Master IB Chemistry R3.1: Proton transfer reactions with notes created by examiners and strictly aligned with the syllabus.
R3.1.1 Brønsted–Lowry acid is a proton donor and a Brønsted–Lowry base is a proton acceptor.
R3.1.2 A pair of species differing by a single proton is called a conjugate acid–base pair.
R3.1.3 Some species can act as both Brønsted–Lowry acids and bases.
R3.1.4 The pH scale can be used to describe the [H+] of a solution: pH = -log10[H+]; [H+] = 10^-pH
An acid is a chemical species that donates a proton, H⁺, to another species in a Brønsted–Lowry reaction. A base is a chemical species that accepts a proton, H⁺, from another species in a Brønsted–Lowry reaction. That version is more useful than the old “sour/slippery/litmus” description, since it tells you what actually moves during the reaction.
The Arrhenius idea still helps: acids form H⁺ in water, and bases form OH⁻ in water. But it has a clear limit. It only deals with aqueous solutions, so it misses bases such as ammonia. Brønsted–Lowry theory covers more cases. In NH₃(g) + HCl(g) → NH₄Cl(s), HCl donates H⁺, so HCl is the acid; NH₃ accepts H⁺, so NH₃ is the base. The definition of acid has evolved over time because better models explain more observations with fewer awkward exceptions.

A bare proton is much too charge-dense to move around freely in water. In aqueous solution, it attaches to water to form H₃O⁺. IB accepts both H⁺(aq) and H₃O⁺(aq) in equations: H⁺(aq) works as convenient shorthand, while H₃O⁺(aq) shows what is actually present.
For example, hydrogen chloride reacting with water can be written as either:
HCl(aq) → H⁺(aq) + Cl⁻(aq)
or, more explicitly,
HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq).
An alkali is a base that is soluble in water and produces hydroxide ions in aqueous solution. Sodium hydroxide is both a base and an alkali. Copper(II) oxide is a base because it reacts with acids, but it is not an alkali because it is not water-soluble. Keep the distinction tidy: all alkalis are bases, but not all bases are alkalis.
To identify the acid and base in a reaction, look for the species that loses H⁺ and the one that gains H⁺. In OH⁻(aq) + H⁺(aq) → H₂O(l), OH⁻ is the Brønsted–Lowry base because it accepts a proton.
A conjugate acid–base pair is a pair of chemical species that differ by exactly one proton. The acid member carries one more H⁺ than the base member. If an acid donates H⁺, it forms its conjugate base; if a base accepts H⁺, it forms its conjugate acid.
For the reaction
HCN(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CN⁻(aq)
the two conjugate pairs are HCN/CN⁻ and H₃O⁺/H₂O. HCN donates H⁺ and becomes CN⁻. H₂O accepts H⁺ and becomes H₃O⁺.

To write a conjugate base, remove one H⁺ and lower the charge by 1: H₂PO₄⁻ → HPO₄²⁻. To write a conjugate acid, add one H⁺ and raise the charge by 1: SO₄²⁻ → HSO₄⁻.
This pattern helps a lot with polyatomic anions. Common conjugate acids include NO₃⁻ → HNO₃, CO₃²⁻ → HCO₃⁻, HCO₃⁻ → H₂CO₃, PO₄³⁻ → HPO₄²⁻, and CH₃COO⁻ → CH₃COOH. Add only one proton at a time. So H₂SO₄ and SO₄²⁻ are not a conjugate pair, since they differ by two protons.
An amphiprotic species is a chemical species that can both donate a proton and accept a proton. Water is the standard example:
H₂O(l) ⇌ H⁺(aq) + OH⁻(aq) water acting as an acid
H₂O(l) + H⁺(aq) ⇌ H₃O⁺(aq) water acting as a base
It can also react with itself:
H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq).
