Several points about pH. The pH scale is logarithmic so a change of 1 pH (from 7 to 6 for example) is a factor of 10 change in the ions that are controlling the pH. The pH is given by log10(1/{activity of H+}). The activity of the H+ ions is approximately equal to the concentration for our purposes. [H+] is the concentration of disassociated hydrogen ions (mostly present as H3O+ hydronium ions or some other bound form). Therefore a pH of 5 indicates that the hydronium concentration is 10^-5.
The other key species for pH of pure water is the hydroxyl ion OH- ion. A first approximation is that [H+] times [OH-] is 10^-14. As a result at a pH of 7 in pure water the concentration of both ions is 10^-7. In soils other negatively charged ions like carbonate are important in determining the pH and will often neutralize hydronium ions.
I found a nice set of examples for calculating the pH of a mixture at ChemTeam: Two solutions of differing pH are mixed. What is the new pH? and the second example is most closely related to our situation of adding an acid to a basic water supply to reduce the pH, generally to a value below 7 (slightly acidic). The method assumes that the acid and base are both fully dissociated and the only ions are H+ and OH- which is a simplification, but it gets us into the right ball park. Here is a summary of the method using the values I had for my water and vinegar.
Acid Mol H+ = volume (liters) * 10^-pH = 1 x 10^-2.72 = 0.00191 mol
Base Mol OH- = volume (liters) * 10^-(14-pH) = 10 x 10^-(14-8.46) = 2.88x10^-5 mol
Excess H+ = mol H+ - mol OH- = 0.00188 mol H+
Concentration of H+ = mol H+ / total volume = 0.00188 / 11 liters = 0.00017
pH mixture = -log10(H+ concentration) = -log10(0.00017) = 3.76
I of course ran the experiment using 10 ml vinegar and 100 ml tap water and got a pH of 3.26 which is a fair bit lower than predicted. However, I could use the calculations to estimate the final pH and then check a mixture to refine the ratio of acid to water in order to obtain a desired pH. That is always a good idea since water chemistry is going to be more complicated because most of the acidic and basic species will only partially disassociate at the pH levels of interest to growing plants. Here are three quick examples at other ratios using my pH 2.72 vinegar and pH 8.46 water.
1 liter vinegar to 1 liter water – predicted pH = 3.02 (very acidic).
1 liter vinegar to 600 liters of water – predicted pH = 6.54 (close to my target of 6.5)
1 liter vinegar to 200 liters of water - predicted pH = 5.18 (probably fairly good for the blueberry patch)
Marty