Its really important to know how much solution will be applied to know if soil will act as a good "antacid". pH changes under dilution because it is the negative log of the hydrogen ion concentration. Thus amount of free ions is relative to the amount solution applied because concentration is amount per unit volume. in this case given in molarity.
So a pH of 4.7 in 1L of solution works out to:
pH = -log[H+] = 4.7
10^-4.7 = [H+] in Molarity = 1.995x10^-5 M
1.995x10^-5 M = 1.995x10^-5 moles/liter (molarity)
1.995X10-5 moles/liter x 1L = 1.995x10^-5 moles
1mol = 6.022x10^23 molecules (Avogadro's number).
So, rounding. 2x10^-5mol x 6x10^23 molecules/mole = 12x10^18 molecules
=1.2x10^19 molecules or 12,000,000,000,000,000,000 which is like 12 bazillion or something.
if you instead put in a pH of 5.3 you end up somewhere around 3x10^18molecules
or 3 with the same amount of zeros. a 4-fold decrease in number.
pH is on a logarithmic scale which means it works in powers of 10, each change in pH of 1 unit is equivalent to one order of magnitude change (or 10-fold).
It's for this reason I've been recently looking into buffering feeding solutions. AN sells something like this I believe, and to my eyes if it's a simple buffer it is waaaaaaay over priced.
