Polyprotic Acid Equilibria and Stepwise Ka

Clinical blood-gas analyzers report pH, $pCO_2$, and calculated bicarbonate using exactly the carbonic-acid stepwise framework from Example 2 — the Henderson-Hasselbalch equation printed on every blood-gas report is literally the first-step equilibrium of a triprotic-adjacent (here diprotic-effective) carbonate system. Municipal water treatment relies on the carbonate ladder to control corrosivity and scaling: operators adjust pH to sit near the $HCO_3^-/CO_3^{2-}$ crossover to precipitate calcium carbonate deliberately (softening) or avoid precipitating it accidentally (pipe scaling). Pharmaceutical formulation chemists choose buffer salts — citrate (triprotic), phosphate (triprotic), succinate (diprotic) — specifically because their pKa ladders let a single salt buffer a chosen physiological pH window, computed exactly as in Example 3's equivalence-point logic. Complexometric titrations with EDTA (formally hexaprotic, $H_6Y^{2+}$ through $Y^{4-}$) use the same dominance-ladder reasoning to decide at what pH the fully deprotonated $Y^{4-}$ — the actual metal-binding form — becomes available, which is why EDTA titrations for water hardness are buffered near pH 10. Fertilizer manufacture and soil chemistry depend on the phosphoric acid ladder to predict whether soil phosphorus exists as agriculturally available $H_2PO_4^-$/$HPO_4^{2-}$ or as unavailable precipitated phosphate salts, governed by exactly the species-fraction diagrams foreshadowed here and developed fully under Fractional Composition of Diprotic Systems. Understanding which species dominates at a given pH is also the first step toward finding the pH of any Amphiprotic Species and Intermediate pH, since amphiprotic ions sit precisely at the boundary between two rungs of this ladder — and the whole apparatus rests on the basic single-step logic developed for Acid Dissociation and Ka.

:::quiz[In the sequential approximation for a diprotic acid, what happens to [A²⁻] once [H⁺] and [HA⁻] are found from the first equilibrium?]

• [A²⁻] must be found by solving a fully simultaneous system with the first equilibrium
• [A²⁻] is simply set equal to C
• [A²⁻] ≈ Ka2, obtained by plugging the already-known [H⁺] ≈ [HA⁻] into the Ka2 expression ✓
• [A²⁻] is always exactly zero

Ask the tutor: How would the sequential approximation change if I were titrating a diprotic acid whose two pKa values differ by less than 2 units, like maleic acid?

A polyprotic acid $H_nA$ releases its protons one at a time, and each release is described by its own acid-dissociation equilibrium:

and so on until the fully deprotonated species $A^{n-}$ is reached. Each of these is a genuine, independent equilibrium with its own equilibrium constant, and — crucially — they are all coupled through the same $[H^+]$, since every proton that leaves ends up in the same solution. This coupling is why you cannot simply solve three unrelated monoprotic problems in parallel; you must recognize that one species's product is the next species's reactant, all sharing a single pH. The reason $K_{a1} > K_{a2} > K_{a3}$ is electrostatic, not some numerical coincidence. Pulling a proton off neutral $H_2A$ separates a positive $H^+$ from a neutral residue — costly, but bounded. Pulling a proton off $HA^-$, however, means separating a positive $H^+$ from a species that is already negatively charged; Coulomb attraction between the departing proton and the residual negative charge makes this dissociation intrinsically less favorable. Each successive deprotonation faces a more negative conjugate base, and the ratio $K_{a1}/K_{a2}$ typically lands in the range $10^4$–$10^5$ (sometimes called the "statistical plus electrostatic" gap). This single number is the reason the entire subject is tractable: it means that at the pH set by the first dissociation, the second dissociation has barely begun, so you can solve the equilibria sequentially rather than simultaneously. The practical consequence is a "dominance ladder" along the pH axis. At very low pH, $H_nA$ (fully protonated) dominates. As pH rises past $pK_{a1}$, $H_{n-1}A^-$ takes over. Past $pK_{a2}$, $H_{n-2}A^{2-}$ dominates, and so on, until at very high pH the fully deprotonated $A^{n-}$ prevails. Each species owns the pH interval between the $pK_a$ values that bracket it, and the crossover points — where two adjacent species are equally abundant — occur exactly at pH $= pK_{ai}$.

