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Chemistry - acids, bases, equilibrium

  1. Feb 25, 2008 #1
    1. The problem statement, all variables and given/known data
    Like all equilibrium constants, Kw varies somewhat with temperature. Given that Kw is 3.49e-13 at some temperature, compute the pH of a neutral aqueous solution at that temperature.

    2. Relevant equations
    Kw = [H+][OH-]
    pH = -log[H+]

    3. The attempt at a solution
    3.49e-13 = [H+]*10e-7
    [H+] = 3.49e-6
    pH = 5.457
  2. jcsd
  3. Feb 26, 2008 #2


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    remember Kw is the ionproduct of water at a given temperature. So {H+}*{OH-}= 3.49e-13 next {H+}*{OH-} can be seen as {H+}e2 so what to do next to gain {H+} from Kw?
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