Polyprotic Acid Equilibria

Introduction

An acid with more than one transferable proton does not usually release every proton in one identical step. Each proton transfer has its own conjugate pair and equilibrium constant. The intermediate species can remain appreciable and may itself act as an acid or a base.

Core explanation

For a diprotic acid H₂A, first dissociation is H₂A + H₂O ⇌ H₃O⁺ + HA⁻ with Ka1. Second is HA⁻ + H₂O ⇌ H₃O⁺ + A²⁻ with Ka2. The species HA⁻ is the conjugate base of H₂A and the conjugate acid of A²⁻. Writing both steps explicitly prevents a mistaken “two hydroniums per H₂A instantly” assumption.

In many common polyprotic systems, Ka1 is substantially larger than Ka2. After losing one proton, the negatively charged intermediate often holds the next proton more strongly relative to the solvent context. This is a broad tendency, not a substitute for measured constants. The actual distribution of H₂A, HA⁻ and A²⁻ depends on pH and both Ka values.

If Ka2 is much smaller than Ka1 and conditions are appropriate, the first step dominates hydronium production in an acid-only solution. A simple first-step weak-acid calculation may then approximate pH. But a problem asking for A²⁻ concentration, buffer behavior or a pH near the second dissociation region must include the second equilibrium. A hierarchy of constants helps choose an approximation, which should be checked.

For phosphoric acid H₃PO₄, three sequential proton donations create H₂PO₄⁻, HPO₄²⁻ and PO₄³⁻, with Ka1, Ka2 and Ka3. Every step changes charge and uses a different starting acid species. The total analytical phosphorus concentration is shared among all these forms, giving a mass-balance equation in advanced calculations.

Polyprotic acid behavior also connects to titration. Distinct steps can create multiple buffer regions and equivalence points if their pKa values are separated enough for the features to resolve. Strong overlap can obscure the stages, so one should not promise a visibly separate titration jump for every proton.

Step-by-step reasoning

1. Remove one proton at a time and write each conjugate pair. 2. Assign Ka1, Ka2 and further constants to their specific steps. 3. Compare magnitudes and pH to decide which forms dominate. 4. Use total-component mass balance if multiple forms are quantitatively important.

Visual explanation

Draw a ladder H₂A → HA⁻ → A²⁻. Put one H₃O⁺ produced beside each downward step and a separate Ka label on each arrow.

Real-world analogy

A person giving away two objects does so in two separate transactions. After the first, the person's situation changes, so the second transaction need not be equally easy.

Real-world example

Phosphate species in aqueous chemistry shift with pH. H₂PO₄⁻ and HPO₄²⁻ form a conjugate pair important in buffer formulations and biological aqueous systems. Their ratio depends on solution pH.

Why?

Why assign two Ka values to H₂A? The second proton is donated by HA⁻, a different chemical species with a different charge and equilibrium balance.

Common misconception

“Diprotic acid means its molarity always produces twice as much hydronium.” The second dissociation may be partial and must be treated separately.

Worked example

Write both dissociations for H₂CO₃. First: H₂CO₃ + H₂O ⇌ H₃O⁺ + HCO₃⁻, governed by Ka1 in a chosen model. Second: HCO₃⁻ + H₂O ⇌ H₃O⁺ + CO₃²⁻, governed by Ka2. If Ka2 is much smaller than Ka1, bicarbonate can dominate over carbonate in an appropriately acidic region; exact fractions require pH and constants.

Quick check

1. Which species donates the proton in the second dissociation of H₂A? Answer: HA⁻, not the original H₂A species directly.

Exam focus

Write each step and conjugate pair before calculation. Do not assume every proton is fully released or every titration step is separately visible.

Advanced insight

Distribution fractions for a diprotic acid can be derived from Ka1, Ka2 and [H₃O⁺] using a shared denominator. These fractions sum to one and connect acid speciation with charge and mass balance.

Summary

Polyprotic acids dissociate one proton at a time, with distinct conjugate pairs and Ka values. Species distribution depends on pH and the relative step constants.

Practice questions

1. What is the intermediate after H₂A loses one proton? Answer: HA⁻. 2. Is Ka2 the constant for H₂A directly losing both protons at once? Answer: No. It describes HA⁻ losing the second proton. 3. Can an intermediate such as HCO₃⁻ act both as acid and base? Answer: Yes. It can donate H⁺ to form CO₃²⁻ or accept H⁺ to form H₂CO₃.