Notice that in this example the two pKa's are separated by 5.05 units — far more than the minimum ~2–3 units usually needed for the sequential approximation to be safe. When the pKa's are close together (as with some dicarboxylic acids, where $\Delta pK_a$ can be under 2), the species overlap more, both dissociations matter simultaneously near the crossover, and the simple sequential method starts to break down; that regime is treated properly under Amphiprotic Species and Intermediate pH.

For a diprotic acid $H_2A$ in water, the two stepwise equilibria are

where: $[H^+]$ is hydronium concentration (mol/L), $[HA^-]$ and $[H_2A]$ are equilibrium concentrations of the mono-anion and neutral acid. Reading: $K_{a1}$ measures how readily the first, most acidic proton leaves the fully protonated form.

where: $[A^{2-}]$ is the fully deprotonated dianion. Reading: $K_{a2}$ measures how readily the second proton leaves the mono-anion — always a harder step, so $K_{a2} < K_{a1}$. For a triprotic acid such as $H_3PO_4$, a third equilibrium extends the ladder:

where: reading directly parallels $K_{a2}$, one further deprotonation step, one further electrostatic penalty, hence $K_{a3} < K_{a2} < K_{a1}$. The overall dissociation constant relating $H_nA$ directly to $A^{n-}$ is the product of the stepwise constants:

where: this is just the equilibrium constant for the net reaction $H_nA \rightleftharpoons nH^+ + A^{n-}$, obtained by adding the stepwise reactions (which multiplies their $K$'s). Reading: this product is rarely used directly for pH work but is essential for computing the fraction of fully deprotonated species at high pH, and for relating polyprotic $K_a$'s to solubility products of salts like $CaC_2O_4$. The practical workhorse, valid whenever $K_{a1}/K_{a2} \gtrsim 10^3$, is the sequential approximation:

where: $C$ is the formal (total) concentration of the acid. Reading: because so little $HA^-$ further ionizes, the first equilibrium alone, treated exactly as a monoprotic weak acid, fixes $[H^+]$; the second and further equilibria are then solved afterward, using that fixed $[H^+]$, to get trace concentrations of $A^{2-}$ (and $A^{3-}$). A companion result, obtained algebraically (derived below), states that for the intermediate amphiprotic species $HA^-$:

where: this holds for the amphiprotic salt $NaHA$ dissolved alone (not the free diprotic acid), and remarkably is nearly independent of concentration $C$. Reading: this is the basis of buffers like $NaHCO_3/Na_2CO_3$-free bicarbonate solutions and phosphate buffers built from $NaH_2PO_4$ or $Na_2HPO_4$ alone.

:::derivation[Deriving [H⁺] ≈ √(Ka1·C) from the exact diprotic mass/charge balance] Goal. Show rigorously that when $K_{a1} \gg K_{a2}$, the exact equilibrium description of a diprotic acid $H_2A$ at formal concentration $C$ collapses to the ordinary monoprotic weak-acid formula, with the second dissociation appearing only as a negligible correction.

Step 1 — Write the exact governing equations. Mass balance on the acid: $C = [H_2A] + [HA^-] + [A^{2-}]$. Charge balance: $[H^+] = [HA^-] + 2[A^{2-}] + [OH^-]$. Two equilibrium expressions, $K_{a1} = [H^+][HA^-]/[H_2A]$ and $K_{a2} = [H^+][A^{2-}]/[HA^-]$, plus $K_w = [H^+][OH^-]$. This is five equations in five unknowns ($[H_2A],[HA^-],[A^{2-}],[H^+],[OH^-]$) — exact, but not directly solvable by hand.

Step 2 — Estimate the size of $[A^{2-}]$ relative to $[HA^-]$. From the $K_{a2}$ expression, $[A^{2-}] = K_{a2}[HA^-]/[H^+]$. If the solution turns out mildly acidic (pH set by $K_{a1}$, typically pH 1–4 for common diprotic acids), and $K_{a2}\ll K_{a1}$ by four to five orders of magnitude, then

because $[H^+]$ from the first dissociation alone is already comparable to or larger than $K_{a1}$, which itself dwarfs $K_{a2}$. Concretely, if $[H^+]\sim 10^{-2}$ M and $K_{a2}\sim 10^{-7}$, the ratio is $\sim 10^{-5}$ — utterly negligible.

