SKY HSC · Mod 8 IQ1
Module 8 · IQ1 · Dot point 2 (M8-IQ1-d2)

Identifying Ions — Flame, Precipitation & Complexation

A diagnostic walk-through for every NESA cation and anion: which test, what you'd see, and the net ionic equation markers reward. Two interactive flowcharts + real HSC model answers.

🧪 8 cations + 8 anions 📊 223 papers · 402 Qs (Second Brain) ⚗ flame · precipitation · complexation

🌱 First time

Read Part 1 principles → walk the cation and anion flowcharts → test yourself on the checkpoints. ~40 min.

🔁 Revision

Skim the TL;DR → drill the two master tables + flashcards → MCQ quiz. ~25 min.

⏰ Exam-eve

Cheat sheet + pre-exam checklist → re-read the model answers. ~12 min.

TL;DR — what every student must know

Three NESA test typesflame test, precipitation, complexation. The question tells you which is "appropriate".
Flame: only 3 usefulBa²⁺ green · Ca²⁺ brick-red · Cu²⁺ blue-green. Mg²⁺, Ag⁺, Fe²⁺/Fe³⁺ give no diagnostic flame; Pb²⁺ isn't flame-tested.
Cation sequenceadd HCl → H₂SO₄ → NaOH, knocking out a group each step (see the flowchart).
Anion sequenceacid (CO₃²⁻) → Ba²⁺ (SO₄²⁻/PO₄³⁻) → litmus → Ag⁺ (halides) / Cu²⁺ (OH⁻/acetate).
Complexation = dissolvingAgCl(s) + 2NH₃ → [Ag(NH₃)₂]⁺. Solubility AgCl(dilute NH₃) > AgBr(conc NH₃) > AgI(insoluble) separates the halides. ⚗
Alwaysstate the reagent, the precise colour, a net ionic equation with states, and run tests in an order that removes interferences.

⚡ The 5-second ID rule

  1. Cation? Add HCl → ppt = Ag⁺/Pb²⁺. No ppt → add H₂SO₄ → ppt = Ba²⁺/Ca²⁺ (flame to split). No ppt → add NaOH and read the precipitate colour.
  2. Anion? Add acid → bubbles = CO₃²⁻. Add Ba²⁺ → ppt = SO₄²⁻/PO₄³⁻ (acid splits). No ppt → litmus (OH⁻/acetate vs halides) → Ag⁺ + ammonia.
  3. Always close with: reagent → observation (colour!) → net ionic equation (with states). That's the full-mark shape. (A net ionic equation shows only the ions that actually react — the spectator ions are left out.)

Part 1 · The principles behind every ion test

1.1 Qualitative vs quantitative

Qualitative analysis answers which ion is present (its identity). Quantitative answers how much (concentration). This dot point is purely qualitative — you identify ions; you don't measure amounts. The three NESA-named qualitative techniques are flame tests, precipitation reactions and complexation reactions.

1.2 Solubility rules — the SKY 3-group version

A precipitation test only works when the salt formed is insoluble. SKY teaches a 3-group version of the solubility rules — more accurate and exam-safe: memorise which group an ion sits in, then the exceptions in brackets.

① Always soluble

Cations: Group 1 (Li⁺ Na⁺ K⁺ Rb⁺ Cs⁺) & ammonium NH₄⁺

Anion: nitrate NO₃⁻

Exceptions: Li₃PO₄ insol (minor)

These ions never form a precipitate.

② Mostly soluble

Cl⁻ Br⁻ I⁻ — except Ag⁺Pb²⁺

SO₄²⁻ — except Ba²⁺Pb²⁺Ag⁺ slCa²⁺ sl

CH₃COO⁻ (acetate) — except Ag⁺Fe³⁺

Lead halides dissolve in hot water; Ag⁺-acetate precipitates only when concentrated.

③ Mostly insoluble

OH⁻ — soluble for Group 1 & NH₄⁺, and Ba²⁺ sol (Ca²⁺ sl)

CO₃²⁻ & PO₄³⁻ — soluble for Group 1 & NH₄⁺ only

⚡ Carbonate & phosphate dissolve in acid (pH < 2) — the key SO₄²⁻-vs-PO₄³⁻/CO₃²⁻ trick.

🔑 How to use the 3 groups

Every prediction is the same two-step lookup: find the group → check the bracket.

  1. A precipitate means the salt is insoluble.
    • Step 1 — find which group the ion is in.
    • Step 2 — check the exceptions listed in the bracket.
    • Worked logic — "Will Ba²⁺ + SO₄²⁻ give a precipitate?"
    • → Sulfate sits in group ② (mostly soluble) …
    • → but the bracket lists Ba²⁺ as an exception …
    • → so BaSO₄ is insoluble = a white precipitate forms ✓
  2. Group ① ions are your "spectator" reagents.
    • Na⁺, K⁺, NH₄⁺ and NO₃⁻ never form a precipitate.
    • So you use them to deliver a test ion without any interference:
    • NaCl → delivers Cl⁻ · Na₂SO₄ → delivers SO₄²⁻ · NaOH → delivers OH⁻ · AgNO₃ → delivers Ag⁺
  3. To identify a cation: add an anion it is insoluble with.
    • (an exception listed in group ②, or any group ③ anion)
    • e.g. SO₄²⁻ pulls down Ba²⁺ and Pb²⁺ · OH⁻ pulls down Cu²⁺, Fe²⁺/Fe³⁺ and Mg²⁺
  4. To identify an anion: add a cation it is insoluble with.
    • e.g. Ag⁺ → AgCl / AgBr / AgI · Ba²⁺ → BaSO₄
  5. The acid trick (group ③).
    • CO₃²⁻ and PO₄³⁻ dissolve in acid; BaSO₄ does not.
    • So add dilute acid to a white Ba precipitate and watch it:
    • • stays (no change) = SO₄²⁻
    • • fizzes (gas given off) = CO₃²⁻
    • • dissolves with no gas = PO₄³⁻
Worked example — "A solution gives a white precipitate with Ba²⁺. Which anion is it?"
  1. List the suspects.
    • Sulfate, carbonate and phosphate are all insoluble with Ba²⁺ — so any of the three could be the white precipitate.
  2. Add dilute HNO₃ and watch the precipitate:
    • Bubbles (a gas is given off) → it was CO₃²⁻ (carbonate).
    • Dissolves, no gas → it was PO₄³⁻ (phosphate).
    • Stays (no change) → it was SO₄²⁻ (sulfate) — because BaSO₄ is acid-insoluble.
Extension — why phosphate dissolves in acid but sulfate doesn't
In acid, phosphate is protonated: PO₄³⁻ + H⁺ ⇌ HPO₄²⁻ lies far right (the reverse ionisation HPO₄²⁻ ⇌ PO₄³⁻ + H⁺ has Ka3 ≈ 4.8×10⁻¹³, i.e. tiny), so at pH < 2 almost no free PO₄³⁻ remains to precipitate Ba₃(PO₄)₂. Sulfate's equilibrium (HSO₄⁻ ⇌ SO₄²⁻ + H⁺, Ka ≈ 1×10⁻²) lies much further right, so plenty of SO₄²⁻ remains → BaSO₄ still precipitates. Raising pH to 10–11 with ammonia frees the PO₄³⁻ again.

