Fractional Composition of Diprotic Systems
Dissolve sodium oxalate in water and add acid dropwise. Then ask a simple question. What fraction of the oxalate exists as the fully protonated diacid H₂A? What fraction exists as the intermediate HA⁻, or the free dianion A²⁻? This depends on the given pH at any moment. The answer is not a guess. It is a closed-form function of pH and the two acid dissociation constants. It is exactly the information needed first. A titration curve, a solubility calculation, or an EDTA complexation problem all require it before anything else can be computed correctly. This matters because almost no real analyte is monoprotic. Oxalic acid, phthalic acid, carbonic acid, and tartaric acid all carry two or more ionizable protons. Amino acids and the ligand forms used in Complex Formation and Stepwise Stability Constants do too. Any calculation that treats such a species as if only one form existed will be wrong. This means ignoring the other species competing for the same total concentration. Such errors can be off by orders of magnitude near a pKa. The fractional composition, or alpha fraction, formalism is the tool that keeps every species properly accounted for. It does so species by species, at every pH. The big picture is this: for a diprotic acid H₂A with stepwise constants Ka1 and Ka2, the total analytical concentration CT splits into three fractions. These are α₀ (H₂A), α₁ (HA⁻), and α₂ (A²⁻). Each is a function of [H⁺] alone. They always sum to unity. These fractions are built directly from the Mass and Charge Balance Equations. They also draw on the equilibrium expressions established in Polyprotic Acid Equilibria and Stepwise Ka. Once you have them, equilibrium concentrations of any species become CTαi. This is a single multiplication. It turns otherwise intractable simultaneous equilibria into arithmetic. This article derives the alpha-fraction expressions from first principles. It builds and reads distribution diagrams. It works through three complete numerical examples. These span a textbook oxalic acid system, a practical buffer-design calculation, and an exam-level problem. The last requires back-solving for pH from a specified fractional composition. Every number is computed explicitly so the algebra can be checked step by step.
- A diprotic acid H₂A distributes among H₂A, HA⁻, and A²⁻; the fractions α₀, α₁, α₂ depend only on [H⁺], Ka1, and Ka2, and always sum to 1.
- Each αi has the same denominator, D = [H⁺]² + Ka1[H⁺] + Ka2Ka1, built from the two stepwise equilibria and the mass balance.
- At pH = pKa1, α₀ = α₁ (crossover); at pH = pKa2, α₁ = α₂; between the two pKa values HA⁻ dominates if pKa2 − pKa1 is large enough.
- Distribution diagrams plot αi vs. pH and instantly show which species predominates, without solving equilibria one at a time.
- Alpha fractions convert total concentration into species concentration: [HAᶻ⁻] = CT·αi, the master multiplication used throughout analytical equilibrium work.
- The same algebra generalizes to any polyprotic or stepwise complexation system, foreshadowing conditional constants in mixed-ligand and mixed-acid systems.
- Key idea: the alpha fraction turns a coupled multi-equilibrium problem into a single algebraic multiplier, CT·αi(pH), for every species in a diprotic (or polyprotic) system.
Where this lives today
Distribution diagrams built from alpha fractions are the working tool behind carbonate system chemistry in oceanography and climate science, where the partitioning among CO₂(aq)/H₂CO₃, HCO₃⁻, and CO₃²⁻ as a function of ocean pH directly determines calcium carbonate saturation state and, therefore, coral reef and shellfish viability under ocean acidification. The same three-species alpha-fraction algebra derived here applies verbatim, with pKa1 ≈ 6.35 and pKa2 ≈ 10.33 (apparent constants at seawater ionic strength). In pharmaceutical formulation, diprotic and polyprotic drug molecules (many amino-acid-derived actives, dicarboxylic acid prodrugs, and phosphate-containing therapeutics) require alpha-fraction calculations to predict what fraction of the drug is charged — and therefore membrane-permeable or not — at physiological pH (≈7.4), directly informing bioavailability and formulation pH choices. In quantitative analysis, alpha fractions underlie the exact calculation of titration curves for polyprotic acids and bases, letting instrument software (potentiometric titrators, automated pH-stat systems) predict equivalence-point pH and buffer capacity without brute-force iterative equilibrium solving at every titrant volume. The same machinery extends to EDTA and other multidentate ligand systems, where the conditional formation constant used in complexometric titrations is built by multiplying the thermodynamic constant by the ligand's alpha fraction at the working pH — the direct bridge to Complex Formation and Stepwise Stability Constants and to the Competing Equilibria and Conditional Constants framework used whenever multiple equilibria compete for the same species. In environmental and soil chemistry, phosphate speciation (H₃PO₄/H₂PO₄⁻/HPO₄²⁻/PO₄³⁻, a triprotic extension of this exact algebra) governs nutrient bioavailability and eutrophication risk, computed routinely from alpha-fraction diagrams as a function of soil or water pH.
