Polyprotic Acids and Stepwise Dissociation

A polyprotic acid can donate more than one proton, and it does so in steps, each with its own dissociation constant (K_a1, K_a2, …), where each later step is much weaker than the one before. In brief: phosphoric acid, H₃PO₄, loses protons in three steps with pK_a values of about 2.15, 7.20 and 12.35; in a 0.100 mol/L solution nearly all the H⁺ comes from the first step, giving pH 1.63.

Sulfuric, carbonic, phosphoric, oxalic and citric acids are polyprotic. Their behavior explains the chemistry of blood, natural waters, fertilizers and food. This article describes stepwise dissociation, tabulates constants, shows how to calculate pH and species concentrations, explains why each step is weaker, describes titrations and distribution of species, works through examples and gives some history.

How does stepwise dissociation work?

Phosphoric acid dissociates in three consecutive equilibria:

1. H₃PO₄ ⇌ H⁺ + H₂PO₄⁻ (K_a1)
2. H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻ (K_a2)
3. HPO₄²⁻ ⇌ H⁺ + PO₄³⁻ (K_a3)

Each step has its own expression, such as K_a2 = [H⁺][HPO₄²⁻] ÷ [H₂PO₄⁻]. The intermediate species H₂PO₄⁻ and HPO₄²⁻ are amphiprotic.

Constants of common polyprotic acids

Why is each step weaker?

It is harder to remove a positive proton from an ion that is already negative, because of electrostatic attraction. K_a2 is usually smaller than K_a1 by a factor of 10⁴ to 10⁶. For phosphoric acid the ratios are 10⁵ between steps 1 and 2 and again between steps 2 and 3.

How do you calculate the pH?

When K_a1 is much larger than K_a2 (by more than about 1,000), treat the acid as monoprotic using K_a1: the second step adds a negligible amount of H⁺. Then [A²⁻] ≈ K_a2. For sulfuric acid, the first step is complete, and the second must be included.

Which species dominate at a given pH?

At pH equal to a pK_a, the acid and its conjugate base are present in equal amounts.

Worked examples

1. 0.100 mol/L H₃PO₄. K_a1 = 7.1 × 10⁻³. Solve x² + 7.1 × 10⁻³x − 7.1 × 10⁻⁴ = 0: x = 0.0233 mol/L, pH 1.63. The second step adds only about 6 × 10⁻⁸ mol/L.
2. 0.100 mol/L H₂CO₃. K_a1 = 4.5 × 10⁻⁷: [H⁺] = √(4.5 × 10⁻⁸) = 2.1 × 10⁻⁴ mol/L, pH 3.67. [CO₃²⁻] ≈ K_a2 = 4.7 × 10⁻¹¹ mol/L.
3. Amphiprotic ion. The pH of NaHCO₃ solution is about (pK_a1 + pK_a2) ÷ 2 = (6.35 + 10.33) ÷ 2 = 8.34.
4. Sulfuric acid, 0.100 mol/L. The first step gives 0.100 mol/L H⁺. The second (K_a2 = 1.0 × 10⁻²) with initial [H⁺] = 0.100: x(0.100 + x) ÷ (0.100 − x) = 0.010 gives x = 0.0084. [H⁺] = 0.1084 mol/L, pH 0.96.
5. Neutralization steps. Titrating 25.0 mL of 0.100 mol/L H₃PO₄ with 0.100 mol/L NaOH, the first equivalence point is at 25.0 mL and the second at 50.0 mL. At the first the main species is H₂PO₄⁻ (pH ≈ (2.15 + 7.20) ÷ 2 = 4.68), and at the second HPO₄²⁻ (pH ≈ (7.20 + 12.35) ÷ 2 = 9.78).

6. Mole ratio. Complete neutralization of 0.0250 mol of H₃PO₄ needs 0.0750 mol of NaOH.

Where are polyprotic acids important?

• Blood: the H₂CO₃/HCO₃⁻ system controls pH.
• Cells: phosphate buffers and ATP (derived from phosphoric acid).
• Food: citric, phosphoric and tartaric acids give flavor and preserve foods.
• Industry: sulfuric and phosphoric acids are among the most produced chemicals.

Thomas Graham distinguished mono-, di- and tribasic acids in the 1830s by their reaction with bases, and the stepwise dissociation constants were measured in the late 19th and early 20th centuries by conductivity and pH methods.

Which mistakes are common?

Mistake 1: Using the total hydrogen content as [H⁺]

Only the first step contributes much.

Mistake 2: Treating sulfuric acid as weak in the first step

Its first ionization is complete.

Mistake 3: Assuming that [A²⁻] equals [H⁺]

In a diprotic acid solution, [A²⁻] is approximately K_a2, not [H⁺].

Mistake 4: Forgetting that amphiprotic ions are both acids and bases

Use the average of the two pK_a values for their pH.