Step 3 — Drop $[A^{2-}]$ from both balances. The charge balance becomes $[H^+] \approx [HA^-] + [OH^-]$, and since the solution is acidic, $[OH^-]\ll[HA^-]$, giving simply

The mass balance becomes $C \approx [H_2A] + [HA^-]$, i.e., $[H_2A] \approx C - [HA^-] \approx C - [H^+]$.

Step 4 — Substitute into the $K_{a1}$ expression.

This is exactly the monoprotic weak-acid quadratic. Solving,

and when $C \gg K_{a1}$ (weak acid, low dissociation), this further simplifies to $[H^+] \approx \sqrt{K_{a1}C}$.

Step 5 — Recover $[A^{2-}]$ as a diagnostic, not a driver. Once $[H^+]$ and $[HA^-]$ are known from Steps 3–4, plug back into $K_{a2}$:

since $[HA^-]\approx[H^+]$ in this regime. This is the elegant payoff: for a diprotic acid dominated by its first ionization, $[A^{2-}] \approx K_{a2}$ directly, independent of $C$ — a fact worth memorizing for quick estimates.

:::derivation[Deriving [H⁺] ≈ √(Ka1·Ka2) for the amphiprotic ion HA⁻ alone in solution] Goal. Find the pH of a solution made by dissolving the salt $NaHA$ (i.e., $HA^-$ as the dominant initial species) at formal concentration $C$, independent of the diprotic-acid derivation above.

Step 1 — Recognize the two competing reactions of $HA^-$. $HA^-$ is amphiprotic: it can act as an acid, $HA^- \rightleftharpoons H^+ + A^{2-}$ (governed by $K_{a2}$), or as a base, $HA^- + H_2O \rightleftharpoons H_2A + OH^-$ (governed by $K_b = K_w/K_{a1}$, since this is the reverse-conjugate reaction of the first acid dissociation). The observed pH is the net result of these two opposing tendencies.

Step 2 — Write mass and charge balances. Mass balance: $C = [H_2A] + [HA^-] + [A^{2-}]$. Charge balance (with Na⁺ present, concentration $C$): $[Na^+] + [H^+] = [HA^-] + 2[A^{2-}] + [OH^-]$, and since $[Na^+] = C$, subtracting the mass balance gives the proton condition

This says: any excess protons beyond neutral pH must have come from converting $HA^-$ back to $H_2A$, and any deficit is balanced by $HA^-$ giving up a proton to form $A^{2-}$.

Step 3 — Express $[H_2A]$ and $[A^{2-}]$ in terms of $[HA^-]$ and $[H^+]$. From the equilibrium expressions, $[H_2A] = [H^+][HA^-]/K_{a1}$ and $[A^{2-}] = K_{a2}[HA^-]/[H^+]$. Substituting into the proton condition:

Step 4 — Solve for $[H^+]^2$. Multiply through by $[H^+]$ and rearrange:

Neglecting $[H^+][OH^-]$ (valid except at very low $C$) and approximating $[HA^-]\approx C$ (valid when $HA^-$ itself dissociates only slightly further either way):

Step 5 — Take the concentration-independent limit. When $C \gg K_{a1}$ — true for essentially any laboratory concentration above micromolar, since $K_{a1}$ for most diprotic acids is $10^{-2}$–$10^{-3}$ — the $C$ in numerator and denominator cancel:

This is why amphiprotic buffers (bicarbonate, monohydrogen phosphate) have a pH that barely shifts with dilution — a strikingly different behavior from ordinary weak-acid solutions, whose pH does shift with $\sqrt{C}$.

Example 1 (Foundational): pH and species distribution of 0.20 M oxalic acid, H₂C₂O₄ Oxalic acid has $K_{a1} = 5.9\times10^{-2}$ ($pK_{a1}=1.23$) and $K_{a2} = 6.4\times10^{-5}$ ($pK_{a2}=4.19$). Note $K_{a1}/K_{a2} \approx 920$ — smaller than the "safe" $10^4$ margin, so we must check the approximation's validity as we go, not just assume it. Step 1 — First dissociation, exact (not simplified) because $K_{a1}$ is not small compared with $C$.

