Polyprotic Acid Titration Curves: Ka1, Ka2, Ka3 and Stepwise Equivalence Points (Worked Examples)
By ChemistryIQ Team · April 18, 2026
Direct Answer: Polyprotic Curves Have Multiple Equivalence Points
A polyprotic acid has more than one ionizable proton. The sequential deprotonation produces a titration curve with multiple equivalence points — one for each acidic proton. Phosphoric acid (H₃PO₄) has three equivalence points because it has three ionizable protons. Carbonic acid (H₂CO₃) and sulfurous acid (H₂SO₃) each have two equivalence points.
Between each equivalence point is a 'buffer region' where pH changes slowly as base is added. The midpoint of each buffer region (the half-equivalence point) is where pH = pKa for that ionization step. This gives you a direct way to measure Ka values from an experimental titration curve.
The successive Ka values typically differ by 4-6 orders of magnitude (Ka1 >> Ka2 >> Ka3). This large separation ensures that each deprotonation step is essentially complete before the next one begins — the curve shows distinct plateaus and sharp equivalence-point transitions rather than blurred overlapping steps.
For H₃PO₄: pKa1 = 2.15, pKa2 = 7.20, pKa3 = 12.35. The three buffer regions center on pH 2.15, 7.20, and 12.35 respectively. The three equivalence points occur at pH values calculated from the species in solution at each stage.
The titration of a polyprotic acid against a strong base is the most common exam format. Every H⁺ added neutralizes with OH⁻ in a 1:1 ratio, so calculating moles at each stage is straightforward — but you must track which species dominate at each pH.
The Stepwise Dissociations and Their Ka Values
Phosphoric acid (triprotic) dissociates in three steps:
Step 1: H₃PO₄ ⇌ H⁺ + H₂PO₄⁻ Ka1 = [H⁺][H₂PO₄⁻] / [H₃PO₄] = 7.1 × 10⁻³ (pKa1 = 2.15)
Step 2: H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻ Ka2 = [H⁺][HPO₄²⁻] / [H₂PO₄⁻] = 6.3 × 10⁻⁸ (pKa2 = 7.20)
Step 3: HPO₄²⁻ ⇌ H⁺ + PO₄³⁻ Ka3 = [H⁺][PO₄³⁻] / [HPO₄²⁻] = 4.5 × 10⁻¹³ (pKa3 = 12.35)
Notice the dramatic decrease: Ka1 / Ka2 ≈ 10⁵, Ka2 / Ka3 ≈ 10⁵. Each successive proton is much harder to remove because: (1) the molecule becomes more negatively charged after losing a proton, making the next deprotonation electrostatically unfavorable, and (2) the remaining protons are bound to oxygen atoms that now bear partial negative charge, strengthening the O-H bond.
Carbonic acid (diprotic): pKa1 = 6.35, pKa2 = 10.33 Sulfurous acid (diprotic): pKa1 = 1.85, pKa2 = 7.20 Oxalic acid (diprotic): pKa1 = 1.25, pKa2 = 4.27 (unusually close) Ascorbic acid (diprotic): pKa1 = 4.10, pKa2 = 11.80
The separation between pKa values determines whether equivalence points on the titration curve are distinct. If pKa1 and pKa2 differ by more than 4 units, you see two clear equivalence points. If they differ by less than 3, the curves begin to merge — H₂SO₄ (both Ka values very large) behaves essentially like a single strong-acid titration in water.
For exam problems: ALWAYS check whether the given pKa values are separated enough for distinct equivalence points. If they differ by less than about 3 units, the analysis gets more complex and requires polynomial equations rather than simplified stepwise treatment.
Worked Example 1: Titration of H₃PO₄ with NaOH
Titrate 50.0 mL of 0.100 M H₃PO₄ with 0.100 M NaOH. Calculate pH at several key points and sketch the curve.
Initial moles of H₃PO₄: 0.0500 L × 0.100 M = 5.00 × 10⁻³ mol = 5.00 mmol.
