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TL;DR Three recurring QA mark-traps cost most students four to six marks across Paper 3. One: Cu²⁺, Zn²⁺ and Al³⁺ all give coloured or white precipitates when you add small amounts of NaOH or NH₃ - the identity is only confirmed by what happens in excess. Two: Fe²⁺ looks pale green but turns brown on standing because of atmospheric oxidation. Three: BaCl₂ gives a white precipitate with both SO₄²⁻ and SO₃²⁻ - only acidified BaCl₂ tells them apart.
When you add a small amount of NaOH or NH₃ to a cation solution, several ions produce precipitates of similar colour. Cu²⁺ and Fe²⁺ both form coloured precipitates; Zn²⁺ and Al³⁺ both form white precipitates. At this stage, the observation alone does not uniquely identify the ion. The mark scheme is structured so that the first step -- the initial precipitate -- typically earns one mark, while the behaviour in excess earns the second and more discriminating mark.
This is the disambiguation step. Examiners weight it heavily because it forces candidates to demonstrate they understand why each reagent is added sequentially, not just what colour appears. A student who writes "white precipitate formed" for both Zn²⁺ and Al³⁺ and nothing more cannot distinguish them. The student who adds "precipitate dissolves in excess NaOH to give a colourless solution" for Zn²⁺, and "precipitate remains insoluble in excess NH₃" for Al³⁺, has made the identification complete.
The practical implication in the exam is that you must always add enough of the reagent to constitute an excess, observe whether the precipitate dissolves or persists, and record both the initial observation and the outcome in excess as two separate lines. Skipping the second line is the single most expensive mark-trap in the QA section.
2 | The full "in excess" decision table
The table below covers the six cations most commonly assessed in O-Level 5073 Paper 3 qualitative analysis. Read across the row for each cation to find the four key observations and the single most diagnostic tell.
Cation
Small NaOH
Excess NaOH
Timings
Weekdays (first slot)
12 noon to 2pm
Weekdays (second slot)
2pm to 4pm
Weekends (first slot)
6pm to 8pm
Weekends (second slot)
8pm to 10pm
Pricing
A-LevelSGD 230per 2-hour session
Small NH₃
Excess NH₃
Key ID tell
Cu²⁺
Blue precipitate
Insoluble (precipitate remains)
Blue precipitate
Soluble; gives deep blue solution
Deep blue in excess NH₃ is unique to Cu²⁺
Zn²⁺
White precipitate
Soluble; gives colourless solution
White precipitate
Soluble; gives colourless solution
Soluble in both excess NaOH and excess NH₃
Al³⁺
White precipitate
Soluble; gives colourless solution
White precipitate
Insoluble (precipitate remains)
Soluble in excess NaOH only, not in excess NH₃
Ca²⁺
White precipitate
Insoluble
No precipitate (or very slight)
No precipitate
No precipitate with NH₃; insoluble in excess NaOH
Fe²⁺
Green precipitate
Insoluble
Green precipitate
Insoluble
Pale green colour; turns brown on standing in air
Fe³⁺
Red-brown precipitate
Insoluble
Red-brown precipitate
Insoluble
Red-brown on first addition; no colour change in excess
Three pairs in this table are the source of the most common errors:
Cu²⁺ vs Fe²⁺ with NaOH: Both give coloured precipitates on adding small NaOH -- blue for Cu²⁺, green for Fe²⁺. The colours are distinct if you know them, but under exam pressure candidates sometimes confuse blue-green shades. The definitive check is NH₃ in excess: only Cu²⁺ dissolves to give the deep blue tetraamminecopper(II) complex.
Zn²⁺ vs Al³⁺ with NaOH: Both give white precipitates, both dissolve in excess NaOH. The test that separates them is excess NH₃: Zn²⁺ precipitate dissolves; Al³⁺ precipitate stays.
Ca²⁺ vs other white-precipitate cations: Ca²⁺ behaves differently from Zn²⁺ and Al³⁺ because its precipitate does not dissolve in excess NaOH, and it gives little or no precipitate with NH₃ at typical exam concentrations.
3 | The Fe²⁺ browning mechanism
Iron(II) hydroxide is pale green. Left standing at room temperature in contact with air, it oxidises to iron(III) hydroxide, which is red-brown. The equation is:
4Fe(OH)2(s)+O2(g)+2H2O(l)→4Fe(OH)3(s)
Examiners treat this as a temporal observation -- an observation that occurs over a short time interval after the initial precipitate forms, not at the moment of addition. The mark scheme typically awards a separate mark for stating "green precipitate turns brown on standing" or "green precipitate that turns reddish-brown in air."
This has two practical consequences for how you write your answer. First, you must record the initial colour (green) and the subsequent colour change (brown) as two distinct observations rather than collapsing them into one. Second, you must attribute the change to standing in air (or exposure to atmospheric oxygen) rather than to the addition of more reagent.
Students who write only "brown precipitate" lose the Fe²⁺ identification mark entirely because red-brown describes Fe³⁺ from the outset. The pale green colour is the Fe²⁺ tell; the browning is the confirmatory secondary observation that demonstrates you recognised the oxidation.
A quick lab habit that helps: when you add NaOH to an Fe²⁺ solution, note the colour immediately and again after 20--30 seconds. If it has shifted towards red-brown, you have confirmed both the identity and the mechanism.
4 | The SO₄²⁻ vs SO₃²⁻ distinction
Both sulfate and sulfite ions form a white precipitate when BaCl₂ solution is added. Many candidates stop there, write "SO₄²⁻ confirmed", and lose the mark for the anion test entirely.
The correct procedure has two steps:
Add acidified barium chloride solution (BaCl₂ acidified with dilute HCl). A white precipitate forms with both SO₄²⁻ and SO₃²⁻.
Add excess dilute hydrochloric acid and observe whether the precipitate dissolves.
If the precipitate dissolves: SO₃²⁻ is present. Barium sulfite (BaSO₃) is slightly soluble in dilute HCl.
If the precipitate does not dissolve: SO₄²⁻ is present. Barium sulfate (BaSO₄) is insoluble in dilute HCl and is one of the most insoluble salts in the SEAB syllabus.
There is a second confirmatory test for SO₃²⁻ that examiners reward: adding acidified potassium manganate(VII) (KMnO₄) solution. The purple KMnO₄ decolourises in the presence of SO₃²⁻ because sulfite is a reducing agent. This test is particularly useful when the BaCl₂ precipitate is ambiguous in quantity or colour.
The reason the BaCl₂ must be acidified is to prevent interference from carbonate ions, which would also produce a white precipitate with unacidified BaCl₂. Acidifying with dilute HCl drives off any CO₂ from carbonate, ensuring the precipitate is specific to sulfate or sulfite.
5 | Observation-language templates
The following phrasings reflect the model answers found in SEAB mark schemes. Reproducing these phrases -- not paraphrasing them -- is the most reliable way to secure PDO marks under time pressure.
Cu²⁺ with NaOH:
"Blue precipitate formed; insoluble in excess aqueous sodium hydroxide."
Cu²⁺ with NH₃:
"Blue precipitate formed; soluble in excess aqueous ammonia to give a deep blue solution."
Zn²⁺ with NaOH:
"White precipitate formed; soluble in excess aqueous sodium hydroxide to give a colourless solution."
Zn²⁺ with NH₃:
"White precipitate formed; soluble in excess aqueous ammonia to give a colourless solution."
Al³⁺ with NaOH:
"White precipitate formed; soluble in excess aqueous sodium hydroxide to give a colourless solution."
Al³⁺ with NH₃:
"White precipitate formed; insoluble in excess aqueous ammonia."
Fe²⁺ with NaOH:
"Green precipitate formed; insoluble in excess aqueous sodium hydroxide; precipitate turns brown on standing."
