IB DP Chemistry Reactivity 3.1: Proton transfer: Kw, Ka, Kb, buffers and titration curves

This page teaches the higher level half of Reactivity 3.1, “Proton transfer reactions”, for IB DP Chemistry. Its numbering follows the Diploma Programme Chemistry guide (February 2023, first assessment 2025), and it works through understandings Reactivity 3.1.9 to 3.1.17 with examples. The guide lists every one of these under additional higher level (AHL), so the entire page is HL only. It applies to examinations from 2025 onwards.

Use it alongside the proton transfer revision notes for fast recall and the proton transfer practice questions for marked answers. The SL half is in Proton transfer: acids, bases and pH. Other units sit on the DP Chemistry hub; mark off understandings on the printable DP Chemistry checklist; and take one of our free diagnostic quizzes if you are unsure where to begin.

Scope of this page

Statement	Skills	Level
Reactivity 3.1.9	Interconvert [H⁺], [OH⁻], pH and pOH	HL only
Reactivity 3.1.10	Compare the strengths of weak acids and bases from Ka, Kb, pKa or pKb	HL only
Reactivity 3.1.11	Derive Ka × Kb = Kw for a conjugate pair and solve problems with it	HL only
Reactivity 3.1.12	Write hydrolysis equations for ions in a salt and predict the pH of its solution	HL only
Reactivity 3.1.13	Interpret pH curves for all four strong/weak acid and base combinations	HL only
Reactivity 3.1.14	Explain indicator colour change using HInd	HL only
Reactivity 3.1.15	Choose a suitable indicator; tell end point from equivalence point	HL only
Reactivity 3.1.16	Describe acidic and basic buffers and explain how they work	HL only
Reactivity 3.1.17	Calculate buffer pH and composition; explain the effect of dilution	HL only

The guide states that the pOH equations, and examples of indicators with their pH ranges, are given in the data booklet. On this page each question supplies the values it needs. All values assume 298 K, where Kw = 1.00 × 10⁻¹⁴, so pH + pOH = 14.00.

Reactivity 3.1.9 The pOH scale

pOH does for hydroxide ions what pH does for hydrogen ions:

• pOH = −log₁₀[OH⁻] and [OH⁻] = 10^(−pOH)
• Taking −log₁₀ of Kw = [H⁺][OH⁻] gives pH + pOH = pKw = 14.00 at 298 K

A solution is neutral when pH = pOH, acidic when pH < pOH and basic when pH > pOH. The value 14.00 holds only at 298 K, because Kw varies with temperature.

Worked example 1: interconverting the four quantities

(a) A potassium hydroxide solution has [OH⁻] = 3.5 × 10⁻³ mol dm⁻³. Find pOH, pH and [H⁺].

• pOH = −log₁₀(3.5 × 10⁻³) = 2.46
• pH = 14.00 − 2.46 = 11.54
• [H⁺] = 10^(−11.54) = 2.9 × 10⁻¹² mol dm⁻³

(b) Rainwater at pH 4.62 has pOH = 9.38 and [OH⁻] = 10^(−9.38) = 4.2 × 10⁻¹⁰ mol dm⁻³.

Reactivity 3.1.10 Describing weak acid and base strength

A weak acid HA only partly ionises: HA(aq) ⇌ H⁺(aq) + A⁻(aq). Its acid dissociation constant is

Ka = [H⁺][A⁻] / [HA], with pKa = −log₁₀Ka

A weak base B accepts a proton from water: B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq). Water is the solvent, so it is left out:

Kb = [BH⁺][OH⁻] / [B], with pKb = −log₁₀Kb

A larger Ka (smaller pKa) means a stronger acid; a larger Kb (smaller pKb) means a stronger base. One pKa unit is a factor of ten in Ka.

Worked example 2: ranking acids

Use these values: hypochlorous acid HOCl, pKa = 7.53; hydrofluoric acid HF, Ka = 6.8 × 10⁻⁴; lactic acid, pKa = 3.86. Rank the acids from strongest to weakest.

pKa(HF) = −log₁₀(6.8 × 10⁻⁴) = 3.17, so the order is HF (3.17) > lactic acid (3.86) > HOCl (7.53).

Reactivity 3.1.11 Ka × Kb = Kw

Take a conjugate pair HA/A⁻. The base reaction of A⁻ is A⁻(aq) + H₂O(l) ⇌ HA(aq) + OH⁻(aq), so

Ka × Kb = ([H⁺][A⁻] / [HA]) × ([HA][OH⁻] / [A⁻]) = [H⁺][OH⁻] = Kw

and so pKa + pKb = pKw = 14.00 at 298 K. The stronger an acid, the weaker its conjugate base.

