pe in redox-unbuffered solutions

[dlparkhurst]

Q. To complete an exercise 10 in solutions (online 2011 course notes, ftp://brrftp.cr.usgs.gov/pub/dlpark/geochem/pc/phreeqc/Geochem2011.tgz), I defined two different solutions – one each with 1mmol/kgw of NaHCO3 (I added the carbon as HCO3) and  1mmol/kgw of Na2CO3. I first charge balanced these two solution with pH and then I mixed them in a proportion of 1:1. My results for pH of the final mixed solution were very similar to yours in the is10 output file (MINE: 10.048  YOURS: 10.107), but were very different in pe (MINE: -4.516  YOURS: 4.00).
 
Conceptually what did I do wrong to get such a different pe and a pH slightly off?

A. The pe is not buffered in these solutions. IMPORTANT: pe should be thought of as the ratio of an oxidized to a reduced species (just like pH should be thought of, in natural waters, as the ratio of HCO3-/CO2(aq), higher ratio, higher pH). In your system there are a few ways to think of it: the ratio of O(0)/H2O, C(4)/C(-4), or H2O/H(0). For each of these couples, one of the concentrations is extremely small (O(0), C(-4), and H(0)). Tiny changes in concentration (< 1e-14) greatly affect the ratios, and the resulting pe.

It is a quirk of PHREEQC that the pe is essentially random for these systems up to a point. I will just tell you what happens. The numerical method will adjust the pe until one of these low concentration species is on the order of 1e-14. It may reduce the pe until H(0) is of that order, or it may increase the pe until O(0) is of that order. In this case, H(0) is about 1e-14, resulting in the pe of -4.516. The key is that, in this system, between pe of say -4 and 12, all the concentrations in solution are the same within 1e-14, and they are all the same to PHREEQC within the numerics of double precision.