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NEET Sample Paper NEET Sample Test Paper-40

  • question_answer
    The dissociation constant 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) \[3.0\times {{10}^{4}}\]

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

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

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

    Correct Answer: A

    Solution :

    \[C{{H}_{3}}COOH\rightleftharpoons C{{H}_{3}}COO{{H}^{-}}+{{H}^{+}}\] \[K{{a}_{1}}=1.5\times {{10}^{-5}}=\frac{[C{{H}_{3}}CO{{O}^{-}}][{{H}^{+}}]}{[C{{H}_{3}}COOH]}\] \[HCN\rightleftharpoons {{H}^{+}}+C{{N}^{-}}\] \[{{K}_{{{a}_{2}}}}=4.5\times {{10}^{-10}}=\frac{[{{H}^{+}}][C{{N}^{-}}]}{[HCN]}\]                 (ii) By (i)(ii) \[K=\frac{{{K}_{{{a}_{1}}}}}{{{K}_{{{a}_{2}}}}}=\frac{[HCN][C{{H}_{2}}CO{{O}^{-}}]}{[C{{N}^{-}}][C{{H}_{3}}COOH]}\] \[=\frac{1.5\times {{10}^{-5}}}{4.5\times {{10}^{-10}}}=3\times {{10}^{4}}\]


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