:::quiz[The common denominator D = [H⁺]² + Ka1[H⁺] + Ka1Ka2 appears in all three alpha-fraction expressions because:]
- it is an approximation valid only at high pH
- it results from clearing denominators in the mass balance after expressing all species in terms of [H₂A] ✓
- Ka1 and Ka2 must be numerically equal for D to apply
- D depends on CT and therefore must be recalculated for each concentration
Ask the tutor: How would the alpha-fraction expressions change for a triprotic acid like citric acid, and where would the extra crossover point fall on the distribution diagram?
The physical picture — thorough prose
Picture oxalic acid, H₂C₂O₄, dissolved in water at some fixed total concentration CT. As soon as it dissolves, it partitions itself among three chemical forms in equilibrium: the neutral diacid H₂A, the monoanion HA⁻, and the dianion A²⁻. Which form dominates depends entirely on the pH of the solution — controlled externally by whatever strong acid or base has been added — because each dissociation step is itself an acid-base equilibrium governed by a fixed Ka. At very low pH, protons are abundant, and the equilibria
are both pushed hard to the left by Le Chatelier's principle: essentially all oxalate exists as H₂A. As pH climbs past pKa1, the first equilibrium tips toward HA⁻. As pH climbs further past pKa2, the second equilibrium tips toward A²⁻. At very high pH, essentially everything is A²⁻. The alpha fractions are simply the mathematically precise version of this qualitative story: they tell you the exact proportion of each species at any pH, not just at the extremes. The crucial insight is that these three equilibria are not independent — they all draw from the same finite pool of total oxalate, CT = [H₂A] + [HA⁻] + [A²⁻]. This is a mass-balance constraint. Combined with the two Ka expressions, it produces three simultaneous algebraic relations in three unknown concentrations, all expressible in terms of [H⁺] alone once CT is fixed. Solving this system yields the alpha fractions as rational functions of [H⁺], Ka1, and Ka2 — with an identical denominator for all three, since they are all normalized against the same total. A distribution diagram plots α₀, α�1, α2 versus pH on the same axes. It is arguably the single most information-dense graphic in analytical acid-base chemistry: one picture answers "what species dominates" at every pH from 0 to 14, replacing dozens of individual equilibrium calculations. The diagram's crossover points sit exactly at pH = pKa1 (where α₀ = α₁) and pH = pKa2 (where α₁ = α₂), a feature that follows directly from the algebra and gives an immediate visual check on any computed diagram.
The core ideas — the governing equations/relationships
For diprotic acid H₂A with total analytical concentration CT, the equilibria are
where:
- Ka1, Ka2 — first and second stepwise acid dissociation constants (Ka1 > Ka2 always, since removing a proton from an already-negative species is harder)
- brackets — equilibrium molar concentrations at the pH of interest
The alpha fractions are defined as the ratio of each species' equilibrium concentration to CT:
where:
- CT — total analytical concentration of all oxalate-containing species, fixed by how much acid or salt was weighed out
- α₀ + α₁ + α₂ = 1 at every pH, since the three species are exhaustive and mutually exclusive
The closed-form expressions, derived below, are
where:
- the denominator D = [H⁺]² + Ka1[H⁺] + Ka1Ka2 is identical across all three fractions
- each numerator is the term in D corresponding to that species — this pattern generalizes directly to triprotic and higher systems
Reading: each numerator term in D is itself proportional to the equilibrium concentration of the corresponding species (up to the common factor [H₂A] used in the derivation), so the fraction of any species is simply "its own term divided by the sum of all terms" — a weighted-average structure that recurs throughout equilibrium chemistry, including in the Complex Formation and Stepwise Stability Constants treatment of metal-ligand distribution.
Goal. Express [H₂A], [HA⁻], [A²⁻] each as a fraction of CT, using only [H⁺], Ka1, Ka2.
Step 1 — Write the mass balance. All oxalate exists as one of the three forms, so
This is the single conservation law tying the three unknown concentrations together at fixed CT.
Step 2 — Express HA⁻ and A²⁻ in terms of H₂A. From the first equilibrium,
From the second equilibrium, substitute this into :
Both minor species are now written purely in terms of [H₂A], [H⁺], Ka1, Ka2 — no other unknowns remain.
