Polyprotic Acids and Salt Hydrolysis

Most of the acids met so far in this course donate a single proton: HCl, HNO₃, CH₃COOH. Many acids of biological and industrial importance, however, can release two or three protons in sequence. Polyprotic acids — sulfuric (H₂SO₄), carbonic (H₂CO₃), phosphoric (H₃PO₄), hydrogen sulfide (H₂S), oxalic (HOOCCOOH) — dissociate stepwise, each step governed by its own Ka. Their conjugate bases include partially-deprotonated species such as HSO₄⁻, HCO₃⁻ and H₂PO₄⁻ — amphiprotic ions that can either lose or gain a further proton. The salts of weak acids and weak bases do not give neutral solutions: a phenomenon called salt hydrolysis shifts the pH systematically. This lesson covers the stepwise-pKa pattern, the four classes of salt-hydrolysis behaviour, quantitative pH prediction for salt solutions, the special algebra of amphiprotic ions, multi-step titration curves, and the biological role of phosphate and bicarbonate buffering. It is the final pH-quantitation lesson before Required Practical 9 in lesson 7.

Spec mapping (AQA 7405): This lesson maps to §3.1.12 (acids and bases — extended treatment), with stepwise dissociation, salt hydrolysis and amphiprotic species developed as the A2 extension of the Brønsted–Lowry framework. It builds directly on lesson 0 (Brønsted–Lowry definitions), lesson 2 (Ka, Kb and Kw quantitative treatment), and lesson 5 (titration curves for monoprotic systems). Group 1 and Group 2 salts in solution are revisited from §3.2 (inorganic chemistry of s-block). Phosphate biological buffering signposts §3.3.13 (amino acids and zwitterions). Refer to the official AQA specification document for the exact wording of §3.1.12.

Assessment objectives: Defining a polyprotic acid, stating that each dissociation step has its own Ka, and explaining what salt hydrolysis means in Brønsted–Lowry terms are AO1 recall items. Predicting the pH of a salt solution (acidic, basic or neutral) from the parent acid and base, and computing pH numerically using Kb = Kw/Ka, are AO2 application skills tested every year on Paper 3. Rationalising why successive pKa values rise (charge-electrostatic argument), predicting the pH of mixed-salt or amphiprotic-ion solutions, and interpreting multi-equivalence-point titration curves test AO3 synthesis.

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What is a Polyprotic Acid?

A polyprotic acid is one that can donate more than one proton per molecule. Each successive deprotonation is a separate equilibrium with its own acid dissociation constant. The standard naming convention:

• Diprotic (two protons): H₂SO₄, H₂SO₃, H₂CO₃, H₂S, H₂C₂O₄ (oxalic acid).
• Triprotic (three protons): H₃PO₄, H₃PO₃ (phosphorous acid, actually only diprotic — see below), citric acid (C₆H₈O₇).
• Polyprotic (more than three): EDTA (six acidic protons across four carboxylates and two amine nitrogens), polyphosphates.

For a generic diprotic acid H₂A, two stepwise equilibria exist:

Step 1: H₂A(aq) + H₂O(l) ⇌ HA⁻(aq) + H₃O⁺(aq), Ka1 = [HA⁻][H₃O⁺] / [H₂A]

Step 2: HA⁻(aq) + H₂O(l) ⇌ A²⁻(aq) + H₃O⁺(aq), Ka2 = [A²⁻][H₃O⁺] / [HA⁻]

The overall equilibrium constant for full dissociation, H₂A ⇌ 2 H₃O⁺ + A²⁻, is the product Ka1 × Ka2. Crucially, Ka1 > Ka2 for every known polyprotic acid: it is harder to remove a proton from a species that already carries negative charge.

The pKa Values You Must Know

Acid	Formula	pKa1	pKa2	pKa3
Sulfuric	H₂SO₄	≈ −3	1.99	—
Sulfurous	H₂SO₃	1.85	7.20	—
Carbonic	H₂CO₃	6.35	10.33	—
Hydrogen sulfide	H₂S	7.00	≈ 19	—
Oxalic	H₂C₂O₄	1.25	4.27	—
Phosphoric	H₃PO₄	2.15	7.20	12.32
Citric	C₆H₈O₇	3.13	4.76	6.40

A-Level questions most commonly use H₂CO₃, H₂SO₄ and H₃PO₄. The full numerical values above are not all memorisation targets — but the pattern (each pKa typically rises by 4–5 units) is examinable.

Key Point: Sulfuric acid's first dissociation has pKa1 ≈ −3, meaning it is essentially complete in dilute aqueous solution. The second dissociation (pKa2 = 1.99) is partial. For pH-calculation purposes at A-Level, H₂SO₄ is treated as fully dissociating to 2 H⁺ + SO₄²⁻ at concentrations above about 0.01 mol dm⁻³, but the more rigorous answer is that [HSO₄⁻] is non-negligible — it features in A* extension questions.

