pH of Weak Acid and Weak Base

When a weak acid is dissolved in water, the acid will undergo ionization. The ionization of a weak acid, HA, in water can be expressed as:

$$ HA+H_2O⇄H^++A^- $$
Four species are present in the solution of the weak acid at equilibrium: HA, A⁻, H⁺, and OH⁻. To calculate the concentrations of the four species in the solution, four equations are needed:

1. Equilibrium: Since the forward and backward reactions are reversible, the equilibrium equation for HA and A⁻ is given by:
   $$ (1)\;\;\;K_a=\frac{[H^+][A^-]}{[HA]} $$
   where K^a is the ionization constant. The water equilibrium is given by:
   $$ (2)\;\;\;K_w=[H^+][OH^-]=10^{-14}\;\;at\;25°C $$
2. Material Balance: Since HA and A⁻ coexist in equilibrium, the sum of the concentration of HA and A⁻ in equilibrium must equal the total concentration of the weak acid initially added to the solution, C_a.
   $$ (3)\;\;\;C_a=[HA]+[A^-] $$
3. Electroneutrality: The three ionic species are composed of one positive charge (H⁺) and two negative charges (OH⁻ and A⁻). The sum of [H⁺] must equal the sum of [OH⁻] and [A⁻] as:
   $$ (4)\;\;\;[H^+]=[OH^-]+[A^-] $$

Substituting Equation (3) for [HA] and Equation (4) for [A⁻] into Equation (1) yields:

$$ (5)\;\;\;K_a=\frac{[H^+]([H^+]-[OH^-])}{C_a-([H^+]-[OH^-])} $$

One can calculate the concentration of H⁺ or OH⁻ in equilibrium by the substitution of Equation (2) into Equation (5). The resulting equation becomes a troublesome cubic one, which cannot be solved easily:

$$ [H^+]+K_a-\frac{C_aK_a+K_w}{[H^+]}-\frac{K_aK_w}{[H^+]^2}=0 $$

One can simplify Equation (5) under certain conditions because some concentration terms are assumed to be negligible in comparison to others:

1. [H⁺]>>[OH⁻]. The concentration of H⁺ is much greater than that of OH⁻ because the ionization of HA produces H⁺. However, this assumption is dependent on how strong or weak the weak acid is. The stronger the acid, the more justifiable the assumption will be. Applying the assumption ([H⁺]>>[OH⁻]) to Equation (5) yields:
   $$ (6)\;\;\;K_a=\frac{[H^+]^2}{C_a-[H^+]} $$
   Substituting Equation (2) into Equation (6) yields a quadratic equation:
   $$ [H^+]^2+K_a[H^+]-C_aK_a=0 $$
   and the concentration of H⁺ in equilibrium is given by:
   $$ [H^+]=\frac{-K_a+\sqrt{K_a^2+4K_aC_a}}{2} $$
2. C_a>>[H⁺] Equation (6) can be further simplified under the assumption that the concentration of H⁺ produced is very small compared to the total concentration of the acid initially added to the solution. When C_a>>[H⁺] is applied, Equation (6) becomes:
   $$ K_a=\frac{[H^+]^2}{C_a} $$
   Or
   $$ [H^+]=\sqrt{K_aC_a} $$
3. C_a>>[H⁺]-[OH^–] When a diluted concentration of the weak acid is used, the concentration of H⁺ produced is very small compared to that of water. In this case, the concentration of the acid is much greater than the concentration difference between H⁺ and OH^– (i.e., C_a>>[H⁺]-[OH^–]). Thus Equation (5) becomes:
   $$ K_a=\frac{[H^+]([H^+]-[OH^-])}{C_a-([H^+]-[OH^-])} $$
   Or
   $$ [H^+]=\sqrt{K_aC_a+K_w} $$

One may decide which equations should be used to calculate the concentration of H⁺. In general, one should proceed with the calculation from the simplest one to the most difficult one.

This same basic approach (equilibrium, material balance, and electroneutrality equations) applies to the calculation of the concentration of OH^– in a solution of a weak base at equilibrium.

Reference:

• Kim, C. (2004). Advanced pharmaceutics : physicochemical principles. London: CRC Press LLC.