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)
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.
| Reaction | Acid and its conjugate base | Base 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 example
Find a pair in an unfamiliar reaction
HCO3- + H2O ⇌ CO32- + H3O+. Which species donates the proton?
- HCO3- loses one H+ and becomes CO32-. Its charge changes from -1 to -2.
- Water gains that proton and becomes H3O+.
- Pair each species with the product differing by one proton.
HCO3- is the acid; CO32- is its conjugate base. H2O is the base and H3O+ its conjugate acid.
Strength describes dissociation, not concentration
Separate how much acid was dissolved from how much becomes ions.
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.
| Comparison | Expected observation | Dissociation explanation |
|---|---|---|
| Dilute HCl versus ethanoic acid, both monoprotic | HCl 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 acids | Hydrogen 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 NH3 | NaOH 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.
Count ions, then take the logarithm
Concentration units and stoichiometric factors come before the calculator.
Worked example
A strong monoprotic acid
Find the pH of 3.20 × 10-3 mol dm-3 HCl at 25 degrees C.
- Assume complete dissociation; one mole HCl provides one mole H+.
- [H+] = 3.20 × 10-3 mol dm-3. Water contributes negligibly at this concentration.
- pH = -log10(3.20 × 10-3) = 2.49485.
pH = 2.495. The pH itself has no unit. Using three decimal places preserves the three significant figures of the concentration.
Worked example
A 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.
- Ca(OH)2(aq) supplies two OH- per formula unit: [OH-] = 5.00 × 10-3 mol dm-3.
- Use Kw: [H+] = (1.00 × 10-14)/(5.00 × 10-3) = 2.00 × 10-12 mol dm-3.
- pH = -log10(2.00 × 10-12) = 11.69897.
pH = 11.699. This uses the stated dissolved concentration; it does not assume every mass of solid calcium hydroxide will dissolve.
Worked example
Neutralise 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.
- Initial H+ amount = 0.0400 × 0.0250 = 1.00 × 10-3 mol.
- Initial OH- amount = 0.0400 × 0.0150 = 6.00 × 10-4 mol.
- The 1:1 neutralisation leaves 4.00 × 10-4 mol H+.
- Total volume = 0.0400 dm3, so [H+] = 0.0100 mol dm-3.
pH = 2.000. Averaging the two starting pH values would ignore both stoichiometry and the logarithmic scale.
Check your understandingA strong acid is diluted tenfold while remaining sufficiently concentrated for water autoionisation to be negligible. How does pH change?Think it through, then reveal the answer
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.
| Indicator | Approximate transition pH | Acid-side / base-side colour |
|---|---|---|
| Methyl orange | 3.1-4.4 | Red / yellow |
| Bromothymol blue | 6.0-7.6 | Yellow / blue |
| Phenolphthalein | 8.2-10.0 | Colourless / pink |
| Titration | Region near equivalence | Usual interpretation |
|---|---|---|
| Strong acid / strong base | Steep change spans acid and alkaline regions at suitable concentrations. | Several indicators may work; inspect the given curve. |
| Weak acid / strong base | Equivalence is alkaline because the conjugate base reacts with water. | Phenolphthalein is commonly suitable. |
| Strong acid / weak base | Equivalence is acidic because the conjugate acid reacts with water. | Methyl orange is commonly suitable. |
| Weak acid / weak base | Often no sufficiently sharp pH change. | A visual indicator is generally unsuitable for a precise endpoint. |
Worked example
Use 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?
- Methyl orange changes well before the steep interval.
- Bromothymol blue also finishes its transition below the stated interval.
- Phenolphthalein changes from 8.2 to 10.0, wholly inside the steep interval.
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.
Keep an acid and its conjugate base available
A buffer removes small additions of acid or base, rather than preventing all pH change.
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.
- Add a little acid
CH3CO2-(aq) + H+(aq) → CH3CO2H(aq). The conjugate base removes the added proton.
- 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 understandingWhy is sodium ethanoate solution alone not the intended two-component acidic buffer?Think it through, then reveal the answer
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.
| Addition | Main buffer reaction | Role |
|---|---|---|
| Acid | CO32-(aq) + H+(aq) → HCO3-(aq) | Carbonate removes H+. |
| Alkali | HCO3-(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 example
Read 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?
- pH is logarithmic, so compare 10-8.0 with 10-8.2.
- Their ratio is 100.2 = 1.5849.
- Both pH values exceed 7 at 25 degrees C, yet the second sample has more H+.
[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 understandingWhy does the existence of an ocean buffer not imply that atmospheric CO2 has no effect on pH?Think it through, then reveal the answer
Quick revision
Revisit the essentials, then return to an explanation when you need it.
| Task | Method |
|---|---|
| Identify conjugate pairs | Species differ by one H+ and one unit of signed charge. |
| Calculate strong-acid pH | Count H+ per formula unit; find concentration; use -log10[H+]. |
| Calculate strong-base pH | Count OH-; use [H+] = Kw/[OH-]; then the logarithm. |
| Mix strong acid and alkali | Neutralise moles first; divide excess by total volume. |
| Choose an indicator | Its transition must lie in the steep supplied pH interval. |
| Explain buffering | Name the component that consumes the added H+ or OH-. |
Scope and references
Learning outcomes and sources
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
3(a) Apply the Arrhenius model.
- Aqueous production of H+ or OH-; neutralisation.
3(b) Apply proton-transfer theory.
- Bronsted-Lowry acids/bases and conjugate pairs.
3(c) Distinguish acid/base strength.
- Qualitative extent of dissociation; strength versus concentration.
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
3(e) Calculate strong-solution hydrogen concentration and pH.
- Strong acids and strong bases; ion factors, volume units, temperature assumptions.
3(f) Choose titration indicators from data.
- Transition ranges, acid/base strengths and sharp pH region.
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
- SEAB H1 Chemistry 8873, 2026 revision
Official scope: section 3, printed pages 13-14. All lettered outcomes and nested requirements checked.
- SEAB H1 Chemistry 8873, 2027
Section 3: same substantive outcomes as the revised 2026 course.
- ASRJC 2025 H2 Acid-Base Equilibria Notes (Grail)
Consulted printed pp. 9-10 for acid/base dissociation and p. 14 for the strength-versus-concentration distinction. Used only the qualitative H1 requirements; no H2 weak-acid or buffer-pH calculations imported.