Polyprotic Acid Titration Curves

Clinical blood-gas analysis rests on the $\mathrm{H_2CO_3}/\mathrm{HCO_3^-}/\mathrm{CO_3^{2-}}$ triprotic-like system (with the first $pK_a$ effectively near 6.1 accounting for dissolved $\mathrm{CO_2}$ equilibrium), where only the first equivalence point is titratable in the physiologically relevant pH window — exactly the "unusable last stage" phenomenon worked out in Example 3. Water-treatment alkalinity titrations exploit the two resolvable carbonate breaks (bicarbonate/carbonate alkalinity) using methyl orange and phenolphthalein endpoints chosen specifically because their color-change pH ranges bracket the two calculated equivalence-point pH values. Food and beverage chemistry relies on titrating polyprotic acids — citric acid in citrus juices and soft drinks, tartaric and malic acids in wine — where total acidity is typically reported using only the last equivalence point because the individual $K_a$'s of these acids are too close together to resolve cleanly, a direct application of the resolution-ratio criterion developed here. Complexometric EDTA titrations (a distinct titration type, but sharing the same amphiprotic-species mathematics) depend on tracking the stepwise protonation states of the hexaprotic EDTA ligand to control which metal-binding species is dominant at a chosen pH. Pharmaceutical quality control uses phosphate-buffer polyprotic equilibria to standardize dissolution testing and formulation pH, and amino acid analysis — the natural next step from this material — extends the same amphiprotic pH formula to compute isoelectric points, foreshadowing Amino Acids and Zwitterion Titrations. Finally, choosing which indicator to use at each stage of a multi-equivalence titration is a direct downstream application of the equivalence-point pH values computed here, the subject taken up in Selecting Indicators for Multi-Equivalence Titrations.

Ask the tutor: How would the calculated pH values change if I titrated phosphoric acid with a strong base at ten times lower concentration — which equivalence point would degrade first, and why?

Consider titrating $\mathrm{H}_3\mathrm{PO}_4$ (a triprotic acid, $pK_{a1}=2.15$, $pK_{a2}=7.20$, $pK_{a3}=12.35$) with NaOH. Each proton is removed by a separate deprotonation reaction:

Because $K_{a1} \gg K_{a2} \gg K_{a3}$ (each step roughly $10^{4.5}$–$10^5$ weaker than the last), the three protons come off sequentially rather than simultaneously — at any given pH, essentially only one deprotonation reaction is "active." This sequential behavior is what carves the titration curve into distinguishable regions: an initial acidic region, a buffer region around $pK_{a1}$, an equivalence point (all acid converted to $\mathrm{H_2PO_4^-}$), a second buffer region around $pK_{a2}$, a second equivalence point ($\mathrm{HPO_4^{2-}}$ dominant), and — in principle — a third buffer/equivalence pair around $pK_{a3}$, though in practice the third equivalence point of phosphoric acid is not titratable in water (it would require pH above 12.35 while consuming a stoichiometric amount of strong base, and the autoionization of water and the weakness of $\mathrm{HPO_4^{2-}}$ as an acid make the break too gentle to detect).

The corresponding species distribution diagram — fraction of each protonation state versus pH — is the mirror image of the titration curve and is often the cleaner way to see why separation succeeds or fails. Where two $\alpha$ curves cross is exactly where $\text{pH}=pK_a$ for that step; where an intermediate species reaches its maximum fraction is exactly the corresponding equivalence point.

The crossing at $\text{pH}=pK_{a1}=2.15$ shows $\mathrm{H_3PO_4}$ and $\mathrm{H_2PO_4^-}$ at equal concentration; the crossing at $pK_{a2}=7.20$ shows $\mathrm{H_2PO_4^-}$ and $\mathrm{HPO_4^{2-}}$ equal; and so on. Because each $pK_a$ is separated from its neighbor by roughly 5 pH units, each species reaches a maximum fraction close to 1 before the next proton starts leaving — that near-total dominance of $\mathrm{H_2PO_4^-}$ near pH 4–5 is exactly what produces a sharp, well-defined first equivalence point.