Worked example 3: the conjugate base of HOCl

pKa(HOCl) = 7.53. pKb(OCl⁻) = 14.00 − 7.53 = 6.47, so Kb = 10^(−6.47) = 3.4 × 10⁻⁷.

pH of a weak acid or weak base

The guide does not expect quadratic equations. Make two assumptions:

1. [H⁺] = [A⁻], because ionisation of water is ignored.
2. [HA] at equilibrium ≈ [HA] initial, because only a small fraction ionises.

Then Ka ≈ [H⁺]² / [HA]initial, so [H⁺] = √(Ka × [HA]initial). For a weak base, [OH⁻] = √(Kb × [B]initial).

Worked example 4: pH of nitrous acid

Find the pH of 0.250 mol dm⁻³ HNO₂ (Ka = 5.6 × 10⁻⁴).

• [H⁺] = √(5.6 × 10⁻⁴ × 0.250) = 1.18 × 10⁻² mol dm⁻³
• pH = −log₁₀(1.18 × 10⁻²) = 1.93

Only about 4.7% ionises, so the approximation is fair.

Worked example 5: pH of dimethylamine

Find the pH of 0.200 mol dm⁻³ (CH₃)₂NH (pKb = 3.27).

• Kb = 10^(−3.27) = 5.37 × 10⁻⁴
• [OH⁻] = √(5.37 × 10⁻⁴ × 0.200) = 1.04 × 10⁻² mol dm⁻³
• pOH = 1.98, so pH = 14.00 − 1.98 = 12.02

Reactivity 3.1.12 pH of salt solutions

A salt holds the conjugate acid of its parent base and the conjugate base of its parent acid. Only an ion from a weak parent reacts with water (hydrolysis).

Ion	Hydrolysis equation	Effect on pH
NH₄⁺	NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)	lowers it
RCOO⁻	RCOO⁻(aq) + H₂O(l) ⇌ RCOOH(aq) + OH⁻(aq)	raises it
CO₃²⁻	CO₃²⁻(aq) + H₂O(l) ⇌ HCO₃⁻(aq) + OH⁻(aq)	raises it
HCO₃⁻	HCO₃⁻(aq) + H₂O(l) ⇌ H₂CO₃(aq) + OH⁻(aq)	raises it slightly

Na⁺, K⁺, Cl⁻ and NO₃⁻ come from strong parents and leave the pH unchanged. So:

• strong acid + strong base (e.g. KNO₃): pH 7
• strong acid + weak base (e.g. NH₄Cl): pH below 7
• weak acid + strong base (e.g. sodium propanoate): pH above 7
• weak acid + weak base: compare Ka of the cation with Kb of the anion

HCO₃⁻ is amphiprotic. With Ka₁ = 4.5 × 10⁻⁷ and Ka₂ = 4.7 × 10⁻¹¹ for carbonic acid, Kb(HCO₃⁻) = Kw ÷ Ka₁ = 2.2 × 10⁻⁸, which beats Ka₂, so sodium hydrogencarbonate solution is mildly basic.

Worked example 6: pH of sodium benzoate

Find the pH of 0.100 mol dm⁻³ sodium benzoate. Ka(benzoic acid) = 6.3 × 10⁻⁵.

• Kb(C₆H₅COO⁻) = 1.00 × 10⁻¹⁴ ÷ 6.3 × 10⁻⁵ = 1.59 × 10⁻¹⁰
• [OH⁻] = √(1.59 × 10⁻¹⁰ × 0.100) = 3.98 × 10⁻⁶ mol dm⁻³
• pOH = 5.40, so pH = 8.60, basic as predicted.

Reactivity 3.1.13 pH curves for the four combinations

The features to read from any curve are the intercept on the pH axis, the equivalence point, the buffer region and the half-equivalence point.

Titration (base added to acid)	Starting pH	Steep section	pH at equivalence
Strong acid + strong base	low (about 1 at 0.1 mol dm⁻³)	long, roughly 3 to 11	7
Weak acid + strong base	higher (about 3)	shorter, above 7	above 7 (salt anion hydrolyses)
Strong acid + weak base	low	shorter, below 7	below 7 (salt cation hydrolyses)
Weak acid + weak base	higher	no sharp vertical section	depends on Ka and Kb

Buffer region and half-equivalence. While strong base runs into a weak acid, the curve flattens early on because HA and A⁻ are both present. At half the equivalence volume, [HA] = [A⁻], so Ka = [H⁺] and pH = pKa. When a weak base is titrated with a strong acid, [B] = [BH⁺] at half-equivalence, so pOH = pKb.

When collecting data, add smaller portions near the equivalence point, where pH changes fastest.

Worked example 7: reading a weak acid curve

20.0 cm³ of a weak acid HX is titrated with 0.0800 mol dm⁻³ NaOH. The equivalence point is at 23.5 cm³, and the pH at 11.75 cm³ is 4.62.

