OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2021: Question 22

16 marks · Hard difficulty · Structured Questions

Calculate the pH of sodium hydroxide, enthalpy changes of neutralisation and enthalpy of reaction, and determine the pH and mass of N2O3 used to form a buffer solution with nitrous acid.

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Question text

22 This question is about acids, bases and buffers.

(a) Sodium hydroxide, NaOH, is a strong base.

Calculate the pH of 0.140 mol dm–3 NaOH(aq) at 298 K.

Give your answer to 2 decimal places.

pH = … [2]

(b) Sulfuric acid reacts with sodium hydroxide as shown in Equation 22.1.

Equation 22.1 H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) ΔH1

This is a neutralisation reaction.

A student carries out an experiment to determine the enthalpy change ΔH1 and uses this

value to deduce the enthalpy change of neutralisation, ΔneutH.

The student measures out two solutions:

• 25.0 cm3 of 1.60 mol dm–3 H SO (aq)

• 55.0 cm3 of 1.50 mol dm–3 NaOH(aq) (an excess).

The temperature of each solution is the same.

The student mixes the two solutions. The temperature increases by 13.0 °C.

(i) Show that NaOH is in excess.

[2]

(ii) Calculate the enthalpy change, ΔH1, for Equation 22.1, and deduce the value for the

enthalpy change of neutralisation, ∆ H, in kJ mol–1.

neut

Assume that the densities of all solutions and the specific heat capacity, c, of the reaction

mixture are the same as for water.

enthalpy change, ΔH = … kJ mol–1

enthalpy change of neutralisation, Δ H = … kJ mol–1 [4]

neut

(iii) The student repeats the experiment using 50.0 cm3 of 1.60 mol dm–3 H SO and

110.0 cm3 of 1.50 mol dm–3 NaOH.

Predict the increase in temperature.

Explain your reasoning.

… [2]

(c)* Nitrous acid, HNO2, is a weak Brønsted–Lowry acid with a pKa value of 3.34 at room

temperature.

HNO2 can be prepared by reacting N2O3 with water.

HNO2 is the only product.

A chemist makes up a buffer solution by the following method.

Step 1 The chemist weighs a sample of N2O3.

Water is then added to form 100 cm3 of 0.500 mol dm–3 HNO (aq).

Step 2 The chemist adds 100 cm3 of 0.150 mol dm–3 NaOH(aq) to the 100 cm3 solution

of 0.500 mol dm–3 solution of HNO (aq).

The resulting solution is made up to 1.00 dm3.

Explain why a buffer solution forms in Step 2. Determine the pH of this buffer solution and the

mass of N2O3 that was used in Step 1.

… [6]

Additional answer space if required.

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

22 (a) FIRST CHECK THE ANSWER ON ANSWER LINE 2 AO2.2

If answer = 13.15 award 2 marks ×2 ALLOW ECF providing pH>7

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

1.00 × 10–14 Calculator: 7.142857143 × 10–14

[H+] = = 7.14….. × 10–14 (mol)

0.140

ALLOW pOH method

pH = –log (7.14….. × 10–14) = 13.15 pOH = –log(0.14) = 0.85………

2 DP required

pH = 14.00 – (0.85……) = 13.15

(b) (i) 25.0 2 AO2.2

n(H2SO4) = 1.60 × 1000 = 0.04(00) (mol)

×2

AND

55.0

n(NaOH) = 1.50 × 1000 = 0.0825 (mol)

0.04(00) mol H2SO4 reacts with 0.08(00) mol NaOH ALLOW 0.0825>0.08

1 mol H2SO4 reacts with 2 mol NaOH

element

(b) (ii) q = mc∆T = 80.0 × 4.18 × 13.0 4 AO2.4 FULL ANNOTATIONS MUST BE USED

= 4347.2 (J) OR 4.3472 (kJ) ×4 -----------------------------------------------------

ALLOW 3 SF up to calculated answer

throughout

4.3472 –1

∆H1 =( –) 0.0400 = (–)108.68 kJ mol ALLOW ECF from q

DO NOT ALLOW division by n(NaOH)