1.3 The three test types

🔥 Flame testHeat the ion in a flame: electrons absorb energy, jump to higher levels, then emit a characteristic colour as they fall back. Only Ba²⁺ (green), Ca²⁺ (brick-red) and Cu²⁺ (blue-green) are exam-useful.
💧 PrecipitationAdd a reagent that forms an insoluble salt with the target ion. The precipitate's colour often identifies the ion (e.g. blue Cu(OH)₂, red-brown Fe(OH)₃).
⚗ ComplexationForm a soluble complex ion — e.g. a silver-halide precipitate dissolving in ammonia: AgCl(s) + 2NH₃(aq) → [Ag(NH₃)₂]⁺(aq) + Cl⁻(aq). Used to separate Cl⁻/Br⁻/I⁻ and to confirm Cu²⁺ and Fe³⁺.
Flame-test reality check: markers penalise students who invent flame colours. Write a flame colour only for Ba²⁺ (green), Ca²⁺ (brick-red) or Cu²⁺ (blue-green). For Mg²⁺, Ag⁺, Fe²⁺, Fe³⁺ (and Pb²⁺) the flame test is not diagnostic — identify these by precipitation instead. The exception: Group 1 cations (Na⁺ yellow, K⁺ lilac, Li⁺ crimson) have only-soluble salts, so a flame test is the only school-lab way to detect them.

🔥 Flame colours — click an ion to see it

These are the colours to memorise. Only Ba²⁺, Ca²⁺ and Cu²⁺ are exam-useful among the syllabus cations; Na⁺/K⁺/Li⁺ are how you detect Group 1.

Pale / apple green
Ba²⁺ — exam-useful

💧 Precipitate colours — click a tube for its equation

The colour of the precipitate is usually what identifies the ion. Memorise these.

Cu(OH)₂blue
Fe(OH)₂green
Fe(OH)₃red-brown
Mg(OH)₂white
Ag₂Obrown
AgClwhite
AgBrcream
AgIyellow
PbI₂golden-yellow
BaSO₄white
👆 Click a test tube to see its net ionic equation.

✅ Checkpoint 1 ⏱ try first — 3 min

Q1. Why is adding NaCl almost useless for identifying an unknown cation? Q2. Which cations give no useful flame colour, and why does that matter?

Model answers
A1. NaCl supplies two ions — Na⁺ and Cl⁻. Na⁺ is group ① (always soluble), so it precipitates nothing. Cl⁻ is group ②, and its only exceptions are Ag⁺ and Pb²⁺ — so chloride only precipitates those two. Therefore, unless the unknown cation is Ag⁺ or Pb²⁺, adding NaCl gives no precipitate and no information. A2. Mg²⁺, Ag⁺ and both irons (Fe²⁺/Fe³⁺) give no diagnostic flame colour — you must identify them by precipitation (the NaOH colour, or Cl⁻/I⁻). Writing a "flame colour" for these loses marks.

Part 2 · Identifying the cations

Eight cations — Ag⁺, Pb²⁺, Ba²⁺, Ca²⁺, Cu²⁺, Fe²⁺, Fe³⁺, Mg²⁺. The flowchart adds reagents that knock out a group at a time (HCl → H₂SO₄ → NaOH). Hover or click any ion to light up its test path.

🧪 How SKY runs the analysis (markers reward this):
  • Order is important — a reagent can precipitate more than one cation, so you must test in the set sequence and remove each group before the next test.
  • Use a fresh sample for every step (a few drops on a spot-test plate); never re-use a tube you've already added reagent to.
  • Minimum ≈ 0.1 mol L⁻¹ — too dilute and a real precipitate is too faint to see, giving a false negative.
  • Stop once a cation is confirmed by two positive results — all the anions are colourless, so any colour in solution is due to the cation.
🔀 Two valid schemes — there is no single "correct" order. Any logical sequence that removes interferences will identify the ion. Below are two SKY methods: Method A uses HCl → H₂SO₄ → NaOH; Method B (the class TB sequence) uses HCl → NH₃ → flame. Learn whichever you prefer, then be consistent in the exam.

Method A — HCl → H₂SO₄ → NaOH

white ppt no ppt dissolves(dilute NH₃) stays white white ppt no ppt green brick-red blue green red-brown white Unknown: Ag⁺ Pb²⁺ Ba²⁺ Ca²⁺ Mg²⁺ Cu²⁺ Fe²⁺ Fe³⁺ Add HCl white ppt → Ag⁺ / Pb²⁺ no ppt → Ba²⁺ Ca²⁺ Mg²⁺ Cu²⁺ Fe²⁺ Fe³⁺ Add NH₃ Add H₂SO₄ Ag⁺⚗ Pb²⁺ white ppt → Ba²⁺ / Ca²⁺ no ppt → Mg²⁺ Cu²⁺ Fe²⁺ Fe³⁺ Flame test Add NaOH Ba²⁺ Ca²⁺ Cu²⁺ Fe²⁺ Fe³⁺ Mg²⁺
green flamebrick-red flame blue pptgreen ppt red-brown pptwhite ppt ⚗ complexation Tap an ion · clear

