H₂S is passed through acidified solution rather than alkaline solution in group II because:

Added H⁺ suppresses the ionisation of H₂S, keeping [S²⁻] low enough that only the least soluble sulphides precipitate

H₂S does not dissolve in alkali

Added H⁺ suppresses the ionisation of H₂S, keeping [S²⁻] low enough that only the least soluble sulphides precipitate

Group II sulphides are soluble in alkali

Acid prevents oxidation of the sulphide

The brown ring in the nitrate test appears only at the junction of the layers because:

Concentrated H₂SO₄ is denser and forms a separate layer, so the reagents meet only at the interface

The complex is unstable elsewhere

Concentrated H₂SO₄ is denser and forms a separate layer, so the reagents meet only at the interface

Nitrate is insoluble in sulphuric acid

Principles of Practical Chemistry — Salt Analysis and Tests

the exam explicitly examines "principles related to practical chemistry", and the questions are unusually predictable: a described observation and four candidate ions, or a named reagent and four possible functional groups. What makes this chapter easy marks is that each test has one diagnostic observation — a colour, a smell, a precipitate that dissolves in excess reagent — and once you attach the observation to the ion, the question answers itself. What makes candidates lose those marks is learning the tests as a list without the chemistry underneath, because then two similar observations become indistinguishable. This chapter deliberately gives the reason alongside each test.

Cation Groups and the Logic of the Sequence

The analytical groups in the order they are tested
Group	Reagent	Ions precipitated
I	Dilute HCl	Pb²⁺, Ag⁺, Hg₂²⁺ — as chlorides
II	H₂S in acidic medium	Cu²⁺, Cd²⁺, Bi³⁺, Sn, As, Sb — least soluble sulphides
III	NH₄OH with NH₄Cl	Fe³⁺, Al³⁺, Cr³⁺ — as hydroxides
IV	H₂S in alkaline medium	Zn²⁺, Mn²⁺, Ni²⁺, Co²⁺ — more soluble sulphides
V	(NH₄)₂CO₃	Ba²⁺, Sr²⁺, Ca²⁺ — as carbonates
VI	No group reagent	Mg²⁺, Na⁺, K⁺, NH₄⁺ — identified by flame or specific tests

🎯 Why groups II and IV both use H₂S but at different pH

This is the single cleverest piece of chemistry in salt analysis, and it is pure common ion effect. H₂S is a weak acid, so in acidic solution the added H⁺ suppresses its ionisation and keeps [S²⁻] very low — low enough that only the sulphides with the smallest K_sp (CuS, CdS, Bi₂S₃) can exceed their solubility product and precipitate. Making the solution alkaline removes H⁺, so H₂S ionises much further, [S²⁻] rises by orders of magnitude, and now the more soluble sulphides (ZnS, MnS, NiS, CoS) precipitate too. So the same reagent separates two sets of ions purely by controlling one concentration. This is also why the order of the groups cannot be rearranged: precipitating group IV first would drag group II down with it and lose the separation entirely.

Confirmatory tests worth knowing outright
Ion	Test and observation
Cu²⁺	Excess NH₄OH gives a deep blue solution, [Cu(NH₃)₄]²⁺
Fe³⁺	KSCN gives blood-red colour; K₄[Fe(CN)₆] gives Prussian blue
Fe²⁺	K₃[Fe(CN)₆] gives Turnbull's blue; no colour with KSCN
Al³⁺	White gelatinous hydroxide that dissolves in excess NaOH
Zn²⁺	White sulphide, and the hydroxide also dissolves in excess NaOH
NH₄⁺	NaOH on warming releases a pungent gas turning litmus blue

🧠 Dissolving in excess reagent is the most informative observation of all

A precipitate that redissolves tells you far more than one that does not, because it means the hydroxide is amphoteric or the cation forms a soluble complex. Al(OH)₃ and Zn(OH)₂ both dissolve in excess NaOH as aluminate and zincate, while Fe(OH)₃ and Mg(OH)₂ do not — that alone separates two pairs of white or off-white precipitates. Then Zn²⁺ dissolves in excess ammonia as [Zn(NH₃)₄]²⁺ but Al³⁺ does not, which separates the remaining pair. So two additions of excess reagent distinguish four cations without any colour at all, and a question that says "the precipitate dissolved on adding more reagent" has already narrowed the field for you.

