AQA A-Level Chemistry Paper 1, 2017: Question 8

Mark scheme

Show the mark scheme

Question Answers Mark Additional Comments/Guidance

BaCl2 / Ba(OH)2 / Ba(NO3)2 / BaX2 or names Ignore acidification but CE = 0/3 if H2SO4

1 If reagent incorrect or blank then CE =0/3

If Ba2+ or wrong formula, lose M1 and mark on

08.1 colourless solution / no (visible) change (nvc) / no ppt / no

1 Ignore nothing happens and no observation

(visible) reaction

white precipitate / white solid

NaOH / sodium hydroxide / other Group 1 hydroxides 1 If reagent incorrect or blank then CE =0/3

If reagent incomplete, lose M1 and mark on

white precipitate / white solid 1

(white) ppt which dissolves in excess (NaOH) If reagent is excess NaOH, allow colourless solution for M3

08.2 Alternative Method

Name or formula of Group 1 carbonate

white precipitate / white solid

(white) precipitate and effervescence 1

Total 6

How to answer it

Ion Tests: Distinguishing Anions and Cations (One Test-Tube Reaction)

What this question tests

Choosing a single suitable qualitative reagent to distinguish ions.
Knowing key observations for sulfate (SO₄²⁻) vs nitrate (NO₃⁻) and for Mg²⁺ vs Al³⁺.
Writing answers in the expected “Reagent + observation for each solution” format to secure method marks.
Using “add dropwise then in excess” where needed (still counts as one test-tube reaction).

Marking style: typically 1 mark for the correct reagent, then 1 mark per correct observation (total 3 marks per part).

Part (a) / Q08.1 — Distinguish KNO₃(aq) from K₂SO₄(aq)

Use one test-tube reaction to tell nitrate from sulfate

✅ Correct answer (what to write)

Prompt	Answer	Marks
Reagent	Add BaCl₂(aq) (or Ba(NO₃)₂(aq) / Ba(OH)₂(aq) / any soluble Ba²⁺ source)	M1
Observation with KNO₃(aq)	No visible change / remains colourless / no precipitate	M2
Observation with K₂SO₄(aq)	White precipitate (BaSO₄(s))	M3

Examiner note: the mark scheme allows BaCl₂ / Ba(OH)₂ / Ba(NO₃)₂ / “BaX₂” (names accepted).

💡 Key knowledge

Sulfate ions form an insoluble white precipitate with Ba²⁺: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
Nitrates are soluble: adding Ba²⁺ to NO₃⁻ gives no precipitate, so “no visible change”.

🧠 Exam technique (how to secure 3/3)

Write in the exact order the question provides: reagent, then observation for each solution.
Use “white precipitate” not just “forms a precipitate” (colour matters).
For the nitrate, say no precipitate / no visible change (don’t leave it blank).

❌ Common errors (seen by examiners)

Using H₂SO₄ as the reagent: the mark scheme explicitly says 0/3 if H₂SO₄ is used (it would introduce sulfate and confuse the test).
Writing just “Ba²⁺” without a reagent (the paper asks for a reagent you would add in a test tube).
Giving “nothing happens” without a clear observation (the scheme credits “no visible change / no ppt”; vague wording may not be credited).
Wrong ion or wrong formula (e.g. suggesting AgNO₃—this tests halides, not sulfate vs nitrate).

Part (b) / Q08.2 — Distinguish MgCl₂(aq) from AlCl₃(aq)

Use one test-tube reaction to tell Mg²⁺ from Al³⁺

✅ Correct answer (main method: NaOH)

Prompt	Answer	Marks
Reagent	Add NaOH(aq) (or another Group 1 hydroxide) dropwise, then in excess	M1
Observation with MgCl₂(aq)	White precipitate of Mg(OH)₂(s) forms (and does not dissolve in excess NaOH)	M2
Observation with AlCl₃(aq)	White precipitate forms, which dissolves in excess NaOH to give a colourless solution	M3

Examiner guidance: If you explicitly say “excess NaOH”, the scheme allows “colourless solution” as the final observation for Al³⁺.

💡 Key knowledge

Both Mg²⁺ and Al³⁺ form white hydroxide precipitates with OH⁻ initially.
Al(OH)₃ is amphoteric: it dissolves in excess OH⁻ to form soluble aluminate.
Mg(OH)₂ is not amphoteric: it stays as a white precipitate in excess OH⁻.

Helpful ionic idea (not required, but good understanding): Al(OH)₃(s) + OH⁻(aq) → [Al(OH)₄]⁻(aq)

🧠 Exam technique (how to hit the mark points)

Make it clear you do the test in one tube per solution using the same reagent.
Say “add NaOH dropwise then in excess” — this is how you unlock the dissolving difference.
Use the phrase “dissolves in excess” for Al³⁺: this is the discriminator.

❌ Common errors (why marks are lost)

Only stating “white precipitate” for both ions without mentioning excess NaOH — you would not distinguish them, so you lose the key mark.
Choosing the wrong reagent (examiner note: if the reagent is incorrect or blank, the response can score 0/3).
Saying Mg(OH)₂ dissolves in excess — it doesn’t.
Not describing the final observation (e.g. not saying it becomes a colourless solution for Al³⁺).

Alternative method (also credited by the mark scheme)

✅ Alternative: Group 1 carbonate

Reagent: a Group 1 carbonate (e.g. Na₂CO₃(aq))
With MgCl₂(aq): white precipitate / white solid
With AlCl₃(aq): white precipitate and effervescence

This is explicitly listed as an “Alternative Method” in the mark scheme.

💡 Why effervescence for Al³⁺?

Al³⁺ solutions are acidic due to hydrolysis; carbonate reacts to release CO₂(g), giving fizzing.
Mg²⁺ gives a carbonate/hydroxide-type precipitate but typically no fizzing in this context.

❌ Trap with this alternative

If you don’t mention effervescence for AlCl₃, you may not distinguish the two solutions.
Don’t call the gas “oxygen” — effervescence here is CO₂.

Quick checklist (full marks approach)

🧠 How to present your answer fast

Name the reagent clearly (correct formula or name).
State the observation for solution 1.
State the observation for solution 2.
Use distinctive words: white precipitate, no visible change, dissolves in excess.

✅ One-line model answers

08.1: Add BaCl₂(aq): KNO₃—no precipitate; K₂SO₄—white precipitate.
08.2: Add NaOH(aq) dropwise then excess: MgCl₂—white ppt insoluble in excess; AlCl₃—white ppt dissolves in excess to colourless solution.

❌ Last-minute “don’ts”

Don’t use H₂SO₄ for 08.1 (kills the question).
Don’t forget “excess” for 08.2.
Don’t write “forms a precipitate” without the colour.

Topics

Inorganic Chemistry · Required Practicals · 3.2.6 Reactions of Ions in Aqueous Solution · Required Practical 4: Carry out simple test-tube reactions to identify Cations and Anions