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Ka of water

  1. Apr 16, 2017 #1
    Sorry if this question has already been asked but I didn't find it in this forum.

    Which is the acidity constant of water in itself?

    Ka of a weak acid HA in water dilute solution is defined considering the aequilibrium:

    HA + H2O ↔ H3O+ + A- (1)
    Ka(HA) = [H3O+]⋅[A-] / [HA]

    this, for what I know, because in dilute solution [H2O] is constant and can then be "incorporated" in the aequilibrium constant Keq:

    Keq. = [H3O+]⋅[A-] / [HA]⋅[H2O]

    Ka = Keq⋅[H2O] = constant1⋅constant2.

    So, my "natural" guess to how Ka(H2O) can be defined, would be to replace HA with H2O in the aequilibrium (1) and then as follow.
    Let's consider the aequilibrium:

    H2O + H2O ↔ H3O+ + OH-


    Ka(H2O) = [H3O+]⋅[OH-] / [H2O] = Kw/[H2O] = 10-14/(1000/18) = 10-15.7.

    But there is a problem: can I here really consider the acid H2O as in "dilute solution"? It seems meaningless...

    How would you define the acidity constant of water in itself, in case is definible?

  2. jcsd
  3. Apr 16, 2017 #2


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    Staff: Mentor

    "Dilute solution" in this context typically means "water concentration can be assumed to be constant".

    Does it hold for water dissociation? (Hint: compare concentration of dissociation products with concentration of water itself).
  4. Apr 16, 2017 #3
    From your answer I deduce that my definition of Ka(H2O) is correct. Is it?
    Now let's see it from the point of view of activities.
    Concerning [H3O+]and [OH_] there is no problem because they are so small to be certainly equal to their activities. But the activity of water is (or it should be) equal to 1. Is it correct?
    But then Ka(H2O) should be equal to Kw = 10-14 and not 10-15.7.

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