Polyprotic Acid Titration Stages

Introduction

When a monoprotic weak acid is titrated with sodium hydroxide, the pH curve shows one buffer region and one steep rise. A polyprotic acid carries more than one ionisable proton, so the curve can show several buffer regions and several steep rises, one for each proton that is removed in turn. Reading such a curve means tracking which species dominates at every stage. This page builds that species-by-species picture using phosphoric acid and the carbonate system.

Core explanation

Protons are removed one at a time. Because the stepwise acid dissociation constants of a polyprotic acid usually differ by several powers of ten, added hydroxide reacts almost entirely with the strongest acid present before it attacks the next one. For phosphoric acid (pKa₁ = 2.15, pKa₂ = 7.20, pKa₃ = 12.35 at 25 °C) the reactions happen in sequence:

- Stage 1: H₃PO₄ + OH⁻ → H₂PO₄⁻ + H₂O - Stage 2: H₂PO₄⁻ + OH⁻ → HPO₄²⁻ + H₂O - Stage 3: HPO₄²⁻ + OH⁻ → PO₄³⁻ + H₂O

Equivalence volumes are equally spaced. Each stage consumes one mole of hydroxide per mole of acid. If the first equivalence point needs a volume V₁ of base, the second arrives at 2V₁ and the third, in principle, at 3V₁. This equal spacing is a useful check on any experimental curve.

Buffer regions sit between equivalence points. Midway through each stage, the acid and its conjugate base for that step are present in equal amounts, so pH ≈ pKa for that step. The curve is flattest here. For phosphoric acid the second half-equivalence point, at 1.5V₁, gives pH ≈ 7.20.

Equivalence points are amphiprotic solutions. At the first equivalence point almost all the phosphorus is present as H₂PO₄⁻, which can both lose and gain a proton. Its pH is approximately the average of the two neighbouring pKa values: pH ≈ ½(pKa₁ + pKa₂) = ½(2.15 + 7.20) ≈ 4.7. At the second equivalence point HPO₄²⁻ dominates and pH ≈ ½(7.20 + 12.35) ≈ 9.8.

Not every stage is visible. The third proton of phosphoric acid is so weakly acidic (pKa₃ = 12.35) that its removal in water gives no clear jump: the pH is already near the region where hydroxide itself sets the pH, so the curve simply flattens towards the pH of the excess base. In practice, phosphoric acid titrated with sodium hydroxide shows two clear equivalence points, not three.

The first half-equivalence point is only approximate. Phosphoric acid's first step is fairly strong, so at 0.5V₁ a significant extra amount of H₂PO₄⁻ comes from dissociation. The measured pH there is usually a few tenths above pKa₁. Half-equivalence estimates are most reliable for the weaker steps.

Formulae

Equivalence volume for stage n: Vₙ = n × V₁. Half-equivalence pH ≈ pKa of that step. Intermediate equivalence point pH ≈ ½(pKaₙ + pKaₙ₊₁).

Step-by-step reasoning

To predict the shape of a polyprotic titration curve:

1. List the stepwise pKa values in order. 2. Calculate V₁ from the moles of acid and the concentration of base; mark V₁, 2V₁ and 3V₁. 3. Mark half-equivalence points at 0.5V₁, 1.5V₁, 2.5V₁ with pH ≈ pKa for each step. 4. Estimate each intermediate equivalence pH as the mean of the neighbouring pKa values. 5. Decide which jumps are large enough to see: a step with pKa near 12 or above, or two steps with close pKa values, will not give a sharp jump.

Visual explanation

Imagine the pH curve drawn above a stacked bar showing species fractions. Across the first plateau the bar changes from H₃PO₄ to H₂PO₄⁻; at V₁ it is almost pure H₂PO₄⁻. Across the second plateau it changes to HPO₄²⁻, which is nearly pure at 2V₁. Each steep rise on the curve lines up with a moment when one species has almost completely replaced another.

Real-world analogy

Think of emptying a building floor by floor with a single lift. The lift clears the top floor completely before starting on the next. Each "floor cleared" is an equivalence point, and the time spent on each floor is a buffer region. A floor that is very hard to reach, like the third proton of phosphoric acid, is never cleared in the time available.

