Derivation of model equations

Before the equivalence point (0⋜ f(i) <1), we have to consider the following chemical equilibria (eq):

HA + H₂O ⇌ A^- + H₃O⁺ for which the acid ionization constant is defined:

Ka=[A^-]_eq [H₃O⁺]_eq/[HA]_eq (1)

and the water auto-ionization

H₂O + H₂O ⇌ H₃O⁺ + OH^- for which the ionic product constant is defined:

Kw=[H₃O⁺]_eq [OH^-]_eq (2)

As the base (NaOH, for example) is strong, it is totally ionized and the OH^-ions from it react with the H₃O⁺ions from the acid. The species electrically charged are A^- , H₃O⁺,OH^- and Na⁺. The total acid concentration in the mixture (AT, see Introduction) is:

AT = [HA]_eq + [A^-]_eq (3)

Additionally, the number of positive charges must be equal to the number of negative ones (electroneutrality condition), so:

BT + [H₃O⁺]_eq =[A^-]_eq + [OH^-]_eq (4)

where BT is the total base concentration in the mixture (see Introduction).

From equations (1), (2), (3) and (4), by simple algebraic manipulation, it is easy to eliminate all [...]_eq, excepting [H₃O⁺]_eq, leading to equation (BEP) of the Introduction:

[H₃O⁺]³ + (Ka+BT) * [H₃O⁺]² + (KA*BT - Ka*AT - Kw)*[H₃O⁺] - Ka*Kw = 0 (BEP)

As for the process at, and after the equivalence point (f(i) ⋝ 1), the salt (NaA), totally ionized, generally undergoes hydrolysis and the base in execess (NaOH) is also assumed completely ionized. Then, the chemical equilibria to be considered are:

A^_- + H₂O ⇌ HA + OH^-

for which the hydrolysis constant is defined:

Kh=[HA]_eq [OH-]_eq/[A-]_eq = Kw/Ka (5)

and, again, the water auto-ionization and respective ionic product constant.

Now, the chemical balances are:

ST=[A-]_eq + [HA]_eq (6)

and

BE+ST+[H3O+]_eq = [OH-]_eq + [A-]_eq (7)

being the last the electroneutrality condition. BE and ST (see Introduction) are, respectively, the concentration of the salt at the equivalence point and the concentration of the base in excess.

Subtracting equations (6) and (7):

[HA]_eq = [OH^-]_eq - [H₃O⁺]_eq - BE (8)

Now, eliminating all [...]_eq but [H₃O⁺]_eq from equations (2), (5), (7) and (8), leads to equation (EPAEP) of the Introduction:

[H₃O⁺]³ + (BE+ST+Ka) * [H₃O⁺]² - (KW-BE *Ka)*[H₃O⁺] - Ka*Kw = 0 (EPAEP)