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Notes/Chemistry/Paper 4/Equilibria
CAIEA2 Level9701§25.1–25.2

Equilibria

Conjugate acid–base pairs, the mathematics of pH, Ka, pKa and Kw, buffers, solubility product and the common ion effect, and partition coefficients — putting numbers on the equilibria that run in every aqueous solution.

260 min read 8 sub-topics
222
question parts
2021–2025 · 37 papers
10 marks
per paper
≈ 10% of the paper
1.7/3
avg difficulty
moderate
#4
most examined
of 15 topics by marks

The AS Equilibria note gave you a qualitative toolkit: dynamic equilibrium, Le Chatelier's principle, and KcK_c for comparing products with reactants at equilibrium. This note bolts numbers onto that picture for the reactions running in every aqueous solution — acids handing protons to bases. What does a pH meter actually measure? Why is ethanoic acid called weak while hydrochloric acid is strong, and what does that difference do to a calculator? How can pure water — neither acid nor alkali — still conduct electricity a little? Each question resolves into a constant: KaK_a scores how far an acid dissociates, KwK_w describes water's own split into ions, and later in the note KspK_{\text{sp}} caps how much of an insoluble salt dissolves while KpcK_{\text{pc}} divides a solute between two solvents. Underneath all these constants sits one structural idea — every acid–base reaction is a proton handover between a conjugate pair — and one exam skill: turning one-line definitions into multi-mark calculations without dropping marks to logs, signs or significant figures.

The route through is: §01 conjugate acids and bases — the pairing rule and amphoteric species, §02 the mathematical definitions of pH, KaK_a and pKa, strong versus weak acids and the weak-acid approximation, §03 KwK_w and the pH of alkaline solutions, §04 buffers: what they are and how they act, §05 calculating buffer pH, §06 solubility product, §07 the common ion effect, and §08 partition coefficients.

Before you start you should be able to

  • Dynamic equilibrium and Le Chatelier's principle: predicting the direction a reversible reaction shifts when conditions change (this subject's own AS Equilibria note)

  • Writing and interpreting Kc expressions for homogeneous equilibria, and judging what a large or small K means (this subject's own AS Equilibria note)

  • Mole and concentration calculations: moles from mass and Ar, n = c × V, and dilution factors (this subject's own Atoms, Molecules and Stoichiometry note)

  • Polarity and electronegativity: why O–H bonds are polarised, and how neighbouring groups strengthen or weaken them (this subject's own Chemical Bonding note)

By the end of this page you can

  • Understand and use the terms conjugate acid and conjugate base, define conjugate acid–base pairs, and identify them in reactions — including amphoteric species such as HCO₃⁻ and H₂PO₄⁻

  • Define mathematically pH, Ka, pKa and Kw, and convert fluently between pH ↔ [H⁺] and pKa ↔ Ka

  • Calculate [H⁺] and pH values for strong acids, strong alkalis and weak acids, deriving and justifying the weak-acid approximation [H⁺] = √(Ka·c)

  • Define a buffer solution, explain how one is made and how it controls pH using chemical equations, and describe its uses, including the role of HCO₃⁻ in controlling blood pH

  • Calculate the pH of buffer solutions from appropriate data, including the half-neutralisation case

  • Understand and use solubility product: write Ksp expressions with correct units, and convert between solubility and Ksp in both directions

  • Explain the common ion effect qualitatively and perform Ksp calculations in solutions containing a common ion

  • State what is meant by a partition coefficient, calculate and use Kpc, and explain how the polarities of solute and solvents affect its value

01

Conjugate acids and bases

Syllabus requirement · §25.1.1–25.1.2

“

understand and use the terms conjugate acid and conjugate base · define conjugate acid–base pairs, identifying such pairs in reactions

”

Every acid reaction is a proton handover

Ask what makes hydrochloric acid an acid and the AS answer was practical: it turns litmus red, it fizzes with carbonates. The Brønsted–Lowry definition cuts underneath all of that: an acid is a proton donor and a base is a proton acceptor, where a proton is simply H+\text{H}^+. Acidity is not a property a substance carries around in isolation — it is a role in a reaction. Hydrochloric acid is only being an acid at the moment it hands H+\text{H}^+ to something willing to receive it.

Watch any such handover closely and a pattern appears. When HCl meets water:

HCl+H2O→Cl−+H3O+\text{HCl} + \text{H}_2\text{O} \rightarrow \text{Cl}^- + \text{H}_3\text{O}^+

HCl loses H+\text{H}^+; water gains it. But look at the products — they are not strangers to the reactants. Cl−\text{Cl}^- is HCl minus exactly one proton, and H3O+\text{H}_3\text{O}^+ is water plus exactly one proton. Each product is the transformed partner of one reactant. Those partnerships — two species differing by a single H+\text{H}^+ — are the organising idea of this entire topic. Buffers (§04–§05), weak-acid calculations (§02) and even water itself (§03) all speak this language, so the pairing rule below is worth perfecting now.

The pairing rule, precisely

The definition, in mark-scheme wording: a conjugate acid–base pair is two species that differ by one proton, H+\text{H}^+.

Which member gets which name follows from who holds the extra proton:

  • the species with the extra H+\text{H}^+ is the conjugate acid;
  • the species without it is the conjugate base.

Every partnership can be written as one equilibrium:

acid⇌base+H+\text{acid} \rightleftharpoons \text{base} + \text{H}^+

Read left to right, the acid loses a proton and becomes its conjugate base; read right to left, the base gains one and becomes its conjugate acid. Which member "is the acid" depends only on which way you read — what is fixed is the pair.

Charge bookkeeping does half the work for you. A proton carries charge +1+1, so:

  • removing H+\text{H}^+ lowers the charge by 1 (acid → conjugate base);
  • adding H+\text{H}^+ raises the charge by 1 (base → conjugate acid).

Nothing else about the species changes: same atoms otherwise, same skeleton. If your conjugate base of CH3COOH\text{CH}_3\text{COOH} comes out as anything other than CH3COO−\text{CH}_3\text{COO}^-, you have altered more than the proton.

conjugate acid

conjugate base

charge change

HCl\text{HCl}

Cl−\text{Cl}^-

0→−10 \rightarrow -1

CH3COOH\text{CH}_3\text{COOH}

CH3COO−\text{CH}_3\text{COO}^-

0→−10 \rightarrow -1

NH4+\text{NH}_4^+

NH3\text{NH}_3

+1→0+1 \rightarrow 0

H3O+\text{H}_3\text{O}^+

H2O\text{H}_2\text{O}

+1→0+1 \rightarrow 0

H2O\text{H}_2\text{O}

OH−\text{OH}^-

0→−10 \rightarrow -1

Every row differs by exactly one H⁺ and exactly one unit of charge. Check both before moving on.

acid⇌base+H+\text{acid} \rightleftharpoons \text{base} + \text{H}^+

A conjugate acid–base pair differs by exactly ONE H⁺.

·

Remove H⁺ → charge falls by 1. Add H⁺ → charge rises by 1. Nothing else changes.

Track the charge, not the vibes

The commonest wrong answer when asked for the conjugate base of HCrO4−\text{HCrO}_4^- is CrO4−\text{CrO}_4^- — right formula, wrong charge. Removing H+\text{H}^+ removes charge +1+1, so the charge MUST fall by one unit: −1−1=−2-1 - 1 = -2, giving CrO42−\text{CrO}_4^{2-}. Write the starting charge down, apply ±1\pm 1, write the new charge. One written line, mark safe.

Finding pairs inside a whole reaction

Exam questions rarely ask about a pair in isolation; they print a full proton-transfer reaction and ask you to identify the pairs hiding inside it. Take ammonia dissolving in water:

NH3+H2O⇌NH4++OH−\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-

One proton moves, but it creates two partnerships at once:

  • NH3\text{NH}_3 gains H+\text{H}^+ and becomes NH4+\text{NH}_4^+: pair (NH3\text{NH}_3, NH4+\text{NH}_4^+);
  • H2O\text{H}_2\text{O} loses H+\text{H}^+ and becomes OH−\text{OH}^-: pair (H2O\text{H}_2\text{O}, OH−\text{OH}^-).

Each pair brackets one reactant with the product it turns into — diagonally across the equation. Mark schemes award one mark per correctly completed pair, so an "identify both pairs" question is really four blanks: two acids, two conjugate bases.

HA₁acid 1+B₂base 2⇌A₁⁻base 1+HB₂⁺acid 2H⁺ donatedH⁺ taken backconjugate pair 2 — B₂ / HB₂⁺ (differ by one H⁺)conjugate pair 1 — HA₁ / A₁⁻ (differ by one H⁺)acid₁ donates H⁺ → its conjugate base A₁⁻ · base₂ accepts H⁺ → its conjugate acid HB₂⁺each pair differs by exactly ONE protonconjugate base = acid − H⁺ · conjugate acid = base + H⁺which way ⇌ sits depends on which pair holds H⁺ more strongly — the stronger base wins it

One proton-transfer reaction split into its two conjugate pairs: HA₁ + B₂ ⇌ A₁⁻ + HB₂⁺. The arrow shows the single H⁺ being donated from HA₁ to B₂; each reactant is bracketed with the product it becomes, one pair per colour.

Amphoteric species — the examiner's favourite twist

Some species carry both pieces of equipment: an H+\text{H}^+ they could donate, AND enough negative charge (or a lone pair) to accept one. These are amphoteric species, and they possess a conjugate acid AND a conjugate base. The rules do not change — you just run the machine twice on the same formula:

  • HCO3−\text{HCO}_3^-: add H+\text{H}^+ → H2CO3\text{H}_2\text{CO}_3; remove H+\text{H}^+ → CO32−\text{CO}_3^{2-}.
  • H2O\text{H}_2\text{O}: add → H3O+\text{H}_3\text{O}^+; remove → OH−\text{OH}^-.
  • HSO4−\text{HSO}_4^-: add → H2SO4\text{H}_2\text{SO}_4; remove → SO42−\text{SO}_4^{2-}.

Water is the everyday example — simultaneously the conjugate acid of OH−\text{OH}^- and the conjugate base of H3O+\text{H}_3\text{O}^+, which is exactly why it can react with acids and alkalis. Hydrogencarbonate's double life is what lets it buffer your blood (§04). The bank's recurring favourites are hydrogen-bearing anions: HCrO4−\text{HCrO}_4^-, H2PO4−\text{H}_2\text{PO}_4^-, HPO42−\text{HPO}_4^{2-} — always good for a cheap mark if your charge bookkeeping is sharp.

Conjugate partners, every time
  1. 1

    Write the species with its charge.

  2. 2

    Conjugate acid: ADD one H+\text{H}^+ to the formula; raise the charge by 1.

  3. 3

    Conjugate base: REMOVE one H+\text{H}^+; lower the charge by 1.

    Both moves touch ONLY the hydrogen count. If any other subscript changed, stop and redo.

  4. 4

    Audit: the pair must differ by exactly one H and exactly one unit of charge — nothing else.

  5. 5

    In a full reaction: bracket each reactant with the product it becomes. Expect TWO pairs per equation.

    One proton moves, but it leaves one species AND arrives at another — both arrivals create a pair.

A clean demonstration: nitric acid meets water

HNO3+H2O→H3O++NO3−\text{HNO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+ + \text{NO}_3^-

Pair 1. HNO3\text{HNO}_3 loses H+\text{H}^+ → NO3−\text{NO}_3^-. Charge check: 0→−10 \rightarrow -1 ✓. So acid = HNO3\text{HNO}_3, conjugate base = NO3−\text{NO}_3^-.

Pair 2. H2O\text{H}_2\text{O} gains H+\text{H}^+ → H3O+\text{H}_3\text{O}^+. Charge check: 0→+10 \rightarrow +1 ✓. So base = H2O\text{H}_2\text{O}, conjugate acid = H3O+\text{H}_3\text{O}^+.

Flip the perspective. In the reverse reaction, H3O+\text{H}_3\text{O}^+ would be the donor (acid) and NO3−\text{NO}_3^- the receiver (base). The roles swap with direction; the pairs themselves never change. That is why examiners ask for pairs rather than "the acid" — the pair is the stable fact.

Amphoteric demo. Take HCO3−\text{HCO}_3^-, charge −1-1. Conjugate acid: add H+\text{H}^+ → H2CO3\text{H}_2\text{CO}_3, charge −1+1=0-1 + 1 = 0. Conjugate base: remove H+\text{H}^+ → CO32−\text{CO}_3^{2-}, charge −1−1=−2-1 - 1 = -2. Two lines, two answers, charges tracked throughout.

Amphoteric behaviour of HCrO₄⁻

9701/44 O/N 2025 Q2(c)(ii)1 mark

HCrO4−\text{HCrO}_4^- can show amphoteric behaviour. State the formula of:

  • the conjugate acid of HCrO4−\text{HCrO}_4^-
  • the conjugate base of HCrO4−\text{HCrO}_4^-
Show full working
  1. 1

    Conjugate acid — add one H+\text{H}^+: HCrO4−+H+→H2CrO4\text{HCrO}_4^- + \text{H}^+ \rightarrow \text{H}_2\text{CrO}_4 Charge: −1+1=0-1 + 1 = 0.

    Only the hydrogen count moved: one Cr, four O, untouched. The added proton brings charge +1, cancelling the −1.

  2. 2

    Conjugate base — remove one H+\text{H}^+: HCrO4−→CrO42−+H+\text{HCrO}_4^- \rightarrow \text{CrO}_4^{2-} + \text{H}^+ Charge: −1−1=−2-1 - 1 = -2.

    The classic error here is writing CrO₄⁻ — right formula, wrong charge. Losing a +1 proton must drop the charge by a full unit.

Answer

Conjugate acid: H₂CrO₄ · Conjugate base: CrO₄²⁻

Amphoteric means run the machine twice: +H⁺ with charge +1, then −H⁺ with charge −1. Never touch the atom skeleton.

Both partners of the hydrogen phosphate ion

9701/42 O/N 2025 Q3(a)(ii)1 mark

Give the formulas of the conjugate acid and the conjugate base of the hydrogen phosphate ion, HPO42−\text{HPO}_4^{2-}.

Show full working
  1. 1

    Conjugate acid — add one H+\text{H}^+: HPO42−+H+→H2PO4−\text{HPO}_4^{2-} + \text{H}^+ \rightarrow \text{H}_2\text{PO}_4^- Charge: −2+1=−1-2 + 1 = -1.

    Start charge −2, add +1, land on −1. Writing the charge arithmetic down is what stops the slip to H₂PO₄²⁻.

  2. 2

    Conjugate base — remove one H+\text{H}^+: HPO42−→PO43−+H+\text{HPO}_4^{2-} \rightarrow \text{PO}_4^{3-} + \text{H}^+ Charge: −2−1=−3-2 - 1 = -3.

    The ion still has one H left after losing this proton — check the product reads PO₄³⁻, not O₄-something. Skeleton untouched.

Answer

Conjugate acid: H₂PO₄⁻ · Conjugate base: PO₄³⁻

Hydrogen-bearing anions (HPO₄²⁻, H₂PO₄⁻, HCrO₄⁻, HCO₃⁻) are the bank's favourite amphoteric species — expect to meet them again in buffers (§04).

In the exam
25 parts · 33 marks · 2021–2025 (question bank)

Conjugate-pair questions are the cheapest marks in the whole topic: across 2021–2025 they made up 25 leaf parts carrying just 33 marks — nearly every question is worth a single mark — and identify-or-define variants are tagged on about a dozen of them. They typically open a long equilibria question as part (a), setting the tone for what follows. Answer them fast and perfectly: one definition sentence, or two formulas with charges tracked.

Common mistakes

  • Saying a conjugate pair differs by one electron

    A conjugate acid–base pair differs by one PROTON, H⁺ — the definition's exact wording is the whole mark.

    'Define conjugate acid–base pair' is a standing 1-mark question; 'one electron' or 'differ in hydrogen content' scores zero.