Method B — HCl → NH₃ → flame (the SKY class TB sequence)

Here ammonia does double duty: it provides OH⁻ to precipitate the hydroxides (and their colours identify Mg/Fe/Cu), while excess NH₃ redissolves Ag⁺ and Cu²⁺ as complexes. Ba²⁺/Ca²⁺ don't precipitate with NH₃ → split them by flame.

white ppt no ppt brown,dissolves in excess white white green brown blue→deep no ppt green brick-red Unknown: Ag⁺ Pb²⁺ Ba²⁺ Ca²⁺ Mg²⁺ Cu²⁺ Fe²⁺ Fe³⁺ Add HCl white ppt → Ag⁺ / Pb²⁺ no ppt → Ba²⁺ Ca²⁺ Mg²⁺ Cu²⁺ Fe²⁺ Fe³⁺ Add dilute NH₃ Add dilute NH₃ Ag⁺⚗ Pb²⁺ Mg²⁺ Fe²⁺ Fe³⁺ Cu²⁺⚗ no ppt → Ba²⁺ / Ca²⁺ Flame test Ba²⁺ Ca²⁺
⚗ complexation — Ag⁺ & Cu²⁺ redissolve in excess NH₃Tap an ion · clear
CationTest & observationNet ionic equationFlameConfirmatory
Ag⁺Method A: HCl → white AgCl, which dissolves in dilute NH₃ (Pb²⁺ stays white).
Method B: dilute NH₃ on a fresh sample → brown Ag₂O ppt that redissolves in excess NH₃ (Pb²⁺ → white Pb(OH)₂).		Ag⁺(aq)+Cl⁻(aq)→AgCl(s)
AgCl(s)+2NH₃(aq)→[Ag(NH₃)₂]⁺(aq)+Cl⁻(aq) ⚗		noneKI → pale-yellow AgI
Pb²⁺HCl → white ppt; NH₃ → stays white.Pb²⁺(aq)+2Cl⁻(aq)→PbCl₂(s)none (toxic)KI → bright golden-yellow PbI₂
Ba²⁺No HCl ppt; H₂SO₄ → white ppt; flame green.Ba²⁺(aq)+SO₄²⁻(aq)→BaSO₄(s)pale/apple greenflame splits Ba/Ca
Ca²⁺No HCl ppt; H₂SO₄ → white ppt (faint); flame brick-red.Ca²⁺(aq)+SO₄²⁻(aq)→CaSO₄(s)brick-redF⁻ → white CaF₂
Cu²⁺NaOH → blue ppt.Cu²⁺(aq)+2OH⁻(aq)→Cu(OH)₂(s)blue-greenexcess NH₃ → deep-blue [Cu(NH₃)₄]²⁺ ⚗
Fe²⁺NaOH → green ppt (darkens to brown in air).Fe²⁺(aq)+2OH⁻(aq)→Fe(OH)₂(s)noneacidified KMnO₄ decolourised (purple→colourless = +ve); read colour promptly
Fe³⁺NaOH → red-brown ppt.Fe³⁺(aq)+3OH⁻(aq)→Fe(OH)₃(s)noneSCN⁻ → blood-red [FeSCN]²⁺ ⚗
Mg²⁺NaOH → white ppt (insoluble in excess).Mg²⁺(aq)+2OH⁻(aq)→Mg(OH)₂(s)nonewhite ppt persists in excess NaOH

✅ Checkpoint 2 ⏱ try first — 3 min

Q1. A solution gives a white ppt with HCl that dissolves in dilute ammonia. Identify the cation and give both equations. Q2. How do you tell Fe²⁺ from Fe³⁺?

Model answers
A1. Ag⁺. Ag⁺(aq)+Cl⁻(aq)→AgCl(s); AgCl(s)+2NH₃(aq)→[Ag(NH₃)₂]⁺(aq)+Cl⁻(aq) (a complexation reaction). Pb²⁺ would stay white. A2. Add NaOH: Fe²⁺ gives a green ppt (Fe(OH)₂), Fe³⁺ gives a red-brown ppt (Fe(OH)₃). Confirm Fe³⁺ with SCN⁻ → blood-red.

Part 3 · Identifying the anions

Eight anions — Cl⁻, Br⁻, I⁻, OH⁻, CH₃COO⁻, CO₃²⁻, SO₄²⁻, PO₄³⁻. Screen out groups one reagent at a time (acid → Ba²⁺ → litmus → Ag⁺/Cu²⁺). Hover or click any ion to light up its path.

🧪 Acidify before the silver test: add a few drops of dilute HNO₃ before AgNO₃ when testing for halides — otherwise carbonate, phosphate (and slightly-soluble Ag₂SO₄) also precipitate with Ag⁺ and give a false halide result. The acid removes CO₃²⁻/PO₄³⁻ first.
🔀 Two valid schemes here too. Method A screens by acid → Ba²⁺ → litmus; Method B (the class TB sequence) splits by litmus first (acidic/neutral vs basic), then tests each branch.

Method A — acid → Ba²⁺ → litmus → Ag⁺/Cu²⁺

bubbling (CO₂) no bubbling no ppt ppt stays red turns blue no change dissolves ppt forms no ppt blue ppt dilute NH₃ conc NH₃ insoluble Unknown: Cl⁻ Br⁻ I⁻ OH⁻ CH₃COO⁻ CO₃²⁻ SO₄²⁻ PO₄³⁻ Add dilute HNO₃ CO₃²⁻ no bubbling → Cl⁻ Br⁻ I⁻ OH⁻ CH₃COO⁻ SO₄²⁻ PO₄³⁻ Add Ba²⁺ no ppt → Cl⁻ Br⁻ I⁻ OH⁻ CH₃COO⁻ ppt → SO₄²⁻ / PO₄³⁻ Red litmus Add HNO₃ to ppt stays red → Cl⁻ Br⁻ I⁻ turns blue → OH⁻ / CH₃COO⁻ SO₄²⁻ PO₄³⁻ Add Ag⁺ (AgNO₃) Add Cu²⁺ Add NH₃ CH₃COO⁻ OH⁻ Cl⁻⚗ Br⁻⚗ I⁻
AgCl whiteAgBr cream AgI yellow⚗ complexation Tap an ion · clear

