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Chemistry: Atoms First 2e · Acid-Base Equilibria

Polyprotic Acids

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On this page 8 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Key takeaway
  6. Check yourself
  7. Study tools
  8. Sources & references

In 30 seconds

An acid that can donate more than one proton is polyprotic. Phosphoric acid (H3PO4) is triprotic: it can give up three protons, one at a time. Carbonic acid (H2CO3) and sulfuric acid (H2SO4) are diprotic. The key idea is that the protons are not all released at once. Each step has its own equilibrium constant — Ka1, Ka2, Ka3 — and each successive constant is much smaller than the one before:

H3PO4(aq) + H2O(l) ⇌ H3O+(aq) + H2PO4-(aq)   Ka1 = 7.5 × 10-3

H2PO4-(aq) + H2O(l) ⇌ H3O+(aq) + HPO42−(aq)   Ka2 = 6.2 × 10-8

HPO42−(aq) + H2O(l) ⇌ H3O+(aq) + PO43−(aq)   Ka3 = 4.2 × 10-13

Removing a proton from an already negatively charged species is progressively harder, which is why Ka1 ≫ Ka2 ≫ Ka3. For most calculations this means the first step dominates: the H3O+ concentration is set almost entirely by Ka1, and later steps contribute so little that they can be ignored. Sulfuric acid is the famous exception — its first proton is strong (complete ionization), and its second step is weak but not negligible.

Why this matters

Polyprotic acids sit at the center of some of the most important natural systems:

  • The body's carbon dioxide transport: Dissolved CO2 forms carbonic acid, which ionizes in two steps to bicarbonate and carbonate. This H2CO3/HCO3−/CO32− family is the blood buffer system and the reason exhaling regulates pH.
  • Phosphate in biology: ATP, DNA, and RNA all contain phosphate groups. The phosphate buffer inside cells uses H2PO4−/HPO42−, and the stepwise constants of phosphoric acid explain which phosphate form exists at body pH.
  • The ocean and the geosphere: The carbonate system controls ocean pH and the dissolution/precipitation of calcium carbonate — the chemistry behind coral reefs, limestone caves, and the shells of marine organisms.
  • Industry: Sulfuric acid, the most-produced chemical worldwide, owes its behavior to being diprotic; the second ionization matters in lead-acid batteries and acid processing.
  • Exams: Polyprotic problems test whether you can set up multiple equilibria, use the "first step dominates" approximation, and recognize when it fails (the H2SO4 case).

The college version

Core Concepts

Stepwise ionization

Each proton leaves in its own step, and each step has its own species and constant. For a diprotic acid H2A:

H2A + H2O ⇌ H3O+ + HA−   Ka1 = [H3O+][HA−][H2A]

HA− + H2O ⇌ H3O+ + A2−   Ka2 = [H3O+][A2−][HA−]

The intermediate HA− is amphiprotic — it can lose another proton or accept one back. Each successive Ka is smaller because removing a proton from a species with a growing negative charge requires more energy: for typical weak diprotic acids Ka2 is 104–105 times smaller than Ka1.

The first step dominates

Because Ka1 ≫ Ka2 ≫ Ka3, nearly all of the H3O+ in solution comes from the first ionization. Two approximations follow for a weak :

  1. [H3O+] ≈ [HA−] — the first step alone fixes the hydronium concentration.
  2. The second and third steps contribute negligible H3O+ and can be ignored when calculating pH.

The second step still matters for one specific question: the concentration of the fully deprotonated ion (e.g., CO32− or PO43−) is governed by the later constants, but for pH purposes the first step is the whole story. This approximation is only valid when the successive constants differ by many orders of magnitude — which they normally do.

Sulfuric acid: the exception

Sulfuric acid's first proton is strong, ionizing completely:

H2SO4(aq) + H2O(l) → H3O+(aq) + HSO4-(aq)

The second proton is weak, with Ka2 = 1.2 × 10-2 for HSO4−. Unlike the tiny Ka2 values of weak diprotic acids, this one is large enough that the second ionization contributes measurably — roughly 10% extra hydronium in a 0.10 M solution. Ignoring it produces a noticeably wrong pH, so H2SO4 problems must include both steps.