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Why Does pKa Rise from Step to Step?

Consider H₃PO₄: pKa1 = 2.15, pKa2 = 7.20, pKa3 = 12.32. Each step is roughly 5 pKa units less favourable than the last, i.e. 10⁵ times weaker as an acid. Two complementary explanations operate:

1. Electrostatic argument. Each deprotonation step generates a species with one extra negative charge. Removing a positively-charged proton from an already-negatively-charged species (e.g. HPO₄²⁻ → PO₄³⁻ + H⁺) requires the H⁺ to be extracted against an attractive Coulombic field. The next departing proton "sees" a more concentrated negative charge, raising the energy cost and lowering Ka.

2. Statistical / symmetry factor. A small additional contribution (typically ~0.6 pKa unit) comes from the changing number of equivalent acidic protons. H₃PO₄ has three equivalent protons to lose; H₂PO₄⁻ has only two; HPO₄²⁻ has only one. With fewer "donor sites" the forward rate of dissociation is statistically reduced. This is a minor effect compared to the electrostatic factor.

The combined effect explains why successive pKa values typically differ by 4–5 units — large enough that, at most pH values, only one or two species out of the full ladder dominate. This separation is what makes polyprotic titration curves show distinct equivalence points.

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Salt Hydrolysis: The Big Idea

A salt is the product of a neutralisation reaction. Dissolved in water, salts dissociate fully into their constituent ions. The cation and anion can independently react with water:

• The conjugate base of a weak acid (e.g. CH₃COO⁻, F⁻, CN⁻, HCO₃⁻) accepts a proton from water, generating OH⁻ — the solution is basic.
• The conjugate acid of a weak base (e.g. NH₄⁺, CH₃NH₃⁺, anilinium C₆H₅NH₃⁺) donates a proton to water, generating H₃O⁺ — the solution is acidic.
• Spectator ions — the conjugate base of a strong acid (Cl⁻, NO₃⁻, ClO₄⁻, Br⁻, I⁻) or the conjugate acid of a strong base (Na⁺, K⁺, Li⁺, alkaline-earth M²⁺ at moderate concentration) — do not measurably react with water. Their parent acid/base is so strong that the reverse reaction is negligible.

This gives the four classical salt categories:

Cation parent	Anion parent	Example	Solution pH
Strong base	Strong acid	NaCl	7 (neutral)
Strong base	Weak acid	NaCH₃COO, Na₂CO₃, K₃PO₄	> 7 (basic)
Weak base	Strong acid	NH₄Cl, C₆H₅NH₃Cl	< 7 (acidic)
Weak base	Weak acid	NH₄CH₃COO, NH₄CN	depends on Ka vs Kb

The fourth case — both parents weak — needs comparison of Ka(cation) and Kb(anion). If Ka > Kb the solution is acidic; if Kb > Ka basic; if equal, neutral. Ammonium acetate (NH₄CH₃COO) is the classic "accidentally neutral" salt because Ka(NH₄⁺) = 5.6 × 10⁻¹⁰ ≈ Kb(CH₃COO⁻) = 5.6 × 10⁻¹⁰.

Worked Example: Sodium Acetate Hydrolysis

Sodium ethanoate (NaCH₃COO) dissolves and dissociates fully:

NaCH₃COO(s) → Na⁺(aq) + CH₃COO⁻(aq)

Na⁺ is a spectator. CH₃COO⁻ is the conjugate base of the weak acid ethanoic acid (Ka = 1.74 × 10⁻⁵, pKa = 4.76). It hydrolyses:

CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq), Kb = ?

Use Kw = Ka × Kb (the conjugate-pair relation, derivable from adding the acid and base equilibria):

Kb(CH₃COO⁻) = Kw / Ka(CH₃COOH) = 1.00 × 10⁻¹⁴ / 1.74 × 10⁻⁵ = 5.75 × 10⁻¹⁰

For a 0.10 mol dm⁻³ solution of NaCH₃COO, set up the ICE table for the hydrolysis:

	CH₃COO⁻	CH₃COOH	OH⁻
Initial	0.10	0	0
Change	−x	+x	+x
Equilibrium	0.10 − x	x	x

Kb = x² / (0.10 − x) ≈ x² / 0.10 (assuming x ≪ 0.10)

x² = 0.10 × 5.75 × 10⁻¹⁰ = 5.75 × 10⁻¹¹

x = [OH⁻] = 7.58 × 10⁻⁶ mol dm⁻³

pOH = −log(7.58 × 10⁻⁶) = 5.12

pH = 14 − pOH = 14 − 5.12 = 8.88

Check assumption: x = 7.58 × 10⁻⁶, which is 0.0076% of 0.10. The approximation holds easily.