Method B — litmus first (the SKY class TB sequence)

Split first by acid–base character: the basic anions (OH⁻, CH₃COO⁻, CO₃²⁻, PO₄³⁻) turn red litmus blue; the rest stay red. Then test each branch. Acidify before AgNO₃ in the halide branch.

stays red turns blue ppt no ppt bubbles no bubbles ppt no ppt blue ppt no ppt dilute NH₃ conc NH₃ insoluble Unknown: Cl⁻ Br⁻ I⁻ OH⁻ CH₃COO⁻ CO₃²⁻ SO₄²⁻ PO₄³⁻ Test with red litmus stays red → Cl⁻ Br⁻ I⁻ SO₄²⁻ turns blue → OH⁻ CH₃COO⁻ CO₃²⁻ PO₄³⁻ Add AgNO₃ (+ HNO₃) Add dilute HNO₃ ppt → Cl⁻ Br⁻ I⁻ SO₄²⁻ Add NH₃ CO₃²⁻ no bubbles → OH⁻ CH₃COO⁻ PO₄³⁻ Add Ba²⁺ PO₄³⁻ no ppt → OH⁻ / CH₃COO⁻ Add Cu²⁺ Cl⁻⚗ Br⁻⚗ I⁻ OH⁻ CH₃COO⁻
⚗ complexation — AgCl/AgBr redissolve in ammoniaTap an ion · clear
AnionTest & observationNet ionic equationConfirmatory
CO₃²⁻Dilute HNO₃ → effervescence (CO₂).CO₃²⁻(aq)+2H⁺(aq)→CO₂(g)+H₂O(l)gas turns limewater milky
SO₄²⁻Add Ba²⁺ → white ppt insoluble in HNO₃.Ba²⁺(aq)+SO₄²⁻(aq)→BaSO₄(s)ppt persists in acid
PO₄³⁻Add Ba²⁺ → white ppt that dissolves in HNO₃.3Ba²⁺(aq)+2PO₄³⁻(aq)→Ba₃(PO₄)₂(s)ammonium molybdate + HNO₃ → yellow ppt
Cl⁻Litmus stays red; Ag⁺ → white ppt; dissolves in dilute NH₃.Ag⁺(aq)+Cl⁻(aq)→AgCl(s)
AgCl(s)+2NH₃(aq)→[Ag(NH₃)₂]⁺(aq)+Cl⁻(aq) ⚗		dissolves in dilute ammonia
Br⁻Ag⁺ → cream ppt; dissolves only in conc NH₃.Ag⁺(aq)+Br⁻(aq)→AgBr(s)
AgBr(s)+2NH₃(aq)→[Ag(NH₃)₂]⁺(aq)+Br⁻(aq) ⚗		dissolves only in conc ammonia
I⁻Ag⁺ → yellow ppt; insoluble in dilute & conc NH₃.Ag⁺(aq)+I⁻(aq)→AgI(s)does not dissolve in ammonia
OH⁻Red litmus → blue; Cu²⁺ → blue ppt (or Ag⁺ → brown Ag₂O).Cu²⁺(aq)+2OH⁻(aq)→Cu(OH)₂(s)strongly alkaline + blue ppt
CH₃COO⁻Litmus turns blue (weak); no ppt with Cu²⁺.CH₃COO⁻(aq)+H⁺(aq)→CH₃COOH(aq) (vinegar smell)weakly basic to litmus + vinegar smell with acid (neutral FeCl₃ → blood-red, beyond syllabus)

✅ Checkpoint 3 ⏱ try first — 3 min

Q1. Three bottles hold Cl⁻, Br⁻, I⁻. Describe a test that separates all three. Q2. Why must you test for (and remove) carbonate before testing for sulfate with Ba²⁺?

Model answers
A1. Add AgNO₃ to each → AgCl white, AgBr cream, AgI yellow. Confirm with ammonia: AgCl dissolves in dilute NH₃, AgBr only in conc NH₃, AgI in neither (AgX(s)+2NH₃→[Ag(NH₃)₂]⁺+X⁻). A2. BaCO₃ is also a white precipitate, so carbonate would give a false positive for sulfate. Acidify first (removes CO₃²⁻ as CO₂); only acid-insoluble BaSO₄ then confirms sulfate.

Part 4 · Revision tools

📌 What to memorise (SKY priority list): (1) the 3-group solubility rules + the colours of the coloured precipitates; (2) two tests for each ion in the syllabus, with balanced net ionic equations; (3) the precipitates that dissolve in acid or in ammonia (the complexation reactions).

Master table — all 16 ions at a glance

IonKey testPositive observationNet ionic equation
Ba²⁺H₂SO₄ then flamewhite ppt; green flameBa²⁺+SO₄²⁻→BaSO₄(s)
Ca²⁺H₂SO₄ then flamewhite ppt; brick-red flameCa²⁺+SO₄²⁻→CaSO₄(s)
Mg²⁺NaOHwhite pptMg²⁺+2OH⁻→Mg(OH)₂(s)
Pb²⁺HCl / KIwhite PbCl₂; golden-yellow PbI₂Pb²⁺+2I⁻→PbI₂(s)
Ag⁺HCl then dilute NH₃white ppt dissolvesAgCl(s)+2NH₃→[Ag(NH₃)₂]⁺+Cl⁻ ⚗
Cu²⁺NaOH (/ NH₃)blue ppt; deep-blue in excess NH₃Cu²⁺+2OH⁻→Cu(OH)₂(s)
Fe²⁺NaOHgreen pptFe²⁺+2OH⁻→Fe(OH)₂(s)
Fe³⁺NaOH / SCN⁻red-brown ppt; blood-red with SCN⁻Fe³⁺+3OH⁻→Fe(OH)₃(s) ⚗
Cl⁻AgNO₃ + NH₃white ppt, dissolves in dilute NH₃Ag⁺+Cl⁻→AgCl(s) ⚗
Br⁻AgNO₃ + NH₃cream ppt, dissolves in conc NH₃Ag⁺+Br⁻→AgBr(s) ⚗
I⁻AgNO₃ + NH₃yellow ppt, NH₃-insolubleAg⁺+I⁻→AgI(s)
OH⁻litmus / Cu²⁺litmus blue; blue Cu(OH)₂Cu²⁺+2OH⁻→Cu(OH)₂(s)
CH₃COO⁻litmus + vinegar smellweakly basic; vinegar smell with acidCH₃COO⁻+H⁺→CH₃COOH
CO₃²⁻dilute acideffervescence; limewater milkyCO₃²⁻+2H⁺→CO₂(g)+H₂O(l)
SO₄²⁻Ba²⁺ then acidwhite ppt, acid-insolubleBa²⁺+SO₄²⁻→BaSO₄(s)
PO₄³⁻Ba²⁺ then acidwhite ppt, dissolves in acid3Ba²⁺+2PO₄³⁻→Ba₃(PO₄)₂(s)