Solving the quadratic: $x = \dfrac{-0.059+\sqrt{0.059^2+4(0.0118)}}{2} = \dfrac{-0.059+\sqrt{0.003481+0.0472}}{2}$.

So $[H^+]\approx[HC_2O_4^-]\approx 0.0831$ M, and $[H_2C_2O_4] = C - x = 0.20-0.0831 = 0.1169$ M. Check: $x/C = 0.0831/0.20=41.6\%$ dissociated — far above the 5% rule, confirming the quadratic (not the square-root shortcut) was mandatory here. pH $= -\log(0.0831) = 1.080$ (3 sig figs, consistent with two-sig-fig $K_{a1}$). Step 2 — Second dissociation, using $[H^+]$ fixed from Step 1.

Notice the algebraic simplification: because $[HC_2O_4^-]\approx[H^+]$ from Step 1, $[C_2O_4^{2-}]$ reduces almost exactly to $K_{a2}$ itself — confirming the "diagnostic" result from Step 5 of the first derivation, here giving $6.4\times10^{-5}$ M, a factor of ~1300 below $[HC_2O_4^-]$. Step 3 — Verify self-consistency. The second dissociation consumed $6.4\times10^{-5}$ M of $HC_2O_4^-$ and produced the same amount of $H^+$ — both utterly negligible next to $0.0831$ M. The sequential treatment is validated post hoc, even though the initial $K_{a1}/K_{a2}$ ratio looked marginal; what matters is not the bare ratio but how much $[H^+]$ the first step generates relative to $K_{a2}$.

Limiting case check: as $C \to 0$ (infinite dilution), the quadratic in Step 1 gives $x \to \sqrt{K_{a1}C} \to 0$, and pH $\to 7$ from below as expected for any acid diluted toward pure water; the sequential approximation for the second step remains valid throughout since $K_{a2}$ stays fixed while $[H^+]$ shrinks even further below it in absolute terms once other buffering from water autoionization is included, though the treatment must switch to the exact charge balance when $[H^+]$ approaches $10^{-6}$ M or so.

Example 2 (Applied): pH of blood plasma bicarbonate buffer and the fraction of species present Human blood plasma at 37°C is buffered near pH 7.40 by the carbonic acid/bicarbonate system, with (at 37 °C, physiological ionic strength) apparent constants $pK_{a1}' = 6.10$ and $pK_{a2}' = 10.25$ for $H_2CO_3^*/HCO_3^-/CO_3^{2-}$ (the star denotes dissolved $CO_2$ lumped with true $H_2CO_3$, standard clinical convention). Total dissolved inorganic carbon in plasma is about $C_T = 25.0$ mmol/L (a typical clinical value). Step 1 — Confirm which species dominate at pH 7.40. Since $pH = 7.40$ lies between $pK_{a1}'=6.10$ and $pK_{a2}'=10.25$ — specifically $1.30$ units above $pK_{a1}'$ and $2.85$ units below $pK_{a2}'$ — the dominant species must be $HCO_3^-$, with $H_2CO_3^*$ present as a significant minority and $CO_3^{2-}$ utterly negligible. This is read directly off the dominance ladder without any calculation. Step 2 — Apply the Henderson–Hasselbalch relation for the first equilibrium (justified because $CO_3^{2-}$ is negligible here).

Step 3 — Partition $C_T$ between the two dominant species.

Step 4 — Verify the neglected species really is negligible.

This is $0.14\%$ of $C_T$ — safely negligible, confirming Step 3's approximation. This is exactly the physiological mechanism by which the lungs (controlling dissolved $CO_2$, hence $[H_2CO_3^*]$) and kidneys (controlling $[HCO_3^-]$ excretion) jointly regulate blood pH: the buffer works precisely because it sits close to, but not at, $pK_{a1}'$, giving a large reserve of $HCO_3^-$ to neutralize incoming acid.

