Polyprotic acids

Result

Polyprotic acids

T-053Home CU-204Threads equilibrium

Statement

Stepwise dissociation and successive constants.

Why it matters

bronsted-lowry and water-autoionisation-ph established single proton-transfer equilibria and the pH scale for simple monoprotic acids; many chemically and biologically important acids — sulfuric, phosphoric, carbonic, amino acids — can donate more than one proton, and polyprotic-acids extends the equilibrium framework to handle these successive, progressively weaker dissociation steps. It is essential groundwork for henderson-hasselbalch's buffer treatment, since many biological buffers (phosphate, carbonate) are polyprotic, and for titration-curves' multi-equivalence-point curves.

Hypotheses

Each successive proton-loss step is treated as its own separate equilibrium, with its own equilibrium constant \(K_{a1}, K_{a2}, K_{a3},\ldots\).Following law-of-mass-action applied stepwise, this is valid because each dissociation is chemically distinct, removing a proton from a progressively more negatively charged, and hence more strongly proton-holding, species. Successive \(K_a\) values decrease substantially at each step, typically \(K_{a1}\gg K_{a2}\gg K_{a3}\).A near-universal empirical pattern, since removing a proton from an increasingly negatively charged species is progressively more electrostatically unfavourable. When successive \(K_a\) values are well separated (commonly by a factor of at least \(\sim1000\)), pH can be calculated from the single dominant equilibrium step relevant at that pH.This decouples what would otherwise be a coupled simultaneous-equilibria problem; the simplification breaks down when successive \(K_a\) values are close together.

Proof

\text{H}_2\text{A} \rightleftharpoons \text{H}^++\text{HA}^-, \ K_{a1}=\frac{[\text{H}^+][\text{HA}^-]}{[\text{H}_2\text{A}]}; \qquad \text{HA}^- \rightleftharpoons \text{H}^++\text{A}^{2-}, \ K_{a2}=\frac{[\text{H}^+][\text{A}^{2-}]}{[\text{HA}^-]}

For a diprotic acid, the two stepwise equilibria are written exactly as for any weak acid, using whatever species is the actual acid at that step (\(\text{H}_2\text{A}\) for the first, \(\text{HA}^-\) for the second). A

K_{a2} \ll K_{a1} \quad (\text{typically by } 10^4\text{–}10^6)

Because \(\text{HA}^-\) is already negatively charged, removing a second proton to form the doubly charged \(\text{A}^{2-}\) is electrostatically more difficult than the first dissociation, so \(K_{a2}\) is essentially always substantially smaller than \(K_{a1}\). A

[\text{H}^+] \approx \sqrt{K_{a1}\cdot C_0} \quad (\text{first-step approximation})

When \(K_{a1}\gg K_{a2}\), essentially all of the \(\text{H}^+\) in solution comes from the first dissociation step, so \([\text{H}^+]\) can be computed to good approximation from \(K_{a1}\) alone, using the standard weak-acid method, treating the acid as effectively monoprotic for this purpose. A

[\text{A}^{2-}] = \frac{K_{a2}[\text{HA}^-]}{[\text{H}^+]}

Once \([\text{H}^+]\) and \([\text{HA}^-]\) are known from Step 3, the fully deprotonated species' concentration follows directly from the second equilibrium expression. A

\text{Speciation: each species dominant over the pH range bracketed by its neighbouring p}K_a\text{ values}

A full speciation diagram, showing the relative fraction of \(\text{H}_2\text{A}\), \(\text{HA}^-\), and \(\text{A}^{2-}\) as a function of pH, follows directly from the two \(K_a\) values by the identical stepwise logic, extended to as many steps as the acid has. A

Result

K_{a1}=\frac{[\text{H}^+][\text{HA}^-]}{[\text{H}_2\text{A}]} \ \gg\ K_{a2}=\frac{[\text{H}^+][\text{A}^{2-}]}{[\text{HA}^-]}

Reading. Each proton is lost via its own, progressively weaker equilibrium; the solution's pH over most of the titration range is set almost entirely by whichever single step is currently dominant.

Scope. A reliable simplification whenever successive \(K_a\) values are well separated, the usual case; acids with closely spaced successive \(K_a\) values require the coupled equilibria to be solved simultaneously rather than stepwise.

Corollaries & converses

henderson-hasselbalch's buffer equation applies separately around each p\(K_a\) of a polyprotic acid, giving it its own set of buffer regions — the phosphate buffer system commonly operates near p\(K_{a2}\), close to physiological pH.
titration-curves for a polyprotic acid show one equivalence point, and one buffer plateau, per dissociation step; a diprotic acid's curve has two equivalence points, directly visualising the successive-\(K_a\) structure derived here.
The same stepwise logic generalises directly to triprotic acids (e.g. \(\text{H}_3\text{PO}_4\), with three \(K_a\) values and four species) by chaining on one further equilibrium step.

Fails without

Apply the single-dominant-step approximation (Hypotheses' third assumption) when \(K_{a1}\) and \(K_{a2}\) are close together, as for oxalic acid: the computed pH is measurably inaccurate, since the second dissociation's contribution to \([\text{H}^+]\) is no longer genuinely negligible.
Substitute the total initial acid concentration, rather than the actual \([\text{HA}^-]\) produced by the first equilibrium, into the second equilibrium expression: gives a badly incorrect \([\text{A}^{2-}]\), since the second step's true reactant concentration is set by the first equilibrium's outcome, not by the original stoichiometric amount of acid dissolved.

