Edexcel A-Level Chemistry Paper 3, June 2022: Question 8
14 marks · Hard difficulty · Calculations
Devise an experiment to determine the acid dissociation constant of ethanoic acid using a pH meter and sodium hydroxide, calculate volumes for a buffer solution of a given pH, and calculate the pH of a mixture of sodium hydroxide and sulfuric acid.
Practise this questionQuestion
Question text
8 This question is about acids and bases.
(a) Devise an experiment to determine the acid dissociation constant, Ka, for a
solution of ethanoic acid, CH3COOH, of unknown concentration.
Assume you have access to a pH meter and a solution of sodium hydroxide of
similar concentration to the acid.
Include how to determine Ka from your results.
(5)
3 *P67095A02136*
(b) 500cm of a buffer solution of pH = 4.70 is required.
Calculate the volume of 0.800 mol dm–3 sodium ethanoate solution and of
0.800 mol dm–3 ethanoic acid needed to make this buffer.
[K for ethanoic acid = 1.74× 10–5 mol dm–3]
a
(3)
(c) Calculate the pH of the solution formed when
51.2 cm3 of 0.927 mol dm–3 NaOH(aq) is mixed with
40.4 cm3 of 0.370 mol dm–3 H SO (aq).
[Ionic product of water K = 1.00 × 10–14 mol2 dm–6]
w
(6)
*P67095A02236*
(Total for Question 8 = 14 marks)
Mark scheme
Show the mark scheme
8(a) An answer that makes reference to the following points: (5)
Titration Stand alone
• titrate (ethanoic acid /weak acid) with strong base / sodium Allow any indication of a titration
hydroxide (1) Allow acid added to base or base added to acid
Then follow the three points for Method 1 or Method 2 In both methods, ignore reference to making a
standard solution / calibration of the pH probe or
meter / practical details of carrying out the titration
Method 1
• measure pH at regular intervals (1)
• plot pH against volume (of strong base) (1) Allow plot a titration / pH curve
• use graph to find pH at half-equivalence point (1) Allow use graph to find pH at volume when half
neutralised
OR
Method 2
• use phenolphthalein indicator to find end-point (1) Allow thymol blue / thymolphthalein indicators
Ignore colour change even if incorrect
• then add same volume of acid to mixture (at end-point) (1) Allow repeat titration (with same volumes but
without indicator) then add original volume of acid to
mixture (at end-point) or use same volume of acid
and half the volume of base
Do not award pH at end point is 7
• measure pH of resultant mixture (with pH meter) (1)
Determining Ka
–pH Stand alone
• (at half neutralisation pH = pKa so) Ka = 10 (1) + –pH +
Allow [H ] = 10 and Ka = [H ]
Question
Answer Additional Guidance Mark
Number
8(b) Example of calculation (3)
EITHER
• calculation of [H+(aq)] (1) [H+(aq)] = 10–4.70 = 1.9953 x 10–5 (mol dm–3)
[acid]/[salt] = 1.9953 x 10–5 / 1.74 x 10–5
• calculation of ratio of [acid]/[salt] or [salt]/[acid] or correct
values substituted into expression for ratio (1) = 1.1467 : 1 / 1 : 0.872
or
[salt]/[acid] = 1.74 x 10–5 / 1.9953 x 10–5
= 0.872 : 1 / 1 : 1.1467
(1.1467 /2.1467) x 500 = 267 cm3 acid
• calculation of volume of acid required and salt required (1)
500 – 267 = 233 cm3 salt
OR
• calculation of log [acid]/[salt] using Henderson-Hasselbalch 4.7595 – 4.70 = 0.05945
(1)
• calculation of ratio of [acid]/[salt] (1) 0.05945
10 = 1.1467 : 1
(1.1467 / 2.1467) x 500 = 267 cm3 acid
• calculation volume of acid required and salt required (1) 3
500 – 267 = 233 cm salt
Allow 270 cm3 acid and 230 cm3 salt
Ignore SF except 1 SF but allow 2 / 2.0 / 2.00 x
10−5 for M1 in ‘EITHER’
Allow TE from M1 throughout
Correct answer with no working scores (3)
Question
Answer Additional Guidance Mark
Number
8(c) Example of calculation (6)
• calculation amount of H2SO4(aq) in mol (1) = (40.4/1000) x 0.370 = 0.014948
• calculation amount of H+(aq) in mol / amount 0.014948 x 2 = 0.029896 (mol)
OH−(aq) needed (1)
• calculation amount of OH–(aq) in mol (1) = (51.2/1000) x 0.927 = 0.047462 (mol)
• calculation amount of excess OH–(aq) in mol (1) = 0.047462 – 0.029896 = 0.017566 (mol)
= 0.017566 / (91.6/1000) = 0.19177 (mol dm−3)
• calculation [OH–] in resultant mixture (1)
[H+] = 1.00 x 10−14/0.19177 = 5.2146 x 10−14 (mol dm−3)
• calculation pH of resultant mixture (1) −14
pH = −log 5.2146 x 10
= 13.3
or
14 – ( – log(0.19177) ) = 13.3
Final answer needs to be to at least 1dp
Allow TE throughout but TE from M5 to M6 must give a
pH > 7
Correct answer with no / some working scores 6 marks
Ignore SF except 1 SF in M1 to M5
(Total for Question 8 = 14 marks)
How to answer it
Acids, Bases, Buffers, and Titration Calculations
This multi-step question assesses core physical chemistry concepts surrounding weak acids and strong bases. It evaluates your ability to design a pH titration experiment to find an acid dissociation constant ( Kₐ ), manipulate buffer preparation calculations using stoichiometry and Kₐ expressions, and execute complex neutralization calculations involving excess reagents and ionic product of water ( K_w ).