Flashcards — click to flip

Ba²⁺ flame?
click to flip ↻
Pale/apple green. (Ca²⁺ = brick-red.)
Cu²⁺ + NaOH?
click to flip ↻
Blue ppt Cu(OH)₂; dissolves in excess NH₃ → deep-blue [Cu(NH₃)₄]²⁺ ⚗
Fe²⁺ vs Fe³⁺ with NaOH?
click to flip ↻
Fe²⁺ → green ppt; Fe³⁺ → red-brown ppt (confirm Fe³⁺ with SCN⁻ → blood-red).
Ag⁺ vs Pb²⁺?
click to flip ↻
Both white with HCl. Ag⁺ dissolves in dilute NH₃; Pb²⁺ stays. Confirm Pb²⁺: KI → golden PbI₂.
Halide colours with Ag⁺?
click to flip ↻
AgCl white · AgBr cream · AgI yellow. NH₃ solubility: Cl(dilute) > Br(conc) > I(none) ⚗
SO₄²⁻ vs PO₄³⁻?
click to flip ↻
Both white with Ba²⁺. Add HNO₃: BaSO₄ stays; Ba₃(PO₄)₂ dissolves.
CO₃²⁻ test?
click to flip ↻
Add acid → effervescence; gas turns limewater milky (CO₂).
What is complexation?
click to flip ↻
Forming a soluble complex ion, e.g. AgCl(s)+2NH₃→[Ag(NH₃)₂]⁺+Cl⁻. One of NESA's 3 test types. ⚗
Which give NO flame?
click to flip ↻
Mg²⁺, Ag⁺, Fe²⁺, Fe³⁺ (and Pb²⁺ isn't flame-tested). Only Ba/Ca/Cu are useful.
Acetate positive test?
click to flip ↻
Neutral FeCl₃ → deep blood-red; or warm with acid → vinegar smell. (Cu²⁺ "no ppt" only eliminates OH⁻.)

Cheat sheet

Cation order (Method A)

Add reagents in sequence: HCl → H₂SO₄ → NaOH.
  • white ppt with HCl = Ag⁺ / Pb²⁺ (NH₃ splits them)
  • white ppt with H₂SO₄ = Ba²⁺ / Ca²⁺ (flame splits them)
  • ppt with NaOH = Cu²⁺ / Fe²⁺ / Fe³⁺ / Mg²⁺ (the colour identifies it)

Anion order

Acid (CO₃²⁻) → Ba²⁺ (SO₄²⁻/PO₄³⁻, acid splits) → litmus → Ag⁺ (Cl/Br/I via NH₃) / Cu²⁺ (OH⁻/acetate).

Flame (only 3)

Ba²⁺ green · Ca²⁺ brick-red · Cu²⁺ blue-green. Everything else: no flame — don't invent one.

NaOH ppt colours

Cu²⁺ blue · Fe²⁺ green · Fe³⁺ red-brown · Mg²⁺ white.

Silver halides ⚗

AgCl white (dilute NH₃) · AgBr cream (conc NH₃) · AgI yellow (insoluble). Dissolving = complexation.

Full-mark shape

reagent → colour observation → net ionic equation with states. Remove interferences (carbonate before sulfate).

Pre-exam checklist

Consolidation only — use this the night before, not as a substitute for the flowcharts above.
  • I can write the cation sequence (HCl → H₂SO₄ → NaOH) and the anion sequence (acid → Ba²⁺ → litmus → Ag⁺/Cu²⁺) from memory.
  • I only ever write a flame colour for Ba²⁺, Ca²⁺, Cu²⁺.
  • I can give a net ionic equation with state symbols for every test.
  • I know the silver-halide colours (white/cream/yellow) and the ammonia-solubility order, and that dissolving is a complexation reaction.
  • I remove carbonate (acidify) before testing sulfate, and I know a second ion can be removed at an earlier step (e.g. Pb²⁺ precipitates as insoluble PbSO₄ at the sulfate step and is filtered off before the lead test).
Sentence starters: "Add … ; if a [colour] precipitate forms, the ion is … , because …" · "This is a complexation reaction: …" · "A flame test is not appropriate here because … gives no characteristic colour."

Part 5 · Past exam questions

📊 Exam frequency (SKY Second Brain · 2019–2025): this dot point appears in 223 papers / 402 questions — one of the most-tested points in Module 8. It is examined as both MCQs and 3–5 mark "outline a sequence of tests" / "evaluate this procedure" items. Q1–Q8 below are real HSC questions; the model answers are official.