108.68 –1

∆neutH = (–) 2 = (–)54.34 kJ mol ALLOW ∆neutH from ∆H1 /2

– sign for ∆H value(s) ALLOW alternative methods

(b) (iii) The same OR 13ºC 2 AO3.1

×2

(Double the moles so) double the energy is spread ALLOW explanation that uses a calculation

over double the volume based on moles, volumes

ALLOW mass for volume

element

(c)* Please refer to the marking instructions on page 4 of this 6 Indicative scientific points may include:

mark scheme for guidance on how to mark this question. 1. Formation of buffer

Level 3 (5–6 marks) AO1.2 • Acid / HNO2 is in excess

Reaches a comprehensive conclusion with most detail and ×2 • + NaOH NaNO + H O

HNO2 2 2

few errors for the formation of the buffer

AND Calculation of the correct buffer pH • Partial neutralisation of HNO2

AND Correct mass of N O . → formation of NO –/ NaNO

23 2 2

There is a well-developed line of reasoning which is clear • Buffer contains HNO AND NO –/NaNO

22 2

and logically structured. The information presented is

relevant and substantiated. 2. Calculation of buffer pH

AO2.6 • n(HNO2) added = 0.0500 (mol)

Level 2 (3–4 marks) ×2

• n(NaOH) added = 0.0150 (mol)

Reaches a sound conclusion with some detail and some –

errors for • n(NO2 ) formed = 0.0150 (mol)

Formation of buffer AND Calculation of the buffer pH • n(HNO2) remaining = 0.0500 – 0.0150

OR = 0.0350 (mol)

Formation of buffer AND Mass of N2O3. • K = 10–3.34 = 4.57… × 10–4 (mol dm–3)

OR 3

• Concentrations = mol (volume 1 dm )

Calculation of the buffer pH AND Mass of N2O3.

4.57… × 10–4 × 0.0350

OR • [H+] =

Partial explanations of formation of the buffer 0.0150

AND buffer pH AND Mass of N O . = 1.0665…. × 10–3 (mol dm–3)

There is a line of reasoning presented with some structure. • pH = 2.97

The information presented is relevant and supported by • pH to 2 dec places

some evidence.

3. Calculation of mass of N2O3

Level 1 (1–2 marks)

AO3.1 • 1 mol N2O3 → 2 mol HNO2

Attempts, with some success, to:

Describe formation of buffer OR Calculate buffer pH ×2 OR N2O3 + H2O → 2HNO2

OR Obtain mass of N2O3. • n(HNO2) = 0.0500 (mol)

There is an attempt at a logical structure with a line of • n(N2O3) = 0.0500/2 = 0.0250 (mol)

reasoning. The information is in the most part relevant. • m(N2O3) = 0.0250 × 76 = 1.9(0) g

0 marks No response or no response worthy of credit.

Total 16

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How to answer it

Acids, Bases and Buffers Study Guide

What this question tests

This comprehensive multi-part question tests your mastery of ionic equilibria and thermochemistry. Key skills include calculating pH for strong bases using ionic product of water (Kw), determining limiting reagents and enthalpy changes of neutralisation from calorimetric data, scaling temperature changes, and constructing rigorous scientific explanations and calculations for buffer systems involving weak acids and salts.

Question Part (a)

Strong Base pH Calculation

✅ Correct Answer

pH = 13.15

💡 Key Knowledge

For strong alkalis, assume complete dissociation: [OH⁻] = [NaOH] = 0.140 mol dm⁻³ .
Use the ionic product of water expression: Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ at 298 K.

📐 Step-by-Step Calculation

Rearrange Kw to find hydrogen ion concentration:
[H⁺] = (1.00 × 10⁻¹⁴) / 0.140 = 7.1428... × 10⁻¹⁴ mol dm⁻³
Calculate pH:
pH = -log(7.1428... × 10⁻¹⁴) = 13.1457...
Round to the requested 2 decimal places: 13.15 .

❌ Common Errors

Calculating the pOH (0.85) and forgetting to subtract it from 14 to find the pH .
Failing to format the final answer strictly to 2 decimal places as requested.

Marks: 2 marks (AO2.2)

Question Part (b)(i)

Proving Limiting vs Excess Reagents

✅ Correct Answer

Initial moles calculated show NaOH is greater than the stoichiometric requirement relative to H₂SO₄.

📐 Step-by-Step Calculation

Calculate moles of H₂SO₄:
25.0 / 1000 × 1.60 = 0.0400 mol
Calculate moles of NaOH:
55.0 / 1000 × 1.50 = 0.0825 mol
Apply stoichiometry: 1 mol of H₂SO₄ requires 2 mol of NaOH. Therefore, 0.0400 mol H₂SO₄ requires 0.0800 mol NaOH. Since 0.0825 mol > 0.0800 mol, NaOH is in excess.