Real-world example

Quality-control laboratories analysing cola drinks and fertiliser solutions for phosphoric acid content often titrate to the first or second equivalence point with a pH meter. Because the two jumps are well separated, the volume to either one gives the phosphate content, and the ratio of the two volumes checks for interfering acids.

Why?

Why does hydroxide react with H₃PO₄ before H₂PO₄⁻? The equilibrium constant for OH⁻ reacting with an acid is Ka/Kw. For H₃PO₄ this is about 7 × 10¹¹, while for H₂PO₄⁻ it is about 6 × 10⁶. With a factor of about 10⁵ between them, the base overwhelmingly chooses the stronger acid until it is almost used up.

Common misconception

"A triprotic acid always shows three equivalence points." The number of visible jumps depends on the pKa values, not the formula. A step that is too weak, or two steps whose pKa values are too close, produce no separate, sharp inflection.

Worked example

Question: 25.0 cm³ of 0.100 mol dm⁻³ H₃PO₄ is titrated with 0.100 mol dm⁻³ NaOH. Give the volumes and approximate pH at the first and second equivalence points, and the pH at 37.5 cm³.

Reasoning: Moles of H₃PO₄ = 0.0250 × 0.100 = 2.50 × 10⁻³ mol, so V₁ = 25.0 cm³ and V₂ = 50.0 cm³. At V₁, pH ≈ ½(2.15 + 7.20) = 4.68. At V₂, pH ≈ ½(7.20 + 12.35) = 9.78. At 37.5 cm³ we are halfway through stage 2, so [H₂PO₄⁻] = [HPO₄²⁻] and pH ≈ 7.20.

Answer: 25.0 cm³ (pH ≈ 4.7); 50.0 cm³ (pH ≈ 9.8); pH ≈ 7.2 at 37.5 cm³.

Quick check

1. In the titration of H₃PO₄ with NaOH, which species dominates at the second equivalence point? Answer: HPO₄²⁻, because two protons per acid molecule have been removed.

Exam focus

Examiners expect you to label buffer regions, half-equivalence points and equivalence points on a polyprotic curve, and to name the dominant species at each. Remember that intermediate equivalence points use the amphiprotic average of pKa values, and explain why a very weak final step gives no visible jump.

Advanced insight

The sharpness of an intermediate jump is governed by the gap between successive pKa values and by concentration. At high dilution, or when ΔpKa is small, the amphiprotic species disproportionates appreciably, e.g. 2H₂PO₄⁻ ⇌ H₃PO₄ + HPO₄²⁻, and the inflection softens. Full treatment uses the distribution fractions and the charge balance rather than the simple average-of-pKa rule.

Summary

A polyprotic acid loses its protons one stage at a time when titrated with strong base, giving equally spaced equivalence volumes V₁, 2V₁, 3V₁. Buffer regions lie midway between them, where pH ≈ pKa of that step. Intermediate equivalence points contain amphiprotic ions with pH ≈ ½(pKaₙ + pKaₙ₊₁). Very weak final steps give no visible jump.

Practice questions

1. A diprotic acid needs 18.0 cm³ of NaOH to reach its first equivalence point. At what volume is the second? Answer: 36.0 cm³, because each proton requires the same amount of base. 2. For carbonic acid (pKa₁ = 6.35, pKa₂ = 10.33), estimate the pH at the first equivalence point. Answer: pH ≈ ½(6.35 + 10.33) ≈ 8.3, where HCO₃⁻ dominates. 3. Why is no third equivalence point seen when phosphoric acid is titrated with aqueous NaOH? Answer: pKa₃ is 12.35, so HPO₄²⁻ is too weak an acid; the pH near the third stage is already controlled by excess hydroxide and no sharp jump appears. 4. What is the approximate pH halfway to the second equivalence point of a phosphoric acid titration, and why? Answer: About 7.2, because equal amounts of H₂PO₄⁻ and HPO₄²⁻ are present, so pH = pKa₂.