OCR A-Level Chemistry AS Depth in chemistry (02), November 2020: Question 1

13 marks · Medium difficulty · Structured Questions

Calculate the volume of hydrochloric acid needed to react with nickel, determine the volume of hydrogen produced, explain the reaction of magnesium, and describe analytical tests to identify nickel(II) bromide, sulfate, and carbonate.

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Question text

1 This question is about nickel and its compounds.

(a) Nickel reacts with dilute hydrochloric acid in a redox reaction.

Ni(s) + 2HCl(aq) NiCl2(aq) + H2(g)

Explain, in terms of the number of electrons transferred, whether nickel is oxidised or reduced.

… [1]

(b) A student completely reacts 0.192 g of nickel with 0.150 mol dm–3 HCl (aq).

(i) Calculate the minimum volume, in cm3, of 0.150 mol dm–3 HCl (aq) that the student needs

for the reaction.

Give your answer to 3 significant figures.

minimum volume of HCl (aq) = … cm3 [3]

(ii) Calculate the volume of H (g), in cm3, that would be produced at RTP.

volume of H (g) = … cm3 [1]

(iii) The student repeats the experiment with 0.192 g of magnesium instead of nickel, using

the same volume of 0.150 mol dm–3 HCl (aq) as in (b)(i).

State and explain whether the volume of H2(g) produced would be greater than, smaller

than, or the same as, the value you have calculated in (b)(ii).

… [2]

(c)* A student is provided with samples of three nickel compounds.

One sample is nickel(II) bromide, another is nickel(II) sulfate and the third is nickel(II)

carbonate. The student doesn’t know which sample is which.

Describe the tests that the student could carry out to identify the anion (negative ion) in each

sample, and write equations for any reactions. [6]

Additional answer space if required

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

1 (a) (i) Oxidised 1 2.1 IGNORE reference to oxidation numbers (even if

AND incorrect)

nickel has lost/donated two electrons

(b) (i) FIRST CHECK ANSWER ON THE ANSWER LINE 3 ALLOW 3.27 × 10–3 up to calculator value of

If answer = 43.6 (cm3) award 3 marks 3.270868825 × 10–3

0.192 ALLOW 6.54 × 10–3 up to calculator value of

n(Ni) = = 3.27… × 10–3 (mol)

58.7 1.1×1 6.541737649 × 10–3

n(HCl) = 3.27 × 10–3 × 2 = 6.54… × 10–3 (mol) 2.4 ×2

6.54 × 10–3

Volume HCl = × 1000 = 43.6 (cm3)

0.15

3 SF required

(ii) Volume H2 = 3.27 × 10–3 × 24000 = 78.5 (cm3) 1 2.4 ×1 ALLOW ECF from incorrect n(Ni) from (b)(i)

ALLOW 78.48 (cm3)

(iii) Volume is the same 2 3.4 ×2

Mg is in excess

Volume of H2 depends on HCl/HCl is limiting reagent

element

(c)* Please refer to the marking instructions on page 4 of this mark 6 1.2 ×2 Indicative scientific points

scheme for guidance on how to mark this question. 2.7 ×2 Tests for anions

Carbonate test:

Level 3 (5–6 marks) 3.4 ×2 Add HNO3(aq)/HCl(aq)/H2SO4(aq)/H+(aq)

The candidate gives a clear description of all three tests with correct fizzing/ forms CO2(g) → Carbonate identified

observations. Sulfate test:

AND Add Ba(NO3)2(aq) OR BaCl2(aq)

Equations are mostly correct. White precipitate → Sulfate identified

AND Bromide test

Some fine detail included in answer. Add AgNO3(aq)

Cream precipitate → Bromide identified

There is a well-developed line of reasoning which is clear and logically Equations (ionic or full)

structured. The information presented is relevant and substantiated. IGNORE state symbols (even if wrong)

Carbonate

Level 2 (3–4 marks) 2H+ + CO 2– → CO + H O

32 2

The candidate describes all three tests with correct observations. OR 2H+ + NiCO → Ni2+ + CO + H O

32 2

OR 2HNO3 + NiCO3 → Ni(NO3)2 + H2O + CO2

OR 2HCl + NiCO3 → NiCl2+ H2O + CO2

OR H2SO4 + NiCO3→ NiSO4 + H2O + CO2

Describes two tests with a few omissions.

Sulfate

AND 2+ 2–

Includes at least one correct equation. Ba + SO4 → BaSO4

OR Ba(NO3)2 + NiSO4 → BaSO4 + Ni(NO3)2

There is a line of reasoning presented with some structure. The OR BaCl2 + NiSO4 → BaSO4 + NiCl2

information presented is relevant and supported by some evidence Bromide

Ag+ + Br– → AgBr

Level 1 (1–2 marks) OR 2AgNO3 + NiBr2 → 2AgBr + Ni(NO3)2

The candidate attempts to describe two tests and observations, but Fine Detail (NOT inclusive)

explanations are incomplete. Sequence of tests on samples

OR Carbonate → Sulfate → Bromide

Gives a thorough description and explanation of one of the tests and Solubility of AgBr

attempts one equation. Soluble in concentrated ammonia

State symbols in ionic or full equations

There is an attempt at a logical structure with a line of reasoning. The e,g.

information is in the most part relevant. • 2H+(aq) + CO 2–(aq) → CO (g) + H O(l)

32 2

OR 2H+(aq) + NiCO (s) → Ni2+(aq) + CO (g) + H O(l)

32 2

0 marks • Ba2+(aq) + SO 2–(aq) → BaSO (s)

No response or no response worthy of credit. + –

• Ag (aq) + Br (aq) → AgBr(s)

How to answer it

Nickel and its Compounds

What this question tests

This question assesses core AS Level physical and inorganic chemistry concepts, including redox definitions via electron transfer, stoichiometry and moles calculations in solutions, gas molar volumes at RTP, limiting reagents, and qualitative analysis (chemical tests for common anions: carbonate, sulfate, and bromide) along with balanced ionic equations.

Question Part (a)

Redox and Electron Transfer

✅ Correct Answer

Oxidised AND nickel has lost/donated two electrons.

💡 Key Knowledge

Remember the acronym OIL RIG (Oxidation Is Loss, Reduction Is Gain of electrons). Elemental nickel goes from an oxidation state of 0 to +2 in NiCl₂, meaning each atom loses 2 electrons.

❌ Common Errors

Students often lose this mark by stating nickel is reduced, or by writing about oxidation numbers instead of focusing strictly on the number of electrons transferred as explicitly requested by the command word.

🎯 1 Mark: 1 mark for stating oxidised and referencing the loss/donation of 2 electrons. (Examiner note: Ignore references to oxidation numbers).

Question Part (b)(i)

Calculation: Minimum Volume of Acid

📐 Step-by-Step Calculation

Step 1: Find moles of Nickel
n(Ni) = mass / molar mass = 0.192 / 58.7 = 3.271 × 10⁻³ mol
Step 2: Use the reacting ratio
From equation Ni + 2HCl → NiCl₂ + H₂ , the ratio is 1:2.
n(HCl) = 3.271 × 10⁻³ × 2 = 6.542 × 10⁻³ mol
Step 3: Calculate volume in cm³
Volume = (moles / concentration) × 1000
Volume = (6.542 × 10⁻³ / 0.150) × 1000 = 43.6 cm³ (to 3 SF)

🧠 Exam Technique & Traps

Significant Figures: The question explicitly demands 3 significant figures. A final answer of 43.6 secures full marks.
Unit Conversions: Don't forget to multiply by 1000 when converting dm³ to cm³.
Rounding early: Keep unrounded values in your calculator until the final step to avoid rounding errors.

🎯 3 Marks: 1 mark for moles of Ni, 1 mark for moles of HCl (using 1:2 ratio), 1 mark for correct volume evaluated to 3 significant figures.

Question Part (b)(ii)

Calculation: Volume of Hydrogen Gas

📐 Step-by-Step Calculation

From the balanced equation, 1 mole of Ni produces 1 mole of H₂.
n(H₂) = n(Ni) = 3.27 × 10⁻³ mol
Volume at RTP = moles × 24000 cm³
Volume = 3.271 × 10⁻³ × 24000 = 78.5 cm³ (Allow ECF from incorrect n(Ni) in b(i), accepting 78.5 to 78.48 cm³).

❌ Common Errors

Forgetting the molar volume constant at RTP ( 24000 cm³ mol⁻¹ vs 24 dm³ mol⁻¹ ) or incorrectly applying the gas stoichiometry ratio.

🎯 1 Mark: 1 mark for correct calculation of H₂ volume using molar gas volume at RTP.

Question Part (b)(iii)

Limiting Reagents: Magnesium vs Nickel

✅ Correct Answer

The volume of H₂ produced is the same because magnesium is in excess (or the volume of H₂ depends strictly on the HCl, making HCl the limiting reagent).

💡 Key Knowledge

Both Mg and Ni react with acids in a 1:2 mole ratio with HCl (Mg + 2HCl → MgCl₂ + H₂). Since the moles of metal (0.192 g / 24.3 g mol⁻¹ = 0.0079 mol for Mg vs 0.00327 mol for Ni) are greater than the nickel moles while using the exact same amount of HCl, acid remains the limiting factor determining product yield.

🎯 2 Marks: 1 mark for stating "the same", 1 mark for explaining that Mg is in excess / HCl is the limiting reagent.

Question Part (c)*

Qualitative Analysis: Identifying Anions

💡 Chemical Tests & Observations

1. Carbonate test: Add dilute acid (e.g., HNO₃ , HCl ). Observation: Effervescence / bubbling / gas turns limewater cloudy ( CO₂ ).
2. Sulfate test: Add aqueous barium ions ( Ba(NO₃)₂ or BaCl₂ ) in the presence of acid. Observation: White precipitate forms ( BaSO₄ ).
3. Bromide test: Add aqueous silver nitrate ( AgNO₃ ), followed by dilute/concentrated ammonia if needed. Observation: Cream precipitate forms ( AgBr ).

🧠 Structuring Top-Level Responses (Level 3)

To achieve top marks (5–6 marks), candidates must provide a logical testing sequence (Carbonate first, because barium carbonate is also a white precipitate and would give a false positive for sulfate!). Include correct observations for all three tests and back them up with valid ionic or full chemical equations.

📝 Required Equations (Any valid ionic or full equation accepted)

Carbonate: 2H⁺(aq) + CO₃²⁻(aq) → H₂O(l) + CO₂In(g) or 2HCl + NiCO₃ → NiCl₂ + H₂O + CO₂
Sulfate: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
Bromide: Ag⁺(aq) + Br⁻(aq) → AgBr(s)

🎯 6 Marks (Level of Response): Assessed via scientific communication, logical sequencing (testing carbonate first), correct observations, and accurate equations.

Topics

Module 2: Foundations in chemistry · Module 3: Periodic table and energy · Module 1: Development of practical skills in chemistry · Practical Activity Groups · PAG 4: Qualitative analysis of ions · 2.1 Atoms and reactions · 3.1 The periodic table

Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.