The fully deprotonated ion is a strong base

The conjugate base left after all protons are removed is the strongest base among the acid's conjugate series. Carbonate ion, CO32−, has Kb = Kw/Ka2 = 1.0 × 10-14/5.6 × 10-11 = 1.8 × 10-4 — strong enough that sodium carbonate solutions are distinctly basic (this connects directly to salt hydrolysis). The see-saw of the previous topic applies at every level of a polyprotic series.

How It Works / Step-by-Step Process

Worked example 1: pH of a weak diprotic acid

Problem. Find the pH of 0.10 M carbonic acid, H2CO3, using Ka1 = 4.3 × 10-7 (and Ka2 = 5.6 × 10-11).

Solution.

  1. Write the dominant step:

H2CO3(aq) + H2O(l) ⇌ H3O+(aq) + HCO3-(aq)   Ka1 = [H3O+][HCO3−][H2CO3]

  1. Let x = [H3O+] = [HCO3−] and [H2CO3] = 0.10 - x; assume x ≪ 0.10:

Ka1 ≈ x20.10 = 4.3 × 10-7   ⇒  x = (4.3 × 10-7)(0.10) = 2.1 × 10-4 M

  1. Check the approximation: 2.1 × 10-4/0.10 = 0.21% < 5%, so it holds.
  1. Convert to pH:

pH = -log(2.1 × 10-4) = 3.68

  1. Verify the second step is negligible: it would add roughly Ka2 = 5.6 × 10-11 M of extra H3O+ — eleven orders of magnitude below the first step's contribution. Dimensional check: all concentrations are molar (M), and Ka1 (dimensionless) times M gives M² under the square root, which returns M.

Worked example 2: sulfuric acid needs both steps

Problem. Find the pH of 0.10 M H2SO4, given Ka2 = 1.2 × 10-2.

Solution.

  1. First step (strong, complete): 0.10 M H2SO4 produces [H3O+] = 0.10 M and [HSO4−] = 0.10 M.
  2. Second step is an equilibrium: HSO4-(aq) + H2O(l) ⇌ H3O+(aq) + SO42−(aq). Let x be the extra hydronium and sulfate formed. Then [H3O+] = 0.10 + x, [SO42−] = x, [HSO4−] = 0.10 - x:

Ka2 = (0.10 + x)(x)0.10 - x = 1.2 × 10-2

  1. Because x is not tiny compared with 0.10, solve the quadratic. Expanding: x2 + 0.10x = 1.2 × 10-3 - 1.2 × 10-2x, so x2 + 0.112x - 1.2 × 10-3 = 0, giving x = 9.9 × 10-3 M.
  2. Total hydronium: [H3O+] = 0.10 + 0.0099 = 0.11 M. The second step added ~10% — clearly not negligible.

pH = -log(0.11) = 0.96

Ignoring the second step would have given pH 1.00 — an error of 0.04 units that matters in precise work.

Worked example 3: when the 5% rule fails

Problem. Estimate the pH of 0.10 M H3PO4 (Ka1 = 7.5 × 10-3). Why must the quadratic be solved?

Solution.

  1. First-step setup with x = [H3O+] = [H2PO4−]:

Ka1 = x20.10 - x = 7.5 × 10-3

  1. Try the shortcut x2/0.10 = 7.5 × 10-3: x = 2.7 × 10-2 M. But that is 27% of 0.10 — the < 5% rule fails, so the shortcut is invalid.
  2. Solve the full quadratic: x2 + 7.5 × 10-3x - 7.5 × 10-4 = 0, giving x = 2.4 × 10-2 M.
  3. Then:

pH = -log(2.4 × 10-2) = 1.62

The second step (Ka2 = 6.2 × 10-8) adds only ~10-6 M of hydronium, so it is still negligible — but the first step itself needed the quadratic because phosphoric acid is a moderately strong weak acid.