Key volumes of NaOH: - First equivalence point: 50.0 mL (5.00 mmol NaOH, converts all H₃PO₄ to H₂PO₄⁻) - First half-equivalence: 25.0 mL (equal H₃PO₄ and H₂PO₄⁻, pH = pKa1) - Second equivalence: 100.0 mL (converts all H₂PO₄⁻ to HPO₄²⁻) - Second half-equivalence: 75.0 mL (equal H₂PO₄⁻ and HPO₄²⁻, pH = pKa2) - Third equivalence: 150.0 mL (converts all HPO₄²⁻ to PO₄³⁻) - Third half-equivalence: 125.0 mL (equal HPO₄²⁻ and PO₄³⁻, pH = pKa3)
pH at initial point (V = 0 mL): H₃PO₄ is mostly undissociated. Apply Ka1: Ka1 = x² / (0.100 - x) ≈ x² / 0.100 (assuming x << 0.100) x² = 0.100 × 7.1 × 10⁻³ = 7.1 × 10⁻⁴ x = [H⁺] = 0.0266 M pH = -log(0.0266) = 1.58
Check: x/0.100 = 0.266 → 26.6% — not negligibly small. Use quadratic: x² + 7.1×10⁻³ x - 7.1×10⁻⁴ = 0 x = [-7.1×10⁻³ + √(5.04×10⁻⁵ + 2.84×10⁻³)] / 2 = 0.0232 pH = -log(0.0232) = 1.63
pH at first half-equivalence (V = 25.0 mL): [H₃PO₄] = [H₂PO₄⁻] → pH = pKa1 = 2.15 (Henderson-Hasselbalch with log(1) = 0)
pH at first equivalence (V = 50.0 mL): All H₃PO₄ → H₂PO₄⁻. H₂PO₄⁻ is amphoteric (can accept or donate proton). pH = (pKa1 + pKa2) / 2 = (2.15 + 7.20) / 2 = 4.68 This is an approximation valid when the acid concentration is not too dilute.
pH at second half-equivalence (V = 75.0 mL): [H₂PO₄⁻] = [HPO₄²⁻] → pH = pKa2 = 7.20
pH at second equivalence (V = 100.0 mL): All H₂PO₄⁻ → HPO₄²⁻. HPO₄²⁻ is also amphoteric. pH = (pKa2 + pKa3) / 2 = (7.20 + 12.35) / 2 = 9.78
pH at third half-equivalence (V = 125.0 mL): [HPO₄²⁻] = [PO₄³⁻] → pH = pKa3 = 12.35
pH at third equivalence (V = 150.0 mL): All HPO₄²⁻ → PO₄³⁻. Now we have a weak base (PO₄³⁻) in solution with Kb = Kw/Ka3 = 10⁻¹⁴/(4.5×10⁻¹³) = 0.022. [PO₄³⁻] = 5.00 mmol / 200.0 mL = 0.0250 M x² / (0.0250 - x) = 0.022 Use quadratic: x = [OH⁻] = 0.0168 M pOH = 1.77, pH = 12.23
(Note: the third equivalence point is difficult to observe experimentally because its pH is close to pKa3, so the transition is gradual rather than sharp.)
Summary table: Volume (mL) | pH 0 | 1.63 25 | 2.15 (pKa1 — first half-equivalence) 50 | 4.68 (first equivalence) 75 | 7.20 (pKa2 — second half-equivalence) 100 | 9.78 (second equivalence) 125 | 12.35 (pKa3 — third half-equivalence) 150 | 12.23 (third equivalence — note this is barely distinguishable)
The Amphoteric Species and the Half-Equivalence Point Trick
At each equivalence point, the dominant species has just received a proton from the previous step and can either accept another proton or donate one further. These amphoteric species give rise to the approximation:
pH at first equivalence point = (pKa1 + pKa2) / 2 pH at second equivalence point = (pKa2 + pKa3) / 2
This approximation assumes: - The amphoteric species concentration is not too low (typically > 0.010 M) - Ka1 >> Ka2 so the species is a clear 'middle state' - The water contribution to proton transfer is negligible
The approximation comes from solving the simultaneous equilibria: H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻ (Ka2, dissociation) H₂PO₄⁻ + H⁺ ⇌ H₃PO₄ (reverse of Ka1)
At equilibrium of a pure H₂PO₄⁻ solution, [H⁺] = √(Ka1 × Ka2), which gives pH = (pKa1 + pKa2) / 2.
At each HALF-equivalence point (between stages), the Henderson-Hasselbalch equation applies: pH = pKa + log([A⁻] / [HA])
At the half-equivalence point: [A⁻] = [HA], so log ratio = 0, and pH = pKa.
This half-equivalence point provides the simplest way to determine pKa experimentally: titrate, find where the volume is exactly half of the next equivalence point, and read the pH directly.
Exam trap: students sometimes apply pH = (pKa + pKa+1)/2 to half-equivalence points. Wrong — it applies to EQUIVALENCE points. Half-equivalence points give pH = pKa directly.
Buffer Regions on Polyprotic Curves
The flat regions of the titration curve around each half-equivalence point are BUFFER regions. Here, pH changes slowly as base is added because the solution is a buffer between two conjugate acid-base species.