Step 3 — Substitute into the mass balance. Replacing [HA⁻] and [A²⁻]:
Factor out [H₂A]:
Step 4 — Clear denominators by multiplying through by [H⁺]². This puts every term over the same common denominator, which is the key simplification:
Define . Solving for [H₂A]:
Step 5 — Divide by CT to get α₀, then back-substitute for α₁ and α₂. By definition α₀ = [H₂A]/CT, so
Using the Step 2 relations, [HA⁻] = Ka1[H₂A]/[H⁺], so
and [A²⁻] = Ka1Ka2[H₂A]/[H⁺]², so
Check: α₀ + α₁ + α₂ = ([H⁺]² + Ka1[H⁺] + Ka1Ka2)/D = D/D = 1 for all [H⁺] — the fractions are automatically normalized, confirming no algebra error crept in.
Go deeper · Derivation B · 1 minLocating the crossover pH values and the maximum of α₁
Goal. Show that α₀ = α₁ exactly at pH = pKa1, that α₁ = α₂ exactly at pH = pKa2, and find where α₁ peaks.
Step 1 — Set α₀ = α₁ and solve. Using the closed forms from the previous derivation,
The common denominator D cancels immediately (it is never zero for real concentrations), leaving
Taking −log of both sides gives pH = pKa1 exactly. This is the same crossover condition familiar from the monoprotic Henderson–Hasselbalch analysis, now shown to survive intact inside the three-species system because the denominator is common to all fractions.
Step 2 — Set α₁ = α₂ and solve analogously.
Both crossovers depend only on the respective Ka, never on CT — a direct consequence of CT canceling out of the ratio α_i/α_j.
Step 3 — Locate the maximum of α₁(pH). Write α₁ explicitly as a function of x = [H⁺]:
Differentiate with respect to x and set the derivative to zero. Using the quotient rule with numerator N = Ka1x and denominator D = x² + Ka1x + Ka1Ka2:
The denominator D² is never zero, so only the numerator must vanish:
Divide through by Ka1 and expand:
Step 4 — Convert to pH. Taking −log of both sides:
The intermediate species HA⁻ peaks exactly midway between the two pKa values on the pH scale — a clean, checkable geometric fact that follows purely from the algebraic structure, with no approximation involved.
Worked examples
Example 1 — Foundational: oxalic acid at pH 3.00. Oxalic acid has pKa1 = 1.25 (Ka1 = 5.6 × 10⁻²) and pKa2 = 4.27 (Ka2 = 5.4 × 10⁻⁵). Total oxalate concentration CT = 0.0500 M. Find α₀, α₁, α₂ and each species' molar concentration at pH 3.00. At pH 3.00, [H⁺] = 1.0 × 10⁻³ M. Compute the three terms of the denominator:
Now the fractions:
Check: 0.0166 + 0.930 + 0.0502 = 0.997 ≈ 1 (rounding in the third decimal accounts for the residual 0.003). Species concentrations follow by multiplying by CT = 0.0500 M:
This confirms the qualitative picture: at pH 3.00 (between pKa1 = 1.25 and pKa2 = 4.27, closer to pKa2), HA⁻ dominates overwhelmingly, exactly as the distribution diagram predicts — pH 3.00 sits well past the first crossover and well before the second.
Example 2 — Applied: buffering an oxalate solution to fix a target ratio for EDTA-style masking. A method requires that [A²⁻]/[HA⁻] = 4.00 in an oxalate buffer used to control free ligand availability (a scenario directly relevant to the conditional-constant treatments in Competing Equilibria and Conditional Constants). Using pKa2 = 4.27, find the required pH, and confirm using the alpha-fraction ratio directly. The ratio α₂/α₁ simplifies because the common denominator D cancels:
This is just the Henderson–Hasselbalch relation for the second dissociation step, recovered automatically from the alpha-fraction framework — a useful consistency check confirming that the diprotic formalism reduces correctly to the familiar monoprotic ratio when only two adjacent species are compared. Setting this ratio to 4.00:
Confirm by computing all three alpha fractions at pH 4.87 ([H⁺] = 1.35 × 10⁻⁵ M):
The α₀ contribution here is negligible ([H⁺]²/D = 4.8 × 10⁻⁵), confirming that at pH 4.87 — nearly a full pH unit above pKa2 — the diacid form has essentially vanished, and the system behaves as an effectively two-species HA⁻/A²⁻ buffer, exactly as the simplified Henderson–Hasselbalch shortcut assumes.
