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J & K CET Engineering J and K - CET Engineering Solved Paper-2007

  • question_answer
    By applying law of mass action, the equilibrium constant,\[K\] for the reaction \[HA+{{H}_{2}}O\rightleftharpoons {{H}_{3}}{{O}^{+}}+\bar{A},\]is given as

    A) \[K=\frac{[HA][{{H}_{2}}O]}{[{{H}_{3}}{{O}^{+}}][\bar{A}]}\]

    B)  \[K=\frac{[{{H}_{3}}{{O}^{+}}][\bar{A}]}{[HA][{{H}_{2}}O]}\]

    C)  \[K=\frac{[{{H}_{3}}{{O}^{+}}][{{H}_{2}}O]}{[\bar{A}][HA]}\]

    D)  \[K=\frac{[HA][\bar{A}]}{[\bar{A}][HA]}\]

    Correct Answer: B

    Solution :

     \[HA+{{H}_{2}}O\rightleftharpoons {{H}_{3}}{{O}^{+}}+{{A}^{-}}\] \[K=\frac{[{{H}_{3}}{{O}^{+}}][{{A}^{-}}]}{[HA][{{H}_{2}}O]}\]


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