Principles Of Qualitative Analysis

Q: Why is ammonium chloride added with ammonium hydroxide in group III analysis?

NH₄Cl supplies NH₄⁺, which suppresses the ionisation of NH₄OH by the common ion effect. The hydroxide ion concentration falls to about 10⁻⁶ M: enough to precipitate Fe(OH)₃, Al(OH)₃ and Cr(OH)₃, which have tiny solubility products, but too low to precipitate Mg(OH)₂ or the hydroxides of later groups.

Q: Why is hydrogen sulphide passed in acid for group II but in ammoniacal solution for zinc and manganese?

In dilute HCl, H⁺ suppresses the ionisation of H₂S, so the sulphide ion concentration is very low and only sulphides with the smallest solubility products, such as CuS, precipitate. Ammonia removes H⁺, raising the sulphide ion concentration enough to precipitate ZnS, MnS, NiS and CoS.

Q: How can chloride, bromide and iodide ions be distinguished?

With silver nitrate, chloride gives white AgCl soluble in dilute ammonia, bromide gives pale yellow AgBr soluble only in concentrated ammonia, and iodide gives yellow AgI insoluble in ammonia. Chlorine water with carbon tetrachloride gives an orange-brown layer for bromide and a violet layer for iodide. Only chloride gives the chromyl chloride test.

ChemistryPrinciples Related to Practical ChemistryFor NEET aspirants

The principles of qualitative analysis explain how an unknown salt is identified from its reactions: the gas it gives with an acid, the precipitate it forms with a group reagent, and the colour that confirms each ion. Two ideas from ionic equilibrium, the solubility product and the common ion effect, sort anions into classes and cations into groups. This page covers the principles of qualitative analysis with dry tests, every anion and cation test with balanced equations, and the reasoning asked in JEE Main, JEE Advanced and NEET.

On this page1Dry tests2Anion classes3Dilute-acid anions4Conc.-acid anions5Class B anions6Cation groups7Why reagents work8Confirmatory tests9Solved examples

Key Formulas - Quick Reference

★ Must learn Precipitation rule: a solid forms only when the ionic product exceeds the solubility product, $Q > K_{s p}$ ; if $Q < K_{s p}$ the ion stays in solution.
★ Must learn Group reagents in order: dil. $HCl$ , $H_{2} S$ in dil. $HCl$ , $N H_{4} Cl$ + $N H_{4} OH$ , $H_{2} S$ in $N H_{4} OH$ , $(N H_{4})_{2} C O_{3}$ , then $N a_{2} HP O_{4}$ .
★ Must learn Common ion effect: $H^{+}$ from $HCl$ lowers $[S^{2 -}]$ (group II); $N H_{4}^{+}$ from $N H_{4} Cl$ lowers $[O H^{-}]$ (group IIIA).
★ Must learn Brown ring: $[Fe (H_{2} O)_{6}]^{2 +}$ + $NO$ $\to$ $[Fe (H_{2} O)_{5} (NO)]^{2 +}$ + $H_{2} O$ (nitrate with conc. $H_{2} S O_{4}$ , nitrite with dilute acid).
★ Must learn Chromyl chloride test: only $C l^{-}$ gives orange-red $Cr O_{2} C l_{2}$ ; it is confirmed as yellow $PbCr O_{4}$ .
Silver halides: $AgCl$ white (dissolves in dil. $N H_{3}$ ), $AgBr$ pale yellow (conc. $N H_{3}$ ), $AgI$ yellow (insoluble).
Iron tests: $F e^{3 +}$ + $[Fe (CN)_{6}]^{4 -}$ gives Prussian blue $F e_{4} [Fe (CN)_{6}]_{3}$ ; $F e^{3 +}$ + $SC N^{-}$ gives blood-red $[Fe (SCN)]^{2 +}$ .
Ammonium: $N H_{4}^{+}$ + $NaOH$ gives $N H_{3}$ ; Nessler's reagent gives a brown precipitate $HgO \cdot Hg (N H_{2}) I$ .
Flame colours: Ca brick red, Sr crimson, Ba apple green, Na golden yellow, K lilac, Cu bluish green.
Other key colours: $N i^{2 +}$ + DMG rosy red; $[Cu (N H_{3})_{4}]^{2 +}$ deep blue; $C u_{2} [Fe (CN)_{6}]$ chocolate brown.

1. What Qualitative Analysis Means

Qualitative analysis finds which ions a substance contains, without measuring how much. A simple salt has one cation (the basic radical) and one anion (the acid radical); a mixture has more.

The purpose of chemical analysis is to establish the composition of natural or manufactured substances. For systematic qualitative analysis, cations are classified into groups on the basis of their behaviour with some reagents. In the same way, anions are classified as class A or class B, depending on how they behave with certain reagents.

The work always follows the same order: preliminary (dry) tests on the solid, anion tests with acids, then the cations group by group from a solution of the salt. Figure 1 shows the whole route.

Figure 1: The order of work in salt analysis. Anions are found first from the gas each one gives with acid; cations are then separated group by group and each ion is confirmed.

1.1 Ions covered

Type	Ions
Cations	$P b^{2 +}$ , $A g^{+}$ , $H g_{2}^{2 +}$ , $H g^{2 +}$ , $C u^{2 +}$ , $C d^{2 +}$ , $B i^{3 +}$ , $A s^{3 +}$ , $S b^{3 +}$ , $S n^{2 +}$ , $F e^{3 +}$ , $A l^{3 +}$ , $C r^{3 +}$ , $Z n^{2 +}$ , $M n^{2 +}$ , $N i^{2 +}$ , $C o^{2 +}$ , $B a^{2 +}$ , $S r^{2 +}$ , $C a^{2 +}$ , $M g^{2 +}$ , $N a^{+}$ , $K^{+}$ , $N H_{4}^{+}$
Anions	$C O_{3}^{2 -}$ , $HC O_{3}^{-}$ , $S O_{3}^{2 -}$ , $S^{2 -}$ , $N O_{2}^{-}$ , $C H_{3} CO O^{-}$ , $S_{2} O_{3}^{2 -}$ , $C l^{-}$ , $B r^{-}$ , $I^{-}$ , $N O_{3}^{-}$ , $S O_{4}^{2 -}$ , $Cr O_{4}^{2 -}$ , $C r_{2} O_{7}^{2 -}$ , $Mn O_{4}^{-}$
Left out	Interfering radicals (oxalate, fluoride, borate, phosphate) and insoluble salts, which need special treatment

1.2 Preparing the solutions

Original solution (O.S.): dissolve the salt in water; if it does not dissolve, try dil. $HCl$ , then conc. $HCl$ . Nitric and sulphuric acids are avoided: $HN O_{3}$ oxidises ions and $H_{2} S O_{4}$ precipitates $B a^{2 +}$ , $S r^{2 +}$ and $P b^{2 +}$ .
Sodium carbonate extract: boiling the salt with $N a_{2} C O_{3}$ solution precipitates most cations as carbonates and leaves the anions in solution as sodium salts. Anion wet tests are done on this extract when the cation would interfere.

2. Preliminary (Dry) Tests

Dry tests on the solid salt take a few minutes and often point straight at the ions present. They never replace confirmatory tests.

2.1 Physical examination

Observation	Inference
Blue	Copper salt
Dark green	Chromium salt
Green	Salts of Fe(II), Ni, Cu or Cr
Light yellow or brown	Salts of Fe(III)
Dark brown	$Pb O_{2}$ , $B i_{2} S_{3}$
Light pink	Salts of Mn
Pink	Salts of Co
Red	$HgO$ , $Hg I_{2}$ , $P b_{3} O_{4}$
Orange red	$S b_{2} S_{3}$
Deliquescent (absorbs moisture and dissolves)	$CaC l_{2}$ , $ZnC l_{2}$ , $MgC l_{2}$ , $MnC l_{2}$ , many nitrites and nitrates
Heavy	Salts of Pb, Hg and Ba
Light, fluffy powder	Carbonates of Mg, Al, Zn, Ca, Sr

2.2 Effect of heating (dry heating test)

A little of the salt is heated in a dry test tube. The table lists what melts, cracks, swells or sublimes, and the colour of the residue.

Observation	Inference
Substance melts	Salts of alkali metals; salts with water of crystallisation
Decrepitates (crackling noise)	$NaCl$ , $KI$ , $Pb (N O_{3})_{2}$ , $Ba (N O_{3})_{2}$
Swells (loses water of crystallisation)	Alums, borates, phosphates
White sublimate	$HgC l_{2}$ , $H g_{2} C l_{2}$ , $N H_{4} X$ , $AlC l_{3}$ , $A s_{2} O_{3}$ , $S b_{2} O_{3}$
Yellow sublimate	$A s_{2} S_{3}$ ; $Hg I_{2}$ (turns red when rubbed with a glass rod)
Blue-black sublimate, violet vapour	Iodides
Residue yellow (hot), white (cold)	$ZnO$
Residue reddish brown (hot), yellow (cold)	$PbO$
Residue black (hot), red (cold)	$HgO$ , $P b_{3} O_{4}$
Residue black (hot), red-brown (cold)	$F e_{2} O_{3}$

Gases evolved on heating. Colourless, odourless gases:

$O_{2}$ rekindles a glowing splinter: alkali nitrates.
$2KN O_{3} Δ 2KN O_{2} + O_{2} ↑$
$C O_{2}$ turns lime water milky: carbonates and oxalates.
$CaC O_{3} Δ CaO + C O_{2} ↑$
$N_{2}$ : ammonium nitrite (also ammonium dichromate).
$N H_{4} N O_{2} Δ N_{2} ↑ + 2 H_{2} O$

Colourless gases with an odour:

$N H_{3}$ turns red litmus blue and mercurous nitrate paper black: ammonium salts.
$(N H_{4})_{2} S O_{4} Δ N H_{4} HS O_{4} + N H_{3} ↑$
$S O_{2}$ smells of burning sulphur and turns acidified $K_{2} C r_{2} O_{7}$ paper green: sulphites and thiosulphates.
$CaS O_{3} Δ CaO + S O_{2} ↑$
$HCl$ has a pungent smell and gives white fumes with ammonia: hydrated chlorides.
$MgC l_{2} \cdot 6 H_{2} O Δ MgO + 2HCl ↑ + 5 H_{2} O$
$H_{2} S$ smells of rotten eggs and turns lead acetate paper black: hydrated sulphides.
$N a_{2} S + 2 H_{2} O Δ 2NaOH + H_{2} S ↑$

Coloured gases:

$N O_{2}$ is brown and turns starch-iodide paper blue: nitrites and nitrates of heavy metals.
$2Cu (N O_{3})_{2} Δ 2CuO + 4N O_{2} ↑ + O_{2} ↑$
$B r_{2}$ is reddish brown; it turns starch paper yellow and starch-iodide paper blue: bromides.
$2CdB r_{2} + O_{2} Δ 2CdO + 2B r_{2} ↑$
$I_{2}$ vapour is violet and turns starch paper blue: iodides.
$2Cd I_{2} + O_{2} Δ 2CdO + 2 I_{2} ↑$
$C l_{2}$ is greenish yellow; it bleaches moist litmus and indigo solution and turns starch-iodide paper blue: some chlorides, such as copper(II) chloride.
$2CuC l_{2} Δ 2CuCl + C l_{2} ↑$

2.3 Flame test

A platinum wire is dipped in conc. $HCl$ , touched to the salt (as a paste with conc. $HCl$ , because chlorides are the most volatile salts) and held in the non-luminous flame. Heat excites electrons of the metal atoms; light of a characteristic wavelength is given out when they fall back.

Metal	Flame colour
Li	Crimson red
Na	Golden yellow
K	Lilac (violet)
Ca	Brick red
Sr	Crimson red
Ba	Apple (grassy) green
Cu	Bluish green

Figure 2: Flame test colours. The salt is made into a paste with conc.

