Edexcel A-Level Chemistry Paper 3, June 2023: Question 7

19 marks · Hard difficulty · Calculations

Calculate the permanent and temporary hardness of a water sample in mg dm-3 using titration data with EDTA and buffer preparation calculations.

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Question

Question text

7 Hardness in water is measured in terms of the concentration of dissolved

calcium compounds.

Titration experiments can be carried out to determine the hardness of

a water sample.

(a) A pipette is used to measure a 50.0 cm3 water sample for titration.

(i) Describe how to remove an air bubble from the tip of the pipette.

(1)

(ii) Calculate the maximum volume that would be obtained by using a

25.0 cm3 pipette twice to measure a total volume of 50.0cm3.

The uncertainty in each 25.0 cm3 pipette measurement is ±0.04 cm3.

(1)

(iii) Compare the percentage uncertainty in using a 25.0 cm3 pipette twice with

using a 50.0 cm3 pipette once to measure 50.0 cm3 of water.

The uncertainty in the 50.0 cm3 pipette measurement is ±0.05 cm3.

(2)

(b) About 2 cm3 of a pH 10 buffer is added to each 50.0 cm3 water sample.

(i) State whether or not a 100 cm3 measuring cylinder is suitable to measure this

volume of buffer solution. Justify your answer.

(1)

(ii) The pH10 buffer can be made by adding solid ammonium chloride to an

aqueous solution of ammonia of concentration 18.1 mol dm–3.

The relevant equation is*P71914A01732*

NH+ NH + H+

K = 5.62 × 10–10 mol dm–3

Calculate the mass of ammonium chloride that must be added to 100 cm3 of

ammonia solution to make the pH10 buffer.

Assume that there is no change in the volume on the addition of

ammonium chloride.

(4)

(iii) State a necessary laboratory precaution, other than wearing a laboratory coat,

gloves and goggles, that must be taken when using concentrated ammonia.

(1)

octahedral complex with the calcium ions in the water sample.

The structure of Eriochrome Black T is shown with a calcium ion.

(i) Complete the diagram to show how Eriochrome Black T forms three

dative covalent or coordinate bonds with the calcium ion.

(1)

NO2

O – +

O Na

18 O

*P71914A01832*OH

Ca2+

(ii) State the number of water molecules needed to complete this complex.

(1)

(d) There are two types of water hardness:

temporary hardness which is removed by boiling as a precipitate forms,

permanent hardness which is unaffected by boiling.

Levels of water hardness are expressed as the concentration of calcium ions

in mg dm–3.

A student carried out a series of experiments to determine the hardness of a

sample of water. 50.0 cm3 samples of the water were titrated with EDTA.

Further 50.0 cm3 samples of water were taken after boiling and then titrated

with EDTA.

(i) Name the process needed before titrating the sample of boiled water.

(1)

(ii) The mean titre of 0.0100*P71914A01932*moldm–3EDTA4–with a 50.0cm3water sample

before boiling was 12.80 cm3.

After boiling the mean titre was 5.15 cm3.

There is a 1 : 1 ratio in the reaction between EDTA4– ions and Ca2+ ions.

Calculate, in this water, the levels of permanent and temporary hardness in

mg dm–3 of calcium ions.

(6)

(Total for Question 7 = 19 marks)

Mark scheme

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Question

Answer Additional Guidance Mark

Number

7(a)(i) An answer that makes reference to the following point: (1)

• expel some solution (to remove the air bubble and suck up again Allow gently tap the side of the pipette to move

with the tip of the pipette in the solution) the air bubble to the top and out of the solution

(fill pipette above the line and) expel some solution Do not award answers referring to opening taps

(to remove the bubble) Do not award inverting the pipette

Question

Answer Additional Guidance Mark

Number

7(a)(ii) An example of calculation (1)

• calculation of maximum volume 25.04 + 25.04 = 50.08 (cm3)

Question

Answer Additional Guidance Mark

Number

7(a)(iii) An example of calculation (2)

• calculation using one 25 cm3 pipette twice (1) % uncertainty = (100 × 0.08 ÷ 50)

= 0.16 %

• calculation using one 50 cm3 pipette % uncertainty = (100 × 0.05 ÷ 50)

and = 0.1(0) %

making a comparison (1) e.g. Difference = 0.16 – 0.10 = 0.06 %

0.16% > 0.10%

TE on (a)(ii)

