AQA Equilibria Part 4: Equilibrium Mixtures & Practical Kc

1. Finding equilibrium composition

If Kc is known but the equilibrium amounts are not, represent the amount reacting by x. Build the initial/change/equilibrium table, substitute into Kc and solve. A valid root must leave all amounts non-negative. Some equal-initial-amount problems simplify using a square root; do not assume every problem avoids a quadratic.

2. Products present initially

Changes follow coefficients, but equilibrium amounts depend on everything present at the start. For H₂ + I₂ ⇌ 2HI, initially 0.500 mol H₂, 0.300 mol I₂ and 0.200 mol HI are present. If a further 0.100 mol I₂ reacts, the final amounts are 0.400 mol H₂, 0.200 mol I₂ and 0.400 mol HI: the original 0.200 mol HI plus 0.200 mol formed. Kc = 0.400²/(0.400 × 0.200) = 2.00.

3. Practical determination of Kc

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

This example uses one homogeneous liquid mixture of ethanoic acid, ethanol, ethyl ethanoate and water, with an acid catalyst. Kc = [ester][water]/([ethanoic acid][ethanol]). All four concentrations refer to the original equilibrium mixture at the stated temperature. The catalyst is not included in this reaction expression.

Prepare known initial amounts. For a pure liquid, mass = density × volume and n = mass/molar mass, with consistent units. Account for purity and any water in solutions, particularly the aqueous catalyst.
Keep the mixture in a closed, suitable vessel at constant temperature. Allow sufficient time for equilibrium; verify unchanged composition in samples taken at suitably separated times. A fixed waiting period alone does not prove equilibrium.
Analyse a measured sample using a validated rapid titration or quench procedure. Cooling or dilution slows reaction but does not freeze the original equilibrium. Analyse promptly so esterification or hydrolysis does not significantly change the composition.
Titrate with standardised NaOH using a suitable indicator such as phenolphthalein. Obtain concordant titres from repeat samples. Use the prescribed first persistent pale-pink endpoint; do not keep adding alkali to chase later fading.
Correct for the acid catalyst using a separate blank with the same amount of catalyst and the same analytical conditions. Scale aliquot results to the whole sample before comparing whole-mixture amounts.

Follow the school risk assessment: eye protection, suitable handling of corrosive acids and alkali, and no naked flame near flammable organic liquids. This is an illustrative Kc investigation, not an additional numbered AQA required practical.

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Equilibrium preparation, sampling and titration

Labels to include:

Stoppered equilibrium vessel in a constant-temperature bath
Known initial acid, alcohol, water and acid-catalyst amounts
Measured sample or quantitative transfer, as specified
Analytical dilution volume and pipetted aliquot labelled separately
Burette of standard NaOH above conical flask and white tile
Separate catalyst blank with matching catalyst amount
Record temperature, sample time, initial/final burette readings

The original equilibrium vessel and the later analytical dilution have different roles. The titration measures acid equivalents; it does not directly measure ester or water.

4. Complete titration calculation

Worked example: catalyst correction and initial water

An illustrative homogeneous mixture initially contains 0.100 mol ethanoic acid, 0.120 mol ethanol, 0.0500 mol water (including water from the catalyst solution), no ester, and 0.00200 mol H₂SO₄ catalyst. After equilibration at constant temperature, the whole mixture is quantitatively diluted to 250.0 cm³. A 25.0 cm³ aliquot requires 22.00 cm³ of 0.200 mol dm⁻³ NaOH. Assume prompt analysis preserves the original acid amount and both catalyst protons are neutralised at the endpoint.

1. Acid equivalents in the aliquot: n(NaOH) = 0.200 × 22.00/1000 = 0.00440 mol.

2. Scale to the whole mixture: 0.00440 × 250.0/25.0 = 0.0440 mol acid equivalents.

3. Remove the catalyst contribution: H₂SO₄ contributes 2 × 0.00200 = 0.00400 mol acid equivalents. Ethanoic acid remaining = 0.0440 − 0.00400 = 0.0400 mol. Do not subtract only one equivalent per H₂SO₄.

4. Determine reaction extent: ethanoic acid used = 0.100 − 0.0400 = 0.0600 mol. All coefficients in the esterification equation are one.