HCl

(chlorides are volatile). Potassium is viewed through blue cobalt glass, which hides the yellow of any sodium impurity.

Magnesium gives no flame colour: its electrons are held so tightly (small atom, high ionisation and excitation energy) that the flame cannot excite them to give visible light.

2.4 Borax bead test

Borax on a platinum loop swells and melts into a clear glassy bead of sodium metaborate and boric anhydride. A trace of a coloured salt fused into the bead forms a metal metaborate with a typical colour.

N a_{2} B_{4} O_{7} \cdot 10 H_{2} O Δ N a_{2} B_{4} O_{7} + 10 H_{2} O

N a_{2} B_{4} O_{7} Δ 2NaB O_{2} + B_{2} O_{3}

CuO + B_{2} O_{3} blue Cu (B O_{2})_{2}

Metal	Oxidising flame (cold bead)	Reducing flame (cold bead)
Cu	Blue	Red, opaque
Fe	Yellow	Bottle green
Cr	Green	Green
Mn	Amethyst (pink-violet)	Colourless
Co	Deep blue	Deep blue
Ni	Reddish brown	Grey

Microcosmic salt, $Na (N H_{4}) HP O_{4} \cdot 4 H_{2} O$ , gives a similar bead of sodium metaphosphate, $NaP O_{3}$ .

2.5 Charcoal cavity and cobalt nitrate tests

The salt is mixed with $N a_{2} C O_{3}$ and heated in a charcoal cavity with a blowpipe flame. A white residue is moistened with a drop of cobalt nitrate solution and heated again; $Co (N O_{3})_{2}$ decomposes to $CoO$ , which combines with the metal oxide.

2Co (N O_{3})_{2} Δ 2CoO + 4N O_{2} + O_{2}

Residue after cobalt nitrate	Compound	Metal
Blue (Thenard's blue)	$CoO \cdot A l_{2} O_{3}$ ( $CoA l_{2} O_{4}$ )	Al
Green (Rinmann's green)	$CoO \cdot ZnO$ ( $CoZn O_{2}$ )	Zn
Pink	$CoO \cdot MgO$	Mg

Hot and cold colours. ZnO is yellow when hot and white when cold;

PbO

goes from reddish brown to yellow. Any identification question with a residue that is yellow while hot and white when cold wants zinc.

Key idea

Dry tests narrow the choice fast: colour, flame and the gas on heating often name the ions before a single solution is made.

3. Classification of Anions

Methods for detecting anions are not as systematic as those for cations. Anions are classified by the process used to detect them.

Class A: anions identified by volatile products obtained with acids. A(i) give gases with dil. $HCl$ or dil. $H_{2} S O_{4}$ ; A(ii) give gases or acid vapours only with conc. $H_{2} S O_{4}$ .
Class B: anions identified by reactions in solution. B(i) precipitation reactions; B(ii) oxidation and reduction in solution.

The classes are tried in order. A salt that gives no gas with dilute acid is tested with conc.

H_{2} S O_{4}

; if that gives nothing either, class B tests are done in solution. Because conc.

H_{2} S O_{4}

also attacks class A(i) anions, the dilute-acid test always comes first.

4. Class A(i): Anions Detected with Dilute Acid

This class includes $C O_{3}^{2 -}$ , $HC O_{3}^{-}$ , $S O_{3}^{2 -}$ , $S^{2 -}$ , $N O_{2}^{-}$ , $C H_{3} CO O^{-}$ and $S_{2} O_{3}^{2 -}$ . Each gives a gas with dil. $HCl$ or dil. $H_{2} S O_{4}$ , and the gas is identified by its own test (Figure 3).

Figure 3: One gas, one test paper. The coloured strips show each paper before and after exposure.

S O_{2}

also turns lime water milky, so the dichromate test must come first when both gases are possible.

4.1 Carbonate

(i) Dilute HCl: brisk effervescence of carbon dioxide.

C O_{3}^{2 -} + 2 H^{+} C O_{2} ↑ + H_{2} O

The gas gives a white turbidity with lime water and with baryta water.

C O_{2} + C a^{2 +} + 2O H^{-} milky CaC O_{3} ↓ + H_{2} O

C O_{2} + B a^{2 +} + 2O H^{-} BaC O_{3} ↓ + H_{2} O

On passing carbon dioxide for a long time the turbidity slowly disappears, because soluble calcium hydrogen carbonate forms.

CaC O_{3} + C O_{2} + H_{2} O Ca (HC O_{3})_{2}

The following tests are done with the aqueous salt solution.

(ii) Barium chloride or calcium chloride solution: white precipitate of barium or calcium carbonate, soluble in mineral acids.

C O_{3}^{2 -} + B a^{2 +} white BaC O_{3} ↓

C O_{3}^{2 -} + C a^{2 +} white CaC O_{3} ↓

(iii) Silver nitrate solution: white precipitate of silver carbonate, soluble in nitric acid and in ammonia. With excess reagent, or on boiling, it turns yellow or brown because silver oxide forms.

C O_{3}^{2 -} + 2A g^{+} white A g_{2} C O_{3} ↓

A g_{2} C O_{3} Δ brown A g_{2} O + C O_{2} ↑

4.2 Bicarbonate (hydrogen carbonate)

$HC O_{3}^{-}$ also effervesces with dilute acid, so the $C O_{2}$ test cannot tell it from carbonate. The difference is that a carbonate solution already contains $C O_{3}^{2 -}$ , while a bicarbonate gives $C O_{3}^{2 -}$ only when heated.

HC O_{3}^{-} + H^{+} C O_{2} ↑ + H_{2} O

2NaHC O_{3} Δ N a_{2} C O_{3} + C O_{2} ↑ + H_{2} O

Mg (HC O_{3})_{2} Δ MgC O_{3} ↓ + C O_{2} ↑ + H_{2} O

Carbonate (e.g.

N a_{2} C O_{3}

)

$MgS O_{4}$ gives a white precipitate in the cold.
$BaC l_{2}$ or $CaC l_{2}$ precipitates at once.
$HgC l_{2}$ gives a reddish-brown precipitate.
Phenolphthalein turns pink.
The dry sodium salt does not decompose on heating.

Bicarbonate (e.g.

NaHC O_{3}

)

$MgS O_{4}$ precipitates only on boiling.
$BaC l_{2}$ or $CaC l_{2}$ precipitates only on boiling.
$HgC l_{2}$ gives no reddish-brown precipitate.
Phenolphthalein stays colourless or faint pink.
The dry salt gives $C O_{2}$ and water on heating.

4.3 Sulphite

(i) Dilute HCl or dilute $H_{2} S O_{4}$ : sulphur dioxide is given off; it has the suffocating smell of burning sulphur.

S O_{3}^{2 -} + 2 H^{+} S O_{2} ↑ + H_{2} O

The following tests are done with the aqueous salt solution or the gas.

(ii) Acidified potassium dichromate: filter paper moistened with acidified $K_{2} C r_{2} O_{7}$ turns green because $C r^{3 +}$ ions form.

3S O_{2} + K_{2} C r_{2} O_{7} + H_{2} S O_{4} K_{2} S O_{4} + green C r_{2} (S O_{4})_{3} + H_{2} O

(iii) Lime water: the gas gives a milky precipitate, which dissolves when more gas is passed because hydrogen sulphite forms.

S O_{2} + Ca (OH)_{2} milky CaS O_{3} ↓ + H_{2} O

CaS O_{3} + S O_{2} + H_{2} O Ca (HS O_{3})_{2}

(iv) Barium chloride or strontium chloride solution: white precipitate of barium or strontium sulphite. Unlike $BaS O_{4}$ , it dissolves in dilute $HCl$ .

S O_{3}^{2 -} + B a^{2 +} white BaS O_{3} ↓

S O_{3}^{2 -} + S r^{2 +} white SrS O_{3} ↓

(v) Acidified potassium permanganate: sulphite decolourises it (carbonate does not).

2Mn O_{4}^{-} + 5S O_{3}^{2 -} + 6 H^{+} 2M n^{2 +} + 5S O_{4}^{2 -} + 3 H_{2} O

4.4 Sulphide

(i) Dilute HCl or dilute $H_{2} S O_{4}$ : a colourless gas smelling of rotten eggs, $H_{2} S$ , is evolved.

S^{2 -} + 2 H^{+} H_{2} S ↑

(ii) The gas turns lead acetate paper black.

(C H_{3} COO)_{2} Pb + H_{2} S black PbS ↓ + 2C H_{3} COOH

The following tests are done with the aqueous salt solution.

(iii) Cadmium carbonate suspension: yellow precipitate of $CdS$ .

N a_{2} S + CdC O_{3} yellow CdS ↓ + N a_{2} C O_{3}

(iv) Silver nitrate solution: black precipitate of silver sulphide, insoluble in cold but soluble in hot dilute nitric acid.

S^{2 -} + 2A g^{+} black A g_{2} S ↓

(v) Sodium nitroprusside solution: purple (violet) colour.

N a_{2} S + N a_{2} [Fe (CN)_{5} NO] purple N a_{4} [Fe (CN)_{5} NOS]

4.5 Nitrite

(i) Dilute HCl or dilute $H_{2} S O_{4}$ : added to a solid nitrite in the cold, it gives a pale blue liquid (free nitrous acid, $HN O_{2}$ , or its anhydride $N_{2} O_{3}$ ) and brown fumes of nitrogen dioxide. Most of the $N O_{2}$ comes from nitric oxide combining with oxygen of the air.

N O_{2}^{-} + H^{+} HN O_{2}

2HN O_{2} N_{2} O_{3} + H_{2} O

3HN O_{2} HN O_{3} + 2NO + H_{2} O

2NO + O_{2} brown 2N O_{2}

The following tests are done with the aqueous salt solution.

(ii) Silver nitrate solution: white crystalline precipitate from concentrated solutions.

N O_{2}^{-} + A g^{+} white AgN O_{2} ↓

(iii) Acidified potassium iodide and starch: the solution (or KI-starch paper) turns blue, because nitrite oxidises iodide to iodine.

2N O_{2}^{-} + 2 I^{-} + 4 H^{+} 2NO + I_{2} + 2 H_{2} O

(iv) Brown ring test: the nitrite solution is added carefully to a concentrated solution of iron(II) sulphate acidified with dilute acetic acid or dilute sulphuric acid. A brown ring (or brown colour) forms at the junction of the two liquids.

N O_{2}^{-} + C H_{3} COOH HN O_{2} + C H_{3} CO O^{-}

3HN O_{2} HN O_{3} + 2NO + H_{2} O

[Fe (H_{2} O)_{6}]^{2 +} + NO brown [Fe (H_{2} O)_{5} (NO)]^{2 +} + H_{2} O

(v) Sulphanilic acid and 1-naphthylamine (Griess test): in acetic acid, nitrite diazotises sulphanilic acid, and the diazonium salt couples with 1-naphthylamine to give a red azo dye. The test is very sensitive.

4.6 Acetate

(i) Dilute sulphuric acid: smell of vinegar.

C H_{3} CO O^{-} + H^{+} C H_{3} COOH

(ii) Neutral iron(III) chloride solution (with the salt solution): deep red colour. On boiling, a reddish-brown precipitate of basic iron(III) acetate separates.

3C H_{3} COONa + FeC l_{3} deep red (C H_{3} COO)_{3} Fe + 3NaCl

(C H_{3} COO)_{3} Fe + 2 H_{2} O Δ reddish brown Fe (OH)_{2} (C H_{3} COO) ↓ + 2C H_{3} COOH

(iii) Ethanol and conc. $H_{2} S O_{4}$ : warming gives the fruity smell of ethyl acetate.

4.7 Thiosulphate

(i) Dilute hydrochloric acid: sulphur separates (the solution turns milky) and sulphur dioxide is given off.