Hydrogencarbonate, HCO₃⁻, is another common one. As an acid:
HCO₃⁻(aq) + H₂O(l) ⇌ CO₃²⁻(aq) + H₃O⁺(aq)
As a base:
HCO₃⁻(aq) + H₂O(l) ⇌ H₂CO₃(aq) + OH⁻(aq).
An amphoteric species is a substance that can react with both acids and bases. All amphiprotic species are amphoteric, but not every amphoteric species is amphiprotic. Zinc oxide, for example, reacts with acids and alkalis. It has no hydrogen atom to donate, so it is amphoteric but not amphiprotic.
Metal oxides are usually basic, while non-metal oxides are usually acidic. Oxides close to the metal–non-metal divide can be amphoteric. That pattern helps explain acid rain from sulfur and nitrogen oxides: these non-metal oxides form acids after reacting with water and oxygen in the atmosphere.

Amino acids are a useful biological example. In water, an amino acid can exist as a zwitterion, with –NH₃⁺ and –COO⁻ groups in the same species. In acid, it accepts H⁺ at the carboxylate end; in base, it donates H⁺ from the ammonium end.
Aqueous hydrogen ion concentrations can cover many powers of ten, which is why chemists use a logarithmic scale. The relationship is:
pH = −log₁₀[H⁺], where pH is the acidity scale value (dimensionless) and [H⁺] is the equilibrium concentration of hydrogen ions in aqueous solution (mol dm⁻³).
Rearranging gives:
[H⁺] = 10⁻ᵖᴴ, where [H⁺] is the equilibrium concentration of hydrogen ions in aqueous solution (mol dm⁻³) and pH is the acidity scale value (dimensionless).
If pH decreases by 1 unit, [H⁺] becomes ten times larger. So pH 2 is not “a bit more acidic” than pH 3; it has ten times the hydrogen ion concentration.
At 298 K, a neutral aqueous solution has equal hydrogen and hydroxide ion concentrations. To identify acidic, neutral and basic solutions, compare [H⁺] and [OH⁻] rather than just memorising pH values:
| solution | condition at 298 K | typical pH |
|---|---|---|
| acidic | [H⁺] > [OH⁻] | less than 7 |
| neutral | [H⁺] = [OH⁻] | 7 |
| basic | [H⁺] < [OH⁻] | greater than 7 |
Universal indicator gives an estimate, since you match its colour to a chart. It’s quick and cheap, but the reading is subjective and usually only precise to about one pH unit. A pH probe gives a more precise digital reading, so it is better for continuous data, small pH changes, coloured solutions, or a titration curve. The probe must be calibrated, usually with buffer solutions of known pH.

A sketch of pH against [H⁺] is a decreasing logarithmic curve: it drops steeply at very small [H⁺], then flattens as [H⁺] increases. That shape is why pH is useful — it compresses a huge concentration range into manageable numbers.

Most water stays as molecules, but a very small amount ionizes:
H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)
For dilute aqueous solutions, the equilibrium is written as:
Kₚ = [H⁺][OH⁻], where Kₚ is the ion product constant of water (mol² dm⁻⁶), [H⁺] is the equilibrium concentration of hydrogen ions (mol dm⁻³) and [OH⁻] is the equilibrium concentration of hydroxide ions (mol dm⁻³).
You’ll usually see this written as Kw. At 298 K, Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶. This value is given in the data booklet.
At a fixed temperature, the product [H⁺][OH⁻] stays constant, so [H⁺] and [OH⁻] change inversely. If [H⁺] increases, [OH⁻] has to decrease. Le Châtelier’s principle explains it: adding acid shifts the water ionization equilibrium to the left.

In pure water at 298 K, [H⁺] = [OH⁻], so each concentration is 1.00 × 10⁻⁷ mol dm⁻³. When temperature rises, water ionizes more, so Kw increases. That’s why “neutral pH” is exactly 7 only with the usual 298 K assumption.
Classify a solution by comparing the two ion concentrations: acidic if [H⁺] > [OH⁻], neutral if equal, and basic if [H⁺] < [OH⁻].