For phosphate buffer around pKa2 = 7.20: pH = 7.20 + log([HPO₄²⁻] / [H₂PO₄⁻])
A 1:1 mixture gives pH = 7.20. A 10:1 mixture (more base form) gives pH = 8.20. A 1:10 mixture gives pH = 6.20. The buffer is most effective when the ratio is within 10:1 to 1:10, corresponding to pH = pKa ± 1.
Phosphate buffer (around pKa2 = 7.20) is widely used in biology because its effective range overlaps physiological pH. Phosphate-buffered saline (PBS), used in countless laboratory protocols, is exactly this buffer system.
Carbonate buffer (around pKa2 = 10.33) is used in higher-pH applications. Bicarbonate buffer (using Ka1, pKa1 = 6.35) is central to blood pH regulation.
Buffer capacity on the titration curve is directly observable: the flatter and longer the plateau, the higher the buffer capacity at that pH. Buffer capacity peaks at exactly pH = pKa (the half-equivalence point) because small changes in acid-base ratio produce minimal pH change.
Clinical connection: blood pH is maintained at 7.40 ± 0.05. The bicarbonate/carbonic acid buffer system (pKa1 = 6.35) has declining effectiveness at pH 7.40 because we're 1.05 pH units above pKa1. However, because the body can actively regulate CO₂ (via respiration) and bicarbonate (via kidneys), this buffer system works effectively despite being on the 'shoulder' of its titration curve. This is 'open system' buffering — different from the closed-system curves we draw in chemistry class.
Identifying Unknown Polyprotic Acids from Titration Data
A common lab exercise: titrate an unknown weak polyprotic acid, measure the titration curve, and identify the acid from its pKa values.
Procedure: 1. Dissolve a known mass of unknown acid in water. 2. Titrate with NaOH of known concentration while measuring pH. 3. Plot pH vs volume of NaOH added. 4. Identify equivalence points as inflection regions (pH changes rapidly). 5. Measure half-equivalence volumes and read pH at each — these are the pKa values. 6. Compare to known pKa values of candidate acids.
Worked example: An unknown triprotic acid is titrated with 0.100 M NaOH. Sharp inflections occur at V = 40.0 mL, 80.0 mL, and 120.0 mL. Half-equivalence points are at 20.0 mL (pH 2.1), 60.0 mL (pH 7.0), and 100.0 mL (pH 12.3).
Identified pKa1 = 2.1, pKa2 = 7.0, pKa3 = 12.3.
Compare to common triprotic acids: - Phosphoric acid (H₃PO₄): pKa 2.15, 7.20, 12.35 — CLOSE MATCH! - Citric acid (H₃C₆H₅O₇): pKa 3.15, 4.77, 6.40 — no match. - Arsenic acid (H₃AsO₄): pKa 2.19, 6.94, 11.50 — somewhat close but pKa3 is off.
Conclusion: the unknown is likely phosphoric acid.
Confirmation via moles: First equivalence at V = 40.0 mL means 4.00 mmol NaOH has been added to reach the monoprotic state. Originally, the acid was triprotic, so moles of acid = 4.00 mmol (each acid molecule neutralized by one NaOH at the first equivalence).
Volume of acid solution should give concentration = 4.00 mmol / initial volume. For a typical 25.0 mL sample, [H₃PO₄] = 4.00 mmol / 25.0 mL = 0.160 M.
Multiply by molar mass (98.0 g/mol for H₃PO₄) to get mass. If the known starting mass was 0.392 g and the computed mass is 0.392 g, identification is confirmed.
This procedure is a staple of quantitative analysis labs. Precise pH electrode, careful titration technique, and accurate volume measurement are essential.
Common Mistakes and Exam Traps
Mistake 1 — Confusing equivalence points with half-equivalence points. Equivalence points: species has been fully converted to next form. pH = (pKa + pKa+1)/2 (approximately, for amphoteric species). Half-equivalence points: species is exactly half-converted. pH = pKa.
Mistake 2 — Forgetting to update the amount of acid remaining after each step. In a triprotic titration, after the first equivalence point, you're working with the monoprotic H₂PO₄⁻. After the second, with HPO₄²⁻. Don't continue to use the original H₃PO₄ concentration.
Mistake 3 — Not accounting for dilution. As you add NaOH, total volume increases. Species concentrations decrease. Dilution matters for quantitative calculations, especially near equivalence points where concentrations can be quite low.
Mistake 4 — Assuming all equivalence points are observable. When pKa values are too close (< 3 units apart), successive equivalence points merge into a single inflection. The theoretical curve might show two equivalence points but experimental detection becomes difficult.