S_{2} O_{3}^{2 -} + 2 H^{+} milky S ↓ + S O_{2} ↑ + H_{2} O

(ii) Iodine solution: decolourised, because iodine is reduced to iodide.

I_{2} + 2 S_{2} O_{3}^{2 -} 2 I^{-} + S_{4} O_{6}^{2 -}

The following tests are done with the aqueous salt solution.

(iii) Barium chloride solution: white precipitate of barium thiosulphate from concentrated solutions; no precipitate with $CaC l_{2}$ .

S_{2} O_{3}^{2 -} + B a^{2 +} white Ba S_{2} O_{3} ↓

(iv) Silver nitrate solution: white precipitate of silver thiosulphate, which is unstable. It turns yellow, brown and finally black on standing, as silver sulphide forms.

S_{2} O_{3}^{2 -} + 2A g^{+} white A g_{2} S_{2} O_{3} ↓

A g_{2} S_{2} O_{3} + H_{2} O black A g_{2} S ↓ + H_{2} S O_{4}

(v) Lead acetate or lead nitrate solution: white precipitate, which turns black on boiling as $PbS$ forms.

S_{2} O_{3}^{2 -} + P b^{2 +} white Pb S_{2} O_{3} ↓

Pb S_{2} O_{3} + H_{2} O Δ black PbS ↓ + 2 H^{+} + S O_{4}^{2 -}

Exam Trick

Two gases turn lime water milky. Both $C O_{2}$ and $S O_{2}$ do. Pass the gas through acidified dichromate first: only $S O_{2}$ turns it green. Carbonate + sulphite mixtures are a favourite trap.

Quick Recall: tap to check

Which dilute-acid gas blackens lead acetate paper?

H_{2} S

, from a sulphide.

Which anion gives a milky solution and

S O_{2}

with dilute

HCl

Thiosulphate,

S_{2} O_{3}^{2 -}

(sulphur separates).

How does sodium nitroprusside respond to a sulphide?

A purple (violet) colour.

Which test tells carbonate from bicarbonate in the cold?

MgS O_{4}

(or

BaC l_{2}

): carbonate precipitates in the cold, bicarbonate only on boiling.

5. Class A(ii): Anions Detected with Conc. $H_{2} S O_{4}$

This class includes $C l^{-}$ , $B r^{-}$ , $I^{-}$ and $N O_{3}^{-}$ . Dilute acid gives no gas with them; conc. $H_{2} S O_{4}$ releases a gas or an acid vapour.

5.1 Chloride

(i) Conc. $H_{2} S O_{4}$ : hydrogen chloride is given off.

C l^{-} + H_{2} S O_{4} HCl ↑ + HS O_{4}^{-}

The gas turns moist blue litmus red, and gives white fumes of $N H_{4} Cl$ when a glass rod dipped in ammonia solution is held at the mouth of the tube.

(ii) Manganese dioxide and conc. $H_{2} S O_{4}$ : a solid chloride heated with $Mn O_{2}$ and conc. $H_{2} S O_{4}$ gives yellowish-green chlorine.

Mn O_{2} + 2 H_{2} S O_{4} + 2C l^{-} M n^{2 +} + C l_{2} ↑ + 2S O_{4}^{2 -} + 2 H_{2} O

The following tests are done with the aqueous salt solution.

(iii) Silver nitrate solution: white, curdy precipitate of $AgCl$ , insoluble in water and dilute nitric acid but soluble in dilute ammonia. Acid brings the precipitate back.

C l^{-} + A g^{+} white AgCl ↓

AgCl + 2N H_{3} [Ag (N H_{3})_{2}] Cl

[Ag (N H_{3})_{2}]^{+} + C l^{-} + 2 H^{+} AgCl ↓ + 2N H_{4}^{+}

(iv) Lead acetate solution: white precipitate of lead chloride (from concentrated solutions), soluble in hot water.

2C l^{-} + P b^{2 +} white PbC l_{2} ↓

(v) Chromyl chloride test: a solid chloride heated with $K_{2} C r_{2} O_{7}$ and conc. $H_{2} S O_{4}$ gives orange-red fumes of chromyl chloride, $Cr O_{2} C l_{2}$ .

K_{2} C r_{2} O_{7} + 4NaCl + 6 H_{2} S O_{4} 2KHS O_{4} + 4NaHS O_{4} + orange-red 2Cr O_{2} C l_{2} ↑ + 3 H_{2} O

The vapour passed into sodium hydroxide gives a yellow solution of sodium chromate, which gives a yellow precipitate of lead chromate with lead acetate.

Cr O_{2} C l_{2} + 4NaOH yellow N a_{2} Cr O_{4} + 2NaCl + 2 H_{2} O

N a_{2} Cr O_{4} + (C H_{3} COO)_{2} Pb yellow PbCr O_{4} ↓ + 2C H_{3} COONa

Figure 4: The chromyl chloride test. Only chloride forms volatile

Cr O_{2} C l_{2}

; the yellow

PbCr O_{4}

at the end proves that chromium really distilled over.

Chlorides of mercury do not respond to the chromyl chloride test, because they are only slightly ionised. Chlorides of lead, silver, antimony and tin are converted only partly.

5.2 Bromide

(i) Conc. $H_{2} S O_{4}$ : reddish-brown bromine vapour comes off with the hydrogen bromide, because conc. $H_{2} S O_{4}$ oxidises part of the $HBr$ .

2KBr + H_{2} S O_{4} K_{2} S O_{4} + 2HBr

2HBr + H_{2} S O_{4} reddish brown B r_{2} ↑ + S O_{2} + 2 H_{2} O

(ii) Manganese dioxide and conc. $H_{2} S O_{4}$ : heating a mixture of the solid bromide, $Mn O_{2}$ and conc. $H_{2} S O_{4}$ gives reddish-brown bromine vapour.

2KBr + Mn O_{2} + 2 H_{2} S O_{4} B r_{2} ↑ + K_{2} S O_{4} + MnS O_{4} + 2 H_{2} O

The following tests are done with the aqueous salt solution.

(iii) Silver nitrate solution: pale yellow precipitate of silver bromide, sparingly soluble in dilute but readily soluble in concentrated ammonia, and insoluble in dilute $HN O_{3}$ .

B r^{-} + A g^{+} pale yellow AgBr ↓

(iv) Lead acetate solution: white crystalline precipitate of lead bromide, soluble in boiling water.

2B r^{-} + P b^{2 +} white PbB r_{2} ↓

(v) Chlorine water: added to the bromide solution shaken with $CC l_{4}$ , it frees bromine, which colours the organic layer orange-brown.

2B r^{-} + C l_{2} B r_{2} + 2C l^{-}

(vi) Potassium dichromate and conc. $H_{2} S O_{4}$ : heating the solid bromide with $K_{2} C r_{2} O_{7}$ and conc. $H_{2} S O_{4}$ and passing the vapour into water gives a yellowish-brown solution of bromine (not chromate, as the chromyl chloride test would).

6KBr + K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} 3B r_{2} + C r_{2} (S O_{4})_{3} + 4 K_{2} S O_{4} + 7 H_{2} O

5.3 Iodide

(i) Conc. $H_{2} S O_{4}$ : violet vapour of iodine.

2 I^{-} + 2 H_{2} S O_{4} violet I_{2} ↑ + S O_{4}^{2 -} + 2 H_{2} O + S O_{2}

The following tests are done with the aqueous salt solution.

(ii) Silver nitrate solution: yellow precipitate of silver iodide, $AgI$ , very slightly soluble in conc. ammonia and insoluble in dilute nitric acid.

I^{-} + A g^{+} yellow AgI ↓

(iii) Lead acetate solution: yellow precipitate of lead iodide. It dissolves in hot water to a colourless solution, which gives golden-yellow plates (spangles) on cooling.

2 I^{-} + P b^{2 +} yellow Pb I_{2} ↓

(iv) Potassium dichromate and conc. $H_{2} S O_{4}$ : iodine is liberated.

6 I^{-} + C r_{2} O_{7}^{2 -} + 7 H_{2} S O_{4} 3 I_{2} + 2C r^{3 +} + 7S O_{4}^{2 -} + 7 H_{2} O

(v) Chlorine water: added drop by drop, it liberates iodine; shaken with $CHC l_{3}$ (or $CC l_{4}$ ) it gives a violet organic layer.

2 I^{-} + C l_{2} I_{2} + 2C l^{-}

(vi) Copper sulphate solution: brown precipitate, a mixture of copper(I) iodide and iodine. Hypo solution removes the iodine and leaves white $CuI$ .

4 I^{-} + 2C u^{2 +} white 2CuI ↓ + brown I_{2}

I_{2} + 2 S_{2} O_{3}^{2 -} 2 I^{-} + S_{4} O_{6}^{2 -}

(vii) Mercury(II) chloride solution: scarlet precipitate of $Hg I_{2}$ , which dissolves in excess $KI$ as the tetraiodomercurate(II) complex.

2 I^{-} + HgC l_{2} scarlet Hg I_{2} ↓ + 2C l^{-}

Hg I_{2} + 2 I^{-} [Hg I_{4}]^{2 -}

Figure 5: Telling the three halides apart. Solubility of the silver halide in ammonia falls from

AgCl

AgI

K_{s p}

falls, and chlorine water frees

B r_{2}

(orange-brown) or

I_{2}

(violet) into the

CC l_{4}

layer.

Chloride vs bromide with conc.

H_{2} S O_{4}

HCl

is a weak reducing agent, so conc.

H_{2} S O_{4}

only displaces it: colourless

HCl

fumes.

Bromide and iodide

HBr

and

HI

are stronger reducing agents; conc.

H_{2} S O_{4}

oxidises them to

B r_{2}

(brown) and

I_{2}

(violet), and is itself reduced to

S O_{2}

5.4 Nitrate

Action of heat. The products depend on the metal.

Nitrate	Products on heating	Equation
Sodium, potassium	Nitrite + $O_{2}$ (glowing splinter relights)	$2NaN O_{3} Δ 2NaN O_{2} + O_{2}$
Ammonium	Dinitrogen oxide + steam	$N H_{4} N O_{3} Δ N_{2} O + 2 H_{2} O$
Noble metals (Ag)	Metal + $N O_{2}$ + $O_{2}$	$2AgN O_{3} Δ 2Ag + 2N O_{2} + O_{2}$
Other metals (Pb, Cu)	Oxide + $N O_{2}$ + $O_{2}$	$2Pb (N O_{3})_{2} Δ 2PbO + 4N O_{2} + O_{2}$

(i) Conc. $H_{2} S O_{4}$ : reddish-brown vapour of nitrogen dioxide on heating. Adding copper turnings makes the brown fumes much denser and the solution turns blue.

4N O_{3}^{-} + 2 H_{2} S O_{4} 4N O_{2} ↑ + 2S O_{4}^{2 -} + 2 H_{2} O + O_{2}

Cu + 4HN O_{3} blue Cu (N O_{3})_{2} + 2N O_{2} ↑ + 2 H_{2} O

The following tests are done with the aqueous salt solution.

(ii) Brown ring test: freshly prepared iron(II) sulphate solution is added to the nitrate solution, and conc. $H_{2} S O_{4}$ is poured slowly down the side of the tube so that it forms a layer underneath. A brown ring forms where the two layers meet (Figure 6).

2N O_{3}^{-} + 4 H_{2} S O_{4} + 6F e^{2 +} 6F e^{3 +} + 2NO + 4S O_{4}^{2 -} + 4 H_{2} O

[Fe (H_{2} O)_{6}]^{2 +} + NO brown ring [Fe (H_{2} O)_{5} (NO)]^{2 +} + H_{2} O

On shaking and warming, the brown colour disappears, nitric oxide escapes and a yellow solution of iron(III) ions remains.

Figure 6: The brown ring test for

N O_{3}^{-}

. The ring forms only at the junction, where conc.

