OCR A-Level Chemistry Unified chemistry (03), June 2024: Question 1

10 marks · Medium difficulty · Structured Questions

Calculate pH values of strong base, diluted strong acid, and buffer solutions, and describe the preparation of a standard solution.

Practise this question

Question

A four-part chemistry question about acids, bases, and salts. Part (a) asks for the pH of 1.00 dm3 of 0.400 mol dm-3 NaOH(aq) at 298 K (2 marks). Part (b) asks for the pH of a diluted HCl(aq) solution produced by adding water to 10.0 cm3 of 0.750 mol dm-3 HCl(aq) to 100 cm3 (1 mark). Part (c) asks for the pH of a buffer solution containing 0.300 mol dm-3 CH3COOH(aq) and 0.100 mol dm-3 CH3COONa(aq) given Ka = 1.75 x 10-5 mol dm-3 (2 marks). Part (d) asks to describe how a student would prepare 100.0 cm3 of a standard solution of Cu(NO3)2·3H2O with a concentration of 0.200 mol dm-3, including quantities, apparatus, and method (5 marks).
Question text

1 This question is about acids, bases and salts.

(a) What is the pH of 1.00 dm3 of 0.400 mol dm–3 of NaOH(aq) at 298 K?

pH = … [2]

(b) Water is added to 10.0 cm3 of 0.750 mol dm–3 HCl (aq) to produce 100 cm3 of diluted HCl (aq).

What is the pH of the diluted HCl(aq)?

Give your answer to 2 decimal places.

pH = … [1]

(c) A solution has concentrations of 0.300 mol dm–3 CH COOH(aq) and 0.100 mol dm–3

CH3COONa(aq).

K for CH COOH = 1.75 × 10–5 mol dm–3 at 298 K.

a 3

What is the pH of the solution at 298 K?

Give your answer to 2 decimal places.

pH = … [2]

(d) A student is provided with hydrated copper(II) nitrate, Cu(NO3)2•3H2O.

The student needs to prepare a standard solution of Cu(NO3)2•3H2O with a concentration of

0.200 mol dm–3. The student has access to usual laboratory apparatus and equipment.

Describe how the student would prepare 100.0 cm3 of this solution, giving quantities, apparatus

and method.

… [5]

Mark scheme

Show the mark scheme A mark scheme showing answers and guidance for question parts (a) to (d). Part (a) requires Kw calculations and awards 2 marks for pH = 13.6(0). Part (b) awards 1 mark for pH = 1.12. Part (c) awards 2 marks for pH = 4.28 using Ka expression. Part (d) awards 5 marks in total: 2 marks for calculating the required mass of 4.83 g, and 3 marks for the practical procedure of dissolving, transferring, rinsing, and making up to the 100 cm3 mark in a volumetric flask.

1 (a) FIRST, CHECK THE ANSWER ON ANSWER LINE 2

IF pH = 13.6(0), award 2 marks

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

K = [H+] 0.400 OR 1.00 10–14 = [H+] 0.400 ALLOW ECF from incorrect [H+] calculated from

w

[OH–] AND K

w

K 1.00 10–14 … for pH > 7 ONLY

+ w + + –14

OR [H ] = 0.400 OR [H ] = 0.400 OR [H ] = 2.5 10 ✓

pH = –log 2.5 10–14 = 13.6(0) ✓ ALLOW method based on pOH:

pOH = –log 0.400 = 0.40 ✓

ALLOW 13.6…… up to calculator value of 13.60205999 correctly Calculator: 0.39794…

rounded pH = 14 – 0.40 = 13.6(0) ✓

(b) FIRST, CHECK THE ANSWER ON ANSWER LINE 1

IF pH = 1.12, award 1 mark

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

pH = –log 0.075 = 1.12 ✓ 2 DP required

Question Answer Marks Guidance

(c) FIRST, CHECK THE ANSWER ON ANSWER LINE 2 COMMON ERRORS

IF pH = 4.28, award 2 marks

---------------------------------------------------------------------- 1 mark for 5.23 inverted [HA] and [A–]

[H+] 0.100 1.75 10–5 0.100

1.75 10–5 = [H+] =

0.300 0.300

1.75 10–5 0.300 OR 5.83…. 10–6

OR [H+] =

0.100 pH = –log 5.83…. 10–6 = 5.23 ✓ ECF

OR [H+] = 5.25 10–5 (mol dm–3) ✓

1 mark for 4.46 [HA] = 0.2 instead of 0.3

pH = –log 5.25 10–5 = 4.28 ✓ 2 DP required 1.75 10–5 0.200

[H+] =

0.100

OR 3.5 10–5

pH = –log 3.5 10–5 = 4.46 ✓ ECF

Other ECF available from ONE transcription error

ONLY, e.g. 1.57 10–5 for K 1.75 10–5

a =

Zero marks for square root approach

[H+]2

e.g. via Ka = 0.300

Zero marks for [A–] : [HA] = 0.1 : 0.1

1.75 10–5 0.100

i.e. [H+] = = 1.75 10–5

0.100

pH = 4.76

ALLOW Henderson-Hasselbalch for both marks:

0.100

e.g. pH = 4.76 + log 0.300

–5 0.100

OR pH = –log(1.75 10 ) + log 0.300 ✓

–5 0.300

OR pH = –log(1.75 10 ) – log 0.100

pH = 4.28 ✓

(d) Calculation 2 marks 5 FULL ANNOTATIONS MUST BE USED

100 ALLOW ECF throughout

n(Cu(NO3)2•3H2O) = 0.200 1000 -----------------------------------------------------

OR 2(.00) 10–2 (mol) OR 0.02(00) ✓

ALLOW ECF from incorrect n(Cu(NO3)2•3H2O)

Mass Cu(NO ) •3H O = 2.00 10–2 241.5 = 4.83 (g) ✓ 4.83 g subsumes 1st mark

32 2

2 or more DP to match balances

Method 3 marks

Dissolve solid in (distilled) water (less than 100 cm3) (in ALLOW small amount/some

DO NOT ALLOW 100 cm3 or more of water

beaker) ✓

IGNORE solvent

Transfer (solution) to volumetric flask

AND ALLOW graduated flask

Wash/rinse (from beaker to flask) ✓

ASSUME that wash/rinse is to a volumetric flask

Make up to mark/up to 100 cm3 with (distilled water)

AND

Invert flask (several times to ensure mixing) ✓ ALLOW swirl/shake

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

ALLOW preparation of solutions > 100 cm3 4 marks

e.g. for 250 cm3

n(Cu(NO3)2•3H2O) = 0.200 1000 OR 0.05 (mol)

Mass Cu(NO3)2•3H2O = 0.05 241.5 = 12.075 (g) ✓

Then method adapted for 250 cm3 volumetric flask

e.g. Make up to 250 cm3 with water

How to answer it

Acids, Bases and Salts Study Guide

What this question tests

This question assesses core quantitative concepts in physical chemistry: calculating pH for strong alkalis using ionic product of water ( Kw ), concentration changes upon dilution, calculating the pH of buffer solutions using acid dissociation constants ( Ka ), and describing practical laboratory techniques for preparing a standard solution from a hydrated salt.

Part (a) [2 Marks]

pH of a Strong Alkali

✅ Correct Answer

pH = 13.60

2 marks awarded for the correct final answer.

📐 Step-by-Step Calculation

  1. State or use the ionic product of water expression: Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ at 298 K.
  2. Rearrange to find [H⁺] = Kw / [OH⁻] = 1.00 × 10⁻¹⁴ / 0.400 = 2.5 × 10⁻¹⁴ mol dm⁻³ .
  3. Calculate pH: pH = -log(2.5 × 10⁻¹⁴) = 13.60 .
  4. Alternative method: Calculate pOH = -log(0.400) = 0.398 , then pH = 14.00 - 0.398 = 13.60 .

