Diprotic Distribution Fractions

Introduction

How much of a diprotic acid is present as H₂A, as HA⁻ and as A²⁻ at a given pH? The answer does not depend on how much acid was added, only on pH and the two acid constants. Expressing each species as a fraction of the total, an alpha fraction , turns the equilibrium expressions into a compact set of formulas that generate the familiar distribution diagrams used for buffers, titrations and precipitation throughout this unit.

Core explanation

Deriving the fractions. Start from the mass balance C = [H₂A] + [HA⁻] + [A²⁻] and write each species in terms of [H₂A] using the Ka expressions (writing h for [H⁺]):

[HA⁻] = Ka1[H₂A] / h and [A²⁻] = Ka1Ka2[H₂A] / h²

Substituting and multiplying through by h² gives C = H₂A / h². So:

α₀ = [H₂A]/C = h² / D

α₁ = [HA⁻]/C = Ka1h / D

α₂ = [A²⁻]/C = Ka1Ka2 / D

where D = h² + Ka1h + Ka1Ka2. The three fractions always sum to 1.

Reading the formulas. Each term in D belongs to one species: h² to H₂A, Ka1h to HA⁻ and Ka1Ka2 to A²⁻. The largest term tells you which species dominates. At low pH, h is large and h² wins; at high pH, h is tiny and the constant term Ka1Ka2 wins.

Key features of the diagram.

- α₀ = α₁ when h = Ka1, that is at pH = pKa1. - α₁ = α₂ when h = Ka2, that is at pH = pKa2. - α₁ has its maximum at pH = ½(pKa1 + pKa2). If the pKa values are far apart, HA⁻ reaches nearly 100% there; if they are close, the maximum is noticeably lower because H₂A and A²⁻ are both present. - Each fraction changes by about a factor of 10 per pH unit on the far side of a crossing point.

Carbonic acid. Using pKa1 = 6.35 and pKa2 = 10.33, H₂CO₃ (dissolved CO₂) dominates below pH 6.35, HCO₃⁻ between 6.35 and 10.33, and CO₃²⁻ above 10.33. Hydrogencarbonate peaks at pH 8.34.

Formulae

D = [H⁺]² + Ka1[H⁺] + Ka1Ka2; α₀ = [H⁺]²/D; α₁ = Ka1[H⁺]/D; α₂ = Ka1Ka2/D; α₀ + α₁ + α₂ = 1; [species] = α × C.

Step-by-step reasoning

1. Convert pH to h = 10⁻ᵖᴴ. 2. Calculate the three terms h², Ka1h and Ka1Ka2. 3. Add them to get D. 4. Divide each term by D to obtain α₀, α₁ and α₂. 5. Multiply by C if actual concentrations are needed, and check that the fractions sum to 1.

Visual explanation

On a distribution diagram, α₀ starts near 1 at low pH and falls; α₂ starts near 0 and rises to 1 at high pH; α₁ is a hill between them. The hill's sides cross the other curves at 0.5 exactly at pKa1 and pKa2 when those values are well separated.

Real-world analogy

Think of a population in three age groups whose proportions change as a single dial, pH, is turned. The proportions always add to 100%, and the dial setting alone, not the size of the population, decides the split.

Real-world example

Blood carries most of its carbon dioxide as hydrogencarbonate. The alpha fractions at pH 7.4 show why: HCO₃⁻ accounts for over 90% of the dissolved inorganic carbon, with dissolved CO₂ a small share and carbonate almost absent.

Why?

Why are alpha fractions independent of C? Every Ka expression is a ratio of two species at the same h. Scaling the total amount scales every species equally, leaving each fraction unchanged. This makes distribution diagrams universal for a given acid.

Common misconception

"At pH = pKa1 the acid is 50% H₂A and 50% HA⁻ exactly." Only approximately: a small amount of A²⁻ is also present, so both fractions are slightly below 0.5 when the pKa values are close together.

Worked example

Question: Calculate α₀, α₁ and α₂ for the carbonic acid system at pH 7.40 (Ka1 = 4.3 × 10⁻⁷, Ka2 = 4.7 × 10⁻¹¹).

Reasoning: h = 10⁻⁷·⁴⁰ = 3.98 × 10⁻⁸. Terms: h² = 1.585 × 10⁻¹⁵; Ka1h = 1.711 × 10⁻¹⁴; Ka1Ka2 = 2.02 × 10⁻¹⁷. D = 1.872 × 10⁻¹⁴.

Answer: α₀ = 0.085, α₁ = 0.914, α₂ = 0.0011. Hydrogencarbonate is dominant, dissolved CO₂ is about 8.5%, and carbonate is about 0.1%.

Quick check

1. At what pH does HA⁻ reach its maximum fraction for an acid with pKa1 = 4.0 and pKa2 = 9.0? Answer: At pH = ½(4.0 + 9.0) = 6.5, midway between the two pKa values.

Exam focus

Be able to derive the expressions from the mass balance, not just quote them. Common questions ask for the species present at a given pH, the pH at which two species are equal, or the concentration of one species given C and pH. Show D explicitly.

Advanced insight

Taking logarithms of the alpha fractions gives straight-line segments with slopes of 0, ±1 or ±2 on a log C–pH diagram. These segments allow quick graphical solutions of equilibrium problems, a technique developed by Sillén and still used in environmental and geochemical modelling.

Summary

The fraction of a diprotic acid in each form depends only on [H⁺], Ka1 and Ka2: α₀ = h²/D, α₁ = Ka1h/D, α₂ = Ka1Ka2/D, with D the sum of the three terms. Adjacent species are equal at the pKa values, and HA⁻ peaks midway between them. Multiplying by C gives actual concentrations.

Practice questions

1. Write α₁ for a diprotic acid in terms of [H⁺], Ka1 and Ka2. Answer: α₁ = Ka1[H⁺] / ([H⁺]² + Ka1[H⁺] + Ka1Ka2). 2. For carbonic acid, which species dominates at pH 11.0 and why? Answer: CO₃²⁻, because pH 11.0 is above pKa2 = 10.33, so the Ka1Ka2 term in D is the largest. 3. Oxalic acid has pKa1 = 1.23 and pKa2 = 4.19. Estimate the pH at which HC₂O₄⁻ is at its maximum fraction. Answer: ½(1.23 + 4.19) = 2.71. 4. A 0.020 mol dm⁻³ carbonate system is at pH 7.40. Using α₁ = 0.914, find [HCO₃⁻]. Answer: [HCO₃⁻] = 0.914 × 0.020 = 0.018 mol dm⁻³.