H_{2} S O_{4}

reduces nitrate to

NO

and

NO

binds to

F e^{2 +}

; shaking or warming destroys it.

(iii) Zinc or aluminium in NaOH: nitrate is reduced to ammonia, which turns red litmus blue. The same reaction lets a nitrate be found after all ammonium has been boiled off with alkali.

N O_{3}^{-} + 4Zn + 7O H^{-} 4Zn O_{2}^{2 -} + N H_{3} ↑ + 2 H_{2} O

(iv) Diphenylamine: a drop of diphenylamine in conc. $H_{2} S O_{4}$ gives a deep blue colour with nitrates (and with other oxidising anions).

Brown ring with nitriteForms with dilute acetic or sulphuric acid, because

HN O_{2}

gives

NO

easily.

Brown ring with nitrateNeeds conc.

H_{2} S O_{4}

as an underlayer; the ring forms only at the junction. A nitrite present must be destroyed first (with urea or sulphamic acid), or it gives a false result.

Key idea

Chloride is the only halide that forms a volatile chromium compound; bromide and iodide are oxidised to the free halogen. Colour of the silver halide plus its solubility in ammonia separates all three.

6. Class B Anions: Tests in Solution

Class B includes anions identified by reactions in solution: precipitation ( $S O_{4}^{2 -}$ ) and oxidation and reduction ( $Cr O_{4}^{2 -}$ , $C r_{2} O_{7}^{2 -}$ , $Mn O_{4}^{-}$ ).

6.1 Sulphate

All sulphates except those of Ba, Pb and Sr are soluble in water; sulphates of calcium and mercury(II) are slightly soluble.

(i) Barium chloride solution: white precipitate of barium sulphate, insoluble in warm dilute hydrochloric acid and in dilute nitric acid, but moderately soluble in boiling conc. hydrochloric acid.

S O_{4}^{2 -} + B a^{2 +} white BaS O_{4} ↓

(ii) Mercury(II) nitrate solution: yellow precipitate of basic mercury(II) sulphate.

S O_{4}^{2 -} + 3H g^{2 +} + 2 H_{2} O yellow HgS O_{4} \cdot 2HgO ↓ + 4 H^{+}

(iii) Lead acetate solution: white precipitate of $PbS O_{4}$ , soluble in excess ammonium acetate.

S O_{4}^{2 -} + P b^{2 +} white PbS O_{4} ↓

PbS O_{4} + 2C H_{3} COON H_{4} (C H_{3} COO)_{2} Pb + (N H_{4})_{2} S O_{4}

6.2 Chromate and dichromate

Metallic chromates give yellow solutions in water. In acid, chromates change into orange-red dichromates; in alkali the change reverses.

yellow 2Cr O_{4}^{2 -} + 2 H^{+} ⇌ orange C r_{2} O_{7}^{2 -} + H_{2} O

C r_{2} O_{7}^{2 -} + 2O H^{-} ⇌ 2Cr O_{4}^{2 -} + H_{2} O

The first equilibrium goes through the hydrogen chromate ion:

2Cr O_{4}^{2 -} + 2 H^{+} ⇌ 2HCr O_{4}^{-} ⇌ C r_{2} O_{7}^{2 -} + H_{2} O

(i) Barium chloride solution: pale yellow precipitate of barium chromate, soluble in dilute mineral acids but insoluble in water and acetic acid.

Cr O_{4}^{2 -} + B a^{2 +} pale yellow BaCr O_{4} ↓

Dichromate gives the same precipitate, but only partly, because a strong acid forms. Adding sodium hydroxide or sodium acetate makes the precipitation complete.

C r_{2} O_{7}^{2 -} + 2B a^{2 +} + H_{2} O ⇌ 2BaCr O_{4} ↓ + 2 H^{+}

(ii) Silver nitrate solution: brownish-red precipitate of silver chromate, soluble in dilute nitric acid and in ammonia, but not in acetic acid. Chloride converts it to $AgCl$ .

Cr O_{4}^{2 -} + 2A g^{+} brick red A g_{2} Cr O_{4} ↓

2A g_{2} Cr O_{4} + 2 H^{+} 4A g^{+} + C r_{2} O_{7}^{2 -} + H_{2} O

A g_{2} Cr O_{4} + 4N H_{3} 2 [Ag (N H_{3})_{2}]^{+} + Cr O_{4}^{2 -}

A g_{2} Cr O_{4} + 2C l^{-} 2AgCl ↓ + Cr O_{4}^{2 -}

A concentrated dichromate solution gives reddish-brown silver dichromate.

C r_{2} O_{7}^{2 -} + 2A g^{+} reddish brown A g_{2} C r_{2} O_{7} ↓

(iii) Lead acetate solution: yellow precipitate of lead chromate, insoluble in acetic acid but soluble in dilute nitric acid.

Cr O_{4}^{2 -} + P b^{2 +} yellow PbCr O_{4} ↓

2PbCr O_{4} + 2 H^{+} ⇌ 2P b^{2 +} + C r_{2} O_{7}^{2 -} + H_{2} O

(iv) Hydrogen peroxide: an acidified chromate solution treated with $H_{2} O_{2}$ turns deep blue because chromium pentoxide forms (it is extracted into ether or amyl alcohol). $Cr O_{5}$ is unstable and decomposes to oxygen and a green $C r^{3 +}$ salt.

Cr O_{4}^{2 -} + 2 H^{+} + 2 H_{2} O_{2} deep blue Cr O_{5} + 3 H_{2} O

6.3 Permanganate

(i) Hydrogen peroxide: decolourises acidified potassium permanganate, with evolution of oxygen.

2Mn O_{4}^{-} + 5 H_{2} O_{2} + 6 H^{+} 5 O_{2} ↑ + 2M n^{2 +} + 8 H_{2} O

(ii) Iron(II) sulphate in dilute sulphuric acid reduces permanganate to manganese(II); the solution turns yellow because iron(III) ions form.

Mn O_{4}^{-} + 5F e^{2 +} + 8 H^{+} 5F e^{3 +} + M n^{2 +} + 4 H_{2} O

(iii) Action of heat: a black residue of potassium manganate and manganese dioxide remains. Extracting with water and filtering gives a green solution of potassium manganate.

2KMn O_{4} Δ green K_{2} Mn O_{4} + Mn O_{2} + O_{2} ↑

Quick Recall: tap to check

Which barium precipitate dissolves in dilute

HCl

BaS O_{4}

BaS O_{3}

BaCr O_{4}

BaS O_{3}

and

BaCr O_{4}

dissolve;

BaS O_{4}

does not.

What turns an acidified chromate deep blue?

H_{2} O_{2}

, forming

Cr O_{5}

Why does a yellow chromate solution turn orange with acid?

H^{+}

shifts the equilibrium to dichromate,

C r_{2} O_{7}^{2 -}

7. Classification of Cations into Groups

For systematic analysis, cations are divided into groups by their behaviour with a few reagents. The basis is whether a cation forms a precipitate with the group reagent or not, which is a difference in solubility. These analytical groups have nothing to do with groups of the periodic table.

Group	Group reagent	Ions	Precipitate and colour
I	dil. $HCl$	$P b^{2 +}$ , $H g_{2}^{2 +}$ , $A g^{+}$	$PbC l_{2}$ , $H g_{2} C l_{2}$ , $AgCl$ : white
IIA (copper group)	$H_{2} S$ in dil. $HCl$	$H g^{2 +}$ , $C u^{2 +}$ , $B i^{3 +}$ , $C d^{2 +}$ , $P b^{2 +}$	Black: $HgS$ , $CuS$ , $PbS$ ; brown: $B i_{2} S_{3}$ ; yellow: $CdS$
IIB (arsenic group)	$H_{2} S$ in dil. $HCl$	$A s^{3 +}$ , $A s^{5 +}$ , $S b^{3 +}$ , $S b^{5 +}$ , $S n^{2 +}$ , $S n^{4 +}$	Yellow: $A s_{2} S_{3}$ , $A s_{2} S_{5}$ , $Sn S_{2}$ ; orange: $S b_{2} S_{3}$ , $S b_{2} S_{5}$ ; brown: $SnS$
IIIA	$N H_{4} OH$ in presence of $N H_{4} Cl$	$F e^{3 +}$ , $A l^{3 +}$ , $C r^{3 +}$	$Fe (OH)_{3}$ reddish brown, $Al (OH)_{3}$ white, $Cr (OH)_{3}$ green
IIIB	$H_{2} S$ in presence of $N H_{4} OH$ and $N H_{4} Cl$ (or $(N H_{4})_{2} S$ )	$N i^{2 +}$ , $C o^{2 +}$ , $M n^{2 +}$ , $Z n^{2 +}$	$ZnS$ white or grey; $CoS$ , $NiS$ black; $MnS$ buff (flesh)
IV	$(N H_{4})_{2} C O_{3}$ in presence of $N H_{4} Cl$ and $N H_{4} OH$	$B a^{2 +}$ , $S r^{2 +}$ , $C a^{2 +}$	$BaC O_{3}$ , $SrC O_{3}$ , $CaC O_{3}$ : white
V	No common group reagent	$M g^{2 +}$ , $N a^{+}$ , $K^{+}$ , $N H_{4}^{+}$	$M g^{2 +}$ with $N a_{2} HP O_{4}$ ; others by their own tests

Mercury(I) exists as the dimeric ion $H g_{2}^{2 +}$ (two Hg atoms joined by a metal-metal bond), which is why its chloride is written $H g_{2} C l_{2}$ .

7.1 The same groups numbered 0 to VI

Some books and question papers number the same groups 0 to VI. The reagents and their order do not change: ammonium ion is tested first on the original salt and called group 0, and groups IIIA and IIIB become III and IV.

Ions	Scheme above	0 to VI numbering
$N H_{4}^{+}$	V	0
$P b^{2 +}$ , $A g^{+}$ , $H g_{2}^{2 +}$	I	I
$C u^{2 +}$ , $P b^{2 +}$ , $A s^{3 +}$ (group II ions)	II	II
$F e^{3 +}$ , $A l^{3 +}$ , $C r^{3 +}$	IIIA	III
$Z n^{2 +}$ , $M n^{2 +}$ , $N i^{2 +}$ , $C o^{2 +}$	IIIB	IV
$B a^{2 +}$ , $S r^{2 +}$ , $C a^{2 +}$	IV	V
$M g^{2 +}$	V	VI

Figure 7: Group reagents are added in a fixed order and each filtrate goes to the next step.

Exam Trick

Which numbering is the question using? If zinc, manganese, nickel or cobalt is called group IV, or calcium and barium group V, the question uses the 0 to VI numbering. The reagent never changes, only the number, so answer from the reagent.

8. Why the Group Reagents Work: Solubility Product and Common Ion Effect

The split of cations into groups rests on the solubility product. To see how it works in analysis, combine it with the law of mass action and the common ion effect: adding a strong electrolyte that shares an ion with a weak electrolyte suppresses the weak electrolyte's ionisation.

For a sparingly soluble salt

AB

in a saturated solution,

K_{s p} = [A^{+}] [B^{-}]

. If the ionic product

Q

(the same product for any solution) exceeds

K_{s p}

, the salt precipitates until

Q

falls back to

K_{s p}

8.1 Group I: chlorides with dilute $HCl$

The metals of group I are precipitated by chloride ions from dil. $HCl$ . Their solubility products at room temperature are $PbC l_{2}$ $1.6 \times 1 0^{- 5}$ , $AgCl$ $1.8 \times 1 0^{- 10}$ and $H g_{2} C l_{2}$ $1.3 \times 1 0^{- 18}$ .