Worked Example: Ammonium Chloride Hydrolysis

NH₄Cl(s) → NH₄⁺(aq) + Cl⁻(aq). Cl⁻ is a spectator (conjugate base of strong HCl). NH₄⁺ is the conjugate acid of the weak base ammonia (Kb = 1.78 × 10⁻⁵):

NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)

Ka(NH₄⁺) = Kw / Kb(NH₃) = 1.00 × 10⁻¹⁴ / 1.78 × 10⁻⁵ = 5.62 × 10⁻¹⁰

For 0.10 mol dm⁻³ NH₄Cl:

Ka = x² / (0.10 − x) ≈ x² / 0.10

x² = 0.10 × 5.62 × 10⁻¹⁰ = 5.62 × 10⁻¹¹

x = [H₃O⁺] = 7.50 × 10⁻⁶ mol dm⁻³

pH = −log(7.50 × 10⁻⁶) = 5.13

Note the symmetry with the acetate case: equal concentrations of NaCH₃COO and NH₄Cl (whose parent acid and base have approximately equal Ka and Kb — both 1.7 × 10⁻⁵) give pH values symmetric about 7. The NaCH₃COO pH 8.88 and NH₄Cl pH 5.13 sum to almost exactly 14.0.

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Amphiprotic Species: A Special Algebra

An amphiprotic ion can either donate or accept a proton. The most important examples are the partially-deprotonated species of polyprotic acids:

• HCO₃⁻ (hydrogencarbonate) — can lose a proton to give CO₃²⁻ or gain one to give H₂CO₃.
• HSO₄⁻ (hydrogensulfate) — loses to give SO₄²⁻, gains to give H₂SO₄ (rarely, since H₂SO₄ is strong).
• H₂PO₄⁻ and HPO₄²⁻ — both intermediate species in the phosphoric acid ladder.
• Amino acids in their zwitterionic form (H₃N⁺−CHR−COO⁻) are also amphiprotic; this is the foundation of the isoelectric-point concept developed in §3.3.13.

For a pure solution of an amphiprotic salt (e.g. NaHCO₃, KH₂PO₄), the simultaneous acid and base reactions interfere. A rigorous derivation from the mass and charge balances gives a remarkably clean result:

pH ≈ ½ (pKa1 + pKa2)

where pKa1 and pKa2 are the dissociation constants of the parent diprotic system on either side of the amphiprotic species in the ladder.

Worked Example: pH of Sodium Hydrogencarbonate (NaHCO₃)

The parent diprotic acid is H₂CO₃, with pKa1 = 6.35 and pKa2 = 10.33.

pH ≈ ½ (6.35 + 10.33) = ½ × 16.68 = 8.34

This is the pH of household baking soda solution — slightly basic, but far less basic than Na₂CO₃ (pH ≈ 11.6 for 0.1 mol dm⁻³). The amphiprotic algebra explains why: HCO₃⁻ both gives H⁺ to water (acting as acid, Ka2 = 10⁻¹⁰·³³) and takes H⁺ from water (acting as base, Kb = Kw/Ka1 = 10⁻⁷·⁶⁵). The two reactions nearly balance, leaving only a small net excess of OH⁻.

Worked Example: pH of NaH₂PO₄ vs Na₂HPO₄

For H₃PO₄: pKa1 = 2.15, pKa2 = 7.20, pKa3 = 12.32.

• NaH₂PO₄ (between pKa1 and pKa2): pH ≈ ½(2.15 + 7.20) = 4.68 (mildly acidic).
• Na₂HPO₄ (between pKa2 and pKa3): pH ≈ ½(7.20 + 12.32) = 9.76 (basic).

These two salts are the active ingredients in many laboratory phosphate buffers at pH 7.4 (physiological) — mixed in the appropriate ratio they buffer near pKa2 = 7.20.

Practical-skills box: A common A-Level error is to predict the pH of NaHCO₃ "from the Na" (basic) or "from the HCO₃" (acidic — students see the H and assume acid). Both intuitions are wrong. The correct reasoning is amphiprotic averaging: pH ≈ ½(pKa1 + pKa2) of the parent acid, giving pH ≈ 8.3. This is precisely the pH of fresh ocean water — and of blood after correction for the open CO₂ system.

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Polyprotic Titration Curves

When a polyprotic acid is titrated with a strong base, the curve shows one equivalence point per dissociable proton, provided the pKa values differ by at least 4 units (the so-called "separation criterion"). If the gap is smaller, the steps overlap and only a single combined equivalence point is observed.

