Acid-Base Titrations

What Is a Titration?

A titration is a quantitative analytical technique used to determine the concentration of a solution (the analyte) by reacting it with a solution of known concentration (the titrant or standard solution). In acid-base titrations, the reaction is a neutralisation:

acid + base → salt + water

The titrant is added from a burette until the equivalence point (also called the stoichiometric point) is reached — the point at which the acid and base have reacted in their exact stoichiometric ratio.

The Practical Procedure

A typical titration follows these steps:

Prepare the standard solution — accurately weigh a primary standard (e.g. anhydrous Na₂CO₃) and dissolve it in a volumetric flask to a known volume
Rinse equipment — the burette with the titrant solution, the pipette with the analyte solution, the conical flask with distilled water (rinsing with analyte would add extra moles)
Pipette a fixed volume of analyte into a conical flask (typically 25.0 cm³)
Add indicator — a few drops of an appropriate indicator
Add titrant from the burette slowly, swirling constantly, until the end point (colour change)
Record the burette reading before and after to find the titre (volume of titrant used)
Repeat until you obtain at least two concordant titres (agreeing within 0.10 cm³)

Why rinse equipment differently? The pipette is rinsed with analyte so that residual water does not dilute the analyte and reduce the moles delivered. The conical flask is rinsed with distilled water because extra water does not change the moles already in the flask — it only affects the volume, which does not matter since you are measuring moles, not concentration.

Standard Solutions and Primary Standards

A standard solution is one of accurately known concentration. It is prepared using a primary standard, which is a substance that:

Has a high purity
Has a known, stable formula
Is not hygroscopic (does not absorb water from the air)
Has a high molar mass (reduces percentage weighing errors)
Is readily available

Common primary standards include anhydrous sodium carbonate (Na₂CO₃) and potassium hydrogen phthalate (KHP).

Why is NaOH not a primary standard? NaOH is hygroscopic — it absorbs water and CO₂ from the air. A weighed sample would contain an unknown mass of H₂O and Na₂CO₃, making its purity uncertain. You must standardise NaOH by titrating it against a primary standard.

The Titration Calculation

The fundamental relationship is:

moles = concentration × volume

n = c × V (where V is in dm³)

For a reaction: acid + base → products, at the equivalence point:

moles of acid × stoichiometric factor = moles of base × stoichiometric factor

Worked Example: 25.0 cm³ of NaOH of unknown concentration requires 22.50 cm³ of 0.100 mol dm⁻³ HCl for neutralisation. Calculate the concentration of NaOH.

The equation is: HCl + NaOH → NaCl + H₂O (1:1 ratio)

Moles of HCl = 0.100 × (22.50/1000) = 2.250 × 10⁻³ mol

Since the ratio is 1:1, moles of NaOH = 2.250 × 10⁻³ mol

Concentration of NaOH = 2.250 × 10⁻³ / (25.0/1000) = 0.0900 mol dm⁻³

Worked Example: Calculating Concentration of H₂SO₄

25.0 cm³ of H₂SO₄ is titrated against 0.100 mol dm⁻³ NaOH. The mean titre is 36.0 cm³. Calculate [H₂SO₄].

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O (1:2 ratio)

Moles of NaOH = 0.100 × (36.0/1000) = 3.60 × 10⁻³ mol

Moles of H₂SO₄ = 3.60 × 10⁻³ / 2 = 1.80 × 10⁻³ mol

[H₂SO₄] = 1.80 × 10⁻³ / 0.0250 = 0.0720 mol dm⁻³

Diprotic Acid Titrations

When a diprotic acid like H₂SO₄ is used:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

The mole ratio is 1:2. One mole of H₂SO₄ reacts with two moles of NaOH.

Worked Example: 25.0 cm³ of 0.050 mol dm⁻³ H₂SO₄ is titrated against NaOH. What volume of 0.10 mol dm⁻³ NaOH is required?

Moles of H₂SO₄ = 0.050 × 0.0250 = 1.25 × 10⁻³ mol

Moles of NaOH needed = 2 × 1.25 × 10⁻³ = 2.50 × 10⁻³ mol

Volume of NaOH = 2.50 × 10⁻³ / 0.10 = 0.0250 dm³ = 25.0 cm³

Back Titrations

A back titration is used when the analyte cannot be titrated directly — for example, if it is an insoluble solid or reacts too slowly.

The procedure is:

Add a known excess of reagent A to the analyte
The analyte reacts with some of reagent A
Titrate the unreacted excess of reagent A with a standard solution of reagent B

Worked Example: 2.00 g of impure CaCO₃ is dissolved in 50.0 cm³ of 1.00 mol dm⁻³ HCl (excess). The excess HCl requires 30.0 cm³ of 0.500 mol dm⁻³ NaOH for neutralisation. Calculate the percentage purity.