Take $AgCl$ . In its saturated solution

AgCl (s) ⇌ A g^{+} + C l^{-}

$K_{s p} (AgCl) = [A g^{+}] [C l^{-}]$ . Adding dil. $HCl$ raises $[C l^{-}]$ , so for a moment $[A g^{+}] [C l^{-}]$ exceeds $K_{s p}$ . $A g^{+}$ and $C l^{-}$ then combine as solid $AgCl$ until the product equals $K_{s p}$ again. Because $K_{s p}$ is constant, the larger $[C l^{-}]$ , the smaller $[A g^{+}]$ left in solution, so a slight excess of reagent removes silver almost completely. $[A g^{+}]$ never becomes exactly zero; it only approaches zero.

In the same way, the extra $C l^{-}$ pushes the ionic products of $P b^{2 +}$ and $H g_{2}^{2 +}$ above the $K_{s p}$ of $PbC l_{2}$ and $H g_{2} C l_{2}$ . The chlorides of later groups are far more soluble, so this $[C l^{-}]$ cannot reach their solubility products and they stay in solution. $PbC l_{2}$ has a fairly large $K_{s p}$ , so its precipitation is incomplete; the lead that stays in solution comes down again as black $PbS$ in group II.

8.2 Group II and group IIIB: sulphides

Group II sulphides are precipitated by $H_{2} S$ in dil. $HCl$ , while group IIIB sulphides need $H_{2} S$ in $N H_{4} OH$ . Take $CuS$ (group II, $K_{s p} \approx 6.3 \times 1 0^{- 36}$ ) and $ZnS$ (group IIIB, $K_{s p} \approx 1.6 \times 1 0^{- 24}$ ). A sulphide precipitates only when $[M^{2 +}] [S^{2 -}]$ exceeds its $K_{s p}$ .

$H_{2} S$ is a weak acid and ionises only feebly:

H_{2} S ⇌ 2 H^{+} + S^{2 -}

HCl H^{+} + C l^{-}

$HCl$ is a strong electrolyte. Its $H^{+}$ is common to both, so it pushes the $H_{2} S$ equilibrium back and cuts $[S^{2 -}]$ sharply. This low $[S^{2 -}]$ is still enough to exceed the tiny $K_{s p}$ of group II sulphides, but not the much larger $K_{s p}$ of group IIIB sulphides:

$[M^{2 +}] [S^{2 -}] > K_{s p} (MS)$ for group II, but $[M^{2 +}] [S^{2 -}] < K_{s p} (MS)$ for group IIIB.

That is why group IIIB sulphides are not precipitated in dil. $HCl$ . In ammoniacal solution the reverse happens:

N H_{4} OH ⇌ N H_{4}^{+} + O H^{-}

H^{+} + O H^{-} H_{2} O

$O H^{-}$ removes $H^{+}$ as un-ionised water, so more $H_{2} S$ ionises and $[S^{2 -}]$ rises. The higher $[S^{2 -}]$ now exceeds the solubility products of the group IIIB sulphides, and they precipitate (Figure 8).

Figure 8: Why

H_{2} S

in acid separates group II from group IIIB.

K_{s p} (ZnS)

is about

1 0^{11}

times

K_{s p} (CuS)

, so the tiny

[S^{2 -}]

in acid precipitates

CuS

but not

ZnS

. Dashed lines are schematic;

K_{s p}

values are approximate and vary between tables.

If $H_{2} S$ is passed through a neutral solution, precipitation of group IIIB is incomplete, because the $HCl$ formed lowers the ionisation of $H_{2} S$ :

MnC l_{2} + H_{2} S ⇌ MnS ↓ + 2HCl

8.3 Group IIIA: hydroxides with $N H_{4} Cl$ and $N H_{4} OH$

Group IIIA hydroxides are precipitated by $N H_{4} OH$ in the presence of $N H_{4} Cl$ . $N H_{4} OH$ is a weak base and dissociates only slightly, while $N H_{4} Cl$ is a strong electrolyte:

N H_{4} OH ⇌ N H_{4}^{+} + O H^{-}

N H_{4} Cl N H_{4}^{+} + C l^{-}

The common $N H_{4}^{+}$ ion suppresses the dissociation of $N H_{4} OH$ : some $O H^{-}$ and $N H_{4}^{+}$ recombine, and $[O H^{-}]$ falls. This lowered $[O H^{-}]$ still exceeds the very small solubility products of the hydroxides of Fe, Al and Cr, so group IIIA precipitates. It is too low for the larger solubility products of the hydroxides of later groups (Zn, Mn, Ni, Co, Mg), so they stay in solution. Without $N H_{4} Cl$ , the undiminished $[O H^{-}]$ would precipitate the hydroxides of groups IIIA, IIIB and IV together.

Figure 9:

N H_{4} Cl

lowers

[O H^{-}]

from

1.3 \times 1 0^{- 3}

1.8 \times 1 0^{- 6} M

Q

for

Mg (OH)_{2}

falls from

1.8 \times 1 0^{- 7}

3.2 \times 1 0^{- 13}

, below its

K_{s p}

(

1.8 \times 1 0^{- 11}

), while

Q = 5.8 \times 1 0^{- 19}

for

M (OH)_{3}

is still far above

K_{s p}

8.4 Group IV: carbonates

$N H_{4} OH$ and $(N H_{4})_{2} C O_{3}$ are added to precipitate the carbonates of Ba, Sr and Ca. With $N H_{4} Cl$ and $N H_{4} OH$ present, conditions keep $[C O_{3}^{2 -}]$ at about $1.6 \times 1 0^{- 3} M$ . The lowest metal-ion concentration that gives an appreciable precipitate is about $1 0^{- 4} M$ , so the ionic product is $1.6 \times 1 0^{- 3} \times 1 0^{- 4} = 1.6 \times 1 0^{- 7}$ . This exceeds the solubility products of $BaC O_{3}$ , $SrC O_{3}$ and $CaC O_{3}$ (about $1 0^{- 9}$ to $1 0^{- 10}$ ) but not the much larger solubility product of $MgC O_{3}$ , so only Ba, Sr and Ca carbonates precipitate in group IV.

$M g^{2 +}$ remains in solution. With $N a_{2} HP O_{4}$ (and ammonia), insoluble magnesium ammonium phosphate, $Mg (N H_{4}) P O_{4}$ , is precipitated.

8.5 Points to remember

Group I cations ( $A g^{+}$ , $P b^{2 +}$ , $H g_{2}^{2 +}$ ) are precipitated as chlorides because $AgCl$ , $PbC l_{2}$ and $H g_{2} C l_{2}$ have much lower solubility products than all other chlorides, which stay in solution.
Group II cations are precipitated as sulphides in dil. $HCl$ . $HCl$ supplies $H^{+}$ , which lowers $[S^{2 -}]$ by the common ion effect; this low $[S^{2 -}]$ is enough only for the group II sulphides, which have the lowest solubility products.
Group IIIA cations are precipitated as hydroxides. $N H_{4} Cl$ suppresses the ionisation of $N H_{4} OH$ , so only the group IIIA hydroxides, with their very low solubility products, precipitate.
$N H_{4} OH$ increases the ionisation of $H_{2} S$ by removing $H^{+}$ as un-ionised water. The extra $S^{2 -}$ makes the ionic products of group IIIB sulphides exceed their solubility products.

Notes on group IIIA:

Excess $N H_{4} Cl$ must be added; otherwise manganese partly precipitates, and air oxidises it to brown hydrated manganese dioxide.
$(N H_{4})_{2} S O_{4}$ cannot replace $N H_{4} Cl$ : the $S O_{4}^{2 -}$ would precipitate barium as $BaS O_{4}$ too early.
$N H_{4} N O_{3}$ cannot replace $N H_{4} Cl$ : $N O_{3}^{-}$ would oxidise $M n^{2 +}$ to $M n^{3 +}$ , and $Mn (OH)_{3}$ would precipitate in group IIIA.
Before group IIIA, $H_{2} S$ is boiled off and a few drops of conc. $HN O_{3}$ are added. This oxidises $F e^{2 +}$ (formed when $H_{2} S$ reduces $F e^{3 +}$ ) back to $F e^{3 +}$ : $Fe (OH)_{2}$ is too soluble to precipitate completely here, while $Fe (OH)_{3}$ precipitates fully.
Of the three group IIIA hydroxides, only $Al (OH)_{3}$ stays dissolved in excess $NaOH$ after boiling (as sodium meta-aluminate); $Fe (OH)_{3}$ and $Cr (OH)_{3}$ remain insoluble.

JEE Advanced

Why the acid strength in group II matters. The gap between $K_{s p} (CdS)$ ( $\approx 1 0^{- 27}$ ) and $K_{s p} (ZnS)$ ( $\approx 1 0^{- 24}$ ) is small, and $[S^{2 -}] \propto 1/ [H^{+}]^{2}$ . The acid is therefore kept near 0.3 M HCl: much stronger acid leaves $CdS$ and $PbS$ incomplete in group II, and much weaker acid lets $ZnS$ come down early. In group IIIA, the ionic product for $M (OH)_{2}$ with $N H_{4} Cl$ (about $3 \times 1 0^{- 13}$ ) still exceeds $K_{s p}$ of $Zn (OH)_{2}$ and $Ni (OH)_{2}$ ; these ions stay dissolved because they form ammine complexes, $[Zn (N H_{3})_{4}]^{2 +}$ and $[Ni (N H_{3})_{6}]^{2 +}$ . $Mn (OH)_{2}$ ( $K_{s p} \approx 1 0^{- 13}$ ) sits close to the limit, which is why a large excess of $N H_{4} Cl$ is needed.

Key idea

Every group reagent is a solubility-product switch: the common ion effect sets

[S^{2 -}]

[O H^{-}]

between the

K_{s p}

of the ions to be precipitated and the

K_{s p}

of the ions to be left behind.

9. Confirmatory Tests for Cations

Each group precipitate is dissolved or treated further, and every ion is confirmed by its own test on the salt solution.

9.1 Group I: $P b^{2 +}$ , $A g^{+}$ , $H g_{2}^{2 +}$

(A) Lead. $PbC l_{2}$ dissolves in hot water. The solution gives a yellow precipitate with $K_{2} Cr O_{4}$ , insoluble in acetic acid but soluble in $NaOH$ .

PbC l_{2} + K_{2} Cr O_{4} yellow PbCr O_{4} ↓ + 2KCl

PbCr O_{4} + 4NaOH N a_{2} [Pb O_{2}] + N a_{2} Cr O_{4} + 2 H_{2} O

(B) With $KI$ , lead gives a yellow precipitate that dissolves in excess $KI$ .

PbC l_{2} + 2KI yellow Pb I_{2} ↓ + 2KCl

Pb I_{2} + 2KI K_{2} [Pb I_{4}]

(C) Silver and mercury(I). $AgCl$ dissolves in $N H_{4} OH$ as a complex, while $H g_{2} C l_{2}$ turns black with $N H_{4} OH$ .

AgCl + 2N H_{4} OH [Ag (N H_{3})_{2}] Cl + 2 H_{2} O

H g_{2} C l_{2} + 2N H_{4} OH white Hg (N H_{2}) Cl ↓ + black Hg ↓ + N H_{4} Cl + 2 H_{2} O

The black colour comes from finely divided mercury mixed with white mercury(II) amidochloride (amino mercuric chloride).

Figure 10: Group I is split by two solubility facts:

PbC l_{2}

dissolves in hot water, and

AgCl

dissolves in

N H_{4} OH

while

H g_{2} C l_{2}

turns black.

9.2 Group IIA: $H g^{2 +}$ , $C u^{2 +}$ , $B i^{3 +}$ , $C d^{2 +}$

(i) Mercury(II): tin(II) chloride gives a white precipitate that turns grey-black.

2HgC l_{2} + SnC l_{2} white H g_{2} C l_{2} ↓ + SnC l_{4}

H g_{2} C l_{2} + SnC l_{2} black 2Hg ↓ + SnC l_{4}

(ii) Copper(II): a deep blue colour with excess $N H_{4} OH$ , and a chocolate-brown precipitate with potassium ferrocyanide (in acetic acid).