Designing an Experiment to Determine Kₐ (5 Marks)
✅ Expected Method & Answer
- Titration Setup: Titrate a known volume of ethanoic acid with sodium hydroxide of similar concentration using a pH meter/probe.
- Method 1 (Graph approach): Measure pH at regular intervals during the addition of the strong base. Plot a pH titration curve against the volume of base added. Use the graph to find the pH at the half-equivalence point.
- Method 2 (Partial neutralization): Add an indicator to find the end-point, or add exactly half the volume of base required for complete neutralization.
- Determining Kₐ: At the half-neutralisation point, [CH₃COOH] = [CH₃COO⁻], meaning Kₐ = [H⁺] and pKₐ = pH (so Kₐ = 10⁻ᵖᴴ ).
💡 Key Knowledge
- The half-equivalence point is the defining feature of weak acid-strong base titrations for determining Kₐ experimentally.
- At this exact midpoint, half of the weak acid has been converted into its conjugate base salt, creating an optimum buffer mixture where concentrations of acid and salt are equal.
🧠 Exam Technique
To secure all 5 marks, ensure you explicitly state: (1) mixing the acid and base via titration, (2) recording pH values as the base is added, (3) plotting a curve or neutralizing half the volume, and (4) explaining the mathematical relationship Kₐ = [H⁺] at half-neutralisation.
❌ Common Errors
- Assuming the equivalence point has a pH of 7 (it is alkaline for weak acid / strong base titrations).
- Stating that Kₐ = pH instead of pKₐ = pH or Kₐ = 10⁻ᵖᴴ .
Buffer Solution Preparation Calculation (3 Marks)
📐 Step-by-Step Calculation
- Find [H⁺] from target pH:
[H⁺] = 10⁻⁴°⁷⁰ = 1.9953 × 10⁻⁵ mol dm⁻³ - Set up the Kₐ expression and find the ratio:
Kₐ = ([H⁺] × [salt]) / [acid]
1.74 × 10⁻⁵ = (1.9953 × 10⁻⁵ × [salt]) / [acid]
Ratio [acid] / [salt] = 1.9953 × 10⁻⁵ / 1.74 × 10⁻⁵ = 1.1467 : 1 - Calculate exact volumes required for 500 cm³ total:
Fraction of acid = 1.1467 / (1.1467 + 1) = 0.5333
Volume of acid = 0.5333 × 500 = 267 cm³
Volume of salt = 500 - 267 = 233 cm³
🧠 Exam Technique & Ratios
Since both the sodium ethanoate and ethanoic acid solutions share the exact same concentration ( 0.800 mol dm⁻³ ), the mole ratio is directly equal to the required volume ratio. Always state your ratios clearly to allow error-carried-forward (TE) marks.
❌ Common Errors
- Inverting the [acid]/[salt] fraction expression.
- Failing to subtract the calculated acid volume from 500 cm³ to find the remaining salt volume.
Neutralization and Excess Reagent pH Calculation (6 Marks)
📐 Step-by-Step Calculation
- Calculate moles of H₂SO₄:
40.4 / 1000 × 0.370 = 0.014948 mol - Calculate moles of H⁺ ions needed (remember H₂SO₄ is diprotic):
0.014948 × 2 = 0.029896 mol H⁺ - Calculate initial moles of NaOH (supplier of OH⁻):
51.2 / 1000 × 0.927 = 0.047462 mol OH⁻ - Determine moles of excess OH⁻:
0.047462 - 0.029896 = 0.017566 mol OH⁻ (in excess) - Calculate concentration of excess OH⁻ in total volume:
Total volume = 51.2 + 40.4 = 91.6 cm³ = 0.0916 dm³
[OH⁻] = 0.017566 / 0.0916 = 0.19177 mol dm⁻³ - Calculate pOH and then final pH:
[H⁺] = K_w / [OH⁻] = (1.00 × 10⁻¹⁴) / 0.19177 = 5.2146 × 10⁻¹⁴ mol dm⁻³
pH = -log(5.2146 × 10⁻¹⁴) = 13.3 (or 14 - (-log(0.19177)) = 13.3 )
💡 Key Knowledge
- Sulfuric acid ( H₂SO₄ ) is a diprotic acid, meaning each mole releases two moles of hydrogen ions ( H⁺ ). Missing the multiplier of 2 is the single most common mark loss in this style of question.
❌ Common Errors
- Treating H₂SO₄ as monoprotic ( 1:1 stoichiometry).
- Forgetting to combine volumes when finding the new concentration of excess ions ( 51.2 + 40.4 = 91.6 cm³ ).
- Calculating pOH instead of converting properly to pH.
Topics
Physical Chemistry · Core Practicals · Core Practical 9: Finding the Ka value for a weak acid · Topic 12: Acid-base Equilibria
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.