Fe³⁺ with NaOH:
"Red-brown precipitate formed; insoluble in excess aqueous sodium hydroxide."
SO₄²⁻ test:
"White precipitate formed with acidified barium chloride solution; precipitate insoluble in excess dilute hydrochloric acid."
SO₃²⁻ test:
"White precipitate formed with acidified barium chloride solution; precipitate dissolves on addition of excess dilute hydrochloric acid."
NH₄⁺ test (included briefly):
On warming with excess aqueous NaOH, a pungent gas is produced that turns damp red litmus paper blue. This identifies NH₄⁺. Note that "pungent smell" alone earns partial credit, but the litmus observation is required for the identification mark.
6 | How to structure a QA answer under time pressure
There is a persistent misconception about QA answer format. Many students believe the correct approach is to state the ion first and then justify it: "Cu²⁺ is present because a blue precipitate formed." This is the ion-first structure, and it is wrong in most contexts.
The SEAB mark scheme defaults to the observation-first structure. You record what you saw, then deduce what ion is present based on that evidence. The examiner awards marks for the observation and then for the inference. If you state the ion before the observation, and your observation language is weak or missing, you receive zero for both lines.
The correct structure for each row in your QA table:
Reagent added
Observation
Inference
Aqueous NaOH (small amount)
Blue precipitate formed
Cu²⁺ may be present
Aqueous NaOH (excess)
Precipitate insoluble in excess
Confirms Cu²⁺ (not Zn²⁺ or Al³⁺)
Aqueous NH₃ (excess)
Precipitate dissolves; deep blue solution formed
Cu²⁺ confirmed
The inference column uses escalating certainty language: "may be present" after an initial test, then "confirmed" after a definitive excess test. Writing "confirmed" after only one test is premature and examiners are trained to spot it.
The one exception to observation-first is when the question explicitly asks "identify the cation in the solution" as a standalone question rather than a table completion. In that case, you state the ion and then give the evidence. But in a table, always observation first.
7 | Common wrong answers that look right
"Precipitate dissolves" without naming the reagent
This phrasing appears in roughly half of all under-marked QA scripts. The observation "precipitate dissolves" is meaningless without specifying which reagent causes the dissolution. "Precipitate dissolves in excess NaOH" earns the mark; "precipitate dissolves" does not. Examiners cannot award marks for ambiguous observations because the dissolution behaviour is the entire basis of the distinction between ions.
"Solution turns blue" with no mention of precipitate disappearance
This error is specific to the Cu²⁺ excess ammonia test. When the blue Cu(OH)₂ precipitate dissolves in excess NH₃, two things happen simultaneously: the precipitate disappears and the solution turns a deep, characteristic blue. Describing only the colour change, without confirming that the precipitate has dissolved, loses the observation mark because the colour alone could refer to the initial precipitate remaining in suspension.
The correct phrasing explicitly links both: "Blue precipitate dissolves; solution turns deep blue."
Writing "brown precipitate" for Fe²⁺
As explained in Section 3, Fe²⁺ initially gives a green (or pale green) precipitate. Writing "brown precipitate" from the outset describes Fe³⁺ and will not earn the Fe²⁺ identification mark.
"White precipitate with BaCl₂" without the acid step
Writing this without the follow-up acid test leaves the anion unidentified. The mark scheme for a complete sulfate identification requires both the precipitate observation and the stability in acid. A white precipitate with unacidified BaCl₂ is not even a reliable sulfate test.
Omitting "in excess" from the inference
Phrasing such as "white precipitate soluble in NaOH" without specifying "excess NaOH" is technically ambiguous because all precipitates are slightly soluble in trace quantities of any reagent. The word "excess" is the qualifier that specifies the test condition.
8 | Mini-drill: five unknown-identification problems
Work through each problem in sequence. Read the question, write your observation and deduction before looking at the reveal.
