Qualitative Analysis — Exam Skills

This guide uses a real CIE exam question (9701/34, Oct/Nov 2020, Question 3) to show exactly how to write observations, carry out further tests, write ionic equations, and avoid the most penalised mistakes. FB 4 is Zn(NO₃)₂(aq) and FB 5 is Na₂SO₃(aq).

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01 — Writing observations

The mark scheme awards marks for specific, precise descriptions of what you see — not what you conclude. Every observation must be written at the correct stage and use the right vocabulary.

Question 3(a) — Test 1 Add an equal volume of aqueous sodium carbonate to FB 4 (Zn(NO₃)₂) and FB 5 (Na₂SO₃). Record your observations.

✕ Wrong

FB 4: zinc carbonate precipitate forms
FB 5: no reaction

✓ Correct

FB 4: white precipitate forms
FB 5: no reaction / no change / no ppt

Zn²⁺(aq) + CO₃²⁻(aq) → ZnCO₃(s)

Why "Zinc carbonate precipitate" is a conclusion, not an observation. The mark scheme awards the star (*) for "white precipitate" — the colour is the observation. "No ppt" alone is sufficient for FB 5. Note: the mark scheme rejects "white ppt soluble in excess" here because Na₂CO₃ does not dissolve precipitates.

Question 3(a) — Test 2 Add aqueous ammonia to FB 4 and FB 5.

✕ Wrong

FB 4: white precipitate
FB 5: no change

✓ Correct

FB 4: white precipitate forms and dissolves / is soluble in excess NH₃
FB 5: no reaction / no change / no ppt

Zn²⁺(aq) + 2NH₃(aq) + 2H₂O(l) → Zn(OH)₂(s) + 2NH₄⁺(aq) Zn(OH)₂(s) + 4NH₃(aq) → [Zn(NH₃)₄]²⁺(aq) + 2OH⁻(aq) (dissolves in excess)

Why The mark scheme requires both — "white precipitate" AND "soluble in excess" — for the full mark. Writing only "white precipitate" misses what makes Zn²⁺ distinctive. The solubility in excess NH₃ is the key distinguishing feature here. Both observations carry a star (*) and two * = 1 mark (round down).

Question 3(b) — NaOH then heat then Al foil Add aqueous sodium hydroxide, warm gently, then add aluminium foil. Record observations at each stage for FB 4 and FB 5. This is a 3-stage test — you must report observations at each stage separately.

Stage	FB 4 (Zn²⁺) — correct observation	FB 5 (SO₃²⁻) — correct observation
Add NaOH	White precipitate, soluble in excess `Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s)` `Zn(OH)₂(s) + 2OH⁻(aq) → [Zn(OH)₄]²⁻(aq) (excess)`	No change / no ppt
Warm	Ignore — mark not available for heating stage alone
Add Al foil	Effervescence / fizzing — gas turns red litmus blue (NH₃) `8Al(s) + 3NO₃⁻(aq) + 5OH⁻(aq) + 2H₂O(l) → 8AlO₂⁻(aq) + 3NH₃(g)` `NH₃ turns damp red litmus blue ✓`	Effervescence / fizzing / bubbling — gas pops with lighted splint (H₂) `2Al(s) + 2OH⁻(aq) + 2H₂O(l) → 2AlO₂⁻(aq) + 3H₂(g)` `H₂ gives squeaky pop with lighted splint ✓`

Why different gases? The anion determines the gas — not the cation. FB 4 contains NO₃⁻, which is reduced by Al in alkaline solution to produce NH₃. FB 5 contains SO₃²⁻, which is not reduced under these conditions — instead Al simply reacts with the excess NaOH to produce H₂. This is why the Al foil test can distinguish NO₃⁻ from SO₃²⁻.

Critical: If you reported gas (NH₃ or any other gas) during the heating stage alone, the gas test observation in the Al foil stage is rejected. Always say "more effervescence" when adding Al foil if you already noted bubbling on heating.

Common mistake: Writing "gas turns litmus blue" for FB 5. FB 5 contains SO₃²⁻ (no nitrogen), so it cannot produce NH₃. The gas from FB 5 + NaOH + Al is H₂ — tested with a lighted splint (squeaky pop), not litmus.

02 — Further tests (part c)

Part (c) asks you to select one further test that distinguishes the anions in FB 4 and FB 5, explain why it works, and record your observations. This is a 3-mark question requiring a reagent, an explanation, and observations.

Question 3(c) Using the information given (nitrite/sulfite easily oxidised; nitrate/sulfate cannot be oxidised), select one further test to identify the anions in FB 4 and FB 5. State the reagent, explain why it works, and record observations.

