Dear Ezzat,

Since H+ is likely to be lognormally distributed in most environments, I believe that the best strategy is to use the geometric mean of H+ (which incidentally turns out to be the same thing as the arithmetic mean of pH), in other words you take the average of the logarithmically transformed H+ activities. This is in line with the IUPAC recommendations (see http://goldbook.iupac.org/G02621.html) and it is also the approach taken in most environmental monitoring programmes.

I don't know the answer, I just wanted to say I never though about that before. I've always used the mean in my measurements, but since pH is measuring H ions on a log scale your approach would seem to make total sense.

Dear Ezzat, I would always go for the anti-log approach. You may want to test it with a drastic example:

The ordinary average of the following pH values: 4.53 / 5.67 / 7.89 is 6.03 but if you transform the pH into H+ concentration this would be 2.95E-05/ 2.14E-06/ 1.23E-08 and the average of this is 1.05E-05. Converted into pH you receive a pH of 4.98.

If your measured pH values are closer together the differences between the two approaches are of course smaller:

mean of pH 4.53/4.67/4.89 = 4.70

mean of H+ conc. 2.95E-05/2.14E-05/1.29E-05 = 2.13E-05 = pH 4.67

Dear Ezzat,

Since H+ is likely to be lognormally distributed in most environments, I believe that the best strategy is to use the geometric mean of H+ (which incidentally turns out to be the same thing as the arithmetic mean of pH), in other words you take the average of the logarithmically transformed H+ activities. This is in line with the IUPAC recommendations (see http://goldbook.iupac.org/G02621.html) and it is also the approach taken in most environmental monitoring programmes.

Dear Jon, that is mean if I used H+ values I should use geometric mean and if I used pH values I should use arithmetic mean? Is it correct? I tried both and I found the final results are the same as you said. It is really interesting. As I know, there is a relationship between the measured pH values and the E values (mV or V) as seen from the known Nernest equation (E = Eo + 0.059 V pH). It means there is a liner relationship of pH value as a function of the electric potential. Therefore, the arithmetic mean is better in such case as pH values here got from electric potential (mV) measured by combined electrode (NOT from measuring H+) and I really confused why some researcher used the average of H+ since they did not measure H+. Thanks again to all of you.

Yes Ezzat, this means that in practice it should be possible to use the arithmetic mean of pH values. I am not sure if the Nernst equation is so relevant here (it is partly a question of cause and effect), i believe the main thing is the expected log-normal distribution of H+.

Dear Jon and Ezzat, I wonder if there is sufficient information available about the distribution of H+ in soils, especially in biologically active zone like the rhizosphere. I know at least one paper where the distribution of the pH was described as normal (this indicates that H+ would have been distributed log-normal, right?). I started to talk with a colleague about this issue and up to now all we would not go for the geometric mean, but we are of course open for discussion.

I forgot to add the example.

I talked with another colleague (very skilled in statistics) and he supported Jon's approach. He also explained to me the reason why the arithmetic mean of the pH is equal to the geometric mean of the H+ concentration.

Thus, thanks to Jon I learned a lot during the last days!

Thanks Stephan for a confirmation. Could you please share why arithmetic mean of pH is equal to the geometric mean of H+. I talked with many colleagues and they gave me different answers but I believe that average pH values is correct as suggested by Jon.

Me too learned from this conservation.

Hi everybody, here is the missing information. I hope that I could present it in a nice and clear way. If not, just let me know.

Thanks Stephan, it is really clear and informative

@ Christoph, I am not sure I understand your question properly, but I would use the geometric standard deviation, a dimension-less number. Those who want to know the boundaries sholud take the geometric mean and divide it by geometric standard deviation (lower boundary) or multiply the geometric mean with geomeric standard deviation (upper boundary).

Thank you guys, 

can we use the mean values in the SPSS output report? 

Hello Everyone!

thank you for helpful discussion  from which  I understand that

arithmeticMean(pH values) = log10(geometricMean(H+ values))

My main confusion is whether IUPAC book is trying to say “the average potential is equivalent to the ARITHMETIC mean concentration”?

The reasons are below:

#1 If arithmeticMean(pH values) = log10(geometricMean(H+ values)) then why to bother with pH transformation to H+ activity?

#2 According to USFDA the correct pH averaging is done by log10(arithmeticMean( H+ values))!

see http://www.fda.gov/ScienceResearch/FieldScience/ucm174295.htm

e.g. R code example

mean(c(4.82,7))  # arithmetic mean of pH values

#[1] 5.91

10^(-c(4.82,7)) # convert pH to H+

#[1] 1.513561e-05 1.000000e-07

-log10((1.513561e-05*1.000000e-07)^(1/2))  # geometric mean of H+ values

#[1] 5.91

-log10(mean(c(1.513561e-05,1.000000e-07))) # arithmetic mean of H+ values

#[1] 5.11817

#3 Can we average bases with acids, or only separate averaging is correct?

Hi Boris, I agree that there is no sense of transformation if

arithmeticMean(pH)=geomean(H+)

The approach I am using is the one described in your link. However, when the measured pH values are very similar (within one pH unit), the difference in averaging is insignificant and this old article is showing some good example:

http://jeb.biologists.org/content/84/1/335.full.pdf

http://jeb.biologists.org/content/84/1/335.full.pdf

Here is a useful tool : http://wgr-sw.com/pH/

I would like a satisfactory answer to this question myself. I think the idea that we would reduce pH to the concentration of hydronium ions is incomplete without also considering the concentration of hydroxide ions. I think it would also depend on how the average is interpreted. I would prefer which ever method is closer to the result of mixing equal portions of two substances being mixed. However in this case we'd also have to consider which substance has the higher buffering capacity. When calculating a weighted average of soil horizons, I prefer an arithmetic average. Sandy soils tend to be acidic but low in cation exchange capacity, whereas the mixing of calcareous loam might be expected to swamp out most of the acidity. On the other hand, if the acidic soil is also high in organic matter, it would take more calcium buffer the pH.

Here is an example of how to calculate it in R.

http://ncss-tech.github.io/stats_for_soil_survey/chapters/4_exploratory_analysis/4_exploratory_analysis.html#61_ph

with example from FAO

https://web.archive.org/web/20170808104311/http://www.fao.org/docrep/field/003/AC175E/AC175E07.htm

"The opinion that pH values must be transformed to [H+] values before averaging appears to be based on the concept of mixing solutions of different pH. In practice, however, the averaging of [H+] values will not provide the correct average pH because buffers present in natural waters have a greater effect on final pH than does dilution alone. For nearly all uses of pH in fisheries and aquaculture, pH values may be averaged directly. When pH data sets are transformed to [H+] to estimate average pH, extreme pH values will distort the average pH. Values of pH conform more closely to a normal distribution than do values of [H+], making the pH values more acceptable for use in statistical analysis."

Boyd, C.E., Tucker, C.S. and Viriyatum, R., 2011. Interpretation of pH, acidity, and alkalinity in aquaculture and fisheries. North American Journal of Aquaculture, 73(4), pp.403-408.

Article Interpretation of pH, Acidity, and Alkalinity in Aquaculture...