🧠 Exam Technique

Always show clear working for both reactants' initial moles. Clearly state the stoichiometric ratio from the balanced equation before comparing values.

Marks: 2 marks (AO2.2)

Question Part (b)(ii)

Enthalpy of Reaction and Neutralisation

✅ Correct Answer

ΔH₁ = -108.68 kJ mol⁻¹
ΔneutH = -54.34 kJ mol⁻¹

📐 Step-by-Step Calculation

Total Volume: 25.0 cm³ + 55.0 cm³ = 80.0 cm³ (mass = 80.0 g).
Heat energy change (q):
q = m c ΔT = 80.0 × 4.18 × 13.0 = 4347.2 J = 4.3472 kJ
Enthalpy change for Equation 22.1 (per mole of limiting reactant H₂SO₄):
ΔH₁ = -4.3472 / 0.0400 = -108.68 kJ mol⁻¹
Enthalpy of neutralisation (per mole of H₂O formed): Since 1 mole of H₂SO₄ produces 2 moles of H₂O, divide by 2:
ΔneutH = -108.68 / 2 = -54.34 kJ mol⁻¹

❌ Common Errors

Dividing by the moles of NaOH (the excess reagent) instead of the limiting reagent (H₂SO₄) will cost you marks. Always link enthalpy changes to the limiting reagent.

Marks: 4 marks (AO2.4)

Question Part (b)(iii)

Scaling Volumes and Temperature Change

✅ Correct Answer

The temperature increase remains the same: 13 °C.

💡 Key Knowledge

When you double the reacting moles, you simultaneously double the total volume of the mixture. Because both heat released ( q ) and mass ( m ) scale proportionally, the ratio q / (mc) remains constant, meaning ΔT does not change.

Marks: 2 marks (AO3.1)

Question Part (c)

Buffer Solutions and Stoichiometry Synthesis

✅ Correct Answer

Buffer pH: 2.97
Mass of N₂O₃: 1.9 g

💡 Key Knowledge: Indicative Scientific Points

Step 1 Reaction: N₂O₃ + H₂O → 2HNO₂
Step 2 Buffer Formation: Partial neutralisation of the weak acid HNO₂ by NaOH creates a buffer containing unreacted HNO₂ and formed salt NaNO₂ (providing NO₂⁻ ).
Mole Calculations:
- Initial HNO₂ added = 0.100 dm³ × 0.500 mol dm⁻³ = 0.0500 mol
- NaOH added = 0.100 dm³ × 0.150 mol dm⁻³ = 0.0150 mol
- Reacted NaOH limits neutralisation, leaving 0.0500 - 0.0150 = 0.0350 mol of HNO₂ and producing 0.0150 mol of NO₂⁻ .
pH Calculation:
- Ka = 10⁻³.34 = 4.57 × 10⁻⁴ mol dm⁻³
- Since concentrations are in the same total volume (1.00 dm³), mole ratios can be used directly:
[H⁺] = Ka × ([HNO₂] / [NO₂⁻]) = 4.57 × 10⁻⁴ × (0.0350 / 0.0150) = 1.0665 × 10⁻³ mol dm⁻³
- pH = -log(1.0665 × 10⁻³) = 2.97 (2 decimal places)
Mass of N₂O₃:
- Total HNO₂ needed = 0.0500 mol.
- From stoichiometry, 1 mol of N₂O₃ produces 2 mol of HNO₂ , so required N₂O₃ = 0.0500 / 2 = 0.0250 mol .
- Molar mass of N₂O₃ = (2 × 14.0) + (3 × 16.0) = 76.0 g mol⁻¹.
- Mass = 0.0250 mol × 76.0 g mol⁻¹ = 1.9 g .

🧠 Examiner Style & Level Descriptors

Top-level responses (Level 3, 5–6 marks) seamlessly link the stoichiometric preparation of the weak acid from its anhydride oxide, track limiting moles during partial neutralisation, correctly use the Ka expression to find [H⁺], and determine the exact starting mass with correct significant figures.

Marks: 6 marks (AO1.2, AO2.6, AO3.1)

Topics

Module 5: Physical chemistry and transition elements · Module 3: Periodic table and energy · Practical Activity Groups · 5.1 Rates, equilibrium and pH · 5.2 Energy · PAG 3: Enthalpy determination · PAG 11: pH measurement

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.