What the redox table tells you: From tests 3(a) and 3(b) you already know FB 4 contains Zn²⁺ (white ppt with NH₃, soluble in excess) and FB 5 contains SO₃²⁻ or SO₄²⁻ (white ppt with Ba²⁺). The redox table is the key — sulfite is easily oxidised, sulfate cannot be.

Approach A — acidified KMnO₄

✕ Wrong

Reagent: KMnO₄
Explanation: it oxidises sulfite
Observation: colour changes in FB 5

✓ Correct

Reagent: acidified potassium manganate(VII) / acidified KMnO₄
Explanation: SO₃²⁻ is easily oxidised and decolourises KMnO₄; NO₃⁻ cannot be oxidised so causes no colour change — this distinguishes the two anions
Observation: FB 5 decolourises the purple KMnO₄; FB 4 no change

5SO₃²⁻(aq) + 2MnO₄⁻(aq) + 6H⁺(aq) → 5SO₄²⁻(aq) + 2Mn²⁺(aq) + 3H₂O(l)

Why the wrong answer fails "KMnO₄" alone is not accepted — it must be acidified. The explanation must link the observation to the redox property in the table: SO₃²⁻ is easily oxidised (decolourises KMnO₄); NO₃⁻ cannot be oxidised (no change). You must describe the contrast between both solutions, not just the positive result for FB 5.

Approach B — acidified K₂Cr₂O₇

✕ Wrong

Reagent: K₂Cr₂O₇
Explanation: dichromate oxidises sulfite
Observation: turns green in FB 5

✓ Correct

Reagent: acidified potassium dichromate(VI) / acidified K₂Cr₂O₇
Explanation: SO₃²⁻ is easily oxidised and reduces Cr₂O₇²⁻ from orange to green; NO₃⁻ cannot be oxidised so no colour change
Observation: FB 5 turns orange solution green; FB 4 no change

3SO₃²⁻(aq) + Cr₂O₇²⁻(aq) + 8H⁺(aq) → 3SO₄²⁻(aq) + 2Cr³⁺(aq) + 4H₂O(l)

Why the wrong answer fails Same issue as KMnO₄ — must specify acidified. The colour change must be stated precisely: orange → green (Cr⁶⁺ → Cr³⁺). Writing "colour changes" without specifying the colours scores no mark for the observation.

Both approaches are accepted by the mark scheme. KMnO₄ and K₂Cr₂O₇ are equally valid. In both cases the key word is acidified — the oxidising power of both reagents depends on H⁺ ions being present.

03 — Golden rules

These are the recurring reasons students lose marks in qualitative analysis questions, drawn directly from mark scheme reject notes.

Observations

✕

Never write conclusions as observations. "Zinc ions present" is a conclusion. "White precipitate forms" is an observation. The mark is for what you see, not what you deduce.
✕

Never omit the solubility in excess. When a precipitate dissolves in excess reagent, you must say so. "White precipitate" alone misses half the observation for many cation tests.
✕

Never report gas during heating if you use it to identify the ion. Gas identity must be confirmed by a separate test (litmus, splint).
✓

Always write "more" effervescence when adding Al foil if bubbling was already seen on heating. The mark scheme requires "more" to distinguish the two stages.
✓

Write observations for both solutions. Even if one shows no change, "no reaction / no change / no ppt" must be stated — it carries a mark star (*) just like a positive result.

Reagents and tests

✕

Never use H₂SO₄ to distinguish SO₄²⁻ from SO₃²⁻. Sulfuric acid introduces SO₄²⁻ ions, creating a false result. Always use HCl or HNO₃.
✕

Never use KMnO₄ without specifying acidified. Acidified potassium manganate(VII) is the correct reagent. Unacidified KMnO₄ is not accepted.
✓

Always give the full reagent name or correct formula including its state — "BaCl₂(aq)" not just "barium" or "Ba²⁺" alone.

Mark scheme vocabulary — words that cost marks

If you write...	Mark scheme says...	Fix
"zinc ions present"	Reject — conclusion not observation	"white precipitate forms"
"white precipitate" (for NH₃ test, no more)	Only half credit — solubility missing	"white precipitate, soluble in excess"
"soluble in excess" for Na₂CO₃ test	Reject — Na₂CO₃ doesn't dissolve ppt	Just "white precipitate"
"add H₂SO₄" to distinguish sulfate/sulfite	Reject — introduces SO₄²⁻	"add dilute HCl or dilute HNO₃"
Ionic equation without state symbols	Zero marks for state symbol component	Add (aq) and (s) to every species