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Chemistry 2e · Acid-Base Equilibria

Polyprotic Acids

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On this page 9 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Check yourself
  8. Study tools
  9. Sources & references

In 30 seconds

The acids you have met so far — HCl, acetic acid, ammonia's conjugate acid — donate or accept exactly one proton. Polyprotic acids can donate more than one: sulfuric acid has two, phosphoric acid has three, carbonic acid has two. The key twist: the protons come off one at a time, each step with its own equilibrium constant. The first step is the "easiest" and has the largest Ka1; later steps fight the growing negative charge on the ion and have much smaller constants. A in water is several simultaneous acid–base equilibria, and knowing which one controls the pH is the heart of this topic.

Why this matters

Polyprotic acids are everywhere in biology and industry. Phosphoric acid is the backbone of DNA and ATP, and phosphate's three pKa values make it a versatile biological buffer. Carbonic acid and bicarbonate control blood pH and drive the release of CO₂ in your lungs. Sulfuric acid is the most-produced industrial chemical in the world, and its two-step behavior changes how it is handled and neutralized. Soft drinks fizz because carbonic acid decomposes; hard water leaves scale because carbonate chemistry shifts with pH. Predicting which species is present at a given pH is a constant skill in biochemistry and environmental chemistry.

The college version

Core Concepts

Stepwise ionization: one proton at a time

Phosphoric acid ionizes in three distinct steps, each with its own constant:

H3PO4 + H2O ⇌ H2PO4- + H3O+   Ka1 = 7.5 × 10-3

H2PO4- + H2O ⇌ HPO42- + H3O+   Ka2 = 6.2 × 10-8

HPO42- + H2O ⇌ PO43- + H3O+   Ka3 = 4.2 × 10-13

Each constant is dramatically smaller than the one before it: removing a proton from a neutral molecule is easier than pulling one off a negatively charged ion, and harder still off a doubly charged one. This pattern — Ka1 ≫ Ka2 ≫ Ka3 — holds for essentially every polyprotic acid. (Exception: sulfuric acid's first step is complete, so only Ka2 = 1.2 × 10-2 is a real equilibrium.)

Which step controls the pH?

For the pure acid, the first ionization dominates [H3O+]. The later steps are suppressed by their tiny constants and by the common-ion effect: the H3O+ already produced pushes them far to the left. For most diprotic acids you can calculate pH using only Ka1; the step-1 species (like H2PO4-) matters for , not for the pH of the acid's own solution.

The special case of sulfuric acid

Sulfuric acid is unique among common acids: the first proton comes off completely (it is a strong acid), leaving HSO4-, which then behaves as a weak acid with Ka2 = 1.2 × 10-2. Calculating the pH of H₂SO₄ requires treating the first step as complete (initial [H+] = [HSO4-] = C) and then solving the second equilibrium for the extra protons. This is the one polyprotic case where you cannot ignore the second step, because Ka2 is large enough to matter.

Speciation: which form dominates at which pH

The fraction of each form (H₃PO₄, H₂PO₄⁻, HPO₄²⁻, PO₄³⁻) depends on pH, not concentration: the acid form dominates below pKa1, the first conjugate base between pKa1 and pKa2, and so on. At blood pH 7.4, phosphate exists mostly as H₂PO₄⁻ and HPO₄²⁻ in nearly equal amounts — exactly the pair that makes a good buffer.

How It Works / Step-by-Step Process

  1. Write the stepwise reactions, each with its own Ka.
  2. For the pure acid, assume the first step sets [H3O+]; ignore later steps unless Ka1 is large (e.g., sulfuric acid).
  3. Set up an ICE table for the first step using Ka1.
  4. Check the 5% rule; if ionization exceeds 5%, solve the quadratic instead of using Ka1 C.
  5. For H₂SO₄ only: treat step 1 as complete, then run a second ICE table with Ka2 for the extra [H3O+].
  6. For speciation, compare pH with pKa1, pKa2, pKa3 to identify the dominant form.

Common Confusions

Do Not ConfuseWithDifference
"One Ka for the whole acid"One Ka per protonEach step has its own constant; H₃PO₄ has three.
Doubling the Ka1 resultAdding step-2 protonsLater steps are usually negligible, not additive — never "double" the first pH.
H₂SO₄ treated like H₃PO₄H₂SO₄'s first step is completeH₂SO₄'s Ka1 is effectively infinite; H₃PO₄'s first step is a real equilibrium (Ka1 = 7.5 × 10-3).
HCO3- as only a baseHCO3- as amphiproticIt can donate a proton (Ka2) or accept one (Kw/Ka1); the bigger constant wins.
Using Ka2 for the acid's pHUsing Ka1For the pure acid solution, Ka1 sets the pH; Ka2 matters for salts of the acid (e.g., NaHCO₃).
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 candy machine that gives out protons one coin at a time. The first coin is easy — the machine is full and happy. After the first proton leaves, the molecule has a negative charge, the machine is less eager, and the second proton is harder to get. Each time one leaves, the next is even harder — which is why the equilibrium constants get smaller with each step.

