Edexcel A-Level Chemistry Paper 3, June 2018: Question 7

11 marks · Hard difficulty · Calculations

Calculate the pH of a buffer solution formed by mixing a weak acid and sodium hydroxide, sketch its titration curve, and explain how to determine Ka from the graph.

Practise this question

Question

Chemistry exam question with two main parts about weak acids. Part (a) asks to calculate the pH of a solution formed by mixing dilute sodium hydroxide with a weak acid HX, given Ka, in 5 marks. Part (b)(i) provides a grid with pH on the y-axis (0 to 14) and volume of sodium hydroxide added on the x-axis (0 to 50 cm3) to sketch a titration curve for propanoic acid, worth 4 marks. Part (b)(ii) asks to explain how to use the graph to obtain the Ka value, worth 2 marks.
Question text

7 This question is about weak acids.

(a) A weak acid, HX, has a K value of 5.25 × 10−5 mol dm−3. A solution was formed by

a

mixing 10.5 cm3 of 0.800 mol dm−3 dilute sodium hydroxide with

25.0 cm3 of 0.920 mol dm−3 HX(aq).

Calculate the pH of the solution formed, showing all your working.

(5)

(b) (i) Propanoic acid, CH3CH2COOH, is a weak acid.

On the grid below, sketch the change in pH during the addition

of 50.0 cm3 of 0.100 mol dm−3 sodium hydroxide solution

to 25.0 cm3 of 0.100 mol dm−3 propanoic acid solution.

(4)

pH 8

0 10 20 30 40 50

Volume of sodium hydroxide added / cm3

(ii) Explain how you would use the graph in (b)(i) to obtain the value of the acid

dissociation constant, Ka, for propanoic acid.

You are not expected to calculate this value.

(2)

… 24

*P52304A02432*(Total for Question 7 = 11 marks)

Mark scheme

Show the mark scheme Mark scheme showing detailed marking points for each part of the question. Part (a) awards 5 marks for calculating moles of reactants, remaining acid, concentrations, substituting into Ka expression, and finding final pH. Part (b)(i) awards 4 marks for starting pH between 2 and 4, correct curve shape ending at pH 12-13, vertical section at 25 cm3, and vertical range between pH 6-7 and 10-12. Part (b)(ii) awards 2 marks for reading pH at half-neutralisation/equivalence point and stating that pH = pKa or Ka = 10^-pH.

Question

Acceptable Answers Additional Guidance Mark

Number

7(a) Example of calculation: (5)

calculates moles of X− / NaOH present in the (moles of X− = mol NaOH = 0.8(00) x 10.5 )

mixture 1000

(1) = 0.0084(0) / 8.4(0) x 10−3 (mol)

calculates moles of HX which remain (moles of HX – mol NaOH = 0.92(0) x 25.0 − 0.0084(0)

unreacted 1000

(1) = 0.023(0) – 0.0084(0))

= 0.0146 / 1.46 x 10−2 (mol)

calculates / shows ratio of [HX] to [X−] OR [HX] = 0.0146 and [X−] = 0.0084(0)

ratio of moles of HX : X− (as total V cancels) 0.0355 0.0355

(1)

= 0.411 and 0.237 (mol dm−3)

Allow use of the ratio of the moles as above (as total V

cancels)

re-arranges K or pK expression correctly and [Η+] = K × [ΗΧ] = 5.25 × 10−5 × 0.411

a a a

substitutes appropriate values (1) [Χ−] 0.237

[Η+] = 9.10443038 × 10−5 (mol dm−3)

final pH to 2 or 3SF (1)

pH = 4.04

Allow use of pH expression to get answer:

pH = pK – log [HX] or pK + log [X-]

a a

[X−] [HX]

ALLOW TE M5 for calculation of pH from any [H+]

Correct answer with no working scores (5)

Question

Acceptable Answers Additional Guidance Mark

Number

7(b)(i) A sketch graph which shows the following: (4)

a starting pH between 2 and 4 (inclusive) (1)

correct general shape and ends at pH = 12-13 (1)

(any) vertical at 25 cm3 (1)

vertical between pH = 6 - 7 and pH = 10 – 12 (1)

Vertical must be no more than 5 pH units within these

ranges

Question

Acceptable Answers Additional Guidance Mark

Number

7(b)(ii) An explanation that makes reference to the following points: (2)

(Read off) pH at half-neutralisation (point) / pH at 12.5 (cm3) May be shown on the sketch graph

OR ALLOW read equivalence vol, add same

pH at half-equivalence (point) (1) volume of (propanoic) acid and measure

pH

As pH = pK / [H+] = K / K = 10-pH (1)

a a a

M2 dependent on mentioning half

equivalent / 12.5 cm3

(Total for Question 7 = 11 marks)

How to answer it

Weak Acids, Buffer Calculations & Titration Curves

What this question tests

This multi-step question assesses your ability to calculate the pH of a buffer solution formed by partial neutralisation of a weak acid by a strong base (5 marks), sketch a weak acid-strong base titration curve with precise inflection points and pH bounds (4 marks), and explain how to graphically determine the acid dissociation constant, Ka (2 marks).

