OCR A-Level Chemistry AS Breadth in chemistry (01), June 2019: Question 22

7 marks · Medium difficulty · Practical Questions

Calculate the concentration of hydrochloric acid from a titration with sodium carbonate, including calculation of titres, mean titre, stoichiometry, and percentage uncertainty.

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Question

Question text

22 A student carries out a titration to determine the concentration of some hydrochloric acid.

The student titrates the hydrochloric acid against a standard solution of sodium carbonate,

Na2CO3. The equation is shown below.

Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

• The student prepares 0.150 mol dm−3 Na CO in a 250.0 cm3 volumetric flask.

• The hydrochloric acid is added to a 50.0 cm3 burette.

• The student pipettes the Na CO (aq) using a 25.0 cm3 pipette.

(a) The student’s burette readings are shown in the table.

The rough titre has been omitted.

(i) Complete the table by adding the titres to the table.

Final reading / cm3 24.60 48.45 34.30

Initial reading / cm3 0.40 24.60 10.00

Titre / cm3

[1]

(ii) Calculate the mean titre of HCl , to the nearest 0.05 cm3, that the student should use for

analysing the results.

mean titre = … cm3 [1]

(b) Calculate the concentration, in mol dm−3, of the hydrochloric acid.

Give your answer to 3 significant figures.

concentration of HCl = … mol dm−3 [3]

• The pipette had an uncertainty of ± 0.04 cm3 in the volume measured.

• The burette had an uncertainty of ± 0.05 cm3 in the volume measured.

Determine whether the volume measured by the pipette or the volume measured by the

burette has the greater percentage uncertainty.

[2]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

22 (a) (i) 2 AO2.4 DO NOT ALLOW 24.2 OR 24.3

Titre/cm3 24.20 23.85 24.30

Correct subtractions to obtain titres to 2 DP

(ii) 24.20 + 22.30 3 AO2.4 DO NOT ALLOW mean of all three titres,

mean titre = 2 = 24.25 (cm ) 24.20 + 23.85 + 22.30

i.e. 3 = 24.10/24.12

i.e. using concordant (consistent) titres

ALLOW ECF from incorrect concordant titres from

22a(i)

(b) FIRST CHECK THE ANSWER ON ANSWER LINE 3 AO2.8 ALLOW 3SF or more throughout

IF answer = 0.309 (mol dm–3) award 3 marks IGNORE trailing zeroes,

×3

--------------------------------------------------------------------– e.g. ALLOW 0.075 for 0.00750

-------------------------------------------------------------------

n(Na2CO3)

25.00 –3

= 0.150 × = 3.75 × 10 (mol)

1000

n(HCl)

= 2 × n(Na CO ) = 7.50 × 10–3 (mol) ALLOW ECF from 2 × incorrect n(Na CO )

23 2 3

[HCl] to 3 SF

1000 ALLOW ECF from incorrect n(HCl),

= n(HCl) × OR from n(Na CO ) if n(HCl) stage omitted

mean titre from b(i) 2 3

–3 1000 –3

= 7.50 × 10 × 24.25 = 0.309 (mol dm ) ALLOW ECF from incorrect mean titre in b(ii)

3 SF required ---------------------------------------------------------

COMMON ERROR for 3 marks

From 24.10 cm3 (mean of all 3 titres in b(ii)),

[HCl] = 0.311 (mol dm–3)

8 AO

element

0.04 ×2 correctly

25.0 × 100 = 0.16 OR 0.2 (%) -------------------------------------------------------------------

Other burette volumes:

0.05 × 2

Burette: (using any of 3 titres or mean titre), e.g. 23.85 × 100 = 0.42 OR 0.4 (%)

0.05 × 2 0.05 × 2

24.20 × 100 = 0.41 OR 0.4 (%) × 100 = 0.41 OR 0.4 (%)

24.30

0.05 × 2

Response does NOT need a statement of whether pipette 24.25 × 100 = 0.41 OR 0.4 (%)

or burette has greater % uncertainty.

ALLOW burette volume of 50 cm3, i.e.