Original equilibrium-mixture mole table
Stage / mol	Ethanoic acid	Ethanol	Ester	Water
Initial	0.100	0.120	0	0.0500
Change	−0.0600	−0.0600	+0.0600	+0.0600
Equilibrium	0.0400	0.0600	0.0600	0.110

5. Calculate Kc: both numerator and denominator have total power two, so the common original equilibrium volume cancels.

Kc = (0.0600 × 0.110) / (0.0400 × 0.0600) = 2.75

Kc = 2.75, no units. The 250.0 cm³ is an analytical dilution volume; do not treat it as the original equilibrium volume. Volume cancellation here removes the need to know that original volume.

5. Errors and assumptions

Avoiding misleading Kc values
Error	Consequence in this example	Improvement
Ignore catalyst acidity	Ethanoic acid appears too high; reaction extent too low; calculated Kc is too low (about 2.11)	Use a matching catalyst blank or correctly subtract acid equivalents
Ignore water initially present	Water is underestimated; Kc is too low (1.50 instead of 2.75)	Include water supplied by all reagents
Forget the aliquot factor	Compare only one tenth of the mixture's acid with whole-mixture initial amounts	Put every amount on the same whole-mixture basis
Delay analysis or chase a fading endpoint	The analytical reaction mixture changes; ester hydrolysis can consume extra alkali	Use the validated prompt endpoint and sampling procedure
Change temperature during equilibration	Measure a different equilibrium constant	Maintain and record the equilibrium temperature

NaOH consumption measures neutralisable acid equivalents. It is not the equilibrium concentration of free H⁺ in a weak-acid mixture. Repeating titres reduces random scatter but does not correct a forgotten catalyst contribution or a systematic sampling error.

Quick checks

Original Finesse practice; indicative solutions, not official AQA mark allocations.

Q1. H₂(g) + I₂(g) ⇌ 2HI(g) has Kc = 4.00. Initially there are 0.800 mol each of H₂ and I₂, with no HI. Find all equilibrium amounts.Show answer

Let x mol of each reactant react. 2x/(0.800 − x) = √4.00 = 2.00. Thus 2x = 1.60 − 2x, so x = 0.400 mol.

H₂ = 0.400 mol, I₂ = 0.400 mol, HI = 0.800 mol. Substitution gives Kc = 4.00.

Q2. A whole mixture requires 0.0300 mol NaOH equivalents. It contains 0.00150 mol H₂SO₄ catalyst. Assuming both catalyst protons are titrated, how much ethanoic acid remains?Show answer

Catalyst contribution = 2 × 0.00150 = 0.00300 mol equivalents. Ethanoic acid = 0.0300 − 0.00300 = 0.0270 mol.

Q3. An aliquot of 20.0 cm³ from a 200.0 cm³ analytical solution contains 0.00320 mol ethanoic acid after catalyst correction. The original mixture had 0.0800 mol acid. How much ester has formed, assuming none initially?Show answer

Whole-mixture acid remaining = 0.00320 × 200.0/20.0 = 0.0320 mol. Acid used = 0.0800 − 0.0320 = 0.0480 mol, so that is also the ester formed.

Q4. An esterification mixture at equilibrium contains 0.0200 mol acid, 0.0300 mol ethanol, 0.0400 mol ester and 0.0600 mol water. Calculate Kc.Show answer

Common volume cancels because total powers are equal. Kc = (0.0400 × 0.0600)/(0.0200 × 0.0300) = 4.00, no units.

Q5. Why are “dilution stops the reaction” and “all the NaOH measures ethanoic acid” unreliable statements?Show answer

Dilution changes concentrations and can disturb equilibrium; it may slow reaction but does not stop it instantly. Use prompt, validated analysis.

NaOH also neutralises the acid catalyst. Correct using a blank or known acid equivalents before calculating ethanoic acid remaining.

Sources

Sources and examiner guidance (reviewed 1 October 2026)

Chemrevise — AQA 1.6 Equilibria revision guide (N. Goalby, Oct 2025 upload) — Equilibrium composition and esterification/titration checklist; catalyst equivalents, initial water and sampling assumptions made explicit.
AQA 7405 specification — 3.1.6 Chemical equilibria, Le Chatelier's principle and Kc — 3.1.6.2: calculating Kc from equilibrium composition.
AQA 7404/1 mark scheme, June 2023 — Q08.1: distinguish amounts remaining from changes.
AQA 7404/1 examiner report, June 2023 — Q08.1: avoid applying coefficient ratios to equilibrium amounts.

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