A strong acid is an acid that ionizes completely in aqueous solution. A weak acid is an acid that ionizes only partially in aqueous solution. In the same way, a strong base is a base that ionizes completely or dissociates completely to produce the basic species in water, whereas a weak base is a base that reacts only partially with water or acids to accept protons.
Use a single arrow for strong acids and bases. Use an equilibrium arrow for weak ones:
HCl(aq) → H⁺(aq) + Cl⁻(aq)
CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq)
The strong acids you need to know are HCl, HBr, HI, HNO₃ and H₂SO₄. Group 1 hydroxides, such as LiOH, NaOH and KOH, are strong bases.
A concentrated solution is a solution that contains a large amount of solute per unit volume. A dilute solution is a solution that contains a small amount of solute per unit volume. Strength tells you about ionization; concentration tells you how much solute has been dissolved. A dilute strong acid is still strong. A concentrated weak acid is still weak.
Compares strength with concentration using strong and weak acid examples.
| Strength | Concentration | Example reagent | Ionization in water | Key idea |
|---|---|---|---|---|
| Strong | Concentrated | 2.0 mol dm⁻³ HCl(aq) | Complete | Many ions because much solute is present and HCl is fully ionized. |
| Strong | Dilute | 0.010 mol dm⁻³ HCl(aq) | Complete | Still a strong acid; dilute only means less solute per volume. |
| Weak | Concentrated | 2.0 mol dm⁻³ CH₃COOH(aq) | Partial | Many acid molecules are present, but only a fraction ionize. |
| Weak | Dilute | 0.010 mol dm⁻³ CH₃COOH(aq) | Partial | Weak and dilute: little solute present and only partial ionization. |
Acid–base equilibria favour the side with the weaker acid and the weaker base — the weaker conjugates. A strong acid has a very weak conjugate base; for example, Cl⁻ barely behaves as a base in water. A weak acid has a conjugate base that accepts H⁺ more readily.
For hydrogen halides, acid strength increases down group 17. The H–X bond gets longer and weaker, making proton donation easier: HF is weak, while HCl, HBr and HI are strong.
In the laboratory, equal-concentration strong and weak acids can be told apart using pH, electrical conductivity, and rate of reaction with a reactive metal or carbonate. The strong acid gives a lower pH, higher conductivity and faster gas production because it has a higher concentration of mobile ions at the start.
A neutralization reaction is a proton transfer reaction in which an acid reacts with a base to form a salt, often with water as a product. A salt is an ionic compound formed when the replaceable hydrogen ion of an acid is replaced by a cation, or when an acid reacts with a base.
You should be able to write these core equations:
| base type | general pattern | example |
|---|---|---|
| metal hydroxide | acid + metal hydroxide → salt + water | HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) |
| metal oxide | acid + metal oxide → salt + water | 2HNO₃(aq) + MgO(s) → Mg(NO₃)₂(aq) + H₂O(l) |
| carbonate | acid + carbonate → salt + carbon dioxide + water | 2HCl(aq) + Na₂CO₃(aq) → 2NaCl(aq) + CO₂(g) + H₂O(l) |
| hydrogencarbonate | acid + hydrogencarbonate → salt + carbon dioxide + water | HCl(aq) + NaHCO₃(aq) → NaCl(aq) + CO₂(g) + H₂O(l) |
The proton transfer is easiest to see in the net ionic equations:
H⁺(aq) + OH⁻(aq) → H₂O(l)
CO₃²⁻(aq) + 2H⁺(aq) → CO₂(g) + H₂O(l)
HCO₃⁻(aq) + H⁺(aq) → CO₂(g) + H₂O(l).
In this topic, bases include ammonia, amines, soluble carbonates and soluble hydrogencarbonates. Organic acids, such as carboxylic acids, act as weak acids. For example:
CH₃COOH(aq) + NH₃(aq) ⇌ CH₃COO⁻(aq) + NH₄⁺(aq).
