Triprotic Acid Speciation

Introduction

A triprotic acid can exist in four protonation states. Citric acid, phosphoric acid and arsenic acid are common examples, and many biological molecules behave similarly. The same logic used for diprotic acids extends naturally: three stepwise equilibria link the four species, and the pH alone fixes how the total is distributed among them. This page builds the general formulas and a fast method for finding the dominant species without a full calculation.

Core explanation

Species and equilibria. For H₃A:

H₃A ⇌ H⁺ + H₂A⁻ (Ka1)

H₂A⁻ ⇌ H⁺ + HA²⁻ (Ka2)

HA²⁻ ⇌ H⁺ + A³⁻ (Ka3)

The mass balance is C = [H₃A] + [H₂A⁻] + [HA²⁻] + [A³⁻].

Alpha fractions. Writing h for [H⁺] and following the same route as for a diprotic acid:

D = h³ + Ka1h² + Ka1Ka2h + Ka1Ka2Ka3

α₀ = h³/D, α₁ = Ka1h²/D, α₂ = Ka1Ka2h/D, α₃ = Ka1Ka2Ka3/D

Each term in D corresponds to one species, and the power of h counts the protons that species still holds.

The ratio shortcut. For any single step, the Henderson–Hasselbalch form gives the ratio of neighbouring species:

[H₂A⁻]/[H₃A] = 10^(pH − pKa1), and similarly for the other steps.

Chaining these ratios gives the relative amounts of all four species, which can then be normalised. This is often quicker than evaluating D directly, and it makes clear why a species is dominant: every ratio pointing towards it is greater than 1.

Dominance ranges. With well-separated pKa values:

- below pKa1: H₃A dominates - between pKa1 and pKa2: H₂A⁻ - between pKa2 and pKa3: HA²⁻ - above pKa3: A³⁻

Citric acid (pKa 3.13, 4.76 and 6.40) is a useful contrast to phosphoric acid because its pKa values are close together. In the pH range 3 to 7, three or even four species coexist in significant amounts, and no single form reaches 90%. That is why citrate buffers work over a broad range.

Average protonation. The mean number of protons still bound per acid unit is 3α₀ + 2α₁ + α₂. It falls smoothly from 3 to 0 as pH rises, and is the quantity followed during a titration.

Formulae

D = h³ + Ka1h² + Ka1Ka2h + Ka1Ka2Ka3; αₙ = (term n)/D; [base]/[acid] = 10^(pH − pKa) for each step; Σα = 1.

Step-by-step reasoning

1. Set the amount of the most protonated species to 1. 2. Multiply by 10^(pH − pKa1) to get the relative amount of H₂A⁻. 3. Multiply that by 10^(pH − pKa2) for HA²⁻, then by 10^(pH − pKa3) for A³⁻. 4. Add all four relative amounts. 5. Divide each by the sum to obtain the fractions.

Visual explanation

A triprotic distribution diagram shows four curves: a falling curve for H₃A, two hills for H₂A⁻ and HA²⁻, and a rising curve for A³⁻. Where pKa values are close, the hills are low and broad and overlap heavily.

Real-world analogy

Picture a relay with four runners passing a baton along a pH track. Each hand-over point is a pKa. If hand-over points are far apart, each runner runs alone for a long stretch; if they are close, several runners are on the track together.

Real-world example

Citric acid is used in soft drinks, jams and effervescent tablets. Its closely spaced pKa values let it buffer across the whole range of fruit acidity, and its partially deprotonated anions bind calcium ions, which is why citrate is also used as an anticoagulant in stored blood.

Why?

Why is the ratio method valid? Each ratio comes directly from one Ka expression at the given h, and neighbouring ratios multiply to link distant species. No approximation is involved, so normalising the chained ratios gives exact alpha fractions.

Common misconception

"A triprotic acid is always found as only one or two species." That is true only when pKa values are widely separated. For citric acid, three species can each exceed 10% at the same pH.

Worked example

Question: Find the fraction of each citric acid species at pH 5.00 (pKa1 = 3.13, pKa2 = 4.76, pKa3 = 6.40).

Reasoning: Set H₃Cit = 1. H₂Cit⁻ = 10^(1.87) = 74.1. HCit²⁻ = 74.1 × 10^(0.24) = 128.8. Cit³⁻ = 128.8 × 10^(−1.40) = 5.13. Sum = 209.0.

Answer: H₃Cit 0.5%, H₂Cit⁻ 35.5%, HCit²⁻ 61.6%, Cit³⁻ 2.5%. HCit²⁻ dominates, but H₂Cit⁻ is also substantial.

Quick check

1. How many terms does the denominator D contain for a triprotic acid, and what does each represent? Answer: Four terms, one for each protonation state, from h³ for H₃A to Ka1Ka2Ka3 for A³⁻.

Exam focus

Examiners expect you to identify the dominant species from pKa values, to calculate a ratio of adjacent species, and occasionally to compute all four fractions. Present the ratio method clearly, with each 10^(pH − pKa) factor shown.

Advanced insight

For molecules such as citric acid, the stepwise constants are macroscopic: they describe how many protons are bound, not which carboxyl group holds them. Different microscopic species with the same charge may coexist, and spectroscopy such as NMR is used to resolve these microscopic constants.

Summary

A triprotic acid has four species linked by three equilibria. Alpha fractions follow from D = h³ + Ka1h² + Ka1Ka2h + Ka1Ka2Ka3, or more quickly from chained ratios 10^(pH − pKa). Widely spaced pKa values give clear dominance ranges; closely spaced values, as in citric acid, give broad regions where several species coexist.

Practice questions

1. Write α₂ for a triprotic acid. Answer: α₂ = Ka1Ka2[H⁺] / ([H⁺]³ + Ka1[H⁺]² + Ka1Ka2[H⁺] + Ka1Ka2Ka3). 2. For citric acid, calculate [HCit²⁻]/[H₂Cit⁻] at pH 4.00. Answer: 10^(4.00 − 4.76) = 0.17. 3. Which citric acid species dominates at pH 8.0? Explain briefly. Answer: Cit³⁻, because pH 8.0 is well above pKa3 = 6.40, so all ratios favour the fully deprotonated form. 4. Explain why citrate buffers work over a wider pH range than a single weak acid buffer. Answer: Its three pKa values lie close together, so overlapping conjugate pairs provide buffering continuously from about pH 2 to 7.5.