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Re: Sato question

Kw = (0.25O2(aq) + H+ + e-)/(0.5H2O)

where Kw = -11.385

-11.385 = .25log(O2) + log(H+) + log(e-) - 0.5log(H2O)

or assuming log(H2O) ~ 0

-11.35 = .25log(O2) - pH - pe

> 1) Is "O2" equal to the molar concentration of O2 calculated from the
Dissolved Oxygen in mg/L?

(O2) represents activity of O2, which is close enough to molality not to
worry about it.

> 2) Is "H+ " equal to pH or the activity of hydrogen ion, or something

(H+) is the activity of hydrogen ion, log(H+) = -pH

3) Is "H2O" equal to the activity of water (in which case it can be assumed
to be 1.0)?


4) Is "e-" equal to pe?

(e-) is the conventional activity of the electron, log(e-) = -pe.

But, my own feeling is that the Sato relation is just a kluge. It has the
effect of lowering the calculated pe several units from what you would
calculate from DO. The main reason it was done was to try to reconcile
platinum electrode measurements, which are not very useful except maybe in
the presence of iron and sulfide.


David Parkhurst (dlpark@xxxxxxxx)
U.S. Geological Survey
Box 25046, MS 413
Denver Federal Center
Denver, CO 80225

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