To find the parent acid and base, split the salt into its ions. Then add H⁺ to the anion and OH⁻ to the cation as needed. For K₂SO₄, the ions are K⁺ and SO₄²⁻, so the parent base is KOH and the parent acid is H₂SO₄. For ammonium salts, the parent base may be written as ammonia, NH₃, or ammonium hydroxide, NH₄OH.
Soluble salts made by neutralization can often be separated by crystallization: gently evaporate some water, let crystals form, then filter and dry them. If the salt is insoluble, filtration is the obvious separation step.
Neutralization reactions are exothermic because the bonds and attractions formed in the products, especially O–H bonds in water and ionic attractions in the salt solution or solid, are energetically favourable compared with the bonds broken. When an acid reacts with a metal to release hydrogen gas, the reaction is not just acid–base. It is also a redox reaction because electrons are transferred to H⁺ to form H₂.
A pH curve is a graph that shows how the pH of a reaction mixture changes as titrant volume is added during a titration. For this SL statement, questions only assess monoprotic strong acid–strong base neutralizations.
When a strong acid analyte is titrated with a strong base, the curve begins at low pH. It rises slowly to start with, climbs very sharply near the equivalence point, then levels out at high pH once base is in excess. If a strong base analyte is titrated with a strong acid, the shape is the same but inverted.

The equivalence point is the point in a titration where reactants have been mixed in the exact stoichiometric ratio shown by the balanced equation. It is therefore also called the stoichiometric point. In a strong acid–strong base titration at 298 K, the equivalence point is at pH 7 because the salt formed does not hydrolyse significantly.
The y-intercept gives the initial pH of the analyte before any titrant has been added. The steep vertical section shows the rapid pH change close to equivalence. Once the curve starts to flatten, the pH is controlled mainly by excess titrant.
For titration calculations, find the amount of solute using n = cV, where n is amount of substance (mol), c is amount concentration (mol dm⁻³) and V is solution volume (dm³). At equivalence, use the ratio from the balanced equation to link the amount of acid with the amount of base. For a 1:1 monoprotic reaction such as HCl + NaOH → NaCl + H₂O, the amounts of HCl and NaOH are equal at equivalence.
pOH is a logarithmic scale value that describes hydroxide ion concentration in aqueous solution. It is defined as:
pOH = −log₁₀[OH⁻], where pOH is the basicity scale value (dimensionless) and [OH⁻] is the equilibrium concentration of hydroxide ions (mol dm⁻³).
Rearrange it and you get:
[OH⁻] = 10⁻ᵖᴼᴴ, where [OH⁻] is the equilibrium concentration of hydroxide ions (mol dm⁻³) and pOH is the basicity scale value (dimensionless).
At 298 K, the pH, pOH and Kw relationships combine to give:
pH + pOH = 14.00, where pH is the acidity scale value (dimensionless) and pOH is the basicity scale value (dimensionless).
You can therefore switch between all four quantities: [H⁺], [OH⁻], pH and pOH. If pH is known, calculate pOH using 14.00 − pH, then calculate [OH⁻] using 10⁻ᵖᴼᴴ. If [OH⁻] is known, calculate pOH first, then pH.
Conversion steps linking [H⁺], pH, pOH and [OH⁻] at 298 K.
| Start quantity | End quantity | Calculation |
|---|---|---|
| [H⁺] / mol dm⁻³ | pH | pH = −log₁₀[H⁺] |
| pH | [H⁺] / mol dm⁻³ | [H⁺] = 10⁻ᵖᴴ |
| pH | pOH | pOH = 14.00 − pH |
| pOH | pH | pH = 14.00 − pOH |
| pOH | [OH⁻] / mol dm⁻³ | [OH⁻] = 10⁻ᵖᴼᴴ |
| [OH⁻] / mol dm⁻³ | pOH | pOH = −log₁₀[OH⁻] |
For a weak acid, written generally as HA:
HA(aq) ⇌ H⁺(aq) + A⁻(aq)
Ka = [H⁺][A⁻]/[HA], where Ka is the acid dissociation constant (mol dm⁻³ for this expression), [H⁺] is the equilibrium concentration of hydrogen ions (mol dm⁻³), [A⁻] is the equilibrium concentration of conjugate base (mol dm⁻³), and [HA] is the equilibrium concentration of undissociated acid (mol dm⁻³).