C u^{2 +} + 4N H_{4} OH deep blue [Cu (N H_{3})_{4}]^{2 +} + 4 H_{2} O

2C u^{2 +} + K_{4} [Fe (CN)_{6}] chocolate brown C u_{2} [Fe (CN)_{6}] ↓ + 4 K^{+}

(iii) Bismuth(III): a solution of $BiC l_{3}$ in $HCl$ turns white and cloudy when diluted with water. Sodium stannite gives a black precipitate of bismuth metal.

BiC l_{3} + H_{2} O white BiOCl ↓ + 2HCl

2BiC l_{3} + 3N a_{2} Sn O_{2} + 6NaOH black 2Bi ↓ + 3N a_{2} Sn O_{3} + 6NaCl + 3 H_{2} O

(iv) Cadmium(II): $NaOH$ gives a white precipitate; $N H_{4} OH$ gives a white precipitate that dissolves in excess. In group II, $C d^{2 +}$ is recognised by its yellow sulphide.

C d^{2 +} + 2NaOH white Cd (OH)_{2} ↓ + 2N a^{+}

C d^{2 +} + 4N H_{4} OH [Cd (N H_{3})_{4}]^{2 +} + 4 H_{2} O

9.3 Group IIB: $A s^{3 +}$ , $S b^{3 +}$ , $S n^{2 +}$

Group IIB sulphides dissolve in yellow ammonium sulphide (as thio salts); group IIA sulphides do not. This separates the two sub-groups.

(v) Arsenic: nitric acid oxidises it to arsenic acid, which gives a yellow precipitate with ammonium molybdate on heating.

H_{3} As O_{4} + 12 (N H_{4})_{2} Mo O_{4} + 21HN O_{3} Δ yellow (N H_{4})_{3} AsM o_{12} O_{40} ↓ + 21N H_{4} N O_{3} + 12 H_{2} O

(vi) Tin(II): $SnC l_{2}$ gives a white precipitate with $HgC l_{2}$ , which turns black on standing.

SnC l_{2} + 2HgC l_{2} SnC l_{4} + white H g_{2} C l_{2} ↓

H g_{2} C l_{2} + SnC l_{2} SnC l_{4} + black 2Hg ↓

(vii) Antimony(III): $SbC l_{3}$ solution gives a white precipitate when diluted with water.

SbC l_{3} + H_{2} O white SbOCl ↓ + 2HCl

9.4 Group IIIA: $A l^{3 +}$ , $F e^{3 +}$ , $C r^{3 +}$

(i) Aluminium: white gelatinous $Al (OH)_{3}$ dissolves in $NaOH$ .

Al (OH)_{3} + NaOH NaAl O_{2} + 2 H_{2} O

Blue lake test: the precipitate is dissolved in dil. $HCl$ , a drop of blue litmus is added, then $N H_{4} OH$ drop by drop. $Al (OH)_{3}$ precipitates and adsorbs the dye, giving a blue ‘lake’ floating in a colourless solution.

(ii) Chromium: $Cr (OH)_{3}$ dissolves in $NaOH$ + bromine water (or $H_{2} O_{2}$ ) as yellow chromate; $BaC l_{2}$ (or lead acetate) then gives a yellow precipitate.

B r_{2} + H_{2} O 2HBr + [O]

2Cr (OH)_{3} + 4NaOH + 3 [O] yellow 2N a_{2} Cr O_{4} + 5 H_{2} O

N a_{2} Cr O_{4} + BaC l_{2} yellow BaCr O_{4} ↓ + 2NaCl

(iii) Iron(III): $Fe (OH)_{3}$ does not dissolve in $NaOH$ . The brown precipitate is dissolved in $HCl$ ; potassium thiocyanate then gives a blood-red colour, and potassium ferrocyanide gives Prussian blue.

Fe (OH)_{3} + 3HCl FeC l_{3} + 3 H_{2} O

FeC l_{3} + 3KSCN blood red Fe (SCN)_{3} + 3KCl

4FeC l_{3} + 3 K_{4} [Fe (CN)_{6}] Prussian blue F e_{4} [Fe (CN)_{6}]_{3} ↓ + 12KCl

The blood-red species is mainly

[Fe (SCN)]^{2 +}

(written

Fe (SCN)_{3}

in older books). Iron(II) gives a deep blue precipitate with potassium ferricyanide (

K_{3} [Fe (CN)_{6}]

), Turnbull's blue; modern work shows it has the same structure as Prussian blue.

9.5 Group IIIB: $N i^{2 +}$ , $C o^{2 +}$ , $M n^{2 +}$ , $Z n^{2 +}$

(i) Nickel and cobalt with $KHC O_{3}$ and bromine water: an apple-green colour shows cobalt; a black precipitate shows nickel.

CoC l_{2} + 6KHC O_{3} K_{4} [Co (C O_{3})_{3}] + 2KCl + 3C O_{2} + 3 H_{2} O

2 K_{4} [Co (C O_{3})_{3}] + 2KHC O_{3} + [O] apple green 2 K_{3} [Co (C O_{3})_{3}] + 2 K_{2} C O_{3} + H_{2} O

NiC l_{2} + 2KHC O_{3} NiC O_{3} + 2KCl + H_{2} O + C O_{2}

2NiC O_{3} + 4NaOH + [O] black N i_{2} O_{3} ↓ + 2N a_{2} C O_{3} + 2 H_{2} O

(ii) Nickel with dimethylglyoxime (DMG): in ammoniacal solution, DMG gives a bright rosy-red precipitate of nickel dimethylglyoximate. This is the standard confirmatory test for nickel.

N i^{2 +} + 2 C_{4} H_{8} N_{2} O_{2} rosy red Ni (C_{4} H_{7} N_{2} O_{2})_{2} ↓ + 2 H^{+}

(iii) Zinc: $NaOH$ gives a white precipitate that dissolves in excess (zinc hydroxide is amphoteric). Potassium ferrocyanide gives a white or bluish-white precipitate.

Z n^{2 +} + 2NaOH white Zn (OH)_{2} ↓ + 2N a^{+}

Zn (OH)_{2} + 2NaOH soluble N a_{2} Zn O_{2} + 2 H_{2} O

2Z n^{2 +} + [Fe (CN)_{6}]^{4 -} bluish white Z n_{2} [Fe (CN)_{6}] ↓

(iv) Manganese: $NaOH$ gives a white (faintly pink) precipitate of $Mn (OH)_{2}$ , which turns brown or black in air or on heating as it is oxidised.

M n^{2 +} + 2NaOH white Mn (OH)_{2} ↓ + 2N a^{+}

Mn (OH)_{2} + [O] brown or black Mn O_{2} ↓ + H_{2} O

Boiling a manganese salt with lead dioxide (or sodium bismuthate) and conc. $HN O_{3}$ gives a purple solution of permanganate. This confirms manganese.

2M n^{2 +} + 5Pb O_{2} + 4 H^{+} purple 2Mn O_{4}^{-} + 5P b^{2 +} + 2 H_{2} O

2M n^{2 +} + 5NaBi O_{3} + 14 H^{+} 2Mn O_{4}^{-} + 5B i^{3 +} + 5N a^{+} + 7 H_{2} O

9.6 Group IV: $B a^{2 +}$ , $S r^{2 +}$ , $C a^{2 +}$

(i) Barium gives (A) a yellow precipitate with $K_{2} Cr O_{4}$ (in acetic acid), (B) a white precipitate with $(N H_{4})_{2} S O_{4}$ , and (C) a white precipitate with $(N H_{4})_{2} C_{2} O_{4}$ . Its flame is apple green.

B a^{2 +} + K_{2} Cr O_{4} yellow BaCr O_{4} ↓ + 2 K^{+}

B a^{2 +} + (N H_{4})_{2} S O_{4} white BaS O_{4} ↓ + 2N H_{4}^{+}

B a^{2 +} + (N H_{4})_{2} C_{2} O_{4} white Ba C_{2} O_{4} ↓ + 2N H_{4}^{+}

(ii) Strontium gives white precipitates with $(N H_{4})_{2} S O_{4}$ and with $(N H_{4})_{2} C_{2} O_{4}$ ; its flame is crimson red.

S r^{2 +} + (N H_{4})_{2} S O_{4} white SrS O_{4} ↓ + 2N H_{4}^{+}

S r^{2 +} + (N H_{4})_{2} C_{2} O_{4} white Sr C_{2} O_{4} ↓ + 2N H_{4}^{+}

(iii) Calcium gives a white precipitate with $(N H_{4})_{2} C_{2} O_{4}$ only (insoluble in acetic acid); its flame is brick red.

C a^{2 +} + (N H_{4})_{2} C_{2} O_{4} white Ca C_{2} O_{4} ↓ + 2N H_{4}^{+}

9.7 Group V: $N H_{4}^{+}$ , $N a^{+}$ , $K^{+}$ , $M g^{2 +}$

(i) Ammonium: every ammonium salt heated with an alkali such as $NaOH$ gives a colourless, pungent gas, $N H_{3}$ .

N H_{4} Cl + NaOH Δ NaCl + N H_{3} ↑ + H_{2} O

(A) The gas gives white fumes with $HCl$ .

N H_{3} + HCl white fumes N H_{4} Cl

(B) Paper soaked in $CuS O_{4}$ solution turns deep blue, because a complex forms.

CuS O_{4} + 4N H_{3} deep blue [Cu (N H_{3})_{4}] S O_{4}

H g_{2} (N O_{3})_{2} + 2N H_{3} black Hg ↓ + Hg (N H_{2}) N O_{3} ↓ + N H_{4} N O_{3}

(D) Nessler's reagent (an alkaline solution of potassium tetraiodomercurate(II)) gives a brown precipitate, the iodide of Millon's base.

N H_{4} Cl + 2 K_{2} [Hg I_{4}] + 4KOH brown HgO \cdot Hg (N H_{2}) I ↓ + 7KI + KCl + 3 H_{2} O

(ii) Potassium salts give a yellow precipitate with sodium cobaltinitrite; the flame is lilac.

N a_{3} [Co (N O_{2})_{6}] + 3KCl yellow K_{3} [Co (N O_{2})_{6}] ↓ + 3NaCl

(iii) Sodium salts give a heavy white precipitate with potassium dihydrogen antimonate; the flame is golden yellow.

K H_{2} Sb O_{4} + NaCl white Na H_{2} Sb O_{4} ↓ + KCl

(iv) Magnesium: $N H_{4} OH$ (without $N H_{4} Cl$ ) or $NaOH$ gives white $Mg (OH)_{2}$ . The confirmatory test is disodium hydrogen phosphate with ammonia, which gives a white crystalline precipitate of magnesium ammonium phosphate. Magneson reagent gives a blue precipitate (lake) in alkaline solution.

M g^{2 +} + 2N H_{3} + 2 H_{2} O white Mg (OH)_{2} ↓ + 2N H_{4}^{+}

M g^{2 +} + HP O_{4}^{2 -} + N H_{3} white Mg (N H_{4}) P O_{4} ↓

Figure 11: The colours examiners expect you to recall. Most identification questions give a colour and ask for the ion, so learn this atlas both ways.

Exam Trick

Iron blues. Ferric with ferro, ferrous with ferri: $F e^{3 +}$ + ferrocyanide gives Prussian blue; $F e^{2 +}$ + ferricyanide gives Turnbull's blue. $F e^{3 +}$ with ferricyanide gives only a brown colour.

Quick Recall: tap to check

Which group I chloride dissolves in hot water?

PbC l_{2}

What turns black when

N H_{4} OH

is poured on the group I residue?

H g_{2} C l_{2}

: it gives

Hg

Hg (N H_{2}) Cl

Which reagent confirms

N i^{2 +}

Dimethylglyoxime in ammonia: rosy-red precipitate.

