AQA A-Level Chemistry Paper 1, 2021: Question 6

16 marks · Medium difficulty · Practical Techniques & Data Analysis

Explain the ionic product of water, evaluate temperature effects and calibrate a pH probe to determine pH changes and calculate mixture pH during a titration.

Practise this question

Question

Question 06 covering pH, ionic product of water Kw, temperature dependence, pH meter calibration curve, titration curve of strong acid with strong base, choice of indicators from Table 6, and a calculation of mixture pH when excess NaOH is added to HCl.
Question text

06 This question is about pH.

Pure water dissociates slightly.

H O(I) ⇌ H+(aq) + OH–(aq) ΔH = +57 kJ mol–1

[H+][OH−]

The equilibrium constant, Kc =

[H2O]

The ionic product of water, K = [H+][OH−]

w

06.1 Explain why [H2O] is not shown in the Kw expression.

[1 mark]

Table 5 shows how Kw varies with temperature.

Table 5

Temperature / °C K / mol2 dm–6

w

10 2.93 × 10−15

20 6.81 × 10−15

25 1.00 × 10−14

30 1.47 × 10−14

50 5.48 × 10−14

06.2 Explain why the value of Kw increases as the temperature increases.

[2 marks]

06.3 Give the expression for pH.

Calculate the pH of pure water at 50 °C

Give your answer to 2 decimal places.

Explain why water is neutral at 50 °C

[4 marks]

Expression

Calculation

pH

Explanation

A pH meter is calibrated using a calibration graph.

To create the calibration, the pH meter is used to measure the pH of separate

solutions, each with a known, accurate pH.

Figure 3 shows the calibration graph.

Figure 3

06.4 Use Figure 3 to give the true pH value when the pH meter reading is 5.6

[1 mark]

06.5 Suggest why the pH probe is washed with distilled water between each of the

calibration measurements.

[1 mark]

06.6 The calibrated pH meter is used to monitor the pH during a titration of

hydrochloric acid with sodium hydroxide.

Explain why the volume of sodium hydroxide solution added between each

*20* pH measurement is smaller as the end point of the titration is approached.

[1 mark]

Figure 4 shows the pH curve for a titration of hydrochloric acid with

sodium hydroxide solution.

Figure 4

Table 6 shows data about some indicators.

Table 6

Indicator pH range Colour at low pH Colour at high pH

Bromocresol green 3.8 – 5.4 yellow blue

Phenol red 6.8 – 8.4 yellow red

Thymolphthalein 9.3 – 10.5 colourless blue

The student plans to do the titration again using one of the indicators in Table 6 to

determine the end point.

06.7 State why all three of the indicators in Table 6 are suitable for this titration.

[1 mark]

06 8 36.25 cm3 of 0.200 mol dm–3 sodium hydroxide solution are added to

.

25.00 cm3 of 0.150 mol dm–3 hydrochloric acid.

*21* Calculate the pH of the final solution at 25 ºC

K = 1.00 × 10–14 mol2 dm–6 at 25 ºC

w

[5 marks]

pH

Mark scheme

Show the mark scheme Mark scheme for Question 06 detailing acceptable points across parts 06.1 to 06.8, including [H2O] constancy, endothermic Le Chatelier shift, pH formula and calculation (pH = 6.63), calibration curve readout (5.55), washing probe to prevent cross-contamination, titration curve steep section match for indicators, and step-by-step excess OH- and pH calculation (pH = 12.76).

Question Answers Additional comments/Guidelines Mark

Allow

[H O] is (very) large in comparison (to [H+] and

[OH-])

[H2O] is (almost) constant or [H2O] is incorporated in Kw

06.1 1

or Kw = Kc[H2O]

or the equilibrium lies very much to the left.

Ignore water has negligible dissociation

Ignore [H2O] = 1 or [H2O] is very small

M1 Equilibrium is endothermic (in forward direction) Ignore more H+ and OH- formed 1

06.2 M2 Equilibrium shifts to the RHS to minimise/oppose

temperature increase

M1 Allow pH = –log[H+]

M1 pH = –log [H+] 1

M2 [H+]2 = 5.48 × 10−14

M2 [H+] = √5.48 × 10−14 (= 2.34 × 10−7)

−7 M3 pH = –log10 M2

06.3 M3 pH = –log10 2.34 × 10 = 6.63 1

M4 Allow equal amounts of H+ and OH-

M4 [H+] = [OH–]

or 1

Dissociation of each water molecule gives one H+

and one OH–

– A-LEVEL CHEMISTRY – – JUNE 2021

06.4 5.55 Allow 5.5 to 5.6 1

pH of previous solution doesn’t contaminate new 1

Different solutions must not contaminate each other solution

or

06.5

To wash off any residual solution/substance (which could interfere Ignore to make neutral/neutralise

with the reading)