Example 3 — Exam-level: back-solving pH from a specified fractional composition, with a quadratic in [H⁺]. A diprotic acid H₂B has Ka1 = 4.5 × 10⁻³ and Ka2 = 1.8 × 10⁻⁷. At what pH is exactly 30.0% of the total analytical concentration present as the fully protonated form H₂B (i.e., α₀ = 0.300)? Set α₀ = 0.300 using the closed form:
Cross-multiply:
Collect terms, moving everything to one side:
Divide through by 0.700 to normalize the leading coefficient, and insert numbers. First compute the needed products:
The quadratic is
where x = [H⁺]. Apply the quadratic formula, , with a = 0.700, b = −1.35 × 10⁻³, c = −2.43 × 10⁻¹⁰:
Taking the positive root (the physically meaningful one, since [H⁺] > 0):
To three significant figures, pH = 2.71. As a sanity check, note that pKa1 = −log(4.5 × 10⁻³) = 2.35, so pH 2.71 sits above pKa1 by 0.36 units — consistent with α₀ being a minority species (30.0%) rather than the dominant one, since α₀ = α₁ occurs exactly at pH = pKa1 = 2.35 and α₀ must fall below 50% for any pH above that point. The full check: recomputing D at [H⁺] = 1.929 × 10⁻³ gives [H⁺]² = 3.72 × 10⁻⁶, Ka1[H⁺] = 8.68 × 10⁻⁶, Ka1Ka2 = 8.10 × 10⁻¹⁰, so D = 1.241 × 10⁻⁵ and α₀ = 3.72 × 10⁻⁶ / 1.241 × 10⁻⁵ = 0.300, matching the target exactly.
Limiting cases, scope & common traps
| Limiting case | Behavior of α₀, α₁, α₂ | Why |
|---|---|---|
| [H⁺] ≫ Ka1 ≫ Ka2 (strongly acidic) | α₀ → 1, α₁, α₂ → 0 | [H⁺]² dominates D; both other terms are negligible by comparison |
| [H⁺] ≪ Ka2 ≪ Ka1 (strongly basic) | α₂ → 1, α₀, α₁ → 0 | Ka1Ka2 dominates D once [H⁺] is small enough |
| pH = pKa1 exactly | α₀ = α₁, and both exceed α₂ if pKa2 − pKa1 is large | Derived directly: [H⁺] = Ka1 makes the first two D-terms equal |
| pH = pKa2 exactly | α₁ = α₂ | Symmetric argument with the second equilibrium |
| pH = (pKa1+pKa2)/2 | α₁ is at its maximum value | Shown by setting dα₁/d[H⁺] = 0 |
| Ka1 ≈ Ka2 (poorly separated) | α₁ never rises much above ~0.3–0.4; no clean plateau | The two equilibria overlap, so a pure HA⁻ region barely exists |
| CT → 0 | Fractions unchanged | α_i depend only on [H⁺], Ka1, Ka2 — never on CT itself |
The chemist behind it
The mathematics of stepwise equilibrium fractions traces to the broader development of ionic dissociation theory pioneered by Svante Arrhenius and refined through mass-action treatments by Wilhelm Ostwald, but the specific alpha-fraction bookkeeping used throughout modern analytical chemistry is most closely tied to the systematic equilibrium formalism developed and popularized in the mid-20th century by physical and analytical chemists working out multi-step protolysis — most visibly consolidated in the pedagogical and research work of Lars Gunnar Sillén, whose work on stepwise stability constants (extending naturally from stepwise acid dissociation to stepwise metal-ligand complexation) established the standard notation and computational machinery still used today, including the direct precursor to the alpha-fraction treatment of complex ion distributions covered in Complex Formation and Stepwise Stability Constants.
Origins · 1 minLars Gunnar Sillén and how this was discovered
Sillén built much of the systematic mathematical apparatus for handling stepwise equilibria — acid-base, complexation, and solubility — as coupled sets of mass-action expressions solvable through consistent fractional bookkeeping. His collaborative compilations of stability constants, produced with Arthur Martell, became a standard reference that made rigorous multi-equilibrium calculation routine rather than exceptional, directly shaping how modern analytical chemistry courses present distribution diagrams and conditional constants.
Why do all three alpha fractions share the same denominator?
The shared denominator D = [H⁺]² + Ka1[H⁺] + Ka1Ka2 arises because all three species concentrations were expressed as multiples of the same reference quantity, [H₂A], during the mass-balance substitution. When the mass balance CT = [H₂A] + [HA⁻] + [A²⁻] is rewritten entirely in terms of [H₂A], factoring pulls out [H₂A]/[H⁺]² as a common factor, and the bracketed sum becomes D. Dividing each species' concentration by CT then produces the same D in every denominator, just with a different numerator term — the numerator that corresponds to that particular species' power of [H⁺]. This structural symmetry is what guarantees α₀ + α₁ + α₂ = 1 automatically, without needing a separate normalization step, and it generalizes cleanly to triprotic or higher polyprotic systems by simply adding more terms to D.
