Edexcel A-Level Chemistry Paper 1, June 2018: Question 10

9 marks · Hard difficulty · Calculations

Analyze an alloy of yellow gold containing copper, silver, and gold through a multi-step chemical procedure involving redox reactions, precipitation, and titration to determine its carat.

Practise this question

Question

Question text

10 Yellow gold is used to make jewellery. It is an alloy of copper, gold and silver.

The purity of gold is measured in carats. The higher the carat, the higher the

percentage of gold in the alloy. Pure gold is 24 carat.

A sample of yellow gold is analysed using the steps below.

Step 1 Excess concentrated nitric acid is reacted with 1.250g of the alloy. The gold

does not react but the copper and silver do react. The half-equations are

Cu(s) → Cu2+(aq) + 2e−

Ag(s) → Ag+(aq) + e−

2HNO (aq) + e− → NO−(aq) + NO (g) + H O(l)

33 2 2

Step 2 The mixture is diluted with distilled water and the gold is filtered off.

Step 3 Excess hydrochloric acid is added to the filtrate. It reacts with the silver ions to

form a precipitate of silver chloride.

Ag+(aq) + Cl−(aq) → AgCl(s)

Step 4 The silver chloride precipitate is filtered off, washed, dried and weighed.

The mass of silver chloride formed is 0.706g.

Step 5 Excess potassium iodide is added to the remaining solution.

A precipitate of copper(I) iodide and a solution of iodine forms.

2Cu2+(aq) + 4I−(aq) → 2CuI(s) + I (aq)

Step 6 The resulting mixture is titrated with 0.100 mol dm−3 sodium thiosulfate solution.

I (aq) + 2S O2−(aq) → 2I−(aq) + S O2−(aq)

22 3 4 6

The titre is 39.40 cm3.

(a) Write the equation for the reaction of copper with concentrated nitric acid,

using the half-equations given in Step 1. State symbols are not required.

(1)

(b) State the indicator used and its colour change at the end-point in the titration

in Step 6.

(2)

*P52302RA02428*CaratPercentage by

mass of gold

9 37.5

10 41.7

14 58.3

18 75.0

Determine, using the experimental data, the carat of the sample of yellow gold

that was analysed.

(6)

(Total for Question 10 = 9 marks)

Mark scheme

Show the mark scheme

Question

Acceptable Answer Additional Guidance Mark

Number

10(a) Examples of equations (1)

correct equation

Cu + 4HNO → Cu2+ + 2NO + 2NO + 2H O

33 2 2

−

Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O

Allow multiples

Allow ⇌ provided equation is written in the

direction shown

Ignore state symbols, even if incorrect

Ignore cancelled electrons

Ignore Ag or Au on both sides

Question Do not award uncancelled electrons

Acceptable Answer Additional Guidance Mark

Number

10(b) (2)

Indicator: starch (1)

Colour change: M2 is conditional on starch or no indicator

Starting colour: blue/black or blue or black

Final colour: colourless (1) Ignore mention of precipitate

Ignore other words to describe colour e.g.

deep / dark

Ignore clear

Question

Answer Additional Guidance Mark

Number

10(c) Example of calculation (6)

calculation of moles of silver chloride (1) moles AgCl = 0.706/(107.9 + 35.5)

= 0.00492329 / 4.92329 x 10−3

calculation of mass of silver (1) mass Ag = 0.00492329 x 107.9 = 0.531223 (g)

calculation of moles of Cu2+ (1) moles S O 2− or moles Cu2+ = 39.40 x 0.100 / 1000

= 0.00394 / 3.94 x 10−3

calculation of mass of copper (1) mass Cu = 0.00394 x 63.5 = 0.25019 (g)

calculation of percentage of gold (1) mass Au = 1.250 – (0.531223 + 0.25019)

= 0.468587 (g)

percentage of gold = 0.468587/1.250 x 100

= 37.5 (%)

percentage of silver = 0.531223/1.250 x 100

= 42.4978 / 42.5 (%)

percentage of copper = 0.25019/1.250 x 100

= 20.0151 / 20 (%)

percentage of gold = 100 – (42.5 + 20) = 37.5 (%)

Allow TE for each step

Allow final answer based on correct rounding at each

stage (36.3 to 37.9%)

Ignore SF except 1 SF in final answer

Correct answer without working scores (5)