Ignore so as not to affect concentrations

To avoid missing the end point 1

Or

06.6

(Very little pH change per cm3 added at start) large change in pH

(near end point)

All have a colour change/pH range within the steep/vertical part of

06.7 Colour change/pH range between pH 3 and 11 1

the titration curve

– A-LEVEL CHEMISTRY – –

M1 Amount of OH– = 36.25 0.200 ÷ 1000 = 7.25 10–3 mol 1

and Amount of H+ = 25.0 0.150 ÷ 1000 = 3.75 10–3 mol

1 21

M2 Amount of excess OH– = 7.25 10–3 – 3.75 10–3

= 3.50 10–3 mol

– –3 –3 –2 M3 [OH–] = (M2) ÷ (61.25 10–3) 1

M3 [OH ] = (3.50 10 ) ÷ (61.25 10 ) (= 5.71 10 mol)

+ –14 –2 –13 M4 [H+] = 1.00 10–14 ÷ M3 1

M4 [H ] = 1.00 10 ÷ 5.71 10 = 1.75 10

06.8 M5 Allow pH = 12.8 1

M5 pH = 12.76

M5 pH = -log10(M4)

Alternative Method

M4 p OH = 1.24

M5 pH = 14 – 1.24 = 12.76

How to answer it

pH, Ionic Product of Water (Kw), and Titration Curves

📌 What this question tests

This question assesses fundamental Acid-Base equilibria and experimental titration skills:

  • Understanding the definition and derivation of Kw and why [H₂O] is omitted.
  • Applying Le Chatelier's Principle to temperature changes for the self-ionisation of water.
  • Calculating the pH of pure water at non-standard temperatures and explaining why neutral does not always mean pH 7.
  • Practical titration techniques: calibrating pH probes, avoiding contamination, and understanding rate of addition near the end-point.
  • Choosing appropriate acid-base indicators using pH titration curves.
  • Multi-step quantitative calculation of pH in strong acid-strong base mixtures with excess base.

Question 06.1: The Kw Expression

Explain why [H₂O] is not shown in the Kw expression. [1 mark]

✅ Correct Answer

[H₂O] is (almost) constant

Also allowed: [H₂O] is extremely large compared to [H⁺] and [OH⁻], [H₂O] is incorporated into Kw (Kw = Kc[H₂O]), or the equilibrium lies very heavily to the left.

❌ Common Errors

  • Saying "[H₂O] = 1" or "[H₂O] is very small" (incorrect; water concentration is ~55.5 mol dm⁻³).
  • Stating "water has negligible dissociation" without mentioning that its concentration remains constant.
Mark scheme: [H₂O] is (almost) constant [1 mark]

Question 06.2: Temperature Dependence of Kw

Explain why the value of Kw increases as the temperature increases. [2 marks]

✅ Correct Answer

Mark 1: The (forward) reaction / dissociation is endothermic (ΔH = +57 kJ mol⁻¹).

Mark 2: Equilibrium shifts to the right-hand side (in the forward direction) to oppose/minimise the increase in temperature.

🧠 Exam Technique: Two-Step Equilibrium Answers

  • Step 1: State whether the forward direction is exothermic or endothermic.
  • Step 2: State the direction the equilibrium shifts and link explicitly to Le Chatelier's principle ("to oppose the temperature rise").
  • Do not just write "more H⁺ and OH⁻ are made" without explaining the shift.
Mark scheme: M1 (Endothermic), M2 (Equilibrium shifts to RHS to oppose temperature increase) [2 marks]

Question 06.3: pH and Neutrality of Pure Water at 50 °C

Give the expression for pH. Calculate pH of pure water at 50 °C (2 d.p.). Explain why it is neutral. [4 marks]

📐 Calculation & Working

1. Expression for pH:
pH = −log₁₀[H⁺] (or −log[H⁺] )

2. In pure water:
[H⁺] = [OH⁻], therefore Kw = [H⁺]²

3. Substitute Kw at 50 °C (5.48 × 10⁻¹⁴ mol² dm⁻⁶):
[H⁺] = √(5.48 × 10⁻¹⁴) = 2.3409 × 10⁻⁷ mol dm⁻³

4. Calculate pH:
pH = −log₁₀(2.3409 × 10⁻⁷) = 6.63 (must be 2 d.p.)

💡 Key Knowledge: Definition of Neutrality

Explanation: Water is neutral because [H⁺] = [OH⁻] (or because dissociation produces equal amounts of H⁺ and OH⁻ ions).

Common Trap: Thinking "neutral" means pH = 7. pH 7 is only neutral at 25 °C! A solution is neutral whenever [H⁺] equals [OH⁻], regardless of the pH value.
Mark scheme: M1 (pH = −log₁₀[H⁺]), M2 ([H⁺] = 2.34 × 10⁻⁷), M3 (pH = 6.63), M4 ([H⁺] = [OH⁻]) [4 marks]

Questions 06.4 – 06.6: Practical pH Meter Calibration & Titration

06.4 Calibration Graph Reading [1 mark]

Find pH meter reading = 5.6 on the y-axis, read across to the calibration line, and read down to the x-axis.

True pH value = 5.55 (allow 5.5 to 5.6)

06.5 Washing the pH Probe [1 mark]

Correct Answer:

  • To prevent solutions from contaminating each other, OR
  • To wash off residual solution that could interfere with the reading.

Ignore: "to make neutral" or "so as not to affect concentrations".

06.6 Adding Smaller Volumes Near End-Point [1 mark]

Correct Answer:

  • To avoid missing the end point, OR
  • Because there is a very large pH change per drop/addition near the end point (compared to very small changes earlier on).

Question 06.7: Selecting Indicators

State why all three indicators in Table 6 are suitable for this titration. [1 mark]

✅ Correct Answer

All three have their complete colour change / pH range within the steep (vertical) section of the titration curve (between approximately pH 3 and 11).

🧠 Exam Technique

Always mention the word "steep" or "vertical" section when explaining why an indicator is suitable for an acid-base titration. For a strong acid – strong base titration, the vertical section is long (typically pH 3 to 11), meaning multiple indicators work.

Mark scheme: All have colour change/pH range within the steep/vertical part of the curve (pH 3–11) [1 mark]

Question 06.8: Mixture pH Calculation (Excess Base)

Calculate the pH of the final solution when 36.25 cm³ of 0.200 mol dm⁻³ NaOH is added to 25.00 cm³ of 0.150 mol dm⁻³ HCl at 25 °C. [5 marks]

📐 Step-by-Step Calculation

  1. Find initial moles of OH⁻ and H⁺:
    Moles of OH⁻ = (36.25 × 0.200) ÷ 1000 = 7.25 × 10⁻³ mol
    Moles of H⁺ = (25.00 × 0.150) ÷ 1000 = 3.75 × 10⁻³ mol
    (1 mark for both values)
  2. Calculate moles of excess OH⁻:
    Excess OH⁻ = (7.25 × 10⁻³) − (3.75 × 10⁻³) = 3.50 × 10⁻³ mol
    (1 mark)
  3. Calculate total volume and [OH⁻]:
    Total volume = 36.25 + 25.00 = 61.25 cm³ = 61.25 × 10⁻³ dm³
    [OH⁻] = (3.50 × 10⁻³) ÷ (61.25 × 10⁻³) = 5.714 × 10⁻² mol dm⁻³
    (1 mark)
  4. Calculate [H⁺] using Kw:
    [H⁺] = Kw ÷ [OH⁻] = (1.00 × 10⁻¹⁴) ÷ (5.714 × 10⁻²) = 1.75 × 10⁻¹³ mol dm⁻³
    (1 mark)
    Alternative: pOH = −log₁₀(5.714 × 10⁻²) = 1.243
  5. Calculate pH:
    pH = −log₁₀(1.75 × 10⁻¹³) = 12.76 (allow 12.8)
    (Alternative: pH = 14 − 1.243 = 12.76)
    (1 mark)

❌ Common Calculation Traps

  • Forgetting total volume: Dividing excess moles by only the volume of NaOH or HCl rather than (36.25 + 25.00 = 61.25 cm³).
  • Not converting cm³ to dm³: Always divide total cm³ by 1000 when computing concentration.
  • Stopping at pOH: Calculating −log₁₀[OH⁻] = 1.24 and writing that as the final pH! A strong alkaline solution must have a pH well above 7.
Mark scheme: M1 (moles OH⁻ and H⁺), M2 (excess OH⁻ = 3.50 × 10⁻³), M3 ([OH⁻] = 5.71 × 10⁻²), M4 ([H⁺] = 1.75 × 10⁻¹³ or pOH = 1.24), M5 (pH = 12.76) [5 marks]

Topics

Physical Chemistry · Required Practicals · 3.1.12 Acids and Bases · 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc · 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 1, 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.