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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Critical Thinking

  • question_answer
    The dissociation constant of a weak acid is \[1.0\times {{10}^{-5}},\] the equilibrium constant for the reaction with strong base is [MP PMT 1990]

    A)                 \[1.0\times {{10}^{-5}}\]              

    B)                 \[1.0\times {{10}^{-9}}\]

    C)                  \[1.0\times {{10}^{9}}\]               

    D)                  \[1.0\times {{10}^{14}}\]

    Correct Answer: C

    Solution :

               HA ⇌\[{{H}^{+}}+{{A}^{-}}\]; \[{{K}_{a}}=\frac{[{{H}^{+}}]\,\,[{{A}^{-}}]}{[HA]}\]           ??(i)            neutralization of the weak acid with strong base is            \[HA+O{{H}^{-}}\]⇌ \[{{A}^{-}}+{{H}_{2}}O\]            \[K=\frac{[{{A}^{-}}]}{[HA]\,\,[O{{H}^{-}}]}\]                                           ??(ii)            dividing (i) by (ii) \[\frac{{{K}_{a}}}{K}=[{{H}^{+}}]\,\,[O{{H}^{-}}]\]\[={{K}_{w}}={{10}^{-14}}\]                                 \[K=\frac{{{K}_{a}}}{{{K}_{w}}}=\frac{{{10}^{-5}}}{{{10}^{-14}}}={{10}^{9}}\].


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