Limiting case check: if $pH$ were instead pushed toward the second pKa (say pH 10.25, as in a strongly alkaline industrial scrubber solution absorbing $CO_2$), the same algebra flips: $[HCO_3^-]=[CO_3^{2-}]$ and $H_2CO_3^*$ becomes negligible — the ladder simply shifts one rung, exactly as the dominance diagram predicts.

Example 3 (Exam/competition level): Triprotic phosphoric acid — locate the pH at the second equivalence point in a titration of 0.100 M H₃PO₄ with 0.100 M NaOH, and find the buffer region width Phosphoric acid: $pK_{a1}=2.15$, $pK_{a2}=7.20$, $pK_{a3}=12.35$ (all well separated, $\Delta pK_a > 5$ each time, so sequential treatment is excellent throughout). Step 1 — Identify what "second equivalence point" means chemically. Titrating $H_3PO_4$ with $NaOH$ proceeds: first equivalence point converts all $H_3PO_4 \to NaH_2PO_4$ (i.e., all to $H_2PO_4^-$); second equivalence point converts all $H_2PO_4^- \to Na_2HPO_4$ (all to $HPO_4^{2-}$). At the second equivalence point, the solution contains only the amphiprotic species $HPO_4^{2-}$ (plus spectator $Na^+$) — precisely the scenario solved in the second derivation. Step 2 — Apply the amphiprotic-pH formula.

Step 3 — Check the validity condition $C \gg K_{a2}$ from Step 5 of the derivation. At the second equivalence point, the solution has been diluted: starting from $V_0$ mL of 0.100 M $H_3PO_4$, two full equivalents of 0.100 M NaOH have been added, so total volume is $3V_0$ and $C = 0.100/3 = 0.0333$ M. Since $K_{a2}=6.3\times10^{-8} \ll C = 0.0333$ M (ratio $\sim 5\times10^{5}$), the concentration-independence condition holds superbly — pH 9.78 would be essentially unchanged even at 10-fold greater dilution. Step 4 — Locate the useful buffering width around the second equivalence point (for context: where is $HPO_4^{2-}/H_2PO_4^-$ a good buffer versus where does the titration curve run steeply?). The buffer region for the $H_2PO_4^-/HPO_4^{2-}$ couple (centered at $pK_{a2}=7.20$, not at the equivalence point pH of 9.78 — a common exam trap, since the equivalence point pH and the buffer pKa are different landmarks) is conventionally taken as $pK_{a2}\pm1$, i.e. pH 6.20 to 8.20. The steep, nearly vertical jump of the titration curve occurs between the buffer region (pH ≈ 6–8) and the equivalence point (pH 9.78), because past pH ≈ 8.2 there is almost no $H_2PO_4^-$ left to consume, so each further drop of NaOH swings the pH rapidly until the next buffer region (centered on $pK_{a3}=12.35$) begins to take hold.

Limiting case check: setting $n=1$ (as if phosphoric acid had only one ionizable proton) collapses this whole three-buffer-region structure to the single, familiar sigmoidal weak-acid/strong-base titration curve with one equivalence point at pH $= 7+\tfrac12(pK_a - \log C)$ — recovering the monoprotic case exactly, confirming the triprotic ladder is a strict generalization, not a different phenomenon.

The quantitative framework for stepwise acid dissociation constants emerged from the broader Brønsted–Lowry revolution in acid-base theory, but it was the Danish biochemist Søren Sørensen and later Niels Bjerrum who gave polyprotic equilibria their modern rigor — Bjerrum in particular developing the mathematical theory of stepwise formation and dissociation constants (originally for metal-ligand complexes, but directly transferable to polyprotic acids) that underlies every distribution diagram used today. The pH scale itself, introduced by Sørensen in 1909 while studying enzyme-catalyzed reactions at the Carlsberg Laboratory in Copenhagen, was motivated by exactly the problem this article addresses: brewing and fermentation depend on carbonic-acid and phosphate buffering of wort, and precise, reproducible acidity measurement was commercially essential before it was theoretically elegant. Sørensen's insistence on defining acidity through a consistent, measurable hydrogen-ion exponent gave later chemists the tool needed to tabulate stepwise $K_a$ values reliably across laboratories.