For a weak base, written generally as B:
B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)
Kb = [BH⁺][OH⁻]/[B], where Kb is the base dissociation constant (mol dm⁻³ for this expression), [BH⁺] is the equilibrium concentration of conjugate acid (mol dm⁻³), [OH⁻] is the equilibrium concentration of hydroxide ions (mol dm⁻³), and [B] is the equilibrium concentration of unprotonated base (mol dm⁻³).
Water does not appear in the expression because it acts as the solvent, so its concentration stays effectively constant in dilute solution.
A larger Ka shows a stronger weak acid. A larger Kb shows a stronger weak base. With p-values, the comparison runs the other way:
pKa = −log₁₀Ka, where pKa is the negative logarithmic acid strength value (dimensionless) and Ka is the acid dissociation constant.
pKb = −log₁₀Kb, where pKb is the negative logarithmic base strength value (dimensionless) and Kb is the base dissociation constant.
So a lower pKa means a stronger acid; a lower pKb means a stronger base. Students often flip this — p-values are negative logs.
How dissociation constants and p-values compare weak acid and base strength.
| Measure / unit | Used for | Stronger when | Weaker when |
|---|---|---|---|
| Ka / mol dm⁻³ | Weak acids | Ka is larger | Ka is smaller |
| pKa / dimensionless | Weak acids | pKa is smaller | pKa is larger |
| Kb / mol dm⁻³ | Weak bases | Kb is larger | Kb is smaller |
| pKb / dimensionless | Weak bases | pKb is smaller | pKb is larger |
For the conjugate pair HA/A⁻:
HA(aq) ⇌ H⁺(aq) + A⁻(aq)
Ka = [H⁺][A⁻]/[HA]
and
A⁻(aq) + H₂O(l) ⇌ HA(aq) + OH⁻(aq)
Kb = [HA][OH⁻]/[A⁻]
When the two expressions are multiplied, [HA] and [A⁻] cancel. That leaves:
Ka × Kb = Kw, where Ka is the acid dissociation constant of an acid, Kb is the base dissociation constant of its conjugate base, and Kw is the ion product constant of water.
At 298 K, the same relationship gives pKa + pKb = 14.00 for a conjugate acid–base pair. Don’t use it for unrelated acids and bases.
The syllabus does not expect quadratic equations. If Ka or Kb is very small, only a small amount ionizes, so the equilibrium concentration of the weak acid or base is usually treated as approximately the same as its initial concentration.
For a weak acid with initial concentration C, if x = [H⁺] at equilibrium, then often:
Ka ≈ x²/C, where x is the equilibrium concentration of H⁺ formed by the weak acid (mol dm⁻³) and C is the initial concentration of the weak acid (mol dm⁻³).
For a weak base:
Kb ≈ y²/Cb, where y is the equilibrium concentration of OH⁻ formed by the weak base (mol dm⁻³) and Cb is the initial concentration of the weak base (mol dm⁻³).
Say what approximation you are making when you use it. The chemical reason is straightforward: weak acids and weak bases mostly remain un-ionized.
Hydrolysis is a reaction in which an ion reacts with water to form its conjugate acid or conjugate base, changing [H⁺] or [OH⁻]. Salt solutions are not automatically neutral. Their pH depends on whether the ions present are conjugates of weak acids or weak bases.