Common errors

Assuming the second (or third) dissociation contributes significantly to \([\text{H}^+]\), when \(K_{a2}\) (and \(K_{a3}\)) are typically far too small relative to \(K_{a1}\) to matter for the overall pH.
Using the total, initial acid concentration incorrectly across multiple steps, rather than recognising that the \(\text{HA}^-\) produced by the first equilibrium is the actual reactant entering the second equilibrium.
Forgetting that every intermediate species genuinely exists in solution at equilibrium, however small its concentration, even when only the first equilibrium step is numerically dominant for pH.
Confusing the number of acidic (dissociable) protons with the total number of hydrogen atoms in the formula — acetic acid, \(\text{CH}_3\text{COOH}\), has four hydrogens but is monoprotic, since only the carboxylic O–H is acidic.

Discussion

The successive-\(K_a\) framework for polyprotic acids was well established by the early-to-mid twentieth century, alongside the broader development of quantitative acid-base equilibrium theory; phosphoric acid and carbonic acid, both biologically central, are the standard textbook examples used to illustrate the full stepwise treatment.

Carbonic acid is a genuinely unusual case: the great majority of dissolved \(\text{CO}_2\) in water exists as hydrated \(\text{CO}_2(aq)\) rather than true \(\text{H}_2\text{CO}_3\), so the "effective" \(K_{a1}\) conventionally tabulated for carbonic acid is actually a composite constant folding in the \(\text{CO}_2\) hydration equilibrium together with the true \(\text{H}_2\text{CO}_3\) dissociation step — a widely used but technically approximate convention worth flagging explicitly.

Common misconception: that a small \(K_{a2}\) means the species \(\text{A}^{2-}\) is essentially absent from solution at equilibrium. Rather, some \(\text{A}^{2-}\) genuinely exists at every pH above p\(K_{a2}\) in appreciable, calculable quantity (Step 4); the small \(K_{a2}\) only means the second dissociation does not significantly affect the pH itself.

Worked examples

0.10\,\text{M H}_2\text{A}, \ K_{a1}=1.0\times10^{-3}, \ K_{a2}=1.0\times10^{-8}

Treating only the first equilibrium (Step 3): \(x^2/(0.10-x)\approx x^2/0.10=1.0\times10^{-3}\Rightarrow x=[\text{H}^+]\approx9.5\times10^{-3}\,\text{M}\), giving \(\text{pH}\approx2.02\). A

[\text{A}^{2-}] = \frac{K_{a2}[\text{HA}^-]}{[\text{H}^+]} \approx \frac{(1.0\times10^{-8})(9.5\times10^{-3})}{9.5\times10^{-3}} = 1.0\times10^{-8}\,\text{M}

Using \([\text{HA}^-]\approx[\text{H}^+]\) from the first step (Step 1's stoichiometry, one \(\text{H}^+\) produced per \(\text{HA}^-\)) and the second equilibrium expression, \([\text{A}^{2-}]\) comes out numerically equal to \(K_{a2}\) itself in this case — small, but nonzero, exactly as the Result's misconception note anticipates. A

\text{pH}\approx2.02; \quad [\text{A}^{2-}]\approx1.0\times10^{-8}\,\text{M}

Reading. The first equilibrium alone sets the pH essentially exactly, while the second equilibrium, evaluated afterward using the first step's results, gives the (much smaller) concentration of the fully deprotonated species.

Scope. This two-stage calculation strategy applies whenever \(K_{a1}\gg K_{a2}\), the standard case for diprotic acids.

Problems

For a diprotic acid \(\text{H}_2\text{B}\) with \(K_{a1}=4.0\times10^{-4}\) and initial concentration \(0.20\,\text{M}\), estimate the pH using only the first equilibrium.
Solution
\(x^2/0.20\approx4.0\times10^{-4}\Rightarrow x^2=8.0\times10^{-5}\Rightarrow x=[\text{H}^+]=8.9\times10^{-3}\,\text{M}\Rightarrow \text{pH}\approx2.05\).
Using the result from Problem 1 and \(K_{a2}=2.0\times10^{-9}\), find \([\text{B}^{2-}]\).
Solution
\([\text{HA}^-]\approx[\text{H}^+]=8.9\times10^{-3}\,\text{M}\) (Step 1's stoichiometry); \([\text{B}^{2-}]=K_{a2}[\text{HA}^-]/[\text{H}^+]=K_{a2}=2.0\times10^{-9}\,\text{M}\) once \([\text{HA}^-]\) and \([\text{H}^+]\) are equal, exactly as in Worked Example 2.
Oxalic acid has \(K_{a1}\) and \(K_{a2}\) differing by only about two orders of magnitude, closer together than many diprotic acids. Explain why this requires more care in approximation than an acid with a very large \(K_{a1}/K_{a2}\) ratio.
Solution
The simplifying approximation of Step 3 (treating pH as set entirely by the first equilibrium) relies on \(K_{a1}\gg K_{a2}\), so that the second dissociation's contribution to \([\text{H}^+]\) is genuinely negligible. When the two constants are closer together, as for oxalic acid, the second dissociation step can contribute a non-negligible extra amount of \(\text{H}^+\), and a more careful treatment — potentially solving the two equilibria simultaneously rather than sequentially — may be needed for an accurate pH.