Buffer can arise when acid and salt (externally added) or acid and conjugate base (salt formed from neutralisation reaction) are present. In books, buffer reaction is simply explained by:
HA + OH- <=> H2O + A-;
A- + H+ <=> HA
in the presence of external salt added.
But when H3PO4 is titrated with KOH, without addition of salt, buffer properties arise, as in the three pKa's of H3PO4:
H3PO4 + 3KOH <=> 3K+ + PO4- + 3H2O where the salt formed from neutralisation reaction completely ionise in aqueous solution and thus buffer is formed in three series of reactions. Since KOH is strong base, H3PO4 is weak base, KOH completely ionise. So potassium can be omitted: H3PO4 + OH- <=> H2PO4- + H2O, H2PO4- + OH- <=> HPO4(2-) + H2O, HPO4(2-) + OH- <=> PO4(3-) + H2O
Everyone knows when protons or hydroxyl ions are added into buffer, they are counter-reacted. I need an advance explanation of why buffer is most resistant to pH when [salt or conjugate base] = [acid] please. Please be aware that I know the henderson-hasselbalch equation, and people tend to use that for explanation instead.
When no external salt is added (neutralisation reaction), I also found that when proton is devoured by the hydroxyl ion, the acid would continue to shift its equilibrium to RHS to dissociate to form more proton, while the conjugate base accumulates. However, in phosphoric acid with pKa's of 7.20 and 12.37, the amount of proton and conjugate base produced is far too little to contribute to proton concentration, thus has little effect on pH. I guess, guess that proton present even in high pH is from H3PO4 which has much higher pKa of 2.15, thus contributing significant amount of proton. Am I right?