A useful class rule: ions from strong parents usually do nothing; ions from weak parents hydrolyse.
| parent acid | parent base | important hydrolysis | effect on pH |
|---|---|---|---|
| strong | strong | none | neutral |
| strong | weak | cation hydrolyses | acidic |
| weak | strong | anion hydrolyses | basic |
| weak | weak | both ions may hydrolyse | depends on relative strengths |
Salt solution pH from parent acid/base strengths and ion hydrolysis.
| Parent acid | Parent base | Ion(s) hydrolysing | Water product | Predicted pH |
|---|---|---|---|---|
| strong | strong | none | neither H₃O⁺ nor OH⁻ | neutral, pH ≈ 7 |
| strong | weak | cation | H₃O⁺ | acidic, pH < 7 |
| weak | strong | anion | OH⁻ | basic, pH > 7 |
| weak | weak | cation and anion | H₃O⁺ and OH⁻ | depends on relative strengths |
You should be comfortable writing these examples:
Ammonium ion, acidic:
NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)
Carboxylate ion, basic:
RCOO⁻(aq) + H₂O(l) ⇌ RCOOH(aq) + OH⁻(aq)
Carbonate ion, basic:
CO₃²⁻(aq) + H₂O(l) ⇌ HCO₃⁻(aq) + OH⁻(aq)
Hydrogencarbonate ion can behave in either direction, so compare the relevant strengths:
HCO₃⁻(aq) + H₂O(l) ⇌ H₂CO₃(aq) + OH⁻(aq)
HCO₃⁻(aq) + H₂O(l) ⇌ CO₃²⁻(aq) + H₃O⁺(aq)
An ion that produces H₃O⁺ lowers pH. An ion that produces OH⁻ raises pH. The acidity of hydrated transition metal ions and Al³⁺(aq) is outside this topic, so don’t use it in these salt-pH predictions.
For monoprotic acid–base titrations, the shape of the curve comes down to the strength of the acid and the strength of the base.
Strong acid with strong base: low initial pH, no buffer region, a large vertical jump, equivalence at pH 7, then a high-pH flattening if base is in excess.
Weak acid with strong base: the initial pH is higher than for a strong acid of the same concentration. There is a buffer region before equivalence, the vertical jump is smaller, and equivalence is above pH 7 because the salt anion hydrolyses to form OH⁻.
Strong acid with weak base: the same idea in reverse — equivalence is below pH 7 because the salt cation hydrolyses to form H₃O⁺.
Weak acid with weak base: buffer behaviour can appear on both sides, but there is no sharp vertical section. That makes the equivalence point hard to locate accurately.

The half-equivalence point is the point in a titration where exactly half of the original weak acid or weak base has been neutralized. In a weak acid–strong base titration, half-equivalence gives a buffer with equal concentrations of HA and A⁻, so pH = pKa.
For a weak base titrated with strong acid, the buffer has equal concentrations of B and BH⁺ at half-equivalence, so pOH = pKb.
The buffer region is the gently sloping part of the curve where both members of a weak conjugate pair are present in appreciable amounts. When you collect data for a pH curve, use smaller additions of titrant where the pH changes quickly — especially near the equivalence point — so you don’t skip over the important features.
An acid–base indicator is a weak acid or weak base whose conjugate acid and conjugate base have different colours. For a weak acid indicator, the equilibrium is written as:
HInd(aq) ⇌ H⁺(aq) + Ind⁻(aq)
KInd = [H⁺][Ind⁻]/[HInd], where KInd is the indicator acid dissociation constant (mol dm⁻³ for this expression), [Ind⁻] is the equilibrium concentration of the deprotonated indicator form (mol dm⁻³), and [HInd] is the equilibrium concentration of the protonated indicator form (mol dm⁻³).
In low pH conditions, the high [H⁺] shifts the equilibrium to the left, so the HInd colour is seen. At high pH, [H⁺] is lower and the equilibrium shifts to the right, so the Ind⁻ colour dominates. The change in colour happens across a range, usually centred close to the indicator pKa.

Universal indicator is a mixture of several acid–base indicators that gives a sequence of colours over a wide pH range. It’s useful for estimating pH, but not for precise titration end points.