Worked example

Example 1: pH of 0.10 M phosphoric acid

Phosphoric acid has Ka1 = 7.5 × 10-3, Ka2 = 6.2 × 10-8, Ka3 = 4.2 × 10-13. Estimate the pH of a 0.10 M solution. Step 1 sets the pH, so use Ka1:

Ka1 = x20.10 - x = 7.5 × 10-3

Try 0.10 - x ≈ 0.10:

x = (7.5 × 10-3)(0.10) = 2.7 × 10-2 M

That is 27% ionization — far above the 5% rule — so the approximation fails and we must solve the quadratic:

x2 + (7.5 × 10-3)x - 7.5 × 10-4 = 0

x = -7.5 × 10-3 + (7.5 × 10-3)2 + 4(7.5 × 10-4)2 = 2.4 × 10-2 M

pH = -log(2.4 × 10-2) = 1.62

The second and third steps contribute almost nothing: Ka2 is a hundred thousand times smaller than Ka1, and the common-ion effect suppresses it further.

Example 2: pH of 0.10 M sulfuric acid

Sulfuric acid's first proton ionizes completely: [H3O+]0 = [HSO4-]0 = 0.10 M. Now treat HSO4- as a weak acid with Ka2 = 1.2 × 10-2:

HSO4- + H2O ⇌ SO42- + H3O+,   Ka2 = [SO42-][H3O+][HSO4-] = 1.2 × 10-2

Let x = [SO42-]. Then [H3O+] = 0.10 + x and [HSO4-] = 0.10 - x:

1.2 × 10-2 = x(0.10 + x)0.10 - x

This rearranges to x2 + 0.112x - 1.2 × 10-3 = 0, giving x = 9.9 × 10-3 M. The total proton concentration:

[H3O+] = 0.10 + 9.9 × 10-3 = 0.11 M,   pH = -log(0.11) = 0.96

Note the contrast: ignoring the second step predicts pH 1.00; including it gives 0.96 — a real difference and a common exam trap.

Example 3: dominant species of carbonic acid at blood pH

Carbonic acid has pKa1 = 6.4 and pKa2 = 10.3. At blood pH 7.4, which form dominates?

Since 7.4 is above pKa1 but below pKa2, the first deprotonation has largely occurred and the second has not: the mixture is mostly HCO3- (bicarbonate) with a substantial fraction of H2CO3. This H₂CO₃/HCO₃⁻ pair is the body's main buffer — see Buffers. CO32- is essentially absent at blood pH.

Key takeaways

  • Protons leave one at a time; each step has its own Ka1, Ka2, Ka3.
  • Ka1 ≫ Ka2 ≫ Ka3: removing a proton from a negatively charged ion is harder.
  • For the pure acid, pH is set almost entirely by Ka1; later steps are negligible in most calculations.
  • Sulfuric acid is the exception: step 1 is complete, then HSO4- ionizes with Ka2 = 1.2 × 10-2 — the second step must be included.
  • Common acids: H₂SO₄, H₃PO₄, H₂CO₃, H₂S, oxalic acid.
  • Amphiprotic intermediates (H₂PO₄⁻, HCO₃⁻) can act as acids or bases; which constant applies decides the outcome.
  • Speciation follows pH: below pKa1 the fully protonated form dominates; between pKa1 and pKa2 the singly deprotonated form does, etc.

Check yourself

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

  1. Why is Ka2 always much smaller than Ka1 for a polyprotic acid?

    Show answer

    Pulling a proton off a negatively charged species is electrostatically harder than off a neutral molecule, and progressively harder with each added charge, so each successive constant is much smaller.

  2. For 0.10 M H₂CO₃ (Ka1 = 4.3 × 10-7), why is it safe to ignore the second ionization when finding pH, and what pH results?

    Show answer

    Ka2 = 5.6 × 10-11 is so tiny its proton contribution is negligible. Using Ka1: x = (4.3 × 10-7)(0.10) = 2.1 × 10-4 M, so pH ≈ 3.68 (0.2% ionization — well within the 5% rule).

  3. Why can't you "ignore the second step" for 0.10 M H₂SO₄ as you can for carbonic acid?

    Show answer

    Its first step is complete, leaving 0.10 M HSO4-; Ka2 = 1.2 × 10-2 is large enough that the second step adds measurably to [H3O+] (pH 0.96 vs. 1.00).

  4. At pH 12, which phosphate species dominates: H₃PO₄, H₂PO₄⁻, HPO₄²⁻, or PO₄³⁻?

    Show answer

    HPO₄²⁻. With pKa2 = 7.2 and pKa3 = 12.4, pH 12 falls between them, so the doubly deprotonated form dominates.

  5. Is HCO3- an acid or a base in water, and how would you decide quantitatively?

    Show answer

    As an acid it uses Ka2 = 5.6 × 10-11; as a base, Kb = Kw/Ka1 = 2.3 × 10-8. Since Kb > Ka, bicarbonate is a net base in water.

Keep learning

Ready to build on this? Continue to the next lesson.

Study tools & related lessonsKey vocabulary · Related

Key vocabulary

polyprotic acid
Acid that can donate more than one proton.
stepwise ionization
Sequential loss of protons, each with its own equilibrium constant.
Kₐ₁, Kₐ₂, Kₐ₃
Equilibrium constants for the first, second, and third proton losses.
amphiprotic ion
Intermediate that can accept or donate a proton (H₂PO₄⁻, HCO₃⁻).
speciation
Distribution of an acid's forms as a function of pH.

Sources & references

  1. openstax.org — Chemistry 2e

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

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