :

     \[\sqrt{\frac{{{K}_{{{a}_{1}}}}}{{{K}_{{{a}_{2}}}}}}=\frac{{{\alpha }_{1}}{{(acid\text{ }strength)}_{1}}}{{{\alpha }_{2}}{{(acid\text{ }strength)}_{2}}}\] \[=\sqrt{\frac{2.7\times {{10}^{-4}}}{3\times {{10}^{-5}}}}=\frac{{{\alpha }_{1}}}{{{\alpha }_{2}}}\] Hence,   \[\frac{{{\alpha }_{1}}}{{{\alpha }_{2}}}=\frac{3}{1}=3:1\]


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