Allow TE for M2

Ignore SF including 1SF

Question

Answer Additional Guidance Mark

Number

7(b)(i) (1)

• unsuitable because the smallest volume you can Allow unsuitable/No because 2 cm3 is too small to measure in a

measure is 10 cm3 100 cm3 measuring cylinder

Allow unsuitable/No because the graduations are too big to

measure such a small volume

Allow unsuitable/No because the (percentage) uncertainty will

be too large

Allow unsuitable/No because the resolution is too low/is not

precise enough

Allow suitable/Yes because the volume doesn’t have to be

accurate and about 2 cm3 can be estimated

Question

Answer Additional Guidance Mark

Number

7(b)(ii) Example of calculation (4)

Method 1

• (expression of K and) [H+] (1) (K = ([NH ] × [H+]) ÷ [NH +]

a a 3 4

and)

[H+] = (inv log −pH =) 1.0 × 10−10 (mol dm−3)

• rearrangement of K expression & [NH +] (1) [NH +] = ((18.1 × 1 × 10−10) ÷ 5.62 × 10−10 =) 3.22 (mol dm−3)

a 4 4

• number of moles of NH4Cl (1) n(NH4Cl) = ((3.22 × (100 ÷ 1000) =) 0.322 (mol)

• mass of NH4Cl (1) m(NH4Cl) = 0.322 × 53.5 = 17.227 / 17.23 / 17.2 / 17 (g)

Final answer with or without working scores (4)

TE at each stage

Ignore SF except 1SF

Method 2 (Use of Henderson-Hasselbalch equation)

pH = pKa + log ( [NH ] ÷ [NH +] )

• expression of pH (1) 3 4

10 = 9.25 + log (18.1 ÷ [NH +] )

[NH +] = (18.1 ÷ 100.75 )= 3.22 (mol dm−3)

• rearrangement of pH expression & [NH +] (1) 4

• number of moles of NH4Cl (1) n(NH4Cl) = ((3.22 × (100 ÷ 1000) =) 0.322 (mol)

• mass of NH Cl (1) m(NH4Cl) = 0.322 × 53.5 = 17.227 / 17.23 / 17.2 / 17 (g)

Question

Answer Additional Guidance Mark

Number

7(b)(iii) An answer that makes reference to the following point: (1)

• use in fume cupboard / fume hood Do not award use mask / well-ventilated room

Question

Answer Additional Guidance Mark

Number

7(c)(i) Example of suitable diagram (1)

• diagram with 3 dative or covalent bonds

Allow lines for arrows

Allow absence of lone pairs

Do not allow dashed lines

Do not award arrows going from the calcium to the N/O

Do not award arrows or line coming from any other atoms to

those shown

Do not award double-headed arrows or curly arrows

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Answer Additional Guidance Mark

Number

7(c)(ii) (1)

• 3 / three

Question

Answer Additional Guidance Mark

Number

7(d)(i) An answer that makes reference to the following point: (1)

• filter (off the precipitate) Allow use of Buchner funnel / suction filtration / filtration

under reduced pressure/ gravity filtration

Ignore decant

Question

Answer Additional Guidance Mark

Number

7(d)(ii) Method 1 An example of calculation (6)

Total Hardness

• (M1) calculation of number of moles of EDTA4− (1) n(EDTA4−) = ((12.80 ÷ 1000) × 0.010 =) 1.28 × 10−4 (mol)

• (M2) calculation of number of moles of calcium ions in 50 cm3 n(Ca2+) = n(EDTA4−) = 1.28 × 10−4 (mol)

in 1 dm3 (1) in 1 dm3 n(Ca2+) = 1.28 × 10−4 × 20 = 2.56 × 10−3 (mol)

• (M3) calculation of mass of calcium ions in 1 dm3 (1) m(Ca2+) = ((2.56 × 10−3 × 40.1 = 0.102656 =) 0.10266 / 0.103 (g)

Total Hardness = (0.103 × 1000 =) 103 (mg dm−3)

• (M4) calculation of total hardness (1)

Permanent and Temporary Hardness (method as above)

• (M5) calculation of permanent hardness (1) Permanent hardness

= ((5.15 ÷ 1000) × 0.010 × 20 × 40.1 × 1000 = 41.303)

= 41 (mg dm−3)

Temporary Hardness = (103 – 41 =) 62 (mg dm−3)

• (M6) calculation of temporary hardness (1)

Final answer without rounding = (102.66 – 41.30 =) 61.36 (mg dm−3)

Final answers without working scores (6)

TE at each stage

Accept M1 – M4 either from the calculation of total or the permanent

hardness

Use of 40 for calcium gives 41.2 and 61.2 which score full marks

Ignore SF except 1SF

Method 2

• (M1) calculation of volume of EDTA4− required for V=(12.80 – 5.15=) 7.65 (cm3)

temporary hardness (1)

• (M2) calculation of number of moles of EDTA4− n(EDTA4−) = ((5.15 ÷ 1000) × 0.010 =) 5.15 × 10−5 (mol)

for permanent hardness (1)

in 50 cm3 n(Ca2+) = n(EDTA4−) = 5.15 × 10−5 (mol)

• (M3) calculation of number of moles of

3 in 1 dm3 n(Ca2+) = 5.15 × 10−5 × 20 = 1.03 × 10−3 (mol)

calcium ions in 1 dm for permanent hardness (1)

• (M4) calculation of mass of calcium ions in 1 dm3 m(Ca2+) = ((1.03 × 10−3 × 40.1 = 0.0413 (g)

for permanent hardness (1)

• (M5) calculation of permanent hardness (1) Permanent hardness =( 0.0413 x 1000=) 41.3 (mg dm−3)

• (M6) calculation of temporary hardness (1) Temporary hardness =((7.65 ÷ 1000) × 0.010 × 20 × 40.1 x 1000=)

= 61.4 (mg dm−3)

Final answers without working scores (6)

TE at each stage

Accept M2 – M5 either from the calculation of permanent or

temporary hardness

Use of 40 for calcium gives 41.2 and 61.2 which score full marks

(Total Question 7 = 19 marks)

How to answer it

Water Hardness, Buffer Calculations & Complexometric Titrations

What this question tests

This question assesses practical chemistry techniques (pipette usage, error analysis, apparatus selection), buffer solution calculations involving Ka and pH, transition metal ligand substitution/dative covalent bonding, and complexometric EDTA titration calculations to determine temporary and permanent water hardness.

Question 7 (a) — Practical Technique & Uncertainties

Pipette Operation and Percentage Uncertainty Analysis

✅ Correct Answers

(i) Expel some solution to remove the air bubble, then suck up again with the tip submerged, or fill past the line and expel solution.
(ii) 25.04 + 25.04 = 50.08 cm³ (or 50.08 )
(iii) % uncertainty for 25.0 cm³ twice = 0.16% . % uncertainty for 50.0 cm³ once = 0.10% . Comparison: using the 25.0 cm³ pipette twice gives a larger percentage uncertainty.

💡 Key Knowledge

Air bubbles in a pipette tip reduce the actual volume delivered, causing inaccurate titration results.
When measurements are repeated, absolute uncertainties accumulate (add together).
Percentage uncertainty formula: (uncertainty × number of uses / measured volume) × 100 .

🧠 Exam Technique

For uncertainty comparisons, calculate both values clearly and end with an explicit comparison statement (e.g., "0.16% > 0.10%").
Do not mention opening taps or tapping the tap area; focus on the pipette tip and meniscus level.

❌ Common Errors

Failing to double the uncertainty when a piece of apparatus is used twice in a calculation.
Vague descriptions of removing bubbles (e.g., just "shaking the pipette").

Total for 7(a): 4 marks

Question 7 (b) — Apparatus Selection & Buffer Calculations

Buffer Solutions and Mass Calculations

✅ Correct Answers

(i) Unsuitable because the smallest volume you can accurately measure is 10 cm³ (or resolution/graduations are too large for 2 cm³).
(ii) Mass of ammonium chloride = 17.2 g (or 17.23 g / 17 g ).
(iii) Use a fume cupboard / fume hood.

📐 Calculation Steps for 7(b)(ii)

Find [H⁺]: pH = 10, so [H⁺] = 10⁻¹⁰ mol dm⁻³.
Rearrange Ka expression for [NH₄⁺]: Ka = ([NH₃][H⁺]) / [NH₄⁺] → [NH₄⁺] = ([NH₃] × [H⁺]) / Ka.
Substitute values: [NH₄⁺] = (18.1 × 1.0 × 10⁻¹⁰) / (5.62 × 10⁻¹⁰) = 3.22 mol dm⁻³.
Calculate moles in 100 cm³: moles = 3.22 × (100 / 1000) = 0.322 mol.
Calculate mass: mass = moles × Mr = 0.322 × 53.5 = 17.23 g.

🧠 Exam Technique

When justifying measuring equipment, always reference the graduation size or resolution relative to the tiny volume being measured (2 cm³).
For safety questions involving volatile toxic gases like concentrated ammonia, always state "fume cupboard"—do not accept generic answers like "mask" or "well-ventilated room".

Total for 7(b): 6 marks

Question 7 (c) — Transition Metals & Ligands

Dative Covalent Bonding in Octahedral Complexes

✅ Correct Answers

(i) Diagram showing 3 dative covalent (coordinate) bonds formed with Ca²⁺ from specific lone-pair donor atoms (arrows pointing from nitrogen/oxygen atoms to Ca²⁺).
(ii) 3 (three water molecules needed to complete the 6-coordinate octahedral geometry alongside the tridentate Eriochrome Black T ligand).

💡 Key Knowledge

An octahedral complex has a coordination number of 6.
Eriochrome Black T is a multidentate (tridentate in this context) ligand providing 3 coordinate bonds; the remaining 3 sites are occupied by water molecules to reach coordination number 6.

❌ Common Errors

Drawing arrows pointing away from the metal ion rather than from the electron-rich donor atom to the Ca²⁺ ion.
Using dashed lines or double-headed curly arrows instead of straight arrows representing coordinate bonds.

Total for 7(c): 2 marks

Question 7 (d) — Water Hardness Titration Calculations

Temporary, Permanent, and Total Hardness via EDTA Titration

✅ Correct Answers

(i) Filter (off the precipitate) / filtration / suction filtration.
(ii) Total hardness = 103 mg dm⁻³ , Permanent hardness = 41 mg dm⁻³ , Temporary hardness = 62 mg dm⁻³ (Accept 102.7, 41.3, 61.4).

📐 Step-by-Step Calculation (Method 1)

Moles of EDTA (Total): (12.80 / 1000) × 0.0100 = 1.28 × 10⁻⁴ mol.
Moles of Ca²⁺ in 50.0 cm³: 1:1 ratio, so moles = 1.28 × 10⁻⁴ mol.
Moles of Ca²⁺ in 1.00 dm³: 1.28 × 10⁻⁴ × (1000 / 50.0) = 2.56 × 10⁻³ mol dm⁻³.
Mass of Ca²⁺ (Total Hardness): 2.56 × 10⁻³ × 40.1 = 0.1026 g = 103 mg dm⁻³.
Permanent Hardness (after boiling): Titre = 5.15 cm³. Moles EDTA = (5.15 / 1000) × 0.0100 = 5.15 × 10⁻⁵ mol. Scaling to 1 dm³ and multiplying by Mr (40.1) gives 41 mg dm⁻³.
Temporary Hardness: Total Hardness − Permanent Hardness = 103 − 41 = 62 mg dm⁻³.

🧠 Exam Technique & Traps

Boiling effect: Boiling precipitates out temporary hardness (as CaCO₃), leaving only permanent hardness ions in solution.
Unit conversions: Remember to convert cm³ to dm³ by dividing by 1000, and grams to milligrams by multiplying by 1000.
Significant figures: Give final answers to 2 or 3 significant figures consistent with data provided.

Total for 7(d): 7 marks | Overall Question Total: 19 marks

Topics

Physical Chemistry · Inorganic Chemistry · Organic Chemistry · Core Practicals · Core Practical 2: Preparation of a standard solution from a solid acid · Core Practical 3: Find the concentration of a solution of hydrochloric acid · Core Practical 11: Find the amount of iron in an iron tablet using redox titration · Topic 15: Transition Metals · Topic 12: Acid-base Equilibria · Topic 5: Formulae, Equations and Amounts of Substance · Topic 2: Bonding and Structure

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.