Anion Tests

Anions by their diagnostic observation
Anion	Test	Observation
CO₃²⁻	Dilute acid	Brisk effervescence; the gas turns limewater milky
SO₃²⁻	Dilute acid	SO₂ with a burning-sulphur smell; decolourises KMnO₄
S²⁻	Dilute acid	H₂S, rotten-egg smell; blackens lead acetate paper
NO₃⁻	FeSO₄ and concentrated H₂SO₄	Brown ring at the junction of the two layers
Cl⁻	AgNO₃	White precipitate, soluble in ammonia
SO₄²⁻	BaCl₂	White precipitate insoluble in dilute HCl

⚠️ Two white precipitates with barium, told apart by acid

Both sulphate and sulphite give a white precipitate with BaCl₂, so the precipitate alone proves nothing. The discriminator is dilute HCl: BaSO₄ is insoluble in it because sulphuric acid is strong and its anion has no affinity for protons, whereas BaSO₃ dissolves with effervescence because sulphurous acid is weak and decomposes to SO₂. The same logic distinguishes the three silver halides by ammonia instead: AgCl dissolves readily, AgBr only in concentrated ammonia, and AgI not at all, because covalent character increases down the halide series and makes the salt progressively harder to break up.

🎯 Why the nitrate test gives a brown ring and not a uniform colour

Concentrated sulphuric acid is denser than the aqueous layer, so when poured carefully down the side of the tube it sinks and forms a separate layer without mixing. The reaction that produces the brown [Fe(H₂O)₅NO]²⁺ complex can therefore only happen at the interface, where nitrate, Fe²⁺ and concentrated acid all meet — which is why the product appears as a sharp ring rather than colouring the whole solution. Shaking the tube destroys the ring, and that is why the standard instruction is to add the acid slowly and not to disturb the tube.

Detecting Functional Groups

One reagent, one functional group
Test	Positive for	Observation
NaHCO₃	Carboxylic acid	Effervescence of CO₂ — phenol gives none
Neutral FeCl₃	Phenol	Violet or blue colouration
2,4-DNP (Brady's reagent)	Aldehyde or ketone	Orange or yellow precipitate
Tollens or Fehling	Aldehyde only	Silver mirror; red Cu₂O precipitate
Iodoform test (I₂/NaOH)	CH₃CO– or CH₃CH(OH)– group	Yellow precipitate of CHI₃
Carbylamine test	Primary amine only	Foul-smelling isocyanide

🧠 Use 2,4-DNP first, then Tollens — the pair identifies the carbonyl type

Brady's reagent reacts with any carbonyl, so an orange precipitate tells you the compound is an aldehyde or a ketone but not which. Tollens' reagent then separates them, because only an aldehyde can be oxidised further to an acid and so only an aldehyde reduces Ag⁺ to the silver mirror. Run the two in that order and you have identified the group in two steps. The one exception worth remembering: formic acid gives a positive Tollens test even though it is an acid, because its structure still contains a hydrogen attached to a carbonyl carbon — which is why HCOOH also decolourises KMnO₄ while acetic acid does not.

Element detection uses the Lassaigne sodium fusion, which converts covalently bound N, S, halogen and P into ionic sodium salts that ordinary tests can find. Nitrogen appears as Prussian blue with ferrous sulphate; sulphur as a violet colour with sodium nitroprusside or a black precipitate with lead acetate; halogens as the silver halides. If both N and S are present the fusion gives sodium thiocyanate, and the ferric thiocyanate that forms is blood red rather than Prussian blue — an examinable subtlety, since the observation changes even though nitrogen is genuinely present.

Key takeaways

Cation groups run I to VI in a fixed order, and the order cannot be rearranged without losing the separation.
H₂S in acid precipitates only group II; in alkali it precipitates group IV too, because [S²⁻] rises sharply.
Al(OH)₃ and Zn(OH)₂ dissolve in excess NaOH; only Zn²⁺ then dissolves in excess ammonia.
BaSO₄ is insoluble in dilute HCl but BaSO₃ dissolves with effervescence — that separates sulphate from sulphite.
The nitrate brown ring forms only at the interface, because the dense acid layer does not mix.
NaHCO₃ distinguishes a carboxylic acid from a phenol; neutral FeCl₃ confirms the phenol.
2,4-DNP detects any carbonyl; Tollens then separates aldehyde from ketone. Formic acid is the exception.
When N and S are both present, Lassaigne gives blood-red ferric thiocyanate instead of Prussian blue.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