❌ Common Errors

  • Directly taking -log(0.400) without converting to [H⁺] first, which results in calculating pOH instead of pH (giving 0.40).
  • Failing to give the answer to 2 decimal places (pH values require decimal places equal to the number of significant figures in the mantissa).
Part (b) [1 Mark]

pH of a Diluted Strong Acid

✅ Correct Answer

pH = 1.12

1 mark awarded for correct evaluation to 2 decimal places.

📐 Step-by-Step Calculation

  1. Calculate moles of HCl initially: 0.0100 dm³ × 0.750 mol dm⁻³ = 7.50 × 10⁻³ mol .
  2. Find new concentration in 100 cm³ ( 0.100 dm³ ): 7.50 × 10⁻³ / 0.100 = 0.0750 mol dm⁻³ .
  3. Since HCl is a strong monoprotic acid, [H⁺] = 0.0750 mol dm⁻³ .
  4. Calculate pH: -log(0.0750) = 1.1249... rounds to 1.12 .

🧠 Exam Technique

Always double-check dilution factors. Diluting 10.0 cm³ to 100 cm³ is a 10-fold dilution, meaning concentration divides by 10 ( 0.750 becomes 0.0750 ).

Part (c) [2 Marks]

pH of an Acid Buffer Solution

✅ Correct Answer

pH = 4.28

2 marks awarded. ECF available from incorrect ratio or transcription errors.

📐 Step-by-Step Calculation

  1. Write the buffer expression for Ka : Ka = ([H⁺][A⁻]) / [HA] .
  2. Rearrange for [H⁺] : [H⁺] = Ka × ([HA] / [A⁻]) .
  3. Substitute values: [H⁺] = (1.75 × 10⁻⁵) × (0.300 / 0.100) = 5.25 × 10⁻⁵ mol dm⁻³ .
  4. Calculate pH: -log(5.25 × 10⁻⁵) = 4.2798... rounds to 4.28 .

❌ Common Errors

  • Inverting the ratio: Putting [A⁻] / [HA] instead of [HA] / [A⁻] gives [H⁺] = 5.83 × 10⁻⁶ leading to pH = 5.23 (1 mark max via ECF).
  • Using square root approximations ( Ka × [HA] ) which are only valid for weak acids alone in water, not buffer mixtures. This scores 0 marks.
Part (d) [5 Marks]

Preparing a Standard Solution

💡 Key Knowledge (Mark Breakdown)

Calculation (2 marks):

  • Moles = 0.200 × (100 / 1000) = 0.0200 mol
  • Molar mass of Cu(NO₃)₂·3H₂O = 63.5 + 2(14.0 + 48.0) + 3(18.015) = 241.6 g mol⁻³ (or 241.5).
  • Mass = 0.0200 × 241.5 = 4.83 g (must be given to 2 or more decimal places to match balance precision).

Method (3 marks):

  • Dissolve solid in distilled water in a beaker using less than 100 cm³ .
  • Transfer solution to a 100 cm³ volumetric flask and rinse beaker/glass rod into the flask.
  • Make up to the mark with distilled water, stopper, and invert to mix thoroughly.

🧠 Examiner Comments & Top Tips

  • Precision matters: Balances read to 2 decimal places, so mass calculations must reflect this ( 4.83 g ).
  • Standard Practical Phrasing: Examiners look for specific safety/procedural keywords: volumetric flask, rinse/washings transferred, invert/shake to mix, and dropwise addition to the graduation mark.
  • Do not state "dissolve in 100 cm³ of water" before adding to the flask, as the total volume would exceed 100 cm³ once the solid is added.

Topics

Module 5: Physical chemistry and transition elements · Module 2: Foundations in chemistry · Practical Activity Groups · 5.1 Rates, equilibrium and pH · 2.1 Atoms and reactions · PAG 2: Acid-base titration

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.