Phosphoric Acid + Sodium Hydroxide

Titrating 0.10 mol dm⁻³ H₃PO₄ (pKa1 = 2.15, pKa2 = 7.20, pKa3 = 12.32) with 0.10 mol dm⁻³ NaOH gives three steps:

pH
14 |                                              ........------
   |                                             /
12 |                                        ____/        ← Eq3 at pH ~12 (incomplete)
   |                                       /
10 |                                  ____/
   |                                 /
 8 |                       ........./            ← Eq2 at pH ~9.8
   |                      /
 7 |                ____/
   |               /
 5 |          ___/                               ← Buffer region around pKa2 = 7.20
   |         /
 4 |    ____/                                   ← Eq1 at pH ~4.7
   |   /
 2 |__/                                          ← Buffer region around pKa1 = 2.15
   |
 1 +----+----+----+----+----+----+----+---->
    0   50  100  150  200  250  300  V(NaOH) / cm³

Three plateau-and-rise features are visible:

1. First buffer region, centred near pH = pKa1 = 2.15, where H₃PO₄/H₂PO₄⁻ are in roughly equal amounts.
2. First equivalence point, pH ≈ ½(pKa1 + pKa2) = 4.68 — solution is essentially pure NaH₂PO₄ (amphiprotic).
3. Second buffer region, centred near pH = pKa2 = 7.20.
4. Second equivalence point, pH ≈ ½(pKa2 + pKa3) = 9.76 — solution is essentially pure Na₂HPO₄ (amphiprotic).
5. Third buffer region, centred near pH = pKa3 = 12.32, but only weakly resolved because pKa3 is close to pKw/2 = 7 + 7 = 14.
6. Third equivalence point at pH > 12, often not fully reachable because hydroxide background swamps the signal.

For carbonic acid (pKa1 = 6.35, pKa2 = 10.33, gap = 4.0 — just barely separable) two equivalence points appear. For sulfurous acid (pKa1 = 1.85, pKa2 = 7.20, gap = 5.4) the two steps are clearly separated.

Indicator Choice for Each Step

• First equivalence point of H₃PO₄ (pH 4.7): methyl orange (3.1–4.4) is too low; bromocresol green (3.8–5.4) works.
• Second equivalence point (pH 9.8): phenolphthalein (8.2–10.0) is the standard textbook choice.
• Third equivalence point (pH > 12): no common indicator changes colour above 12; in practice the third step is followed potentiometrically with a glass electrode rather than visually.

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Biological Relevance: Why Polyprotic Acids Matter

Phosphate buffer system. Intracellular fluid maintains pH ≈ 7.2 via the H₂PO₄⁻/HPO₄²⁻ pair, buffering near pKa2 = 7.20. The Henderson–Hasselbalch equation gives [HPO₄²⁻]/[H₂PO₄⁻] = 10^(7.2 − 7.2) = 1 in pure buffer at pKa2; in actual cells the ratio is ~1.6:1 because intracellular pH is ~7.4.

Carbonic acid / bicarbonate (blood plasma). Blood pH 7.35–7.45 is maintained by the H₂CO₃ / HCO₃⁻ pair, buffering near pKa1 = 6.35. The ratio [HCO₃⁻]/[H₂CO₃] = 10^(7.4 − 6.35) = 11.2 — bicarbonate is the dominant species. The system is open because CO₂ is exchanged with the lungs: ventilation rate adjusts [H₂CO₃] within seconds, while kidneys adjust [HCO₃⁻] within hours. Acidosis (pH < 7.35) and alkalosis (pH > 7.45) are clinically classified by which arm of this buffer is disturbed.

Amino acids as zwitterions. An amino acid like glycine carries a carboxylic acid (pKa1 ≈ 2.3) and an amine (pKa2 ≈ 9.6); in neutral solution it exists predominantly as the zwitterion H₃N⁺−CH₂−COO⁻. Its isoelectric point (pI) is the pH at which the net charge is zero: pI = ½(pKa1 + pKa2) = 5.95 for glycine — the same amphiprotic algebra. This concept is developed in §3.3.13.

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Summary of Predictive Rules

Salt class	Cation behaviour	Anion behaviour	Solution pH
Strong-strong	Spectator	Spectator	7
Strong base / weak acid	Spectator	Accepts H⁺	> 7 (basic)
Weak base / strong acid	Donates H⁺	Spectator	< 7 (acidic)
Weak base / weak acid	Donates H⁺	Accepts H⁺	Depends on Ka vs Kb
Amphiprotic ion	Both donor and acceptor	—	½(pKa1 + pKa2)

For numerical pH calculation, the algorithm is: