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8873 / 2027
Theories of Acids and Bases

Full chapter

Theories of Acids and Bases

Follow protons, calculate strong-solution pH and explain how buffers respond to change.

A-Level 8873, revised syllabus (2026-2027)

01

Follow what the acid and base actually do

One model focuses on water; another follows proton transfer.

In the Arrhenius model, an acid produces H+(aq) in water and a base produces OH-(aq). HCl(aq) provides H+ and Cl-; NaOH(aq) provides Na+ and OH-. Neutralisation removes H+ and OH- as water: H+(aq) + OH-(aq) → H2O(l).

The Bronsted-Lowry model follows a proton: an acid donates H+, and a base accepts it. A conjugate acid-base pair differs by exactly one proton. Losing that proton also reduces the signed charge by one; gaining it increases the signed charge by one.

Track the transferred proton
ReactionAcid and its conjugate baseBase and its conjugate acid
HCl + H2O → H3O+ + Cl-HCl / Cl-H2O / H3O+
NH3 + H2O ⇌ NH4+ + OH-H2O / OH-NH3 / NH4+
CH3CO2H + H2O ⇌ CH3CO2- + H3O+CH3CO2H / CH3CO2-H2O / H3O+

Water accepts a proton from HCl but donates one to NH3; a substance capable of either role is amphiprotic. In aqueous equations H+(aq) is convenient shorthand for the hydrated proton. Writing H3O+ makes the transfer to water visible.

Worked exampleFind a pair in an unfamiliar reaction

HCO3- + H2O ⇌ CO32- + H3O+. Which species donates the proton?

Try it first. Then show the worked stepsHide the worked steps
  1. HCO3- loses one H+ and becomes CO32-. Its charge changes from -1 to -2.
  2. Water gains that proton and becomes H3O+.
  3. Pair each species with the product differing by one proton.
Answer

HCO3- is the acid; CO32- is its conjugate base. H2O is the base and H3O+ its conjugate acid.

Check your understanding

In NH3 + H2O <=> NH4+ + OH-, which species acts as the Bronsted-Lowry acid?

  1. H2O, because it donates a proton to NH3
  2. OH-, because it makes the solution alkaline
  3. NH3, because it contains hydrogen atoms
Show the answer

H2O, because it donates a proton to NH3

A Bronsted-Lowry acid donates H+. Water gives a proton to ammonia, forming OH-. The conjugate pairs are H2O/OH- and NH4+/NH3.

02

Strength describes dissociation, not concentration

Separate how much acid was dissolved from how much becomes ions.

Make a guess. You are not marked.

Two acids are both 0.100 mol dm-3: hydrochloric acid and ethanoic acid. Which solution has the higher pH?

  1. Hydrochloric acid, because it has more H atoms
  2. Ethanoic acid, because it only partly ionises
  3. Ethanoic acid, because it has more carbon atoms to dilute it
Show the answer

Ethanoic acid, because it only partly ionises

Ethanoic acid is weak and only partly ionises, so its [H+] is much lower than the 0.100 mol dm-3 from fully ionised HCl.

A strong acid is essentially fully dissociated in dilute aqueous solution. A weak acid establishes an equilibrium in which only a proportion is dissociated. Strong and weak bases differ similarly in the extent to which they generate OH- in water. "Concentrated" and "dilute" instead describe amount per volume. A concentrated weak acid is still weak.

Equal analytical concentrations, different dissociation

Strong acid is mostly ions; weak acid is mostly undissociated HA alongside some ions. Water and exact particle counts are omitted.

Qualitative sketch: the number of printed symbols is illustrative, not a measured dissociation fraction.
Compare equal analytical concentrations at the same temperature
ComparisonExpected observationDissociation explanation
Dilute HCl versus ethanoic acid, both monoproticHCl has lower pH and usually greater electrical conductivity.HCl supplies a much larger concentration of H+ and other mobile ions; most ethanoic acid remains as neutral molecules.
Equal Mg pieces in excess of these two acidsHydrogen is normally released faster initially in HCl.The larger initial H+ concentration gives more frequent effective collisions at the same exposed metal surface. Control temperature and surface area.
Fully dissolved NaOH versus aqueous NH3NaOH has higher pH at the same analytical concentration.NaOH dissociates essentially completely to give OH-; NH3 reacts only partially with water to form NH4+ and OH-.

A weak acid can still react completely with enough strong base. As H+ is removed, more weak acid dissociates. Equal volumes of equal-concentration monoprotic strong and weak acids therefore require the same amount of NaOH for complete neutralisation, even though their initial pH and initial reaction rates differ. Comparing pH alone cannot establish acid strength unless concentration and the number of ionisable protons are also controlled.

For a monoprotic weak acid HA(aq) ⇌ H+(aq) + A-(aq), Ka = [H+][A-]/[HA] at equilibrium. At a fixed temperature, a larger Ka means a greater tendency to donate a proton. For a base B + H2O ⇌ BH+ + OH-, Kb = [BH+][OH-]/[B]. Pure liquid water is omitted from these expressions.

pH = -log10[H+(aq)], with the concentration expressed in mol dm-3 for these calculations. One pH unit corresponds to a tenfold change in hydrogen-ion concentration. Kw = [H+][OH-] is the ionic product of water. At 25 degrees C, Kw = 1.00 × 10-14 mol2 dm-6.

Check your understanding

Which solution has the lowest pH?

  1. 1.0 mol dm-3 ethanoic acid
  2. 0.10 mol dm-3 hydrochloric acid
  3. 0.10 mol dm-3 ethanoic acid
  4. 0.0010 mol dm-3 hydrochloric acid
Show the answer

0.10 mol dm-3 hydrochloric acid

HCl dissociates fully, so [H+] = 0.10 mol dm-3 and pH = 1.0. Ethanoic acid only partly dissociates, so even 1.0 mol dm-3 has [H+] well below 0.10.

Exam-style question

Separate 25.0 cm3 samples of 0.100 mol dm-3 hydrochloric acid and 0.100 mol dm-3 ethanoic acid are compared at 25 °C.
(a) Explain why the hydrochloric acid has the lower pH. [2]
(b) Equal pieces of magnesium are added. Predict and explain the difference in the initial rate. [1]
(c) Explain why both acids need the same volume of 0.100 mol dm-3 NaOH for neutralisation. [1] [4]

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03

Count ions, then take the logarithm

Concentration units and stoichiometric factors come before the calculator.

Worked exampleA strong monoprotic acid

Find the pH of 3.20 × 10-3 mol dm-3 HCl at 25 degrees C.

Try it first. Then show the worked stepsHide the worked steps
  1. Assume complete dissociation; one mole HCl provides one mole H+.
  2. [H+] = 3.20 × 10-3 mol dm-3. Water contributes negligibly at this concentration.
  3. pH = -log10(3.20 × 10-3) = 2.49485.
Answer

pH = 2.495. The pH itself has no unit. Using three decimal places preserves the three significant figures of the concentration.

Worked exampleA base with two hydroxide ions

A fully dissolved Ca(OH)2 solution has concentration 2.50 × 10-3 mol dm-3. Find its pH at 25 degrees C.

Try it first. Then show the worked stepsHide the worked steps
  1. Ca(OH)2(aq) supplies two OH- per formula unit: [OH-] = 5.00 × 10-3 mol dm-3.
  2. Use Kw: [H+] = (1.00 × 10-14)/(5.00 × 10-3) = 2.00 × 10-12 mol dm-3.
  3. pH = -log10(2.00 × 10-12) = 11.69897.
Answer

pH = 11.699. This uses the stated dissolved concentration; it does not assume every mass of solid calcium hydroxide will dissolve.

Worked exampleNeutralise before finding pH

Mix 25.0 cm3 of 0.0400 mol dm-3 HCl with 15.0 cm3 of 0.0400 mol dm-3 NaOH. Assume additive volumes and 25 degrees C.

Try it first. Then show the worked stepsHide the worked steps
  1. Initial H+ amount = 0.0400 × 0.0250 = 1.00 × 10-3 mol.
  2. Initial OH- amount = 0.0400 × 0.0150 = 6.00 × 10-4 mol.
  3. The 1:1 neutralisation leaves 4.00 × 10-4 mol H+.
  4. Total volume = 0.0400 dm3, so [H+] = 0.0100 mol dm-3.
Answer

pH = 2.000. Averaging the two starting pH values would ignore both stoichiometry and the logarithmic scale.

Try it step by step

Work it out in steps: [H+] or [OH-], then pH.

  1. Problem 1. Find the pH of 0.0250 mol dm-3 nitric acid, a strong acid, at 25 degrees C. Give pH to 2 decimal places.
  2. Problem 2. Find the pH of 0.00400 mol dm-3 Ba(OH)2, fully dissolved, at 25 degrees C. Use Kw = 1.00 x 10-14 mol2 dm-6. Give pH to 2 decimal places.
  3. Problem 3. Find the pH of 0.00500 mol dm-3 sodium hydroxide at 25 degrees C. Use Kw = 1.00 x 10-14 mol2 dm-6. Give [OH-] to 3 significant figures and pH to 2 decimal places.
Show the answer

Problem 1.

[H+] or [OH-]: HNO3 ionises completely and gives one H+ each, so [H+] = 0.0250 mol dm-3.

pH: pH = -log(0.0250) = 1.60.

Problem 2.

[H+] or [OH-]: Each Ba(OH)2 gives two OH-, so [OH-] = 2 x 0.00400 = 0.00800 mol dm-3.

pH: [H+] = 1.00 x 10-14 / 0.00800 = 1.25 x 10-12 mol dm-3, so pH = 11.90.

Problem 3.

[H+] or [OH-]: NaOH gives one OH- each, so [OH-] = 0.00500 mol dm-3.

pH: [H+] = 1.00 x 10-14 / 0.00500 = 2.00 x 10-12 mol dm-3, so pH = 11.70.

Check your understanding

A strong acid, not extremely dilute, is diluted tenfold. How does its pH change?

  1. It rises tenfold, so a pH of 1 becomes a pH of 10.
  2. It falls by 1, because the acid is now spread through ten times the water.
  3. It rises by 1, because [H+] falls by a factor of ten.
Show the answer

It rises by 1, because [H+] falls by a factor of ten.

pH = -log[H+], so a tenfold fall in [H+] raises pH by one. In extremely dilute acid, H+ from water matters and this rule breaks down.

04

Choose a colour change inside the sharp rise

The equivalence point and the observed endpoint are related but not identical.

At equivalence, the acid and base have reacted in the stoichiometric proportion. An indicator changes colour over a pH range; its endpoint should occur within the steep portion of the titration curve, where a very small added volume causes the whole indicator transition. Choose using the supplied curve or transition data, not merely a memorised colour.

Useful indicator ranges
IndicatorApproximate transition pHAcid-side / base-side colour
Methyl orange3.1-4.4Red / yellow
Bromothymol blue6.0-7.6Yellow / blue
Phenolphthalein8.2-10.0Colourless / pink
How strengths affect the choice
TitrationRegion near equivalenceUsual interpretation
Strong acid / strong baseSteep change spans acid and alkaline regions at suitable concentrations.Several indicators may work; inspect the given curve.
Weak acid / strong baseEquivalence is alkaline because the conjugate base reacts with water.Phenolphthalein is commonly suitable.
Strong acid / weak baseEquivalence is acidic because the conjugate acid reacts with water.Methyl orange is commonly suitable.
Weak acid / weak baseOften no sufficiently sharp pH change.A visual indicator is generally unsuitable for a precise endpoint.

Worked exampleUse given data rather than the name alone

A weak-acid/strong-base titration changes sharply from pH 7.8 to 11.0 near equivalence. Which listed indicator is suitable?

Try it first. Then show the worked stepsHide the worked steps
  1. Methyl orange changes well before the steep interval.
  2. Bromothymol blue also finishes its transition below the stated interval.
  3. Phenolphthalein changes from 8.2 to 10.0, wholly inside the steep interval.
Answer

Choose phenolphthalein. The endpoint then needs very little extra titrant to complete the colour change.

For reliable observations, add titrant dropwise near the endpoint, swirl to mix, and use a consistent faint permanent colour. A darker final colour after extra titrant is not "more complete": it means the endpoint has been overshot.

Check your understanding

Ethanoic acid is titrated with sodium hydroxide. The steep part of the curve runs from about pH 7 to pH 11. Which indicator suits it?

  1. Methyl orange, which changes over pH 3.1-4.4
  2. Phenolphthalein, which changes over pH 8.2-10.0
  3. Any indicator, since every titration ends at pH 7
Show the answer

Phenolphthalein, which changes over pH 8.2-10.0

Its whole colour-change range sits inside the steep rise, so it changes sharply at the end-point. The equivalence pH is above 7 because ethanoate is a weak base.

05

Keep an acid and its conjugate base available

A buffer removes small additions of acid or base, rather than preventing all pH change.

Make a guess. You are not marked.

A small amount of sodium hydroxide is added to a buffer of ethanoic acid and sodium ethanoate. What happens to the pH?

  1. It rises sharply, as it would in water
  2. It falls, because the buffer makes more acid
  3. It rises only very slightly
Show the answer

It rises only very slightly

The added OH- reacts with ethanoic acid to form ethanoate and water, so very little free OH- remains.

An acidic buffer contains appreciable amounts of a weak acid and its conjugate base, for example CH3CO2H and dissolved sodium ethanoate. Added H+ is consumed by ethanoate; added OH- is consumed by the weak acid. Because most added strong acid/base is removed, the pH changes much less than it would in water.

Two complementary responses
  1. Add a little acid

    CH3CO2-(aq) + H+(aq) → CH3CO2H(aq). The conjugate base removes the added proton.

  2. Add a little alkali

    CH3CO2H(aq) + OH-(aq) → CH3CO2-(aq) + H2O(l). The weak acid removes hydroxide.

A basic buffer uses a weak base and its conjugate acid, such as NH3/NH4+. NH3 consumes H+; NH4+ consumes OH-. Buffers help keep enzyme conditions suitable and prevent unwanted pH shifts in formulations. Their capacity is finite: a large acid/base addition can use up one component.

Check your understanding

Why is sodium ethanoate solution on its own not an acidic buffer?

  1. It has ethanoate to remove added acid, but no weak acid to remove added alkali.
  2. It is a salt of a strong acid, so its ions do not react with added acid or alkali.
  3. Its solution is neutral, and a buffer must start with a pH below 7 to work.
Show the answer

It has ethanoate to remove added acid, but no weak acid to remove added alkali.

A buffer needs both a weak acid and its conjugate base in useful amounts, so it can resist both acid and alkali.

Exam-style question

Paper 2 Section B style

This question draws on the Mole Concept and Stoichiometry, and Theories of Acids and Bases.

An antacid tablet of mass 1.20 g contains calcium carbonate and an unreactive binder. A student added the tablet to 50.0 cm3 of 1.00 mol dm-3 hydrochloric acid, an excess. When the fizzing stopped, the solution was boiled for a minute, cooled and made up to 250 cm3 with water. A 25.0 cm3 portion needed 34.60 cm3 of 0.100 mol dm-3 sodium hydroxide for neutralisation. Mr(CaCO3) = 100.1; Kw = 1.0 x 10-14 mol2 dm-6.

  1. (a) Define a Bronsted-Lowry acid and a Bronsted-Lowry base. [2]
  2. (b) Write an equation, with state symbols, for the reaction of calcium carbonate with hydrochloric acid. [1]
  3. (c) Calculate the pH of the 1.00 mol dm-3 hydrochloric acid and of the 0.100 mol dm-3 sodium hydroxide. [3]
  4. (d) Calculate the percentage by mass of calcium carbonate in the tablet. [5]
  5. (e) The pH changes sharply from about 3 to about 11 at the end point of this titration. Explain why phenolphthalein (colour change range pH 8.3 to 10.0) is a suitable indicator. [2]
  6. (f) Suggest why the solution was boiled before the titration, and how the percentage in (d) would change if it had not been boiled. [2]
  7. (g) Explain the difference between a strong acid and a concentrated acid. [2]
  8. (h) Blood contains a buffer made of carbonic acid, H2CO3, and hydrogencarbonate ions, HCO3-. Explain, with an equation, how this buffer keeps the pH of blood nearly constant when a small amount of acid enters it. [3]
 [20]

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06

Apply the same proton bookkeeping to the ocean

Acidification is a fall in pH, even while seawater remains alkaline.

The required ocean pair is CO32-/HCO3-. Carbonate can accept a proton to form hydrogencarbonate; hydrogencarbonate can donate a proton to react with added hydroxide. This is one part of the wider seawater carbonate system.

The ocean pair responds in both directions
AdditionMain buffer reactionRole
AcidCO32-(aq) + H+(aq) → HCO3-(aq)Carbonate removes H+.
AlkaliHCO3-(aq) + OH-(aq) → CO32-(aq) + H2O(l)Hydrogencarbonate removes OH-.

More atmospheric CO2 promotes more dissolved CO2. Its reaction with water can be represented overall as CO2(aq) + H2O(l) ⇌ H+(aq) + HCO3-(aq). The extra H+ is partly taken up by carbonate, increasing hydrogencarbonate and reducing carbonate availability. Buffering limits the change; it cannot cancel a sustained large input, so ocean pH falls.

Worked exampleRead a pH change as a concentration change

A water sample changes from pH 8.2 to 8.0. Has [H+] increased by only 0.2?

Try it first. Then show the worked stepsHide the worked steps
  1. pH is logarithmic, so compare 10-8.0 with 10-8.2.
  2. Their ratio is 100.2 = 1.5849.
  3. Both pH values exceed 7 at 25 degrees C, yet the second sample has more H+.
Answer

[H+] is about 1.58 times as large, a roughly 58% increase. The sample has acidified without becoming acidic in the pH-below-7 sense.

Check your understanding

The ocean is buffered. Why can rising atmospheric CO2 still change its pH?

  1. CO2 does not dissolve in seawater, so it cannot change the pH of the ocean.
  2. A steady CO2 input uses up the buffer's carbonate, so the pH slowly falls.
  3. The buffer fixes the pH exactly, so any change must come from pollution instead.
Show the answer

A steady CO2 input uses up the buffer's carbonate, so the pH slowly falls.

A buffer resists change only within its capacity. Continued CO2 input shifts the carbonate and hydrogencarbonate balance, so the pH drifts down.

Quick revision

Revisit the essentials, then return to an explanation when you need it.

Exam definitions: learn these in mark-scheme words
TermDefinition to write
Arrhenius acid and baseAn Arrhenius acid produces H+ ions in aqueous solution; an Arrhenius base produces OH- ions in aqueous solution.
Brønsted-Lowry acid and baseAn acid is a proton (H+) donor; a base is a proton acceptor.
Conjugate acid-base pairTwo species that differ by one proton, such as CH3COOH and CH3COO-.
Strong acidAn acid that dissociates completely in water.
Weak acidAn acid that dissociates only partially in water.
pHpH = -lg[H+].
Ionic product of water, KwKw = [H+][OH-] = 1.00 × 10-14 mol2 dm-6 at 25 °C.
Buffer solutionA solution that resists a change in pH when a small amount of acid or base is added.
Acid-base decisions
TaskMethod
Identify conjugate pairsSpecies differ by one H+ and one unit of signed charge.
Calculate strong-acid pHCount H+ per formula unit; find concentration; use -log10[H+].
Calculate strong-base pHCount OH-; use [H+] = Kw/[OH-]; then the logarithm.
Mix strong acid and alkaliNeutralise moles first; divide excess by total volume.
Choose an indicatorIts transition must lie in the steep supplied pH interval.
Explain bufferingName the component that consumes the added H+ or OH-.

Scope

Learning outcomes

3. Theories of Acids and Bases (8873, 2026 revision; examinations 2026 and 2027). Use the outcome map to find the explanation for a particular syllabus requirement.

See the learning outcome map

  1. 3(a) Apply the Arrhenius model.

    • Aqueous production of H+ or OH-; neutralisation.

    Follow what the acid and base actually do

  2. 3(b) Apply proton-transfer theory.

    • Bronsted-Lowry acids/bases and conjugate pairs.

    Follow what the acid and base actually do

  3. 3(c) Distinguish acid/base strength.

    • Qualitative extent of dissociation; strength versus concentration.

    Strength describes dissociation, not concentration

  4. 3(d) Explain pH and equilibrium terms.

    • pH, Ka, Kb and Kw; KaKb relation not required.

    Strength describes dissociation, not concentrationCount ions, then take the logarithm

  5. 3(e) Calculate strong-solution hydrogen concentration and pH.

    • Strong acids and strong bases; ion factors, volume units, temperature assumptions.

    Count ions, then take the logarithm

  6. 3(f) Choose titration indicators from data.

    • Transition ranges, acid/base strengths and sharp pH region.

    Choose a colour change inside the sharp rise

  7. 3(g) Explain buffers and their uses.

    • (i) Consumption of added acid/base by buffer components.
    • (ii) CO3 2-/HCO3- ocean buffer and increased atmospheric CO2 causing ocean acidification.

    Keep an acid and its conjugate base availableApply the same proton bookkeeping to the ocean