  • Writing the conjugate base of HCrO₄⁻ as CrO₄⁻

    Removing H⁺ removes charge +1, so the charge falls a full unit: CrO₄²⁻.

    Formula right but charge untracked is the most common loss on amphoteric-partner questions.

  • Changing other subscripts while adding or removing H⁺ (e.g. H₂CrO₃)

    Only the hydrogen count changes — same skeleton, charge ±1.

    The pairing rule is mechanical precisely so it needs no chemical imagination; improvising breaks it.

  • Assuming amphoteric species must be neutral, like water

    Amphoteric means able to donate AND accept H⁺ — HCO₃⁻, HCrO₄⁻, H₂PO₄⁻ and HPO₄²⁻ all do it while charged.

    The bank asks for partners of charged amphoterics far more often than of water itself.

Your turn

One definition, two identifications inside reactions, and a family drill — the exact shapes the bank uses.

  1. 19701/42 O/N 2025 Q3(a)(i)1 mark

    Define the term conjugate acid–base pair.

    Stuck? Show hint

    Two species, one proton — say both halves.

    Show solution
    1. 1

      Quote the mark scheme's wording: two species that differ by one proton (H+\text{H}^+).

      The word 'one' is load-bearing — differing by two protons, or by an electron, scores nothing. An illustration such as HCl/Cl⁻ is welcome but the sentence itself carries the mark.

    Answer

    Two species that differ by one proton (H⁺).

  2. 29701/42 M/J 2025 Q5(a)(i)1 mark

    In the reaction between hypochlorous acid and sodium hydroxide,

    HClO+NaOH→NaClO+H2O\text{HClO} + \text{NaOH} \rightarrow \text{NaClO} + \text{H}_2\text{O}

    identify the two conjugate acid–base pairs, naming the acid and its conjugate base in each.

    Stuck? Show hint

    Follow the single proton: who lost it, who gained it.

    Show solution
    1. 1

      Strip the spectator first: Na+\text{Na}^+ rides through unchanged, so the proton handover is HClO+OH−⇌ClO−+H2O\text{HClO} + \text{OH}^- \rightleftharpoons \text{ClO}^- + \text{H}_2\text{O}

      The paper prints the molecular form (NaOH, NaClO); removing the spectator ion lines up every conjugate-pair member where you can see it.

    2. 2

      Pair I: HClO\text{HClO} loses H+\text{H}^+ → ClO−\text{ClO}^-. Acid I = HClO\text{HClO}, conjugate base = ClO−\text{ClO}^-. Charge check: 0→−10 \rightarrow -1 ✓.

    3. 3

      Pair II: OH−\text{OH}^- gains H+\text{H}^+ → H2O\text{H}_2\text{O}; viewed from the acid side, H2O\text{H}_2\text{O} loses H+\text{H}^+ → OH−\text{OH}^-. Acid II = H2O\text{H}_2\text{O}, conjugate base = OH−\text{OH}^-. Charge check: 0→−10 \rightarrow -1 ✓.

      The mark scheme fills four blanks in order: HClO, ClO⁻, H₂O, OH⁻. Water plays the conjugate-acid partner here — students who decide 'water is only ever a base' miss pair II entirely.

    Answer

    Pair I: HClO / ClO⁻ · Pair II: H₂O / OH⁻

  3. 39701/43 O/N 2024 Q1(a)(i)1 mark

    Disodium phosphate, (Na+)2(HPO42−)(\text{Na}^+)_2(\text{HPO}_4^{2-}), reacts with an acid to form monosodium phosphate, Na+(H2PO4−)\text{Na}^+(\text{H}_2\text{PO}_4^-).

    Identify the ions that are a conjugate acid–base pair in this reaction.

    Stuck? Show hint

    Follow the proton the acid donates — which two species end up differing by exactly one H⁺?

    Show solution
    1. 1

      Write the proton transfer: the acid hands H+\text{H}^+ to the base HPO42−\text{HPO}_4^{2-}: HPO42−+H+→H2PO4−\text{HPO}_4^{2-} + \text{H}^+ \rightarrow \text{H}_2\text{PO}_4^- Charge check: −2+1=−1-2 + 1 = -1 ✓.