Does the alpha-fraction formalism assume activity coefficients equal 1?
Yes, as presented here, Ka1 and Ka2 are treated as concentration-based constants, implicitly assuming ideal dilute solution behavior (activity coefficients ≈ 1). At appreciable ionic strength, thermodynamic constants must be corrected using the Davies equation or extended Debye–Hückel treatment to obtain "apparent" or "conditional" constants valid at that ionic strength, and the alpha-fraction algebra is then applied using those conditional Ka values instead. This is exactly the situation in seawater carbonate chemistry, where apparent pKa1 and pKa2 for the carbonate system differ measurably from their infinite-dilution values because of the high ionic strength of seawater, yet the same alpha-fraction formulas apply unchanged once the correct apparent constants are substituted in.
How does this extend to triprotic acids like phosphoric acid?
The same substitution method extends directly: express [H₂PO₄⁻], [HPO₄²⁻], and [PO₄³⁻] all in terms of [H₃PO₄] using the three stepwise Ka values, substitute into the four-term mass balance, and clear denominators to get a common D with four terms: [H⁺]³, Ka1[H⁺]², Ka1Ka2[H⁺], and Ka1Ka2Ka3. Each of the four alpha fractions is then that species' corresponding term divided by this four-term D. The crossover logic also extends: α₀ = α₁ at pH = pKa1, α₁ = α₂ at pH = pKa2, and α₂ = α₃ at pH = pKa3, exactly analogous to the diprotic case, since each crossover condition only ever involves canceling the shared denominator between two adjacent terms.
Can alpha fractions be negative or exceed 1?
No — each αi is a ratio of a physically real, non-negative equilibrium concentration to a positive total concentration, so 0 ≤ αi ≤ 1 always, and the three (or more, for polyprotic systems) fractions sum to exactly 1 at every pH. If a computed alpha fraction falls outside [0, 1], the almost certain cause is an arithmetic slip in computing D or in matching the correct numerator to the correct species — not a genuine physical possibility. This makes the sum check (Σαi = 1, to within rounding) an extremely reliable and fast way to catch calculation errors before proceeding to convert fractions into species concentrations.
Why is Ka1 always larger than Ka2 for a diprotic acid?
Removing a proton from H₂A produces HA⁻, an anion; removing a second proton from that anion to form A²⁻ requires pulling a positive proton away from an already negatively charged species, which is electrostatically less favorable. This raises the free energy of the second dissociation relative to the first, making Ka2 smaller (weaker acid) than Ka1 essentially always, with the ratio Ka1/Ka2 typically ranging from about 10² for the two well-separated steps of a system like phosphoric acid up to only about 10¹ for closely spaced steps like the two carboxylic groups of a short-chain dicarboxylic acid, where the negative charge from the first ionization is far enough from the second site that the electrostatic penalty is smaller.
How do alpha fractions connect to titration curve shape?
At any point during a titration of H₂A with strong base, the charge and mass balance equations that determine pH are written most cleanly in terms of alpha fractions: the total base added equals CT(α₁ + 2α₂) plus corrections for water autoionization, since HA⁻ contributes one negative charge equivalent and A²⁻ contributes two. Buffer regions on the titration curve correspond to pH ranges where one alpha fraction dominates and changes slowly with added base; equivalence points correspond to pH values where a particular alpha fraction crosses a specific threshold (for instance, the first equivalence point occurs near where α₁ is maximal). This is precisely why titration curve calculations for polyprotic acids are built from the alpha-fraction machinery rather than solved as isolated single equilibria at each titrant volume.
What real difference does it make whether pKa1 and pKa2 are close together or far apart?
When pKa2 − pKa1 is large (roughly greater than 4, as in phosphoric acid, pKa1 = 2.15, pKa2 = 7.20), α₁ rises to nearly 1.0 at its maximum, producing a broad, well-resolved buffer plateau for HA⁻ and two clearly separated inflection points on a titration curve. When pKa2 − pKa1 is small (as in oxalic acid, 4.27 − 1.25 = 3.02, or smaller still for some dicarboxylic acids), α₁ never approaches 1, the two equivalence points on a titration curve blur together, and potentiometric titration to resolve the first equivalence point becomes experimentally difficult or impossible — a limitation with direct consequences for analytical method design, since it determines whether a diprotic acid can be titrated stepwise at all.