Question 7 (a)

Buffer Calculation (5 Marks)

Calculate the pH of the solution formed by mixing 10.5 cm³ of 0.800 mol dm⁻³ NaOH with 25.0 cm³ of 0.920 mol dm⁻³ HX(aq). Ka = 5.25 × 10⁻⁵ mol dm⁻³

📐 Step-by-Step Calculation

  1. Moles of NaOH added:
    (10.5 / 1000) × 0.800 = 0.00840 mol
  2. Initial moles of HX:
    (25.0 / 1000) × 0.920 = 0.0230 mol
  3. Moles of unreacted HX remaining:
    0.0230 - 0.00840 = 0.0146 mol
  4. Moles of salt (X⁻) formed:
    Equal to moles of NaOH reacted = 0.00840 mol
  5. Calculate concentrations (or use mole ratio directly as total volume cancels):
    Total Volume = 10.5 + 25.0 = 35.5 cm³
    [HX] = 0.0146 / 0.0355 = 0.411 mol dm⁻³
    [X⁻] = 0.00840 / 0.0355 = 0.237 mol dm⁻³
  6. Rearrange Ka expression & solve for [H⁺]:
    [H⁺] = Ka × ([HX] / [X⁻])
    [H⁺] = 5.25 × 10⁻⁵ × (0.411 / 0.237) = 9.104 × 10⁻⁵ mol dm⁻³
  7. Calculate final pH:
    pH = -log(9.104 × 10⁻⁵) = 4.04 (to 3 SF)

❌ Common Errors & Traps

  • Forgetting limiting reagents: Assuming all the weak acid reacts completely instead of recognising a buffer mixture is created.
  • Volume errors: Calculating concentrations incorrectly by dividing by individual volumes instead of the combined total volume ( 35.5 cm³ ). Examiner tip: Because total volume appears in both numerator and denominator for [HX]/[X⁻], using mole ratios directly is a great shortcut to avoid arithmetic slips!
  • Rounding too early: Rounding intermediate [H⁺] values leading to final pH boundary errors. Keep full calculator display until the final pH step.
Mark Breakdown (5): M1: Moles of X⁻/NaOH calculated | M2: Moles of unreacted HX calculated | M3: Ratio/concentrations of HX to X⁻ shown | M4: Correct substitution into rearranged Ka expression | M5: Final pH to 2 or 3 SF (4.04).
Question 7 (b)(i)

Titration Curve Sketch (4 Marks)

Sketch the change in pH during the addition of 50.0 cm³ of 0.100 mol dm⁻³ NaOH to 25.0 cm³ of 0.100 mol dm⁻³ propanoic acid.

💡 Key Features Required on Sketch

  • Starting pH: Must start between pH 2 and 4 (reflecting a weak acid).
  • General Shape: Gradual upward curve initially (buffer region), followed by a steep vertical inflection, leveling off towards a high pH.
  • Equivalence Point Volume: The vertical inflection must be centred exactly at 25.0 cm³ of NaOH added (since concentrations are equal and acid volume is 25.0 cm³).
  • Inflection Bounds: The vertical section must occur strictly between pH 6 - 7 at the bottom and pH 10 - 12 at the top, ending at pH 12–13 overall.

🧠 Exam Technique for Sketches

Examiners look for clear, distinct geometric landmarks. Do not draw a lazy diagonal line. Ensure the curve starts smoothly, rises gently, curves sharply upward near 25 cm³, and flattens out well below pH 14 (around 12-13 due to excess 0.100 mol dm⁻³ NaOH).

Mark Breakdown (4): M1: Starting pH between 2 and 4 inclusive | M2: Correct general shape ending at pH 12–13 | M3: Vertical section positioned at exactly 25 cm³ | M4: Vertical section constrained between pH 6–7 and pH 10–12.
Question 7 (b)(ii)

Graphical Determination of Ka (2 Marks)

Explain how you would use the graph in (b)(i) to obtain the value of the acid dissociation constant, Ka, for propanoic acid.

✅ Correct Answer & Marking Points

  • Point 1: Read off the pH value at half-neutralisation / half-equivalence. For 25.0 cm³ equivalence point, this is at 12.5 cm³ of NaOH added.
  • Point 2: State the governing relationship: at this specific halfway point, [H⁺] = Ka , therefore pH = pKa (or use Ka = 10⁻ᵖᴴ ).

🧠 Top-Level Response Guidance

Top students explicitly state both the volume coordinate (12.5 cm³) where half-neutralisation occurs and the theoretical justification ( [acid] = [salt] , simplifying the Ka expression so Ka = [H⁺] ).

Mark Breakdown (2): M1: Read off pH at half-neutralisation / half-equivalence / at 12.5 cm³ | M2: State that at this point pH = pKa or [H⁺] = Ka (M2 is dependent on referencing half-equivalence/12.5 cm³).

Topics

Physical Chemistry · Topic 12: Acid-base Equilibria

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.