Problem (a): Unknown solution containing Cu²⁺
Question: You add aqueous NaOH dropwise to solution A until it is in excess. Describe what you observe at each stage and state your deduction.
Observation: Blue precipitate forms on addition of small NaOH. On adding excess NaOH, the precipitate remains; it does not dissolve.
Deduction: Cu²⁺ is present. The precipitate is Cu(OH)₂. Insolubility in excess NaOH distinguishes Cu²⁺ from Zn²⁺ and Al³⁺, both of which dissolve. To confirm, add excess aqueous ammonia: the blue precipitate should dissolve to give a deep blue solution.
Problem (b): Distinguishing Zn²⁺ from Al³⁺
Question: Solution B and solution C both give white precipitates with aqueous NaOH and both dissolve in excess NaOH. Describe a single further test that distinguishes them and state the expected observations for each.
Test: Add aqueous ammonia in excess.
Observation for Zn²⁺: White precipitate forms; dissolves in excess aqueous ammonia to give a colourless solution.
Observation for Al³⁺: White precipitate forms; insoluble in excess aqueous ammonia (precipitate remains).
Deduction: If the precipitate dissolves in excess NH₃, Zn²⁺ is present. If it remains, Al³⁺ is present.
Problem (c): Fe²⁺ with the browning observation
Question: You add aqueous NaOH to solution D and observe a green precipitate. The precipitate is insoluble in excess NaOH. Thirty seconds later, the precipitate has changed colour. Identify the cation and explain the colour change.
Observation: Green precipitate formed on addition of aqueous NaOH; insoluble in excess; precipitate turns brown (reddish-brown) on standing in air.
Deduction: Fe²⁺ is present. The green precipitate is iron(II) hydroxide, Fe(OH)₂. It oxidises in atmospheric oxygen to form iron(III) hydroxide, Fe(OH)₃, which is red-brown. The equation is:
4Fe(OH)2(s)+O2(g)+2H2O(l)→4Fe(OH)3(s)
Problem (d): SO₄²⁻ vs SO₃²⁻
Question: Both solution E and solution F give a white precipitate when acidified barium chloride solution is added. Describe how you would distinguish between them using one additional step.
Test: Add excess dilute hydrochloric acid to each precipitate and observe.
Observation for SO₄²⁻: White precipitate remains; insoluble in excess dilute HCl.
Observation for SO₃²⁻: White precipitate dissolves in excess dilute HCl.
Alternative test for SO₃²⁻: Add acidified aqueous KMnO₄. Purple colour of KMnO₄ decolourises in the presence of SO₃²⁻ (which acts as a reducing agent). No decolourisation with SO₄²⁻.
Problem (e): Combined cation and anion -- unknown salt solution containing Fe²⁺ and SO₄²⁻
Question: Solution G is a pale green solution. You are given aqueous NaOH, aqueous NH₃, and acidified BaCl₂. Describe how you would identify both the cation and the anion, giving expected observations and inferences at each step.
Step 1 -- Cation test: Add aqueous NaOH to a portion of solution G.
Observation: Green precipitate forms; insoluble in excess NaOH; precipitate turns brown on standing.
Inference: Fe²⁺ is present (iron(II) hydroxide forms, then oxidises to iron(III) hydroxide).
Step 2 -- Confirm cation: Add aqueous NH₃ to a fresh portion.
Observation: Green precipitate forms; insoluble in excess NH₃; again turns brown on standing.
Inference: Fe²⁺ confirmed (insoluble in both excess NaOH and excess NH₃; rules out Cu²⁺, Zn²⁺, Al³⁺).
Step 3 -- Anion test: Add acidified BaCl₂ to a fresh portion.
Observation: White precipitate forms; insoluble in excess dilute HCl.
Inference: SO₄²⁻ is present. Salt is iron(II) sulfate, FeSO₄.
9 | Where this fits
This post addresses the in-excess mark-trap in detail. For the complete QA reference table covering all common cations and anions, see the O-Level Chemistry Qualitative Analysis Toolkit.
