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VMMC VMMC Medical Solved Paper-2007

  • question_answer
    The dissociation constant of two acids \[\text{H}{{\text{A}}_{\text{1}}}\] and \[\text{H}{{\text{A}}_{2}}\] are \[3.0\times {{10}^{-4}}\] and \[1.8\times {{10}^{-5}}\] respectively. The relative strengths of the acids is

    A) 1 : 16                                     

    B) 1 : 4                       

    C) 4 : 1                       

    D)        16 : 1

    Correct Answer: C

    Solution :

    Key Idea: \[\frac{strength\,of\,acid\,H{{A}_{1}}}{Strength\,of\,acid\,H{{A}_{2}}}=\sqrt{\frac{{{K}_{1}}}{{{K}_{2}}}}\] Given: dissociation constant for \[H{{A}_{1}}\] \[=3.0\times {{10}^{-4}}\] Dissociation constant for \[H{{A}_{2}}=1.8\times {{10}^{-5}}\] \[\therefore \] \[\frac{\text{stregth}\,\text{ofacid}\,\text{H}{{\text{A}}_{\text{1}}}}{\text{strength}\,\text{of}\,\text{acid}\,\text{H}{{\text{A}}_{\text{2}}}}\text{=}\sqrt{\frac{3.0\times {{10}^{-4}}}{1.8\times {{10}^{-5}}}}\]                                 \[=\sqrt{\frac{16.66}{1}}=\frac{4}{1}\] \[\therefore \] strength of acid\[\text{H}{{\text{A}}_{\text{1}}}\text{:}\]   strength of acid \[\text{H}{{\text{A}}_{2}}=4:1\]


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