H₂S is passed through acidified solution rather than alkaline solution in group II because:
1. H₂S does not dissolve in alkali
2. Added H⁺ suppresses the ionisation of H₂S, keeping [S²⁻] low enough that only the least soluble sulphides precipitate
3. Group II sulphides are soluble in alkali
4. Acid prevents oxidation of the sulphide
Show answer
Answer: B — Added H⁺ suppresses the ionisation of H₂S, keeping [S²⁻] low enough that only the least soluble sulphides precipitate
H₂S is a weak acid, so the common ion effect from added H⁺ keeps its sulphide-ion concentration very low. At that low [S²⁻] only CuS, CdS and Bi₂S₃, which have the smallest solubility products, can exceed K_sp and precipitate, leaving group IV in solution. Making the medium alkaline later raises [S²⁻] by orders of magnitude and brings ZnS, MnS, NiS and CoS down as group IV.
A white precipitate with BaCl₂ dissolves in dilute HCl with effervescence. The anion is:
1. SO₄²⁻
2. SO₃²⁻
3. Cl⁻
4. NO₃⁻
Show answer
Answer: B — SO₃²⁻
Both sulphate and sulphite give a white barium precipitate, so the precipitate alone is not diagnostic; the behaviour in acid is. BaSO₃ dissolves with effervescence because sulphurous acid is weak and breaks down to SO₂, whereas BaSO₄ is insoluble in dilute HCl since sulphuric acid is strong. Chloride and nitrate give no precipitate with barium at all.
A white gelatinous precipitate with NaOH dissolves in excess NaOH but not in excess ammonia. The cation is:
1. Zn²⁺
2. Al³⁺
3. Fe³⁺
4. Mg²⁺
Show answer
Answer: B — Al³⁺
Al(OH)₃ is amphoteric and dissolves in excess NaOH as the aluminate ion, but aluminium forms no ammine complex so it stays precipitated in excess ammonia. Zinc behaves the same way with NaOH but does dissolve in excess ammonia as [Zn(NH₃)₄]²⁺, so the ammonia step is exactly what separates the two. Fe(OH)₃ is brown and Mg(OH)₂ dissolves in neither reagent.
Which reagent distinguishes a carboxylic acid from a phenol?
1. Neutral FeCl₃
2. NaHCO₃ solution
3. Tollens' reagent
4. 2,4-DNP
Show answer
Answer: B — NaHCO₃ solution
A carboxylic acid is strong enough to displace carbon dioxide from bicarbonate and gives visible effervescence, while a phenol is too weakly acidic to do so. Neutral FeCl₃ also distinguishes the pair but from the other side, by giving a violet colour with the phenol. Tollens and 2,4-DNP both test for carbonyl compounds and would be negative for either of these.
A compound gives an orange precipitate with 2,4-DNP but no silver mirror with Tollens' reagent. It is:
1. An aldehyde
2. A ketone
3. A carboxylic acid
4. An alcohol
Show answer
Answer: B — A ketone
2,4-DNP reacts with any carbonyl, so the orange precipitate establishes an aldehyde or a ketone. Only an aldehyde can be oxidised further to a carboxylic acid, so only an aldehyde reduces Tollens' reagent to metallic silver; a negative Tollens test therefore points to a ketone. Alcohols and acids would fail the 2,4-DNP test in the first place, since neither has a free carbonyl.
In the Lassaigne test, a compound containing both nitrogen and sulphur gives:
1. Prussian blue
2. A blood-red colouration
3. A violet colouration
4. No colour at all
Show answer
Answer: B — A blood-red colouration
When both elements are present, sodium fusion produces sodium thiocyanate rather than separate cyanide and sulphide, and ferric thiocyanate is blood red. Prussian blue appears only when nitrogen is present without sulphur. The observation therefore changes even though nitrogen genuinely is present, which is exactly why this case is examined — a candidate looking only for blue would wrongly report nitrogen absent.
The brown ring in the nitrate test appears only at the junction of the layers because:
1. The complex is unstable elsewhere
2. Concentrated H₂SO₄ is denser and forms a separate layer, so the reagents meet only at the interface
3. Nitrate is insoluble in sulphuric acid
4. The reaction needs light
Show answer
Answer: B — Concentrated H₂SO₄ is denser and forms a separate layer, so the reagents meet only at the interface
Poured carefully down the side of the tube, the denser concentrated acid sinks and stays as a distinct layer, so nitrate, Fe²⁺ and concentrated acid come together only where the two liquids touch — and the brown nitrosyl complex forms there as a sharp ring. Shaking mixes the layers and destroys the ring, which is why the procedure insists on not disturbing the tube.
Which of the three silver halides is insoluble even in concentrated ammonia?
1. AgCl
2. AgBr
3. AgI
4. All three dissolve
Show answer
Answer: C — AgI
Covalent character increases from chloride to iodide because the large, easily polarised iodide is strongly distorted by the silver ion, so AgI is held together too tightly for ammonia to break up. AgCl dissolves readily in dilute ammonia and AgBr only in concentrated ammonia, which makes the sequence a clean way to identify which halide is present from a single reagent.