Sodium hydroxide reacts with phosphoric acid to remove sequentially two acidic hydrogens, as shown below. Although some PO43– is present beyond the second equivalence point, OH– is not a strong enough base to remove the acidic hydrogen completely. Reactions (1) and (2) are both product-favoured, and therefore the major phosphorus-containing species depends on how much OH– has been added.
The Ka (or Ka's) in which the major species in solution appear are used to calculate the concentrations of minor species at various points during the titration described above (including H3O+ and therefore pH).
*For solutions containing a single solute which has an amphiprotic anion,
(1) H3PO4 + OH– 
H2PO4– + H2O
(2) H2PO4– + OH– HPO42– + H2O
The Ka expressions for the three conjugate pairs derived from phosphoric acid are shown below. H2PO4– + H2O(2) H2PO4– + OH–
HPO42– + H2O| K1 = | [H2PO4–][H3O+] | K2 = | [HPO42–][H3O+] | |||
| [H3PO4] | [H2PO4–] |
| K3 = | [PO43–][H3O+] | |
| [HPO42–] |
The Ka (or Ka's) in which the major species in solution appear are used to calculate the concentrations of minor species at various points during the titration described above (including H3O+ and therefore pH).
| major species | use to calculate pH | |
At start: | H3PO4 acid only | pK1(H3PO4) |
| first buffer region: | H3PO4/H2PO4–, Na+ acid/conj base | pK1(H3PO4) |
| first equivalence: | H2PO4–, Na+ amphiprotic | ½(pK1 + pK2)* |
| second buffer region: | H2PO4–/HPO42–, Na+ acid/conj base | pK2(H3PO4) |
| second equivalence: | HPO42–, Na+ amphiprotic | ½(pK2 + pK3)* |
*For solutions containing a single solute which has an amphiprotic anion,
the pH of the aqueous solution is independent of the concentration of the solute and is given by
pH = ½(pK1 + pK2)
pK1 and pK2 are pKa for the two conjugate pairs involving the amphiprotic species