AQA A-Level Chemistry Paper 3, 2017: Question 4
9 marks · Medium difficulty · Practical Techniques & Data Analysis
Analyse a weak acid-strong base titration curve to deduce indicator range, calculate acid concentration, initial pH, half-neutralisation pH, and complete the plotted titration curve.
Practise this questionQuestion
Question text
A 0.100 mol dm−3 solution of sodium hydroxide was gradually added to 25.0 cm3 of a
solution of a weak acid, HX, in the presence of a suitable indicator.
A graph was plotted of pH against the volume of sodium hydroxide solution, as shown
in Figure 1.
The first pH reading was taken after 20.0 cm3 of sodium hydroxide solution had been
added.
The acid dissociation constant of HX, K , = 2.62 × 10−5 mol dm−3
a
Figure 1
04.1 The pH range of an indicator is the range over which it changes colour.
Suggest the pH range of a suitable indicator for this titration.
[1 mark]
04.2 Give the expression for the acid dissociation constant of HX.
[1 mark]
Ka =
04.3 Calculate the concentration of HX in the original solution.
*13* [2 marks]
Concentration mol dm−3
04.4 Calculate the pH of the solution of HX before the addition of any sodium hydroxide.
(If you were unable to calculate a value for the concentration of HX in Question 4.3
you should use a value of 0.600 mol dm−3 in this calculation. This is not the correct
value.)
[2 marks]
pH of HX
04.5 Calculate the pH of the solution when half of the acid has reacted.
[1 mark]
pH of solution
04.6 Plot your answers to Questions 4.4 and 4.5 on the grid in Figure 1.
Use these points to sketch the missing part of the curve between 0 and 20 cm3 of
NaOH solution added.
[2 marks]
Section B
Answer all questions in the spaces provided
Only one answer per question is allowed.
For each answer completely fill in the circle alongside the appropriate answer.
CORRECT METHOD WRONG METHODS
If you want to change your answer you must cross out your original answer as shown.
If you wish to return to an answer previously crossed out, ring the answer you now wish to select as
shown.
You may do your working in the blank space around each question but this will not be marked.
Do not use additional sheets for this working.
Mark scheme
Show the mark scheme
![Mark scheme for Question 04 listing 9 total marks. 04.1 awards 1 mark for any pH range within 7–10.2. 04.2 awards 1 mark for Ka = [H+][X-]/[HX]. 04.3 awards 2 marks for calculating amount of NaOH = 2.40 x 10^-3 mol and conc HX = 0.0960 mol dm^-3. 04.4 awards 2 marks for calculating [H+] = 1.59 x 10^-3 mol dm^-3 and pH = 2.80. 04.5 awards 1 mark for pH = -log(2.62 x 10^-5) = 4.58. 04.6 awards 2 marks: 1 mark for plotting both points correctly and passing through them, and 1 mark for a line steeper at start then levelling off to show buffering.](/.netlify/images?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fexamquestions-bf477.firebasestorage.app%2Fo%2FmarkSchemes%252FQ17.L3.04-1790631424758.png%3Falt%3Dmedia&w=1000)
Question Answers Mark Additional Comments/Guidance
04.1 7–10.2 1 any range (i.e. 2 values) within this range
ALLOW H O+ for H+ and A for X
K = [H+][X–] IGNORE [H+]2/[HX]
04.2 a 1
[HX] must be square brackets
IGNORE state symbols
Amount NaOH = (24.0 x 0.100)/1000 = 2.40 x 10–3 mol 1
04.3 (= amount HX)
Conc HX = 2.40 x 10–3/0.025 = 0.0960 mol dm–3 1 ecf for M1/0.025
(K = 2.62 x 10–5 = [H+]2/0.0960) ecf from 04.3 [H+] = (2.62 x 10–5 x ans to 04.3)
a
From alternative data
[H+] = (2.62 x 10–5 x 0.0960) (= 1.59 x 10–3 mol dm–3) 1 [H+] = (2.62 x 10–5 x 0.600) (= 3.96 x 10–3 mol dm–3)
04.4
(pH = –log 1.59 x 10–3 =) 2.80 (must be 2 or more dp) 1 pH = 2.40 (must be 2 or more dp)
M2 dependent on a calculation of [H+]
(pH at half-neutralisation = pKa)
04.5
= –log 2.62 x 10–5 = 4.58 (must be 2 or more dp) 1 ALLOW 1dp if already penalised in 04.4
Both points plotted correctly and line touches both points 1 ecf from 04.4 and 04.5 within 1 small square
04.6
Line steeper at start then levels (to show buffering) 1 Mark independently
Total 9
How to answer it
Weak Acid Titration Curves, Ka & Buffer Regions
This question assesses core skills in aqueous equilibrium, acid-base titrations, and graphical analysis from Section 3.1.12 of the AQA specification:
- Interpreting pH curves to locate equivalence points and choosing suitable indicators based on the steep equivalence region.