Buffer-region pH (each step). In the region between the start and the $i$-th equivalence point, once enough base has been added that both the conjugate acid $\mathrm{H}_{n-i+1}\mathrm{A}^{(i-1)-}$ and conjugate base $\mathrm{H}_{n-i}\mathrm{A}^{i-}$ coexist in comparable amounts, and provided the neighboring equilibria are far enough away to be ignored:

where: $pK_{a,i} = -\log K_{a,i}$ is the acidity constant for the $i$-th proton loss, and the log ratio is the mole ratio of conjugate base to conjugate acid present at that point in the titration (computed exactly as in a monoprotic weak-acid titration).

Half-equivalence shortcut. At exactly half the volume needed to reach the $i$-th equivalence point (measured from the $(i-1)$-th equivalence point), the ratio is 1 and

Intermediate (amphiprotic) equivalence-point pH. At the $i$-th equivalence point (for $1 \le i \le n-1$), the dominant species $\mathrm{H}_{n-i}\mathrm{A}^{i-}$ is amphiprotic — it can both donate ($K_{a,i+1}$) and accept ($K_{a,i}$, in reverse) a proton. The standard approximation, valid when the amphiprotic species' concentration $C$ is not too small relative to $K_{a,i}$ and $K_{a,i+1}$, is:

so that

where: $C$ is the formal (analytical) concentration of the amphiprotic species at that point in the titration (diluted by the added titrant volume), and $K_w = 1.0\times10^{-14}$ is water's autoionization constant. The strikingly simple final form says the intermediate equivalence-point pH is independent of concentration to leading order — a genuinely useful and testable prediction.

First-equivalence-point sanity range. The approximation $\text{pH}\approx\frac12(pK_{a1}+pK_{a2})$ is reliable provided $K_{a1}C \gg K_w$ (the solution isn't so dilute that water dominates) and $C \gg K_{a2}$ isn't wildly violated (the amphiprotic species doesn't dissociate so much that the "just average the two pK_a's" logic breaks down).

Final equivalence point (full deprotonation). At the last equivalence point, all the acid has been converted to its fully deprotonated conjugate base $\mathrm{A}^{n-}$, which hydrolyzes as a simple monoprotic weak base:

where: $C'$ is the analytical concentration of $\mathrm{A}^{n-}$ at that titration volume (original moles of acid divided by total solution volume, since every mole of $\mathrm{H}_n\mathrm{A}$ produces one mole of $\mathrm{A}^{n-}$).

Resolution criterion. Two adjacent equivalence points are resolvable as distinct breaks on a practical titration curve only if consecutive $K_a$'s differ enough. A widely used rule of thumb, consistent with Skoog's treatment of feasibility, is

with the required ratio increasing as the analytical concentration $C$ decreases (more dilute acids need more separation between $K_a$'s to still show a visible break, exactly as a single weak acid needs a larger $K_a \cdot C$ product to show a feasible single break).

Example 1 — Foundational: diprotic oxalic acid, full curve. Titrate 40.00 mL of 0.0800 M oxalic acid ($\mathrm{H_2C_2O_4}$, $K_{a1}=5.9\times10^{-2}$, $K_{a2}=6.4\times10^{-5}$) with 0.1000 M NaOH. Find the pH (a) at the start, (b) at the first half-equivalence volume, (c) at the first equivalence point, (d) at the midpoint between the two equivalence points, (e) at the second equivalence point. Moles of acid $=0.04000\text{ L}\times0.0800\text{ M}=3.200\times10^{-3}$ mol. Volume to first equivalence point: $V_1 = (3.200\times10^{-3}\text{ mol})/(0.1000\text{ M}) = 32.00$ mL. Volume to second equivalence point: $V_2 = 64.00$ mL (double, since the second proton needs an equal number of moles of NaOH). (a) Start (0 mL added). $K_{a1}=5.9\times10^{-2}$ is large enough that oxalic acid is not a "weak-in-the-usual-sense" acid; solve the quadratic exactly rather than approximating:

(b) First half-equivalence (16.00 mL added). Moles NaOH added $=0.01600\text{ L}\times0.1000\text{ M}=1.600\times10^{-3}$ mol, exactly half of the original acid — so $[\mathrm{H_2C_2O_4}]=[\mathrm{HC_2O_4^-}]$ (equal moles, same volume, ratio =1):

