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