How is

M n^{2 +}

confirmed?

Boiling with

Pb O_{2}

(or

NaBi O_{3}

) and conc.

HN O_{3}

gives purple

Mn O_{4}^{-}

Which reagent precipitates

M g^{2 +}

after group IV?

N a_{2} HP O_{4}

with

N H_{4} OH

: white

Mg (N H_{4}) P O_{4}

Key idea

Confirmatory tests are chosen to be specific: a colour or precipitate that no other ion left at that stage can give.

Figure 12: The whole chapter on one page. Revise it branch by branch before attempting the solved examples.

10. Solved Examples

Solved Example 1

You are given two test tubes, one with an aqueous solution of a carbonate and the other with a sulphite. How will you identify them?

Solution:

Add acidified $KMn O_{4}$ to each. The sulphite decolourises it; the carbonate does not react, so its solution stays pink.

2Mn O_{4}^{-} + 5S O_{3}^{2 -} + 6 H^{+} 2M n^{2 +} + 5S O_{4}^{2 -} + 3 H_{2} O

Acidified dichromate can be used in the same way: sulphite turns it green, carbonate does not.

Solved Example 2

A colourless salt, on heating, forms a white sublimate on the cooler sides of the test tube. The gas evolved is colourless and has a pungent smell. It gives dense white fumes when a rod dipped in conc.

HCl

is brought near the mouth of the tube. Identify the basic radical.

Solution:

A white sublimate plus a pungent gas that fumes with $HCl$ (that is, $N H_{3}$ ) shows the ammonium ion, as in $N H_{4} Cl$ . On heating, the salt splits into $N H_{3}$ and $HCl$ , which recombine on the cooler part of the tube as a white sublimate.

N H_{4} Cl ⇌ Δ N H_{3} ↑ + HCl ↑

N H_{3} + HCl white fumes N H_{4} Cl

Solved Example 3

When silver nitrate solution is added to a solution of a salt A, a black precipitate forms. Salt A also decolourises acidified permanganate solution. Identify the acid radical in A.

Solution:

A black silver salt and a reducing anion point to sulphide.

S^{2 -} + 2A g^{+} black A g_{2} S ↓

5 S^{2 -} + 2Mn O_{4}^{-} + 16 H^{+} 5S ↓ + 2M n^{2 +} + 8 H_{2} O

Solved Example 4

A salt decrepitates on heating. Heated with manganese dioxide and conc.

H_{2} S O_{4}

, it gives a greenish-yellow gas that turns moist blue litmus red and then colourless. In the flame test the salt gives a crimson-red flame. Identify the cation and the anion.

Solution:

The greenish-yellow gas is chlorine, so the anion is chloride. Chlorine with water gives an acid (litmus turns red) and nascent oxygen, which bleaches the dye.

2C l^{-} + Mn O_{2} + 2 H_{2} S O_{4} MnS O_{4} + 2 H_{2} O + C l_{2} ↑ + S O_{4}^{2 -}

C l_{2} + H_{2} O 2HCl + [O]

A crimson-red flame shows strontium, so the salt is $SrC l_{2}$ . (Lithium also gives a crimson flame, but it is not among the cations analysed here.)

Solved Example 5

A salt solution gives a yellow precipitate with

BaC l_{2}

, soluble in dil.

HCl

. On adding acid to the salt solution, it turns orange. Identify the acid radical.

Solution:

The radical is chromate: yellow $BaCr O_{4}$ dissolves in mineral acid, and acid turns yellow chromate into orange dichromate.

Cr O_{4}^{2 -} + B a^{2 +} yellow BaCr O_{4} ↓

2Cr O_{4}^{2 -} + 2 H^{+} ⇌ orange C r_{2} O_{7}^{2 -} + H_{2} O

Solved Example 6

With dil.

HCl

the salt gives a colourless gas that turns brown in air. The salt solution gives a brown colour with

FeS O_{4}

solution. Identify the acid radical.

Solution:

The radical is nitrite. Nitrous acid gives colourless $NO$ , which air turns into brown $N O_{2}$ ; $NO$ with $F e^{2 +}$ gives the brown nitrosyl complex even with dilute acid.

N O_{2}^{-} + H^{+} HN O_{2}

3HN O_{2} HN O_{3} + 2NO + H_{2} O

2NO + O_{2} brown 2N O_{2}

[Fe (H_{2} O)_{6}]^{2 +} + NO brown [Fe (H_{2} O)_{5} (NO)]^{2 +} + H_{2} O

Solved Example 7

Though

Mg (OH)_{2}

is insoluble,

M g^{2 +}

is not precipitated in group III (IIIA). Explain.

Solution:

The group reagent is $N H_{4} OH$ with plenty of $N H_{4} Cl$ . $N H_{4} Cl$ is a strong electrolyte; its $N H_{4}^{+}$ suppresses the dissociation of the weak base $N H_{4} OH$ . $[O H^{-}]$ becomes so low that the ionic product for $Mg (OH)_{2}$ stays below its comparatively large $K_{s p}$ , so no precipitate forms (see Solved Example 23 for the numbers).

Solved Example 8

When a chloride is heated with conc.

H_{2} S O_{4}

HCl

gas is evolved, but a bromide gives

B r_{2}

. Explain.

Solution:

Both first give the hydrogen halide:

C l^{-} + H_{2} S O_{4} HCl + HS O_{4}^{-}

B r^{-} + H_{2} S O_{4} HBr + HS O_{4}^{-}

$HBr$ is a much stronger reducing agent than $HCl$ . Conc. $H_{2} S O_{4}$ oxidises it to bromine and is reduced to $S O_{2}$ ; it cannot oxidise $HCl$ .

2HBr + H_{2} S O_{4} B r_{2} + S O_{2} + 2 H_{2} O

Solved Example 9

A salt solution gives a brown solution with neutral

FeC l_{3}

; on heating, a brown precipitate forms. With

KI

solution the salt gives a straw-yellow precipitate. Identify the salt.

Solution:

The $FeC l_{3}$ test shows acetate: deep red iron(III) acetate, which gives brown basic iron(III) acetate on boiling.

3C H_{3} CO O^{-} + FeC l_{3} (C H_{3} COO)_{3} Fe + 3C l^{-}

(C H_{3} COO)_{3} Fe + 2 H_{2} O Δ brown Fe (OH)_{2} (C H_{3} COO) ↓ + 2C H_{3} COOH

With $KI$ , copper(II) gives white $CuI$ and free iodine, so the precipitate looks straw yellow. The cation is copper(II) and the salt is copper(II) acetate.

2C u^{2 +} + 4 I^{-} white 2CuI ↓ + I_{2}

Solved Example 10

Passing

H_{2} S

through a salt solution gives a brown precipitate. Adding water to an acidic solution of the salt makes it turbid. Identify the cation.

Solution:

A brown sulphide in group II means $B i^{3 +}$ or $S n^{2 +}$ . Turbidity on dilution (hydrolysis to $BiOCl$ ) confirms bismuth(III).

2B i^{3 +} + 3 H_{2} S brown B i_{2} S_{3} ↓ + 6 H^{+}

BiC l_{3} + H_{2} O white BiOCl ↓ + 2HCl

Solved Example 11

A colourless, water-soluble solid X on heating gives equimolar amounts of Y and Z. Y gives dense white fumes with

HCl

, and Z does so with ammonia. Y gives a brown precipitate with Nessler's reagent, and Z gives a white precipitate with the nitrates of

A g^{+}

P b^{2 +}

and

H g_{2}^{2 +}

. What is X?

Solution:

Y is $N H_{3}$ and Z is $HCl$ , so X is ammonium chloride.

N H_{4} Cl Δ Y N H_{3} + Z HCl

Solved Example 12

Gradual addition of

KI

solution to

Bi (N O_{3})_{3}

solution first gives a dark brown precipitate, which dissolves in excess

KI

to give a clear yellow solution. Write equations for these reactions.

Solution:

Iodide precipitates bismuth(III) iodide, which is black-brown. Excess iodide turns it into the soluble tetraiodobismuthate(III) complex, which is yellow-orange.

Bi (N O_{3})_{3} + 3KI dark brown Bi I_{3} ↓ + 3KN O_{3}

Bi I_{3} + KI yellow K [Bi I_{4}]

Solved Example 13

A colourless solid A gives a brown gas B on acidification, a colourless alkaline gas C with

NaOH

, and a colourless unreactive gas D on heating. On continued heating, A disappears completely. Identify A to D.

Solution:

An alkaline gas with $NaOH$ ( $N H_{3}$ ) shows an ammonium salt; an unreactive gas on heating ( $N_{2}$ ) with no residue shows a nitrite. A is ammonium nitrite; B = $N O_{2}$ , C = $N H_{3}$ , D = $N_{2}$ .

N H_{4} N O_{2} Δ D N_{2} ↑ + 2 H_{2} O

N H_{4} N O_{2} + NaOH NaN O_{2} + H_{2} O + C N H_{3} ↑

N H_{4} N O_{2} + HCl N H_{4} Cl + HN O_{2}

3HN O_{2} HN O_{3} + 2NO + H_{2} O

2NO + O_{2} B 2N O_{2}

Solved Example 14

A compound X of S, Cl and O has a vapour density of 67.5. It reacts with water to form two acids, and with

KOH

to form two salts Y and Z. Y gives a white precipitate with

AgN O_{3}

; Z gives a white precipitate with

BaC l_{2}

. Identify X.

Solution:

Y is a chloride and Z a sulphate. Molar mass of X = $2 \times 67.5 = 135$ . Sulphuryl chloride, $S O_{2} C l_{2}$ , has $M = 32 + 32 + 71 = 135$ , so X is $S O_{2} C l_{2}$ .

S O_{2} C l_{2} + 2 H_{2} O H_{2} S O_{4} + 2HCl

S O_{2} C l_{2} + 4KOH Z K_{2} S O_{4} + Y 2KCl + 2 H_{2} O

KCl + AgN O_{3} AgCl ↓ + KN O_{3}

K_{2} S O_{4} + BaC l_{2} BaS O_{4} ↓ + 2KCl

Solved Example 15

A white powder A, strongly heated, gives a colourless, odourless gas B that turns lime water milky because C forms; more gas makes the milkiness disappear, giving solution D. The residue E is yellow when hot and white when cold. Identify A to E with equations.

Solution:

A gas that turns lime water milky is $C O_{2}$ , so A is a carbonate; a residue yellow when hot and white when cold is $ZnO$ . A = $ZnC O_{3}$ , B = $C O_{2}$ , C = $CaC O_{3}$ , D = $Ca (HC O_{3})_{2}$ , E = $ZnO$ .

ZnC O_{3} Δ ZnO + C O_{2} ↑

C O_{2} + Ca (OH)_{2} CaC O_{3} ↓ + H_{2} O

CaC O_{3} + H_{2} O + C O_{2} Ca (HC O_{3})_{2}

Solved Example 16

An aqueous solution of a gas X (i) turns red litmus blue; (ii) in excess, gives a deep blue colour with copper sulphate solution; (iii) gives a brown precipitate with

FeC l_{3}

solution, soluble in dilute nitric acid. Identify X and write equations for (ii) and (iii).

Solution:

X is ammonia.

CuS O_{4} + 4N H_{4} OH deep blue [Cu (N H_{3})_{4}] S O_{4} + 4 H_{2} O

FeC l_{3} + 3N H_{4} OH brown Fe (OH)_{3} ↓ + 3N H_{4} Cl

Fe (OH)_{3} + 3HN O_{3} soluble Fe (N O_{3})_{3} + 3 H_{2} O

Solved Example 17

An unknown inorganic compound X loses its water of crystallisation on heating. Its aqueous solution (i) gives a white turbidity with dil.

