Titrating Weak and Polyprotic Acids

Introduction

Weak acids create buffer regions during titration, and polyprotic acids can create more than one. The curve records both stoichiometry and equilibrium. A half-equivalence pH can estimate a pKₐ, while the spacing of dissociation constants determines whether separate equivalence points can be observed. Treating every polyprotic acid as though all protons behaved identically loses that information.

Core explanation

For a monoprotic weak acid HA titrated by strong base, added OH⁻ converts HA to A⁻. Before equivalence, both forms coexist and form a buffer. The Henderson–Hasselbalch approximation gives pH = pKₐ + log([A⁻]/[HA]) when activities and other acid–base reactions can be neglected. At half-equivalence, roughly half of the starting HA is converted, so [A⁻] ≈ [HA] and pH ≈ pKₐ. The equivalence volume remains determined by initial acid moles, not by the acid's strength.

A diprotic acid H₂A loses protons in steps: H₂A ⇌ H⁺ + HA⁻ with Kₐ₁, then HA⁻ ⇌ H⁺ + A²⁻ with Kₐ₂. Usually Kₐ₁ > Kₐ₂, because removing the second proton from an already negative ion is less favourable. As strong base is added, a first buffer region can involve H₂A/HA⁻, followed by a second involving HA⁻/A²⁻. If the dissociation constants are sufficiently separated and the solution sufficiently concentrated, two distinct pH jumps or inflection regions may be visible. If the constants are close or the sample dilute, the stages can overlap and separate visual endpoints may be poor.

The first stoichiometric equivalence for a diprotic acid occurs after one mole of OH⁻ per mole H₂A; the second after two. At the first, HA⁻ may be amphiprotic, able to donate or accept a proton, so pH is not automatically 7. At the second, A²⁻ is present and may hydrolyse. Different indicators may be needed for the two points, or potentiometric analysis may identify them more reliably.

An acid with several hydrogen atoms is not necessarily polyprotic in a given titration. Acetic acid has four hydrogens in its formula but only the carboxyl hydrogen is appreciably acidic in ordinary aqueous acid–base titration. Count chemically titratable protons at the selected endpoint, not all hydrogens in a molecular formula.

Step-by-step reasoning

1. List stepwise acid equilibria and their Kₐ values or relative strengths. 2. Calculate initial acid moles and the titrant volumes for each stoichiometric proton equivalent. 3. Identify which conjugate pair dominates each buffer region. 4. At half-equivalence, use pH ≈ relevant pKₐ only when the stages are sufficiently distinct. 5. Select a suitable endpoint method and check whether distinct jumps are observable.

Visual explanation

Draw a rising pH curve for a diprotic acid with two possible steep regions. Label first and second equivalence volumes V and 2V. Mark half-way to each equivalence stage as a buffer region associated with pKₐ₁ or pKₐ₂. Draw a second smoother curve with overlapping regions to show that two theoretical proton transfers do not guarantee two visually resolvable endpoints.

Real-world analogy

A diprotic acid is like a person carrying two keys, one easy to hand over and one held tightly. Base removes the first under milder conditions and the second later. If both keys are surrendered over similar conditions, an observer may see one broad transition rather than two neat events.

Real-world example

Phosphoric acid is triprotic, but its titration curve and usable endpoints depend on concentration, titrant and pH range. In quality control, an analyst must specify which proton-equivalent endpoint defines the reported acidity. A volume to the first endpoint and a volume to a later endpoint do not represent the same chemical quantity.

Why?

Why can pH at half-equivalence estimate pKₐ? The reaction has converted approximately equal amounts of weak acid and conjugate base, so their activity ratio is near one. The logarithm of one is zero, leaving pH close to pKₐ. The approximation is less reliable when other dissociation steps or strong dilution substantially perturb the simple pair.

Common misconception

“A diprotic acid always yields two sharp endpoints” ignores overlapping equilibria and measurement limits. Another error is using the second-equivalence base volume as though it corresponds to one OH⁻ per acid molecule; it corresponds to two when both protons are titrated.

Worked example

Start with 25.00 mL of 0.1000 mol L⁻¹ ideal diprotic H₂A and titrate with 0.1000 mol L⁻¹ NaOH. Initial H₂A amount is 0.002500 mol. One equivalent of base takes 0.002500 mol, or 25.00 mL, giving the first stoichiometric equivalence. Two equivalents take 0.005000 mol, or 50.00 mL total, giving the second. The predicted volumes do not prove the pH curve has two easily visible jumps.

Quick check

1. If a simple weak acid is halfway to its first equivalence point, what pH relation is often useful? Answer: The conjugate base and acid amounts are approximately equal, so pH is often close to that acid step's pKₐ, provided other equilibria do not dominate.

Exam focus

Separate chemical proton count, stoichiometric equivalence volumes and observable endpoints. Show one base equivalent per proton removed. Explain buffer regions from conjugate pairs and avoid assigning pH 7 to amphiprotic or weak-base-containing equivalence solutions without calculation.

Advanced insight

For a polyprotic system, pH at a given volume is governed by mass balance, charge balance and all relevant equilibria. Simple Henderson–Hasselbalch formulas can be accurate in well-separated buffer regions but fail near overlapping dissociations or very dilute endpoints. Numerical equilibrium solving becomes useful when high accuracy is needed.

Summary

Weak-acid titration creates a buffer region and a half-equivalence pH near pKₐ. Polyprotic acids undergo stepwise proton loss, leading to multiple stoichiometric equivalence points, though not always separate visible endpoints. Balanced equivalents, dissociation constants and curve shape must be considered together.

Practice questions

1. How much 0.200 mol L⁻¹ NaOH is needed for the second equivalence of 10.00 mL of 0.100 mol L⁻¹ H₂A? Answer: H₂A amount is 0.00100 mol; two equivalents need 0.00200 mol NaOH, or 0.0100 L = 10.0 mL.

2. Why can an ion at first equivalence have pH different from 7? Answer: An intermediate such as HA⁻ may be amphiprotic, acting as both acid and base in water. Its acid and base equilibria determine pH.

3. Why is acetic acid treated as monoprotic despite four hydrogen atoms? Answer: Only the carboxyl-group hydrogen is appreciably donated in ordinary aqueous acid–base titration. The hydrogens bonded to carbon are not titrated under those conditions.