Acid–base indicators and redox indicators both use a visible colour change to show a stage in a titration. What differs is the equilibrium involved: acid–base indicators change colour through proton transfer, while redox indicators change colour through electron transfer or, in some titrations, the titrant is self-indicating.
The end point is the point in a titration where the indicator shows its chosen colour change. The equivalence point is the point where acid and base have reacted in the exact stoichiometric ratio. In a careful titration, you try to get these two points as close together as possible. They’re related, but they are not the same thing.
Choose an indicator with a transition range that falls within the steep section of the pH curve around the equivalence point. The salt formed helps you estimate the equivalence pH:

For example, sodium ethanoate at equivalence points back to weak ethanoic acid and strong sodium hydroxide. That puts the equivalence point above pH 7, so an indicator that changes in the basic region, such as phenolphthalein, is suitable. With ammonium chloride, the parent acid is strong hydrochloric acid and the parent base is weak ammonia, so you’d choose an acidic-range indicator instead.
Use only a few drops of indicator. Indicators are weak acids or bases themselves, and too much of one adds a small but real systematic error.
A buffer solution is an aqueous solution that resists pH change when small amounts of acid or alkali are added. To do this, it needs appreciable amounts of both members of a weak conjugate acid–base pair.
An acidic buffer is a buffer made from a weak acid and its conjugate base. A typical example is CH₃COOH/CH₃COO⁻.
When acid is added:
H⁺(aq) + CH₃COO⁻(aq) → CH₃COOH(aq)
When alkali is added:
OH⁻(aq) + CH₃COOH(aq) → CH₃COO⁻(aq) + H₂O(l)
A basic buffer is a buffer made from a weak base and its conjugate acid. A typical example is NH₃/NH₄⁺.
When acid is added:
H⁺(aq) + NH₃(aq) → NH₄⁺(aq)
When alkali is added:
OH⁻(aq) + NH₄⁺(aq) → NH₃(aq) + H₂O(l)

Strong acids and strong bases do not form useful buffers with their conjugates, since those conjugates are too weak to remove added acid or alkali. In a working buffer, added strong acid gets converted into a weak acid, while added strong base gets converted into a weak base. Le Châtelier’s principle describes the same behaviour: adding H⁺ or OH⁻ disturbs the conjugate equilibrium, so the buffer components react to oppose that disturbance.
A buffer has limited capacity. Once either conjugate component has been used up, the pH changes rapidly.
For an acidic buffer containing HA and A⁻, begin with:
Ka = [H⁺][A⁻]/[HA]
Rearrange it:
[H⁺] = Ka[HA]/[A⁻]
Taking negative logarithms gives the Henderson–Hasselbalch form:
pH = pKa + log₁₀([A⁻]/[HA]), where pH is the acidity scale value (dimensionless), pKa is the negative logarithmic acid strength value (dimensionless), [A⁻] is the concentration of conjugate base in the buffer (mol dm⁻³), and [HA] is the concentration of weak acid in the buffer (mol dm⁻³).
For a basic buffer, it’s usually simplest to find pOH first:
pOH = pKb + log₁₀([BH⁺]/[B]), where pOH is the basicity scale value (dimensionless), pKb is the negative logarithmic base strength value (dimensionless), [BH⁺] is the concentration of conjugate acid in the buffer (mol dm⁻³), and [B] is the concentration of weak base in the buffer (mol dm⁻³).
Then convert using pH + pOH = 14.00 at 298 K.

The ratio does the work here. If [A⁻] = [HA], then log₁₀(1) = 0, so pH = pKa. When the conjugate base concentration is larger than the acid concentration, pH is above pKa; when it is smaller, pH is below pKa.
Dilution usually changes buffer pH very little because it lowers both buffer concentrations by the same factor, so their ratio stays the same. It does reduce buffer capacity, though: fewer moles of HA and A⁻ remain to remove added OH⁻ or H⁺. At extreme dilution, the solution tends towards the behaviour of water at that temperature.