Common Confusions

Do Not ConfuseWithDifference
All protons released at onceStepwise releaseProtons leave one at a time; each step has its own Ka, and intermediate ions exist in solution.
H2SO4 treated like a weak diprotic acidIts two weak stepsThe first proton is strong (complete); only the second is weak, and it still contributes ~10% hydronium.
Ignoring later steps alwaysIgnoring them only for pHLater steps are negligible for pH but control the concentration of the fully deprotonated ion (e.g., PO43−).
First-step shortcut x ≈ Ka1CAlways validFails when ionization exceeds ~5% (e.g., 0.10 M H3PO4); then solve the quadratic.
H2CO3 and H2SO4 behaving alikeBoth being "diprotic"Both are diprotic, but carbonic acid's first step is weak while sulfuric acid's is strong — very different pH behavior.
Ka2 of the acidKb of the fully deprotonated ionCO32−'s basicity comes from Kb = Kw/Ka2, not from Ka2 itself.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

A polyprotic acid is like a person carrying several heavy boxes up a staircase, setting one down at each landing. The first box comes off easily — the person was carrying it with two hands. Each next box is harder because the person is getting more and more loaded down (and more negative). So the acid gives up its first proton readily, the second reluctantly, and the third almost never. For most questions you only need to watch the first landing; sulfuric acid is the show-off that drops its first box instantly and then still manages to set down a second.

Key takeaways

  • Polyprotic acids ionize in steps: Ka1 ≫ Ka2 ≫ Ka3; each step is harder because the species becomes more negatively charged.
  • For weak polyprotic acids, pH is controlled by the first step: [H3O+] ≈ Ka1 × [H2A]0 when ionization is < 5%.
  • The intermediate ions (HCO3−, H2PO4−, HSO4−) are amphiprotic — they can act as acids or bases.
  • H2SO4 is special: first proton strong (complete), second weak but not negligible (Ka2 = 1.2 × 10-2); include both steps for accurate pH.
  • The fully deprotonated ion is the strongest base in the series (e.g., CO32− with Kb = 1.8 × 10-4).
  • Common values at 25 °C: H2CO3: Ka1 = 4.3 × 10-7, Ka2 = 5.6 × 10-11; H3PO4: 7.5 × 10-3, 6.2 × 10-8, 4.2 × 10-13.
  • Later steps set the concentration of the fully deprotonated anion, but rarely affect pH.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

  1. Why is each successive ionization constant of a polyprotic acid smaller than the previous one?

    Show answer

    Removing a proton from an increasingly negative species requires more energy, so each step is less favorable — hence Ka1 ≫ Ka2 ≫ Ka3.

  2. Write the two ionization steps of carbonic acid, H2CO3, with their constants' symbols.

    Show answer

    H2CO3 + H2O ⇌ H3O+ + HCO3− (Ka1); HCO3− + H2O ⇌ H3O+ + CO32− (Ka2).

  3. For a 0.10 M weak diprotic acid with Ka1 = 1.0 × 10-4, why can you ignore Ka2 when finding pH?

    Show answer

    Because Ka2 is typically 104–105 times smaller than Ka1; the second step adds a negligible amount of H3O+, so pH is set by the first step alone.

  4. Why must both steps be considered for 0.10 M H2SO4?

    Show answer

    The first proton is strong (complete ionization giving 0.10 M H3O+), and the second step's Ka2 = 1.2 × 10-2 is large enough to add ~10% more hydronium; ignoring it misreports the pH.

  5. Which species is the strongest base in the series H2CO3, HCO3−, CO32−? How would you find its Kb?

    Show answer

    CO32−, the fully deprotonated ion, is the strongest base; Kb(CO32−) = Kw/Ka2 = 1.0 × 10-14/5.6 × 10-11 = 1.8 × 10-4.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

polyprotic acid
An acid that can donate more than one proton
monoprotic / diprotic / triprotic
Donates one / two / three protons
stepwise ionization
Protons released one at a time, each with its own Ka
Kₐ₁, Kₐ₂, Kₐ₃
Successive ionization constants, each much smaller than the last
amphiprotic intermediate
Ion (e.g., HSO4−, HCO3−) that can donate or accept a proton
first-step approximation
Ignoring Ka2 and Ka3 for pH calculations

Sources & references

  1. openstax.org — Chemistry Atoms First 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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