- Writing rigorous equilibrium expressions for the acid dissociation constant (Ka) with correct notation.
- Stoichiometric titration calculations linking titre volumes to molar concentration.
- Calculating the pH of a weak acid before neutralisation using Ka approximations.
- Applying the half-neutralisation principle (pH = pKa) to buffer solutions.
- Accurate graph plotting and understanding the characteristic initial steep rise and buffering plateau of weak acid-strong base curves.
Indicator Selection for Weak Acid-Strong Base Titration
Suggest the pH range of a suitable indicator for this titration.
✅ Correct Answer
Any range completely contained within 7.0 to 10.2.
Examples of accepted ranges: 8.3 – 10.0 (phenolphthalein) or any two values such as 7.5 – 9.5 or 8.0 – 10.0.
🧠 Exam Technique
Read the vertical section of the graph directly from Figure 1:
- The vertical region begins at approx. pH 6.8 – 7.0 and ends at approx. pH 10.2.
- A suitable indicator must change colour completely within this vertical section.
- The mark scheme requires a range (i.e. two numbers), not a single pH value.
❌ Common Errors
- Suggesting methyl orange (pH 3.1 – 4.4) – this would change colour far too early.
- Giving a single number (e.g. "8.5") instead of an operational range.
- Extending the range below 7.0 (e.g. 6.0 – 9.0), which would cross into the non-vertical buffer region.
[1 mark]: Any range (2 values) completely within 7.0 – 10.2.
Writing the Acid Dissociation Constant Expression
Give the expression for the acid dissociation constant of HX.
✅ Correct Answer
Ka = [H⁺][X⁻] / [HX]
Also allowed: [H₃O⁺][X⁻] / [HX] or using A instead of X .
💡 Key Knowledge
For any weak monobasic acid HX(aq) ⇌ H⁺(aq) + X⁻(aq) :
- Concentrations must be enclosed in square brackets [ ].
- State symbols are ignored, but curved brackets ( ) are penalized.
❌ Common Errors
- Writing the simplified calculation form: [H⁺]² / [HX] . This is an approximation for working, not the definition of Ka.
- Omitting charges on ions (writing [H][X] ).
- Using round brackets (H⁺)(X⁻) instead of square brackets.
[1 mark]: Fully correct expression with square brackets.
Calculating the Concentration of the Weak Acid
Calculate the concentration of HX in the original solution.
📐 Step-by-Step Calculation
- Read the equivalence volume from Figure 1:
The midpoint of the vertical inflection occurs at V(NaOH) = 24.0 cm³. - Calculate the amount (moles) of NaOH used at equivalence:
Amount = concentration × volume (in dm³)
Amount of NaOH = 0.100 mol dm⁻³ × (24.0 / 1000) dm³ = 2.40 × 10⁻³ mol1st mark awarded here - Use stoichiometry to find concentration of HX:
The reaction equation is: HX + NaOH → NaX + H₂O (1:1 molar ratio).
Amount of HX = 2.40 × 10⁻³ mol in 25.0 cm³.
Concentration of HX = Amount / Volume (in dm³)
Conc(HX) = (2.40 × 10⁻³ mol) / (25.0 / 1000 dm³) = 0.0960 mol dm⁻³ (or 0.096 mol dm⁻³)2nd mark awarded here
❌ Common Errors
- Misreading the equivalence volume from the graph (e.g. reading 20.0 or 25.0 instead of 24.0 cm³).
- Forgetting to divide volumes by 1000 to convert cm³ to dm³.
- Dividing by the titre volume (24.0 cm³) instead of the original acid volume (25.0 cm³).