Continued on next page

deduction that alloy is 9 carat gold (1) Conditional on some correct working to show the

percentage of gold

If calculated % is not 37.5, allow:

calculated value of carat (24 x their percentage/100)

or ‘less than 9 carat gold’ if calculated % is less than

37.5%

or nearest carat value from table

or a (rough) interpolated carat value

or between the two relevant carat values

(Total for Question 10 = 9 marks)

How to answer it

Analysis of Yellow Gold Alloy

What this question tests

This multi-step synoptic problem assesses redox half-equation combination, redox titrations involving iodine-thiosulfate, gravimetric analysis via precipitation, stoichiometry, percentage composition calculations, and interpretation of alloy purity standards (carats).

Part (a)

Combining Redox Half-Equations

✅ Correct Answer

Combine the copper oxidation half-equation and the nitric acid reduction half-equation, balancing the electrons:

Cu + 4HNO₃ → Cu²⁺ + 2NO₃⁻ + 2NO₂ + 2H₂O

or ionic form: Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

💡 Key Knowledge

Oxidation: Cu(s) → Cu²⁺(aq) + 2e⁻
Reduction: HNO₃(aq) + e⁻ → ... (derived from given half-equation multiplied by 2 to balance electrons).
State symbols are not required as per the question stem.

🎯 1 Mark: Awarded for a correctly balanced equation. Multiples are accepted.

Part (b)

Titration Indicator and Colour Change

✅ Correct Answer

Indicator: Starch
Starting Colour: Blue-black (or blue / black)
Final Colour: Colourless

🧠 Exam Technique

Be precise with iodine titration colour changes. Starch is added near the end-point when the solution turns straw-yellow (reacting with remaining iodine to form a deep blue-black complex). Titrating until the colour disappears leaves a colourless final solution.

🎯 2 Marks: 1 mark for identifying starch; 1 mark for correct starting and final colours (M2 is conditional on using starch or no indicator stated).

Part (c)

Determining the Carat of Yellow Gold (Multi-Step Calculation)

📐 Step-by-Step Calculation

Step 1: Find moles of AgCl formed

Molar mass of AgCl = 107.9 + 35.5 = 143.4 g mol⁻¹

Moles AgCl = 0.706 / 143.4 = 0.004923 mol

Step 2: Find mass of Silver (Ag)

1 mol AgCl contains 1 mol Ag.

Mass Ag = 0.004923 × 107.9 = 0.5312 g

Step 3: Find moles of Cu²⁺ from titration

Moles S₂O₃²⁻ = (39.40 × 0.100) / 1000 = 0.00394 mol

From equation 2: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻

From equation 1: 2Cu²⁺ + 4I⁻ → 2CuI + I₂

Ratio: 2 moles Cu²⁺ produce 1 mole I₂, which reacts with 2 moles S₂O₃²⁻. Therefore, moles Cu²⁺ = moles S₂O₃²⁻ = 0.00394 mol .

Step 4: Find mass of Copper (Cu)

Mass Cu = 0.00394 × 63.5 = 0.2502 g

Step 5: Calculate percentage of Gold (Au)

Total alloy mass = 1.250 g

Mass Au = 1.250 - (0.5312 + 0.2502) = 0.4686 g

% Gold = (0.4686 / 1.250) × 100 = 37.5%

Step 6: Deduce Carat

Comparing 37.5% to the given table shows it corresponds exactly to 9 carat gold.

❌ Common Errors & Traps

Stoichiometry confusion: Forgetting the 1:1 mole ratio link between Cu²⁺ and S₂O₃²⁻ via I₂. Always write out or trace intermediate molar ratios carefully.
Significant figures: Final answers should be given to 2 or 3 significant figures matching input data. Avoid rounding intermediate steps too early.
Missing conclusion: Calculating the percentage correctly (37.5%) but failing to explicitly state that the alloy is 9 carat loses the final mark.

🎯 6 Marks: 1 mark for each logical numerical stage (moles AgCl, mass Ag, moles Cu²⁺, mass Cu, percentage Au) and 1 final mark for deducing the carat rating based on the percentage calculated.

Topics

Physical Chemistry · Inorganic Chemistry · Core Practicals · 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 3: Redox I · Topic 5: Formulae, Equations and Amounts of Substance · Topic 4: Inorganic Chemistry and the Periodic Table · Topic 15: Transition Metals

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.