      Converting the paper's salt formulas into the ions that did the reacting turns one sentence of prose into one line of chemistry.

    2. 2

      The two species in that handover differ by exactly one proton: H2PO4−\text{H}_2\text{PO}_4^- (conjugate acid) and HPO42−\text{HPO}_4^{2-} (conjugate base). Both are ions, as the question demands.

      The mark scheme awards for H₂PO₄⁻ AND HPO₄²⁻ — both names needed.

    Answer

    H₂PO₄⁻ and HPO₄²⁻ (both required).

  4. 4

    For each species, state its conjugate acid and its conjugate base — or explain briefly why no partner exists on that side:

    (i) HSO4−\text{HSO}_4^- (ii) SO42−\text{SO}_4^{2-} (iii) H2SO4\text{H}_2\text{SO}_4

    Stuck? Show hint

    Run +H⁺ and −H⁺ mechanically — and look at what the formula actually contains.

    Show solution
    1. 1

      (i) HSO4−\text{HSO}_4^- (charge −1-1): add H+\text{H}^+ → H2SO4\text{H}_2\text{SO}_4 (charge 00); remove H+\text{H}^+ → SO42−\text{SO}_4^{2-} (charge −2-2). Both partners exist — HSO4−\text{HSO}_4^- is amphoteric.

    2. 2

      (ii) SO42−\text{SO}_4^{2-}: add H+\text{H}^+ → HSO4−\text{HSO}_4^- ✓. Remove H+\text{H}^+? The formula contains no hydrogen at all, so there is nothing to donate: SO42−\text{SO}_4^{2-} has no conjugate base.

      Every species has at least one partner — the side it can reach by ±H⁺. Only species with BOTH a removable H and the capacity to accept one are amphoteric.

    3. 3

      (iii) H2SO4\text{H}_2\text{SO}_4: remove H+\text{H}^+ → HSO4−\text{HSO}_4^- ✓. Adding another H+\text{H}^+ would need sulfuric acid to accept a proton onto an already proton-saturated −OH group — not a species recognised at this level, so H2SO4\text{H}_2\text{SO}_4 has no conjugate acid.

    Answer

    (i) acid H₂SO₄, base SO₄²⁻ (amphoteric) · (ii) acid HSO₄⁻, no conjugate base (no H in the formula) · (iii) base HSO₄⁻, no conjugate acid.

Practise conjugate acid–base pairsReal past-paper questions · Conjugate acid-base pairs

The rest of this note

Checking your access…

Can you do all of these?

  • I can define a conjugate acid–base pair and identify both pairs in any proton-transfer reaction, including pairs where both partners are ions

  • I can write the conjugate acid and the conjugate base of amphoteric species (H₂O, HCO₃⁻, HCrO₄⁻, H₂PO₄⁻, HPO₄²⁻) by adding or removing exactly one H⁺, tracking the charge

  • I can define pH, Ka, pKa and Kw mathematically, and interconvert pH ↔ [H⁺] and pKa ↔ Ka without sign slips

  • I can calculate the pH of a strong acid, a weak acid (via [H⁺] = √(Ka·c), checking the approximation) and a strong alkali (via Kw or pOH), quoting pH to 2 dp

  • I can explain how a buffer is prepared (weak acid + its salt, or excess weak acid partially neutralised), track the moles mixed, and write the equations showing it intercepting added H⁺ and added OH⁻

  • I can calculate a buffer's pH with Henderson–Hasselbalch in either form, including moles-based ratios, backwards composition questions and the half-neutralisation shortcut pH = pKa

  • I can convert between solubility and Ksp in both directions for any ion ratio (MX, MX₂, M₂X, MX₃), deriving the power law from the expression and stating correct units

  • I can calculate solubilities and ion concentrations in a solution sharing an ion with the salt, justifying why the common ion's concentration is treated as unchanged

  • I can define a partition coefficient, use Kpc to calculate masses extracted and left behind, and predict qualitatively how solute and solvent polarity change Kpc

Now do the questions
222 real Paper 4 parts from 2021–2025, sorted by difficulty, with mark schemes