Monoprotic acids, like hydrofluoric acid, contain a single ionizable proton. In contrast, polyprotic acids contain two or more ionizable protons. For example, sulfurous acid has two ionizable protons, and phosphoric acid has three.

A polyprotic acid loses each of its protons sequentially, and each reaction has its own K_a.

It is easier to remove a proton from a neutral molecule than a negatively charged one because the negative charge increases the strength of the bond between the proton and the anion.

Therefore, for phosphoric acid, the K_a for the removal of the first proton, K_a1, is higher than the second one, K_a2, which is higher than K_a3.

The pH of a polyprotic acid can be estimated using only the first reaction if K_a1 is at least one thousand times larger than its subsequent K_a’s.

For example, the pH of a 0.050 molar ascorbic acid solution can be determined using its K_a1 and an ICE table.

When dissolved in water, ascorbic acid dissociates into hydronium and ascorbate ions. The K_a1 for this reaction is 8 × 10⁻⁵, and it is equal to the concentration of hydronium times the concentration of ascorbate monoanion, divided by the concentration of ascorbic acid.

An ICE table can be prepared for this reaction with the initial and equilibrium concentrations. Due to the small value of x, 0.050 minus x is approximately equal to 0.050.

Substituting these values into the expression for K_a1, the value for x is equal to 0.0020 molar, which is only 4% of the initial concentration of ascorbic acid. Thus, the approximation is valid.

The pH of the solution equals 2.70.

The concentration of ascorbate dianion formed in the second step of the reaction can also be calculated using K_a2 and an ICE table.

K_a2 is equal to 1.6 × 10⁻¹², and it can be expressed as the concentration of hydronium times the concentration of ascorbate dianion divided by the concentration of ascorbate monoanion.

For this reaction, the ICE table is filled in with the initial concentration of ascorbate monoanion and hydronium from the first reaction, 0.002 molar. Due to the small value of x, it can be omitted from the equilibrium concentrations of ascorbate monoanion and hydronium.

After substituting these values in the K_a2 expression and solving, x is 1.6 ×10⁻¹² molar.

Since x is less than 5% of 0.002 molar, the approximation is valid.

As the concentration of hydronium ions that were generated during the second step of ascorbic acid dissociation is negligible, the first ionizable proton determines the pH of the solution.