• n(NaOH) = 0.0800 × 0.0235 = 1.88 × 10⁻³ mol, so [HX] = 1.88 × 10⁻³ ÷ 0.0200 = 0.0940 mol dm⁻³
• 11.75 cm³ is half-equivalence, so pKa = 4.62 and Ka = 10^(−4.62) = 2.4 × 10⁻⁵

Reactivity 3.1.14 Indicators

An indicator is a weak acid, HInd, whose conjugate base has a different colour:

HInd(aq) ⇌ H⁺(aq) + Ind⁻(aq) colour A ⇌ colour B

• Add acid: [H⁺] rises, the equilibrium shifts left, colour A shows.
• Add alkali: OH⁻ removes H⁺, the equilibrium shifts right, colour B shows.
• The colour changes when [HInd] ≈ [Ind⁻]. Then Ka(HInd) = [H⁺], so the end point pH ≈ pKa(HInd).

Universal indicator mixes many indicators, so its colour shifts gradually over a wide pH range: good for estimating pH, poor for spotting an end point.

Reactivity 3.1.15 Choosing an indicator

At the equivalence point, acid and base have been mixed in exactly the stoichiometric ratio. The end point is the volume at which you see the indicator switch colour. A good indicator makes the two coincide: its colour change range must lie inside the steep part of the curve, around the equivalence pH.

Worked example 8: picking an indicator

Use these ranges: methyl red 4.4–6.2; bromothymol blue 6.0–7.6; phenolphthalein 8.3–10.0.

• Ammonia with nitric acid: NH₄NO₃ is an acidic salt, so methyl red.
• Propanoic acid with sodium hydroxide: sodium propanoate is basic, so phenolphthalein.
• Hydrochloric acid with potassium hydroxide: KCl, pH 7, long jump, so any of the three.
• Weak acid with weak base: no sharp jump, so no indicator gives a clear end point.

Reactivity 3.1.16 Buffer solutions

Adding a little acid or alkali to a buffer barely moves its pH.

• Acidic buffer: weak acid plus a soluble salt containing its conjugate base (lactic acid with sodium lactate).
• Basic buffer: weak base plus a soluble salt containing its conjugate acid (ammonia with ammonium chloride).

In the lactic acid buffer (HA/A⁻), added H⁺ is removed: A⁻ + H⁺ → HA. Added OH⁻ is removed: HA + OH⁻ → A⁻ + H₂O. Both reservoirs are large, so [A⁻]/[HA] barely changes and nor does the pH.

A strong acid cannot do this: its conjugate base does not take protons back. You can also make an acidic buffer by adding strong base to excess weak acid.

Reactivity 3.1.17 Calculating buffer pH

Making [H⁺] the subject of Ka gives:

[H⁺] = Ka × [HA] / [A⁻], or pH = pKa + log₁₀([A⁻] / [HA])

So buffer pH depends on pKa and on the ratio of the two concentrations. For a basic buffer use pOH = pKb + log₁₀([BH⁺] / [B]).

Worked example 9: pH from composition

A buffer contains 0.150 mol dm⁻³ lactic acid and 0.250 mol dm⁻³ sodium lactate (pKa = 3.86).

pH = 3.86 + log₁₀(0.250 ÷ 0.150) = 4.08

To make a pH 4.50 buffer instead, you need [A⁻]/[HA] = 10^(4.50 − 3.86) = 4.4.

Worked example 10: buffer by partial neutralisation

50.0 cm³ of 0.200 mol dm⁻³ lactic acid is mixed with 25.0 cm³ of 0.120 mol dm⁻³ NaOH.

• n(HA) initially = 0.0100 mol; n(OH⁻) = 0.00300 mol
• After reaction: n(A⁻) = 0.00300 mol, n(HA) = 0.00700 mol
• Both share one volume, so use the mole ratio: pH = 3.86 + log₁₀(0.00300 ÷ 0.00700) = 3.49

Worked example 11: a basic buffer

A solution has [NH₃] = 0.120 and [NH₄⁺] = 0.0800 mol dm⁻³ (pKb = 4.75).

pOH = 4.75 + log₁₀(0.0800 ÷ 0.120) = 4.57, so pH = 9.43

Dilution. Diluting lowers [HA] and [A⁻] by the same factor, so the ratio and the pH stay almost unchanged. With fewer moles per cm³, though, the buffer absorbs less added acid or alkali.

Official syllabus

This guide is based on the Diploma Programme Chemistry guide (International Baccalaureate Organization), first assessment 2025, published February 2023, sub-topic Reactivity 3.1, understandings 3.1.9 to 3.1.17 (additional higher level).