Mistake 5 — Applying Henderson-Hasselbalch in the wrong buffer region. Between V=25 mL and V=75 mL, you're in the Ka1/Ka2 buffer region. Between V=75 mL and V=125 mL, you're in the Ka2/Ka3 buffer region. Use the right Ka for the right region.
Mistake 6 — Ignoring quadratic requirements for initial concentration calculations. At the start of titration, if the weak acid concentration is not dilute, don't use the approximation x << C. Use the quadratic formula to avoid systematic error.
Mistake 7 — Forgetting that the last equivalence point for triprotic acids is often poorly defined. For H₃PO₄, the third equivalence point (pH ≈ 12.2) is barely different from the buffer region around pKa3 (pH 12.35). Drawing a sharp curve through this region is misleading — it's often a gentle slope.
Mistake 8 — Mis-reading direction of approximation. pH at first equivalence = (pKa1 + pKa2)/2 works AS AN APPROXIMATION. For precise values, solve the full amphoteric equilibrium, which can differ by 0.1-0.3 pH units depending on concentration and Ka ratios.
Mistake 9 — Using Ka values instead of pKa when doing pH calculations directly. Keep pKa values (positive numbers) in the mental framework; convert to Ka when explicitly needed.
Mistake 10 — Confusing polyprotic acid titration with polybasic base titration. H₃PO₄ + NaOH goes acid-to-base. A polybasic base like Na₃PO₄ titrated with HCl goes the opposite direction, but with the same set of equivalence points in reverse order.
FAQs
Common questions about polyprotic acid titration curves
Each acidic proton is removed in a separate reaction step with its own equilibrium constant (Ka1, Ka2, Ka3...). As base is added, the first proton is removed first (converting H3A to H2A⁻). Once that conversion is complete, the first equivalence point is reached. Additional base then removes the second proton (converting H2A⁻ to HA²⁻), creating the second equivalence point. And so on. The large separation between Ka values ensures each step is essentially complete before the next begins, producing distinct equivalence points rather than a single smooth transition.
The half-equivalence point for each step of a polyprotic acid titration is the midpoint of that step's buffer region — where exactly half of the species at that stage has been converted to the next form. At this point, the Henderson-Hasselbalch equation gives pH = pKa for that step, since the ratio [conjugate base]/[acid] = 1 and log(1) = 0. Half-equivalence points are the easiest way to read pKa values off an experimental titration curve.
Because pKa3 = 12.35 is very close to the pH of pure NaOH solution (around 12-13). When you add NaOH to the HPO4²⁻ solution, there is no sharp inflection — the pH simply rises gradually as excess OH⁻ accumulates. The third equivalence point, in theory, occurs at approximately pH 12.2, but the titration curve is too gentle in this region to identify precisely. In practice, H3PO4 titrations often stop at the second equivalence point (pH ≈ 9.8) and measure pKa3 separately using other methods (like conductometric titration or spectrophotometry).
Oxalic acid has pKa1 = 1.25 and pKa2 = 4.27 — separated by only 3 pH units, which is on the borderline for clear separation. The first equivalence point is less sharp than for acids with larger pKa separation. When pKa values differ by less than 2.5 units, the equivalence points begin to merge into a single gradual transition. Full mathematical treatment requires solving coupled equilibria rather than the simplified stepwise approach. H2SO4's two Ka values are essentially the same (both very large) — so it titrates as if it were monoprotic with a single sharp equivalence point at pH 7 (for strong-acid/strong-base titration).
The body uses multiple buffer systems stacked to cover the physiological pH range. The most important are: (1) bicarbonate/carbonic acid (pKa1 = 6.35, covering pH 6-7); (2) phosphate (pKa2 = 7.20, covering pH 7.2 — closest to physiological); and (3) proteins (multiple pKa values of side chains, covering broad range). The bicarbonate system is especially effective because it's an 'open' system — the body can actively adjust CO2 through respiration (respiratory compensation) and HCO3- through renal regulation (renal compensation). These dynamic adjustments allow the bicarbonate system to function effectively even though physiological pH (7.4) is above its optimal buffering pH.
Yes. Snap a photo of any titration curve or polyprotic acid problem and ChemistryIQ identifies which stage of the titration applies (initial, buffer region, equivalence point), applies Henderson-Hasselbalch for buffer regions, uses the (pKa_n + pKa_(n+1))/2 approximation for equivalence points with amphoteric species, and walks through the dilution and mole-balance calculations. It also handles unknown polyprotic acid identification from titration data — matching measured pKa values against standard tables.