Moles of HCl initially = 1.00 × 0.0500 = 0.0500 mol

Moles of excess HCl = moles of NaOH used = 0.500 × 0.0300 = 0.0150 mol

Moles of HCl that reacted with CaCO₃ = 0.0500 - 0.0150 = 0.0350 mol

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Moles of CaCO₃ = 0.0350 / 2 = 0.0175 mol

Mass of CaCO₃ = 0.0175 × 100.1 = 1.75 g

Percentage purity = (1.75 / 2.00) × 100 = 87.5%

Finding Molar Mass from a Titration

A 0.500 g sample of an unknown monoprotic acid HA requires 25.0 cm³ of 0.200 mol dm⁻³ NaOH for complete neutralisation. Calculate the molar mass.

Moles of NaOH = 0.200 × 0.0250 = 5.00 × 10⁻³ mol

Since HA is monoprotic, moles of HA = 5.00 × 10⁻³ mol

Mr = mass / moles = 0.500 / 5.00 × 10⁻³ = 100.0 g mol⁻¹

Common Exam Calculation Types

Type	Method
Simple titration	moles = c × V, use stoichiometry
Diprotic acid	Remember the 1:2 or 2:1 ratio
Back titration	Initial moles - excess moles = reacted moles
Finding Mr	Calculate moles from titration, then Mr = mass/moles
Percentage purity	Calculate moles of pure substance, convert to mass, compare with impure mass

Concordant Titres and Data Processing

Concordant titres are results that agree within 0.10 cm³. They demonstrate reliability. When processing results:

Discard any rough titre or outliers
Average the concordant values
Use this average for calculations

Example: Titres: 23.50, 23.45, 23.55, 26.00 cm³. Discard 26.00 (outlier). Mean = (23.50 + 23.45 + 23.55) / 3 = 23.50 cm³.

Common Exam Mistakes in Titration Calculations

Mistake	Correction
Using cm³ instead of dm³ in n = cV	Always convert: divide cm³ by 1000
Ignoring the stoichiometric ratio	Check the balanced equation for the mole ratio
Using the wrong titre (not concordant)	Average only concordant titres
Forgetting H₂SO₄ is diprotic	1 mol H₂SO₄ reacts with 2 mol NaOH
Including the rough titre in the mean	Always discard the rough and any outliers

Worked Example: Finding Molar Mass with a Diprotic Acid

A 0.750 g sample of an unknown diprotic acid H₂A requires 25.0 cm³ of 0.400 mol dm⁻³ NaOH. Calculate the molar mass.

H₂A + 2NaOH → Na₂A + 2H₂O

Moles of NaOH = 0.400 × 0.0250 = 0.0100 mol

Moles of H₂A = 0.0100 / 2 = 0.00500 mol (1:2 ratio)

Mr = mass / moles = 0.750 / 0.00500 = 150.0 g mol⁻¹

The ratio is critical here. Forgetting the 1:2 stoichiometry would give Mr = 75.0, which is incorrect.

Precision and Accuracy in Titrations

Burette precision: A standard burette reads to ±0.05 cm³. The total uncertainty in a titre (two readings: initial and final) is ±0.10 cm³.

Percentage uncertainty: For a titre of 25.00 cm³, percentage uncertainty = (0.10/25.00) × 100 = 0.40%. Using a larger titre reduces percentage uncertainty.

Reducing errors: Use a white tile behind the conical flask to see colour changes more clearly. Add titrant dropwise near the end point. Use a burette funnel when filling but remove it before titrating to avoid drips.

A-Level Deep Dive: Acid-Base Titrations

Spec mapping

Edexcel 9CH0 specification, Topic 12: Acid-base equilibria, sub-strand 12.5 requires you to perform acid-base titration calculations from titre and known concentration; identify the equivalence point (stoichiometric end of reaction) and recognise its dependence on the strong/weak nature of the species; choose a suitable indicator for a given titration based on its pKₐ relative to the pH at equivalence (refer to the official specification document for exact wording). Examined in Paper 2 (numerical, Topic 12) and Paper 3 (Core Practical 2 and CP13 directly). Stoichiometry (Topic 5) and mole/concentration calculations (Topic 1) are prerequisites.

Worked example with full mark scheme

Question (8 marks):

A student titrates a 25.0 cm³ aliquot of an unknown weak monoprotic acid HA against 0.100 mol dm⁻³ NaOH(aq). The mean titre is 22.50 cm³.

