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    Weak Electrolyte undergoes Ionization

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    Water is a very weak electrolyte that undergoes the ionization-

    H2O= H+(aq) + OH-(aq)

    If k1= 2.4x 10-5 s-1 and k-1 = 1.3 x 10 to the eleventh/M*s

    What is the equilibrium constant K where K= [H=] [OH-]/[H2O] ?

    What is the product [H+][OH-] ?

    What is [H+] ?

    What is [OH-] ?

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    Solution Preview

    H2O -------> H+(aq) + OH-(aq)

    The rate of the forward reaction = k1[H2O]

    The rate of the reverse reaction = k-1[H+][OH-]

    At equilibrium, the two rates must be equal.

    Therefore, at equilibrium:

    k1[H2O] = k-1[H+][OH-]

    Rearranging, we get:

    k1/k-1 = [H+][OH-]/[H2O] = Kd

    Kd is the ...

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