MCQ quiz — 12 questions

Progress: 0 / 12 · 0 correct

1. Which pair of ions can be distinguished using a flame test in the school laboratory?[2022 HSC Q5]
B. Ba²⁺ (green) and Ca²⁺ (brick-red) give distinct flames. Ag⁺/Mg²⁺ give none; Br⁻/Cl⁻ are anions (not flame-tested); Fe²⁺/Fe³⁺ give no distinct flame.
2. Which pair of ions produce different colours in a flame test?[2024 HSC Q8]
C. Cu²⁺ (blue-green) vs Ca²⁺ (brick-red) — different flame colours. The others are anions or non-flame ions.
3. Which ion can be detected using a precipitation reaction with silver nitrate?[2021 HSC Q2]
B. Ag⁺ + Cl⁻ → white AgCl(s). Mg²⁺ and NO₃⁻ form no precipitate with silver nitrate; Ag⁺ is the reagent.
4. Four tests on a dilute solution: (1) red litmus stays red; (2) Ba²⁺ → white ppt; (3) OH⁻ → brown ppt; (4) Cl⁻ → white ppt. Which compound fits?[2023 HSC Q13]
A. Not basic (rules out carbonate); SO₄²⁻ present (Ba²⁺ ppt); cation gives brown ppt with OH⁻ and white ppt with Cl⁻ → Ag⁺ (Ag₂O brown; AgCl white). So silver sulfate.
5. Two colourless solutions contain Pb²⁺ and Na⁺. Which ion, added to both, identifies them?[2025 HSC Q4]
A. I⁻ gives bright-yellow PbI₂ with Pb²⁺; NaI stays soluble. NH₄⁺/NO₃⁻/CH₃COO⁻ form no precipitate with either.
6. [Cu(H₂O)₆]²⁺ is blue and [CuCl₄]²⁻ is green. Which technique best distinguishes these two complexes?[2025 HSC Q7]
C. The two complex ions are different colours, so they absorb visible light differently — UV-vis measures that. AAS gives total Cu only; IR/¹³C NMR suit organics.
7. An unknown may contain Mg²⁺, Pb²⁺ and/or Ba²⁺ (as nitrates). Which solid, dissolved in water, is the best first reagent?[2021 Homebush Q19]
C. Sulfate precipitates BaSO₄ and PbSO₄ (white) but not Mg²⁺ — separating the three in one step. AgCl is insoluble (won't dissolve); KCl only precipitates Pb²⁺; CuSO₄ adds an interfering coloured ion.
8. A mixture of KI and KCl is titrated with AgNO₃. Why is this valid for quantifying each anion?[2023 HSC Q16]
B. AgI has the smallest Ksp (least soluble), so it precipitates first; AgCl follows — two separate equivalence points let you quantify each.
9. AgCl, AgBr and AgI are all pale precipitates. Which observation, using ammonia, identifies Cl⁻?[SKY practice · NESA style]
C. AgCl dissolves in dilute NH₃ (AgCl + 2NH₃ → [Ag(NH₃)₂]⁺ + Cl⁻, a complexation reaction). AgBr needs conc NH₃; AgI dissolves in neither; silver halides don't fizz with acid.
10. Adding NaOH to a solution gives a red-brown precipitate. The cation is:[SKY practice · NESA style]
D. Fe(OH)₃ is red-brown. Cu(OH)₂ is blue, Mg(OH)₂ white, Fe(OH)₂ green. Confirm Fe³⁺ with SCN⁻ → blood-red.
11. A gas released on adding dilute acid turns limewater milky. The anion is:[SKY practice · NESA style]
B. Acid + carbonate → CO₂, which turns limewater milky (CaCO₃). The other anions give no gas with dilute acid.
12. Which is a complexation reaction used in ion identification?[SKY practice · NESA style]
D. A complex ion [Ag(NH₃)₂]⁺ forms. A and C are precipitations; B is an acid–carbonate (gas) reaction.
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Model answers — real HSC items with official marking

2020 HSC Q22 — Identify an unknown salt (cation + anion) 5 marks

Question (NESA, word-for-word): "A 0.1 mol L⁻¹ solution of an unknown salt is to be analysed. The cation is one of magnesium, calcium or barium. The anion is one of chloride, acetate or hydroxide. Outline a sequence of tests that could be performed in a school laboratory to confirm the identity of this salt solution. Include expected observations and a balanced chemical equation in your answer."

✍ Full-mark answer — write this Cation: Perform a flame test on a fresh sample. A pale-green flame = Ba²⁺; a brick-red flame = Ca²⁺; no characteristic colour = Mg²⁺.
Anion: To a fresh sample add copper(II) nitrate solution — a pale-blue precipitate confirms hydroxide: Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s). If no precipitate forms, add silver nitrate to another fresh sample — a white precipitate confirms chloride (Ag⁺(aq) + Cl⁻(aq) → AgCl(s)); no precipitate = acetate.

How the 5 marks are earned:

  • 1 — a valid cation test (flame) with the three expected colours
  • 1 — a valid anion test that distinguishes the three anions
  • 2 — the correct expected observation paired with each test
  • 1 — a correct balanced net-ionic equation with state symbols
Official NESA mark bands — what each score needs (verbatim)
  • 5 — Outlines a sequence of suitable tests with expected observations • Includes a balanced chemical equation
  • 4 — Outlines a sequence of suitable tests and most of the expected observations • Includes a substantially correct balanced chemical equation
  • 3 — Provides suitable tests that can identify cation(s) and anion(s) present • Includes some expected observations and/or a balanced chemical equation
  • 2 — One test + observation for a cation OR anion; OR tests for cations and anions; OR an equation + one valid test
  • 1 — Provides some relevant information

The answer above hits the top band: a full test sequence, an expected observation for every test, and a balanced equation with states. NESA's own sample answer also accepts a pH test for the anion (Cl⁻ neutral · CH₃COO⁻ slightly basic · OH⁻ very basic).

Marker note: pair every test with its observation and give at least one balanced equation with states. A pH/indicator test (OH⁻ very basic, acetate slightly basic, Cl⁻ neutral) is also accepted for the anion.
Source: 2020 HSC Chemistry Q22 question paper + NESA marking guidelines (transcribed verbatim).
2024 HSC Q27 — Why a Pb²⁺/Ba²⁺ test procedure fails 4 marks

Question (NESA, word-for-word): "The following procedure is proposed to test for the presence of lead(II) and barium ions in water at concentrations of 0.1 mol L⁻¹. 1. Add excess 0.1 mol L⁻¹ sodium sulfate solution. If a precipitate is produced, then barium ions are present.
2. Filter any precipitate produced.
3. Add excess 0.1 mol L⁻¹ sodium bromide solution to the filtrate. If a precipitate is produced, then lead(II) ions are present. Explain why this procedure gives correct results when only barium ions are present, but not when both barium and lead(II) ions are present. Include ONE balanced chemical equation in your answer."

✍ Full-mark answer — write this When only barium is present, step 1 precipitates it as BaSO₄ — Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) — so the conclusion "barium present" is correct. Step 3 then gives no precipitate, correctly showing no lead is present.
When both ions are present, lead(II) also forms an insoluble sulfate, PbSO₄, so it too precipitates at step 1 and is removed by filtration. This leaves insufficient Pb²⁺ in the filtrate to give a precipitate at step 3 — so the procedure wrongly concludes that only barium is present, when lead(II) is in fact also present.

How the 4 marks are earned:

  • 1 — a balanced equation with states for the BaSO₄ precipitation
  • 1 — recognise the procedure works correctly when only Ba²⁺ is present
  • 2 — explain PbSO₄ is also insoluble → Pb²⁺ removed by filtration at step 1 → too little left → false "barium only" verdict at step 3
Official NESA mark bands — what each score needs (verbatim)
  • 4 — Demonstrates a thorough understanding of the procedure and qualitative ion testing • Includes a balanced chemical equation, including states
  • 3 — Demonstrates a sound understanding of the procedure and qualitative ion testing
  • 2 — Demonstrates some understanding of the procedure and qualitative ion testing
  • 1 — Provides some relevant information
Marker note: the discriminator is recognising that Pb²⁺ is lost at the sulfate step (PbSO₄ is insoluble too) — not a vague "selectivity" statement. The equation must carry state symbols for the top band.
Source: 2024 HSC Chemistry Q27 question paper + NESA marking guidelines (transcribed verbatim).
2021 HSC Q30 — Evaluate an ion-identification procedure 5 marks

Question (NESA, word-for-word): "A student was trying to identify the ions present in a dilute aqueous solution. The solution contained ions of barium, calcium or magnesium, and ions of hydroxide or acetate. The student performed the following tests and recorded their observations. A fresh sample of the solution was used for each test. • When aqueous sodium chloride was added, no visible reaction was observed.
• When aqueous silver nitrate was added, brown precipitate was produced. The precipitate dissolved when dilute hydrochloric acid was added.
• When concentrated aqueous sodium sulfate was added, white precipitate was produced. Evaluate this procedure as a method of identifying the ions."

✍ Full-mark answer — write this Test 1 (NaCl) gives no information: sodium and chloride form no precipitate with any of the possible ions, so it is a wasted step.
Test 2 (AgNO₃) correctly identifies the anion: adding silver nitrate gives no precipitate with acetate (silver acetate is soluble) but a brown Ag₂O precipitate with hydroxide (which dissolves in acid) — so the brown precipitate confirms the anion is OH⁻, not acetate. (It also rules out Mg²⁺ as the cation: if OH⁻ were present with Mg²⁺, insoluble Mg(OH)₂ would already have precipitated.)
Test 3 (concentrated sulfate) cannot identify the cation: both Ba²⁺ and Ca²⁺ form a precipitate with concentrated sulfate, so it does not distinguish them.
Judgement: the procedure is insufficient — it identifies the anion (OH⁻) but not the cation. A flame test is needed (pale-green = Ba²⁺, brick-red = Ca²⁺).

How the 5 marks are earned:

  • 2 — outline the limited/useless steps (Test 1 gives no info; Test 3 can't split Ba²⁺/Ca²⁺)
  • 2 — outline what does work (Test 2 identifies OH⁻; also eliminates Mg²⁺)
  • 1 — an informed judgement: insufficient → a flame test is needed (state this explicitly)
Official NESA mark bands — what each score needs (verbatim)
  • 5 — Shows a comprehensive understanding of the procedure • Outlines positive and negative aspects of the procedure • Makes an informed judgement
  • 4 — Outlines some positive and negative aspects of the procedure • Makes a judgement
  • 3 — Outlines some positive and/or negative aspects of the procedure
  • 2 — Identifies positive and/or negative aspects; OR outlines a positive OR negative aspect
  • 1 — Provides some relevant information
Marker note: "Evaluate" needs both positives and negatives and a clear verdict — the 5th mark is lost without the explicit judgement.
Source: 2021 HSC Chemistry Q30 question paper + NESA marking guidelines (transcribed verbatim).
2023 HSC Q30 — Anion sequence: Br⁻ and/or CO₃²⁻ 4 marks

Question (NESA, word-for-word): "A water sample contains at least one of the following anions at concentrations of 1.0 mol L⁻¹. • bromide (Br⁻)
• carbonate (CO₃²⁻) Outline a sequence of tests that could be performed in a school laboratory to confirm the identity of the anion or anions present. Include expected observations and TWO balanced chemical equations in your answer."

✍ Full-mark answer — write this Step 1 — add dilute nitric acid to a sample. Effervescence of a colourless gas that turns limewater (calcium hydroxide solution) milky — cloudy white from CaCO₃ — confirms carbonate, and removes it before the next test: 2H⁺(aq) + CO₃²⁻(aq) → CO₂(g) + H₂O(l).
Step 2 — to a fresh, acidified sample add silver nitrate. A cream precipitate confirms bromide: Ag⁺(aq) + Br⁻(aq) → AgBr(s).

How the 4 marks are earned:

  • 1 — correct carbonate test + observation (effervescence / limewater)
  • 1 — balanced equation (with states) for carbonate + acid
  • 1 — correct bromide test + observation (cream ppt)
  • 1 — balanced equation (with states) for AgBr
Official NESA mark bands — what each score needs (verbatim)
  • 4 — Demonstrates a thorough understanding of anion testing in an appropriate sequence with expected observations • Includes TWO balanced chemical equations including states
  • 3 — Demonstrates a sound understanding of anion testing with expected observation(s) and/or a correct chemical equation
  • 2 — Demonstrates some understanding of anion testing
  • 1 — Provides some relevant information

NESA also accepts an alternative: add excess silver nitrate → precipitate; then add dilute nitric acid — bubbles + a brown precipitate dissolving = carbonate was present, a cream precipitate remaining = bromide.

Marker note: test carbonate first (acid removes it) so AgBr isn't confused with Ag₂CO₃; both equations need state symbols.
Source: 2023 HSC Chemistry Q30 question paper + NESA marking guidelines (transcribed verbatim).
2020 Independent Q36 — Distinguish Cl⁻, Br⁻ and I⁻ 3 marks

Question (word-for-word): "Bottles A, B and C contain aqueous solutions of chloride (Cl⁻), bromide (Br⁻) and iodide (I⁻) ions respectively. Describe chemical tests that could identify which bottle contains which anion." (diagram of three bottles A · B · C)

✍ Full-mark answer — write this Add silver nitrate to a separate sample of each bottle. All three give a precipitate, and the colour identifies the anion: white = chloride (AgCl), cream = bromide (AgBr), pale-yellow = iodide (AgI).
Confirm by adding ammonia: the AgCl precipitate dissolves in dilute ammonia, AgBr dissolves only in concentrated ammonia, and AgI does not dissolve in either: AgX(s) + 2NH₃(aq) → [Ag(NH₃)₂]⁺(aq) + X⁻(aq) ⚗ complexation.

How the 3 marks are earned:

  • 1 — identify Cl⁻ (white AgCl / dissolves in dilute NH₃)
  • 1 — identify Br⁻ (cream AgBr / dissolves only in conc NH₃)
  • 1 — identify I⁻ (pale-yellow AgI / insoluble in NH₃)
Official mark bands — what each score needs (verbatim)
  • 3 — Describes chemical tests that clearly identify the three separate anions
  • 2 — Describes chemical tests that clearly identify TWO separate anions
  • 1 — Describes a chemical test that clearly identifies an anion OR makes a correct statement about how the ions could be separately identified

The marking guideline's own answer distinguishes them by precipitate colour alone (white / cream / pale-yellow with AgNO₃). The ammonia-solubility step above is a stronger confirmation and also scores the top band.

Source: 2020 Independent Trial Chemistry Q36 + marking guidelines (answer is SKY-original; question quoted for study).
Why AgCl dissolves in ammonia but AgI does not SKY practice · 3 marks

Question: Explain, using complexation, why AgCl dissolves in ammonia but AgI does not.

✍ Full-mark answer — write this Ammonia forms a soluble diamminesilver(I) complex ion: AgX(s) + 2NH₃(aq) ⇌ [Ag(NH₃)₂]⁺(aq) + X⁻(aq). Whether the silver halide dissolves depends on its solubility (Ksp, the solubility product — the smaller Ksp is, the more insoluble the salt). AgCl has the largest Ksp (most soluble), so dilute ammonia supplies enough NH₃ to shift this equilibrium right and dissolve it. AgI has the smallest Ksp (least soluble), so even concentrated ammonia cannot provide enough NH₃ to dissolve it; AgBr is intermediate and dissolves only in concentrated ammonia.

How the 3 marks are earned:

  • 1 — the complexation equation forming [Ag(NH₃)₂]⁺
  • 1 — link dissolving to Ksp / solubility order (AgCl > AgBr > AgI)
  • 1 — conclude AgCl dissolves (dilute NH₃) but AgI does not (smallest Ksp)
This is the ⚗ complexation logic that powers the Cl⁻/Br⁻/I⁻ split in the anion flowchart.

Exam formats you must be able to read — flowcharts & tables

Markers love presenting ion-ID as a flowchart or a results table. Same chemistry — you just have to read the diagram. Here are two, rebuilt in SKY style so they read clearly in light and dark mode.

① Flowchart question. A solution contains three cations Ba²⁺, Mg²⁺ and Ag⁺. The flow chart shows a student's plan to confirm each ion. Which row correctly identifies ion 1, ion 2 and ion 3?

precipitate filtered precipitate filtered Solution: Ba²⁺, Mg²⁺, Ag⁺ Add H₂SO₄ ion 1 Add HCl ion 2 Add NaOH ion 3
A.  ion 1 = Ba²⁺  ·  ion 2 = Ag⁺  ·  ion 3 = Mg²⁺
B.  ion 1 = Mg²⁺  ·  ion 2 = Ag⁺  ·  ion 3 = Ba²⁺
C.  ion 1 = Ag⁺  ·  ion 2 = Mg²⁺  ·  ion 3 = Ba²⁺
D.  ion 1 = Ba²⁺  ·  ion 2 = Mg²⁺  ·  ion 3 = Ag⁺
Answer + reasoning
✓ Answer: A Read the reagents in order — each one removes a different cation:
Add H₂SO₄ → only Ba²⁺ gives an insoluble sulfate (BaSO₄), so ion 1 = Ba²⁺: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).
Add HCl → of the two ions left, only Ag⁺ gives an insoluble chloride (AgCl), so ion 2 = Ag⁺: Ag⁺(aq) + Cl⁻(aq) → AgCl(s).
Add NaOH → the remaining Mg²⁺ gives a white hydroxide, so ion 3 = Mg²⁺: Mg²⁺(aq) + 2OH⁻(aq) → Mg(OH)₂(s).
Why this order works: each reagent is chosen so that only one of the ions still in solution precipitates — that is the whole logic of a separation flowchart. (Sulfate first because BaSO₄ is far less soluble than Ag₂SO₄.)
Question format from 2023 Independent Trial Q4 · rebuilt and answered by SKY.

② Results-table question. Four solutions were each given a flame test and then tested with silver nitrate (after acidifying with HNO₃). Which row correctly matches the observations to the salt?

Flame colourWith AgNO₃ / HNO₃Salt
Aapple-greenwhite precipitateBaCl₂
Bbrick-redpale-yellow precipitateCaCl₂
Cno colourcream precipitateMgCl₂
Dlilacwhite precipitateKBr
Answer + reasoning
✓ Answer: A Check both the flame colour and the halide colour in each row:
A ✓ — Ba²⁺ burns apple-green and Cl⁻ gives a white AgCl precipitate. Both correct → BaCl₂.
B ✗ — the flame (brick-red = Ca²⁺) is right, but Cl⁻ gives a white precipitate, not pale-yellow (pale-yellow = iodide).
C ✗ — no flame colour does fit Mg²⁺, but Cl⁻ is white, not cream (cream = bromide).
D ✗ — lilac flame fits K⁺, but Br⁻ gives a cream precipitate, not white.
Trap: table questions put one correct detail next to one wrong detail. Verify every cell before you commit — the silver-halide colours (Cl⁻ white · Br⁻ cream · I⁻ yellow) are the usual catch.
SKY practice · NESA results-table format.

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