Roger Rowlett wrote:
> No. Kw = [H3O+][OH-] = 1 x 10^-14 at 25 deg C.
>
> So at pH 7.0, you have 10^-7 M each at equilibrium no matter how you slice it or whatever
> else is in solution. If equilibrium [H3O+] goes up [OH-] goes down commensurately.
>
> The "pKa" of water as an acid is based on Kw and water's effective concentration of 55 M
> in pure water. This "pKa" is used to compare the instrinsic acidity of water to other weak
> acids. Water is an exceptionally weak acid or base.
>
>
And note that Kw, like other physical constants, depends on temperature,
ionic strength, etc. Therefore neutrality, defined as the concentration
where H+ = OH- , which is half of pKw*, is not exactly 7.0, but varies.
A lot of students come out of first-year chemistry with the idea that
pH 7.0000 is by definition neutral.
(*That is using the old Kw definition where {H2O} is taken as 1)
The pK at 14 (or 14+log(55)) is for H2O <-> OH-, H+
i.e. pH where H2O = OH-
The pK at 0 (or -log 55)) is for H3O+ <-> H2O, H+
i.e. pH ph when H3O+ = H2O
But isn't H+ = H30? then when H3O+ = H2O, [H3O] = 55/2,
pH would be -log (55/2)
pH is log of "activity of water, or whatever the glass electrode measure.
Or- assuming [H2O] always unity, when H2O = H3O+, H3O+ = 1, pH=0
assuming [H2O] always 55, extrapolate to where H3O+ = H2O while keeping H20=55,
then pH would be -log 55
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