🧠 Exam Technique
Always inspect the grid carefully: each major grid block represents 5 cm³, divided into 5 small squares (each small square = 1 cm³). The vertical line lies directly on 24.0 cm³.
Calculating Initial pH of the Weak Acid
Calculate the pH of the solution of HX before the addition of any sodium hydroxide.
📐 Step-by-Step Calculation
- State the weak acid approximation:
Before neutralisation, dissociation is negligible so [HX]eq ≈ [HX]initial = 0.0960 mol dm⁻³, and [H⁺] ≈ [X⁻].
Ka = [H⁺]² / [HX] ⟹ [H⁺]² = Ka × [HX] - Calculate [H⁺]:
[H⁺] = √(2.62 × 10⁻⁵ × 0.0960) = √(2.5152 × 10⁻⁶) = 1.586 × 10⁻³ mol dm⁻³Mark 1: Correct [H⁺] expression and value - Calculate pH:
pH = −log₁₀[H⁺] = −log₁₀(1.586 × 10⁻³) = 2.80Mark 2: Correct pH (must be given to 2 or more decimal places)
💡 Alternative Data Note
If you used the fallback value provided in the paper ([HX] = 0.600 mol dm⁻³):
[H⁺] = √(2.62 × 10⁻⁵ × 0.600) = 3.96 × 10⁻³ mol dm⁻³
pH = −log₁₀(3.96 × 10⁻³) = 2.40 (Full marks awarded via error carried forward).
❌ Significant Figure Trap
Rule: In A-Level Chemistry, all calculated pH values must be given to at least 2 decimal places. Writing pH = 2.8 loses the final mark!
pH at Half-Neutralisation (Buffer Point)
Calculate the pH of the solution when half of the acid has reacted.
✅ Correct Answer
pH = 4.58
(Must be to 2 or more decimal places, unless already penalised in 04.4 where 4.6 is allowed).
💡 The Half-Neutralisation Shortcut
When half of a weak acid has been neutralised:
[HX] = [X⁻]
Substitute this into the Ka expression:
Ka = [H⁺][X⁻] / [HX] = [H⁺]
Taking negative logs of both sides:
pH = pKa
Therefore: pH = −log₁₀(2.62 × 10⁻⁵) = 4.58
🧠 Examiner Insight
This is a 1-mark question intended to take less than 30 seconds. Do not carry out long equilibrium or mole calculations. Recognise the term "half of the acid has reacted" immediately as the half-neutralisation point where pH = pKa .
Graph Plotting and Sketching the Buffer Region
Plot your answers to 4.4 and 4.5 on Figure 1 and sketch the missing curve between 0 and 20 cm³.
📍 Points to Plot
- Point 1 (Initial pH from 04.4):
At 0.0 cm³ of NaOH, plot at pH = 2.80 (within 1 small square: 2.7 to 2.9). - Point 2 (Half-neutralisation from 04.5):
Since equivalence is at 24.0 cm³, half-neutralisation is at 12.0 cm³ of NaOH.
Plot at (12.0 cm³, pH = 4.58) (within 1 small square: 4.5 to 4.7).
✏️ Shape of the Curve (Buffer Action)
The curve must display the standard shape for a weak acid-strong base titration:
- Initial steep rise: Rises steeply from pH 2.8 over the first 1–2 cm³.
- Buffer plateau: Levels out between ~4 cm³ and 18 cm³ as an acid-salt buffer system forms ([HX] and [X⁻] coexist).
- Smooth continuation: Meets the provided curve cleanly at 20.0 cm³ (where pH is ~5.5).
❌ What Lost Marks
- Plotting the half-neutralisation point at 10.0 cm³ (half of 20 cm³) instead of 12.0 cm³ (half of the true 24.0 cm³ equivalence volume).
- Drawing a straight diagonal line from (0, 2.8) to the curve at 20 cm³ without showing the initial steep rise followed by the buffer plateau.
- Missing the plotted points with the drawn line (the curve must pass directly through the plotted points).
Topics
Physical Chemistry · Required Practicals · 3.1.12 Acids and Bases · 3.1.2 Amount of Substance · Required Practical 9: Investigate how pH changes when a weak acid reacts with a strong base and when a strong acid reacts with a weak base
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.