(a) Calculate the concentration of HA. (3)

(b) The pH at the half-equivalence point is 4.32. Calculate Ka and pKa for HA. (3)

(c) Suggest, with reasoning, a likely identity of HA from the following table of common weak acids: methanoic (pKa = 3.75), ethanoic (4.76), propanoic (4.87), benzoic (4.20). (2)

Solution with mark scheme:

(a) Reaction: HA + NaOH → NaA + H₂O (1:1 stoichiometry).

Moles NaOH at equivalence = 0.02250 × 0.100 = 2.25 × 10⁻³ mol.

Moles HA in aliquot = 2.25 × 10⁻³ mol (1:1).

[HA] = 2.25 × 10⁻³ / 0.0250 = 0.0900 mol dm⁻³.

M1 — moles NaOH from titre and concentration. M1 — 1:1 mole ratio applied. A1 — concentration = 0.0900 mol dm⁻³ (3 s.f.).

(b) At half-equivalence, [HA] = [A⁻] (half the HA has been neutralised), so by Henderson-Hasselbalch pH = pKa.

pKa = 4.32. Ka = 10⁻⁴·³² = 4.79 × 10⁻⁵ mol dm⁻³.

M1 — recognises pH = pKa at half-equivalence. M1 — correct conversion pKa → Ka. A1 — Ka = 4.79 × 10⁻⁵ mol dm⁻³ (3 s.f., units).

M1 — comparison with table. A1 — benzoic acid (acceptable: ethanoic acid as alternative if reasoning given, but benzoic is closer).

Total: 8 marks (M5 A3, split as shown).

Specimen question modelled on the Edexcel 9CH0 Paper 2 format

Question (6 marks): A student standardises a sodium hydroxide solution by titrating against a primary-standard solution of potassium hydrogen phthalate (KHP, M = 204.2 g mol⁻¹). She dissolves 0.510 g of KHP in 100 cm³ of distilled water, then titrates 25.0 cm³ of this solution against the NaOH(aq); the mean titre is 23.85 cm³.

(a) Write a balanced equation for the reaction. (KHP is a 1:1 monoprotic acid for this titration.) (1)

(b) Calculate the concentration of the NaOH solution. (4)

Mark scheme decomposition by AO:

(a)

B1 (AO1.1) — KHC₈H₄O₄(aq) + NaOH(aq) → NaKC₈H₄O₄(aq) + H₂O(l) (or analogous correctly balanced equation).

(b)

M1 (AO2.6) — moles KHP in 100 cm³ flask = 0.510 / 204.2 = 2.498 × 10⁻³ mol.
M1 (AO2.6) — moles KHP in 25.0 cm³ aliquot = 2.498 × 10⁻³ × (25.0/100) = 6.244 × 10⁻⁴ mol.
M1 (AO2.6) — moles NaOH at equivalence = 6.244 × 10⁻⁴ mol (1:1).
A1 (AO1.1) — [NaOH] = 6.244 × 10⁻⁴ / 0.02385 = 0.0262 mol dm⁻³ (3 s.f.).

(c)

B1 (AO2.5) — KHP is a weak acid (pKa₂ ≈ 5.4); titration with strong base NaOH gives a weakly basic equivalence pH (~9), suiting phenolphthalein (pKa ≈ 9.3, range 8.2–10.0).

Total: 6 marks split AO1 = 2, AO2 = 4. A clean stoichiometry-then-indicator-choice structure typical of Paper 2.

Synoptic links

Connects to:

Topic 5 — Stoichiometry: every titration calculation reduces to mole bookkeeping. Aliquot, dilution, stoichiometric ratio, titre — all Year 1 ideas applied at higher precision.
Topic 12.4 — Buffers (Lesson 5–6): the half-equivalence-point pH = pKa observation links titration data to Ka determination. Lesson 7 is the technique; Lesson 5 is the theory.
Topic 12.5 — Titration curves (Lesson 8): Lesson 7 produces a single titre value (the equivalence point); Lesson 8 maps the whole curve. The two are complementary.
Topic 12.5 — Indicators (Lesson 9): indicator choice is governed by the pH at equivalence, which depends on the strong/weak classification of acid and base. Lesson 9 explains why certain indicators suit certain titrations; Lesson 7 uses the indicator.
Analytical chemistry skills: primary standards, replicate titres, concordance criteria (typically ±0.10 cm³), uncertainty propagation — Lesson 7 is the canonical practical application of these skills.

Mark-scheme literacy

Titration questions on 9CH0 split AO marks as follows: