Polyprotic Acids

Core Idea

Polyprotic acids can donate more than one proton per molecule, ionizing in sequential steps. Each successive ionization has a smaller Ka (Ka₁ >> Ka₂ >> Ka₃) because removing a proton from an increasingly negative species requires more energy. For most polyprotic acids, the pH is determined almost entirely by the first ionization — the second and third contribute negligibly to [H⁺]. Sulfuric acid is a notable exception: its first ionization is strong (complete), so Ka₂ must be used for the second proton. Intermediate species (like HCO₃⁻ or H₂PO₄⁻) are amphoteric — they can act as either acid or base.

How It's Best Learned

Solve the first ionization as a standard weak acid ICE table, then verify that the second ionization's contribution to [H⁺] is negligible (typically [H⁺] from Ka₂ ≈ Ka₂ itself when Ka₁ >> Ka₂). For the pH of an amphoteric intermediate, use the formula pH ≈ ½(pKa₁ + pKa₂).

Common Misconceptions

• Students often try to solve all ionizations simultaneously. Because Ka₁ >> Ka₂, the ionizations can be treated as independent sequential equilibria — the first is solved, and its result becomes the initial condition for the second.
• The concentration of the doubly-deprotonated species is approximately equal to Ka₂ (not dependent on initial concentration), a counterintuitive result that follows directly from the algebra of sequential equilibria.

Explainer

From weak acid ionization, you know how to set up an ICE table for a monoprotic acid like acetic acid: it partially ionizes, and Ka tells you the equilibrium ratio of products to reactant. A polyprotic acid is simply an acid with more than one ionizable proton — diprotic acids like H₂SO₄ and H₂CO₃ can donate two protons, and triprotic acids like H₃PO₄ can donate three. The essential new idea is that these protons come off one at a time, in sequential equilibria, each with its own Ka.

The reason for sequential ionization is electrostatic: after the first proton leaves, the remaining species carries a negative charge. Removing a second proton from a negatively charged ion is harder — you are pulling a positive charge away from something that is already pulling it inward. This is why Ka₁ is always much larger than Ka₂, which is much larger than Ka₃. For phosphoric acid, the ratios are dramatic: Ka₁ = 7.5 × 10⁻³, Ka₂ = 6.2 × 10⁻⁸, Ka₃ = 4.2 × 10⁻¹³. Each successive ionization is roughly 100,000 times weaker than the one before it.

This enormous drop in Ka values leads to a practical simplification: the first ionization dominates the pH calculation. When you dissolve H₃PO₄ in water, the first ionization produces H⁺ and H₂PO₄⁻. You solve this exactly as you would for a monoprotic weak acid using Ka₁ and an ICE table. The H⁺ produced by the first step suppresses the second ionization (Le Chatelier's principle), and since Ka₂ is already tiny, the second ionization contributes a negligible amount of additional H⁺. You can verify this: after solving the first equilibrium, plug the results into the Ka₂ expression and confirm that the additional [H⁺] is insignificant. A useful shortcut emerges from this algebra — the concentration of the doubly-deprotonated species (like HPO₄²⁻) is approximately equal to Ka₂ regardless of the initial acid concentration.

Sulfuric acid is the important exception. Its first ionization is strong (complete dissociation: H₂SO₄ → H⁺ + HSO₄⁻), so Ka₁ is effectively infinite. This means you cannot ignore the second ionization the way you normally would — you must use Ka₂ (1.2 × 10⁻²) to calculate how much additional H⁺ the bisulfate ion contributes, especially in dilute solutions where it matters most. The intermediate species of polyprotic acids — HSO₄⁻, HCO₃⁻, H₂PO₄⁻ — are amphoteric: they can donate a proton (acting as an acid) or accept one (acting as a base). To find the pH of a solution of an amphoteric intermediate, a clean approximation is pH ≈ ½(pKa₁ + pKa₂), which averages the two equilibria that the species participates in.