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NEET Sample Paper NEET Sample Test Paper-39

  • question_answer
    The dissociation constants for acetic acid and HCN at \[25{}^\circ C\] are\[1.5\times {{10}^{-5}}\]and \[4.5\times {{10}^{-10}},\] respectively. The equilibrium constant for the equilibrium. \[C{{N}^{-}}+C{{H}_{3}}COOH\rightleftharpoons HCN+C{{H}_{3}}CO{{O}^{-}}\]would be

    A) \[3.0\times {{10}^{5}}\]

    B) \[3.0\times {{10}^{-5}}\]

    C) \[3.0\times {{10}^{-4}}\]

    D) \[3.0\times {{10}^{4}}\]  

    Correct Answer: D

    Solution :

     Given, \[C{{H}_{3}}COOH\rightleftharpoons C{{H}_{3}}O{{O}^{-}}+{{H}^{+}};\] \[{{K}_{a1}}=1.5\times {{10}^{-5}}\] (i) \[{{K}_{a2}}=4.5\times {{10}^{-10}}\] (ii) \[HCN\rightleftharpoons {{H}^{+}}+C{{N}^{-}};\] \[C{{N}^{-}}+C{{H}_{3}}COOH\rightleftharpoons \] \[HCN+C{{H}_{3}}CO{{O}^{-}}\] \[K=?\] On subtracting Eq. (ii) from Eq. (i), we get \[C{{H}_{3}}COOH+C{{N}^{-}}\rightleftharpoons HCN+C{{H}_{3}}CO{{O}^{-}}\] \[K=\frac{{{K}_{a1}}}{{{K}_{a2}}}=\frac{1.5\times {{10}^{-5}}}{4.5\times {{10}^{-10}}}=\frac{{{10}^{5}}}{3}=3.33\times {{10}^{4}}\]


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