(Note $K_{a1}$ is large enough that the simple Henderson–Hasselbalch equality is only approximate here too, but is standard practice at this level; a rigorous treatment would re-solve the mass-balance quadratic, giving a value close to 1.3.) (c) First equivalence point (32.00 mL added, total volume $=72.00$ mL). Formal concentration of $\mathrm{HC_2O_4^-}$:

Apply the amphiprotic approximation:

Check the approximation's validity: $C=0.0444$ M is not $\gg K_{a1}=0.059$ M — they're actually comparable, which is a warning sign for oxalic acid specifically (its unusually large $K_{a1}$ makes the simple square-root formula shakier here than for phosphoric acid). A more careful treatment using the full Step-5 formula from the first derivation gives:

This more rigorous value (pH 2.89) versus the quick estimate (pH 2.71) is a realistic illustration of how the simple $\sqrt{K_{a1}K_{a2}}$ shortcut can be off by a few tenths of a pH unit when $K_{a1}$ is unusually large — exactly the kind of asymmetric, "not both nice round numbers" case that exposes whether the formula is being applied correctly rather than just pattern-matched. (d) Midpoint between equivalence points (48.00 mL added, second half-equivalence). Here moles of NaOH added past $V_1$ equal half of $(V_2-V_1)=16.00$ mL worth, i.e. $[\mathrm{HC_2O_4^-}]=[\mathrm{C_2O_4^{2-}}]$:

(e) Second equivalence point (64.00 mL added, total volume $=104.0$ mL). All acid converted to $\mathrm{C_2O_4^{2-}}$:

The full picture: pH climbs from 1.38 through a shallow buffer plateau near pH 1.2–2, breaks at pH 2.71–2.89 (first equivalence), climbs through a second, much more pronounced buffer plateau near pH 4.2, and reaches a mildly basic second equivalence point at pH 8.34 — a genuinely useful endpoint for phenolphthalein.

Example 2 — Applied: does the second equivalence point of oxalic acid actually resolve from the first? Using $K_{a1}/K_{a2} = (5.9\times10^{-2})/(6.4\times10^{-5}) = 922$, apply the resolution criterion.

This falls below the $10^3$–$10^4$ threshold quoted above. In practice, titrating oxalic acid does show two breaks, but the first one is visibly less sharp and more curved than the corresponding first break in phosphoric acid ($K_{a1}/K_{a2}\approx10^{5.05}$), because the buffer region for $K_{a2}$ starts encroaching on the tail end of the first equivalence-point jump. This is exactly why oxalic acid is a textbook example used to illustrate marginal resolvability rather than a showcase acid like phosphoric or citric acid — quantitative analysts titrating oxalate historically preferred redox methods (permanganate titration) over acid–base titration for exactly this reason: the acid–base endpoints are real but not crisp enough for high-precision volumetric work. Cross-check with the non-overlap bound from Step 2 of the second derivation: $pK_{a2}-pK_{a1} = 4.19-1.23 = 2.96 > 2$, so the bare non-overlap criterion is satisfied (breaks don't literally overlap), consistent with the fact that two breaks are indeed observed experimentally — but the stricter $10^3$–$10^4$ ratio criterion correctly predicts that the first break will be noticeably softer than in a well-separated system like phosphoric acid. Both criteria agree in direction; they simply set the bar for "detectable" versus "sharp and precisely usable" at different heights.

Example 3 — Exam-level: triprotic acid, locate every stage precisely and assess the feasibility of the third break. A triprotic acid $\mathrm{H_3X}$ has $K_{a1}=7.5\times10^{-3}$, $K_{a2}=1.8\times10^{-7}$, $K_{a3}=4.0\times10^{-12}$. A 20.00 mL sample of 0.0500 M $\mathrm{H_3X}$ is titrated with 0.1000 M NaOH. Determine (a) the volumes at each equivalence point, (b) the pH at the second equivalence point, (c) whether the third equivalence point will be experimentally usable, and (d) the pH at 5.00 mL past the first equivalence point (still within buffer region 1–2 crossover, testing whether the simple Henderson–Hasselbalch treatment is even appropriate there). (a) Volumes. Moles of acid $=0.02000\text{ L}\times0.0500\text{ M}=1.000\times10^{-3}$ mol. Each equivalence point requires the same molar amount of NaOH as the original acid (one proton per stage):

(b) pH at the second equivalence point (20.00 mL added, total volume $=40.00$ mL). Dominant species is $\mathrm{HX^{2-}}$, formal concentration:

Check whether $C\gg K_{a2}$ and $K_{a2}C\gg K_w$: $K_{a2}=1.8\times10^{-7}\ll C$, comfortably satisfied; $K_{a2}C = 4.5\times10^{-9} \gg K_w=10^{-14}$, satisfied. The simplified formula applies cleanly:

Equivalently, $\text{pH} \approx \tfrac12(pK_{a2}+pK_{a3}) = \tfrac12(6.745+11.398) = 9.07$ — matches exactly, confirming the shortcut. (c) Feasibility of the third break. Test the ratio between the second and third dissociation constants:

This comfortably clears the $10^3$–$10^4$ threshold on ratio grounds alone. But there is a second, independent feasibility condition (as in a single weak acid, per Effect of Concentration and Ka on Titration Feasibility): the third step's own $K_{a3}\cdot C$ product must be large enough to produce a break at all. At the third equivalence point the total volume is $0.02000+0.03000=0.05000$ L, giving formal $\mathrm{X}^{3-}$ concentration $C'''=1.000\times10^{-3}/0.05000=0.02000$ M, so:

Compare to the conventional feasibility floor of roughly $K_a C \gtrsim 10^{-8}$ for a clearly visible break (below this, the jump is smaller than about 2 pH units and gets lost in the noise of a real electrode/indicator). Here $K_{a3}C'''=8\times10^{-14}$ is six orders of magnitude below that floor — so even though $K_{a2}/K_{a3}$ is large enough for separation from the second break, the third break itself is far too weak an acid at this concentration to titrate at all. Practically, the solution near $V_3$ is already so close to being an extremely weak base solution (conjugate base of an acid with $pK_{a3}=11.40$) that adding more strong base barely changes the tiny residual $\mathrm{HX^{2-}}/\mathrm{X^{3-}}$ ratio compared to the swamping effect of $\mathrm{OH^-}$ itself — the third jump is smeared into the general rise of pH from excess strong base and is not detectable as a distinct equivalence point. This mirrors exactly why phosphoric acid's third equivalence point is unusable in practice. (d) pH at 5.00 mL past the first equivalence point (15.00 mL total added, i.e., exactly the second half-equivalence point). At 15.00 mL, moles of NaOH added $=1.500\times10^{-3}$ mol; this consumed all $1.000\times10^{-3}$ mol for the first proton and $0.500\times10^{-3}$ mol toward the second — exactly half of the second stage. So $[\mathrm{HX^{2-}}]=[\mathrm{H_2X^-}]$ and:

This is a clean check that "half-equivalence between stage $i$ and stage $i+1$" always lands exactly on $pK_{a,i+1}$, regardless of concentration — the same logic as part (b) of Example 1.

The systematic language of stepwise dissociation constants and buffer regions used throughout this article descends from the acid–base equilibrium framework built up in the early twentieth century, most influentially by the Danish chemist Søren Peter Lauritz Sørensen, who introduced the pH scale itself in 1909 while working on enzyme kinetics at the Carlsberg Laboratory in Copenhagen. Sørensen needed a workable, reproducible way to express and control the acidity of protein solutions for brewing and biochemical studies — precisely the kind of multi-component, buffered system where polyprotic acids like phosphoric and carbonic acid dominate real physiological and industrial chemistry. His collaborators, including his wife Margrethe Høyrup Sørensen, extended this into detailed studies of amino acid titration behavior, directly anticipating the amphiprotic-species mathematics used above. The formalism of stepwise ionization constants ($K_{a1}, K_{a2}, \ldots$) and the systematic treatment of polyprotic equilibria were consolidated later by physical chemists working on buffer theory — most notably through the widespread adoption of the Henderson–Hasselbalch relationship (Lawrence Joseph Henderson's physiological buffer work, formalized mathematically by Karl Albert Hasselbalch in 1916–1917) into a general stepwise tool applicable to any acid with multiple ionizable protons, from carbonic acid in blood to phosphoric acid in soil chemistry to citric acid in food science.