0.05 × 2

50 × 100 = 0.2%

ALLOW ECF from incorrect titre in 22(a)

IF BOTH calculations are ‘correct’ but ×100 is

omitted BOTH times, ALLOW 1 mark

Total 7

How to answer it

Titration Calculations & Practical Uncertainties

OCR AS Level Chemistry • Core Practical Skills

What this question tests

This question assesses fundamental titration techniques and quantitative chemistry skills: processing raw burette readings, identifying concordant titres to calculate a mean, performing multi-step concentration calculations using stoichiometry, and calculating equipment percentage uncertainties in practical chemistry.

Question 20 (a) — Titration Table & Mean Titre

✅ Correct Answers

(i) Titres:
Col 1: 24.20 cm³
Col 2: 23.85 cm³
Col 3: 24.30 cm³
(ii) Mean Titre:
24.25 cm³

💡 Key Knowledge

Burette readings must always be recorded to 2 decimal places, ending in 0 or 5 .
Concordant titres are those within 0.10 cm³ of each other. Here, titres 1 ( 24.20 ) and 3 ( 24.30 ) are concordant.

🧠 Exam Technique

Never include the rough titre in your mean calculation.
Only average the concordant titres: (24.20 + 24.30) / 2 = 24.25 cm³ .
Always state the final mean to 2 decimal places matching the precision of the burette.

❌ Common Errors

Averaging all three titres (including 23.85 ), which results in an incorrect mean of 24.12 cm³ and penalises subsequent calculation marks.
Writing burette values to 1 decimal place (e.g., 24.2 ).

Marks available: 2 marks total (1 for table completion, 1 for correct mean calculation).

Question 20 (b) — Concentration Calculation

📐 Step-by-Step Calculation

Find moles of sodium carbonate ( Na₂CO₃ ) in the pipette:
n(Na₂CO₃) = concentration × volume / 1000
n = 0.150 × 25.0 / 1000 = 3.75 × 10⁻³ mol
Use the stoichiometric equation to find moles of HCl :
Equation: Na₂CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) + CO₂(g)
Ratio Na₂CO₃ : HCl is 1 : 2 .
n(HCl) = 2 × 3.75 × 10⁻³ = 7.50 × 10⁻³ mol
Calculate the concentration of HCl :
Concentration = (moles / volume in cm³) × 1000
[HCl] = (7.50 × 10⁻³ / 24.25) × 1000 = 0.30927... mol dm⁻³
Round to 3 significant figures: 0.309 mol dm⁻³

🧠 Exam Technique & ECF

Error Carried Forward (ECF): If you used an incorrect mean titre from part (a)(ii), examiners apply ECF so you can still pick up credit for correct calculation steps.
Always check your final answer matches the requested 3 significant figures.

❌ Common Errors

Forgetting to multiply by 2 for the mole ratio.
Inverting the volume conversion factor (multiplying by 1000 instead of dividing, or vice-versa).

Marks available: 3 marks (1 for moles of carbonate, 1 for moles of acid, 1 for final concentration to 3 SF).

Question 20 (c) — Percentage Uncertainty

📐 Step-by-Step Calculation

Pipette Percentage Uncertainty:
A pipette has a single measurement uncertainty, but manufacturers account for both filling and delivery by doubling it, OR use the formula:
(uncertainty / volume measured) × 100
(0.04 / 25.0) × 100 = 0.16% (or 0.2% )
Burette Percentage Uncertainty:
A burette involves two readings (initial and final), so the total instrument uncertainty is ±0.05 × 2 = ±0.10 cm³ .
Using the mean titre ( 24.25 cm³ ):
(0.10 / 24.25) × 100 = 0.41% (or 0.4% )
Conclusion: The burette has the greater percentage uncertainty ( 0.41% vs 0.16% ).

💡 Key Knowledge

Burette rule: Always multiply the stated uncertainty by 2 because it is read twice (initial and final volume).
Pipettes only involve filling to a single graduation line once.

❌ Common Errors

Failing to multiply the burette uncertainty by 2.
Forgetting to multiply the final fraction by 100 to convert into a percentage.

Marks available: 2 marks (1 for calculating pipette uncertainty, 1 for calculating burette uncertainty and identifying which is larger).

Topics

Module 1: Development of practical skills in chemistry · Module 2: Foundations in chemistry · Practical Activity Groups · Module 6: Organic chemistry and analysis · 6.3 Analysis · PAG 2: Acid-base titration · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.