HCl

; (ii) decolourises iodine in

KI

; (iii) gives a white precipitate with

AgN O_{3}

that turns black on standing; (iv) gives a golden-yellow flame. Identify X and give equations for (i) to (iii).

Solution:

Decolourising iodine, sulphur with acid and the white-to-black silver salt all point to thiosulphate; the flame shows sodium. X is sodium thiosulphate, $N a_{2} S_{2} O_{3} \cdot 5 H_{2} O$ .

N a_{2} S_{2} O_{3} \cdot 5 H_{2} O Δ N a_{2} S_{2} O_{3} + 5 H_{2} O

N a_{2} S_{2} O_{3} + 2HCl 2NaCl + H_{2} O + S O_{2} ↑ + S ↓

2N a_{2} S_{2} O_{3} + I_{2} N a_{2} S_{4} O_{6} + 2NaI

N a_{2} S_{2} O_{3} + 2AgN O_{3} white A g_{2} S_{2} O_{3} ↓ + 2NaN O_{3}

A g_{2} S_{2} O_{3} + H_{2} O black A g_{2} S ↓ + H_{2} S O_{4}

Solved Example 18

The gas from a mixture of two salts heated with

NaOH

gives a reddish-brown precipitate with an alkaline solution of

K_{2} Hg I_{4}

. The aqueous solution of the mixture gives, with

BaC l_{2}

, a white precipitate sparingly soluble in conc.

HCl

. Heating the mixture with

K_{2} C r_{2} O_{7}

and conc.

H_{2} S O_{4}

gives red vapours of A. The solution gives a deep blue colour B with potassium ferricyanide. Identify the radicals and write balanced equations for the formation of A and B.

Solution:

Nessler's reagent turns brown: $N H_{4}^{+}$ .
Red vapours with dichromate and conc. $H_{2} S O_{4}$ (chromyl chloride): $C l^{-}$ .
White $BaS O_{4}$ : $S O_{4}^{2 -}$ .
Deep blue with ferricyanide (Turnbull's blue): $F e^{2 +}$ .

The mixture contains $N H_{4}^{+}$ , $F e^{2 +}$ , $S O_{4}^{2 -}$ and $C l^{-}$ (for example $N H_{4} Cl$ and $FeS O_{4}$ ).

4NaCl + K_{2} C r_{2} O_{7} + 3 H_{2} S O_{4} A 2Cr O_{2} C l_{2} + K_{2} S O_{4} + 2N a_{2} S O_{4} + 3 H_{2} O

3F e^{2 +} + 2 K_{3} [Fe (CN)_{6}] B F e_{3} [Fe (CN)_{6}]_{2} + 6 K^{+}

Solved Example 19

An orange crystalline compound A, heated with common salt and conc.

H_{2} S O_{4}

, gives an orange-yellow gas B. B passed into caustic soda gives a yellow solution C. (i) With

AgN O_{3}

, C first gives a white precipitate that then turns red; 0.155 g of B needs 2.0 mmol of

AgN O_{3}

to give the first trace of red. (ii) Acidified with dil.

H_{2} S O_{4}

, C gives an orange solution (chromium +6) that liberates iodine from

KI

, leaving a green chromium(III) solution; 0.155 g of B liberates 1.5 mmol of iodine. Deduce A, B and C.

Solution:

A is $K_{2} C r_{2} O_{7}$ , B is chromyl chloride, $Cr O_{2} C l_{2}$ ( $M = 155$ ) and C is sodium chromate (with NaCl).

K_{2} C r_{2} O_{7} + 4NaCl + 6 H_{2} S O_{4} 2KHS O_{4} + 4NaHS O_{4} + 2Cr O_{2} C l_{2} + 3 H_{2} O

Cr O_{2} C l_{2} + 4NaOH N a_{2} Cr O_{4} + 2NaCl + 2 H_{2} O

The 2 $C l^{-}$ from each $Cr O_{2} C l_{2}$ precipitate white $AgCl$ first; red $A g_{2} Cr O_{4}$ appears only after that. So 155 g of B needs 2 mol $AgN O_{3}$ , and 0.155 g ( $1 0^{- 3}$ mol) needs $2 \times 1 0^{- 3}$ mol = 2 mmol, as observed.

N a_{2} Cr O_{4} + 2AgN O_{3} red A g_{2} Cr O_{4} ↓ + 2NaN O_{3}

2N a_{2} Cr O_{4} + H_{2} S O_{4} N a_{2} C r_{2} O_{7} + N a_{2} S O_{4} + H_{2} O

C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 I^{-} 2C r^{3 +} + 7 H_{2} O + 3 I_{2}

Two $Cr O_{2} C l_{2}$ give one $C r_{2} O_{7}^{2 -}$ , which frees 3 $I_{2}$ . So $1 0^{- 3}$ mol of B frees $1.5 \times 1 0^{- 3}$ mol = 1.5 mmol of iodine, as observed.

Solved Example 20

A white amorphous powder A, when heated, gives a colourless gas B (which turns lime water milky, the milkiness dissolving in excess B) and a residue C, yellow while hot and white when cold. C dissolves in dil.

HCl

, and the solution gives a white precipitate with potassium ferrocyanide. A dissolves in dil.

HCl

giving the same gas B. This solution gives a white precipitate D with

N H_{4} Cl

, excess

N H_{4} OH

and

H_{2} S

. Another portion gives a white precipitate E with

NaOH

, which dissolves in excess;

H_{2} S

passed into this solution again gives a white precipitate. The white precipitate heated with dil.

H_{2} S O_{4}

gives a gas used in group analysis. Identify A to E.

Solution:

A = $ZnC O_{3}$ , B = $C O_{2}$ , C = $ZnO$ (yellow hot, white cold), D = $ZnS$ , E = $Zn (OH)_{2}$ (amphoteric).

ZnC O_{3} Δ ZnO + C O_{2} ↑

C O_{2} + Ca (OH)_{2} CaC O_{3} ↓ + H_{2} O

CaC O_{3} + C O_{2} + H_{2} O Ca (HC O_{3})_{2}

ZnO + 2HCl ZnC l_{2} + H_{2} O

2ZnC l_{2} + K_{4} [Fe (CN)_{6}] white Z n_{2} [Fe (CN)_{6}] ↓ + 4KCl

ZnC O_{3} + 2HCl ZnC l_{2} + H_{2} O + C O_{2} ↑

ZnC l_{2} + H_{2} S + 2N H_{4} OH D ZnS ↓ + 2N H_{4} Cl + 2 H_{2} O

ZnC l_{2} + 2NaOH E Zn (OH)_{2} ↓ + 2NaCl

Zn (OH)_{2} + 2NaOH N a_{2} Zn O_{2} + 2 H_{2} O

N a_{2} Zn O_{2} + H_{2} S ZnS ↓ + 2NaOH

ZnS + H_{2} S O_{4} ZnS O_{4} + H_{2} S ↑

$H_{2} S$ is the reagent for group II (in acid) and for group IIIB (in ammoniacal solution).

Solved Example 21

Which reagent distinguishes a solid chloride from a solid bromide?
(A) Dilute

HN O_{3}

alone
(B)

K_{2} C r_{2} O_{7}

and conc.

H_{2} S O_{4}

, vapour passed into

NaOH

H_{2} S O_{4}

(D)

BaC l_{2}

solution

Solution:

Answer: (B). Only the chloride gives $Cr O_{2} C l_{2}$ , so the $NaOH$ turns yellow (chromate). A bromide gives free $B r_{2}$ and the $NaOH$ stays colourless. Neither halide reacts with dilute acids or $BaC l_{2}$ .

Solved Example 22

In group IIIA,

N H_{4} Cl

is added before

N H_{4} OH

mainly to
(A) raise

[O H^{-}]

so that

Fe (OH)_{3}

precipitates completely
(B) lower

[O H^{-}]

by the common ion effect so that only group IIIA hydroxides precipitate
(C) oxidise

F e^{2 +}

F e^{3 +}

(D) convert

Z n^{2 +}

into

ZnS

Solution:

Answer: (B). $N H_{4}^{+}$ suppresses the ionisation of $N H_{4} OH$ . The low $[O H^{-}]$ still exceeds $K_{s p}$ for $Fe (OH)_{3}$ , $Al (OH)_{3}$ and $Cr (OH)_{3}$ , but not for $Mg (OH)_{2}$ and the later groups. Oxidation of $F e^{2 +}$ is done by conc. $HN O_{3}$ , not by $N H_{4} Cl$ .

Solved Example 23

A solution is 0.1 M in

M g^{2 +}

and 0.1 M in

N H_{3}

. Will

Mg (OH)_{2}

precipitate (a) with

N H_{3}

alone, (b) after adding

N H_{4} Cl

to 1 M?

K_{b} (N H_{3}) = 1.8 \times 1 0^{- 5}

K_{s p} (Mg (OH)_{2}) = 1.8 \times 1 0^{- 11}

Solution:

(a) $[O H^{-}] = K_{b} c = 1.8 \times 1 0^{- 5} \times 0.1 = 1.34 \times 1 0^{- 3} M$ . $Q = [M g^{2 +}] [O H^{-}]^{2} = 0.1 \times (1.34 \times 1 0^{- 3})^{2} = 1.8 \times 1 0^{- 7}$ . Since $Q > K_{s p}$ , $Mg (OH)_{2}$ precipitates.

(b) With the buffer, $[O H^{-}] = K_{b} \frac{[ N H _{3} ]}{[ N H _{4}^{+} ]} = 1.8 \times 1 0^{- 5} \times \frac{0.1}{1} = 1.8 \times 1 0^{- 6} M$ . $Q = 0.1 \times (1.8 \times 1 0^{- 6})^{2} = 3.2 \times 1 0^{- 13}$ . Since $Q < K_{s p}$ , no precipitate: magnesium passes on to the last group (Figure 9).

Solved Example 24

After dil.

HCl

is added in group I,

[C l^{-}] = 0.1 M

. How much

A g^{+}

and

P b^{2 +}

stay in solution?

K_{s p}

AgCl

1.8 \times 1 0^{- 10}

PbC l_{2}

1.6 \times 1 0^{- 5}

. Why is lead found again in group II?

Solution:

$[A g^{+}] = \frac{1.8 \times 1 0 ^{- 10}}{0.1} = 1.8 \times 1 0^{- 9} M$ : silver is removed almost completely.

$[P b^{2 +}] = \frac{1.6 \times 1 0 ^{- 5}}{( 0.1 ) ^{2}} = 1.6 \times 1 0^{- 3} M$ . For a typical 0.01 M lead solution, about 16% of the lead stays dissolved. $H_{2} S$ in the next step precipitates it as black $PbS$ ( $K_{s p} \approx 8 \times 1 0^{- 28}$ ), so $P b^{2 +}$ appears in both group I and group II.

Solved Example 25

A green salt dissolves in water. It gives no precipitate with dil.

HCl

, with

H_{2} S

in acid, or with

N H_{4} Cl

N H_{4} OH

H_{2} S

passed into the ammoniacal solution gives a black precipitate. The precipitate, dissolved in aqua regia and made ammoniacal, gives a rosy-red precipitate with dimethylglyoxime. The salt solution gives a white precipitate with

BaC l_{2}

, insoluble in dil.

HCl

. Identify the salt.

Solution:

A black sulphide in group IIIB is $NiS$ or $CoS$ ; the rosy-red DMG complex proves nickel. White $BaS O_{4}$ insoluble in acid proves sulphate. The salt is nickel(II) sulphate, $NiS O_{4} \cdot 7 H_{2} O$ (green).

N i^{2 +} + H_{2} S + 2N H_{3} black NiS ↓ + 2N H_{4}^{+}

N i^{2 +} + 2 C_{4} H_{8} N_{2} O_{2} rosy red Ni (C_{4} H_{7} N_{2} O_{2})_{2} ↓ + 2 H^{+}

B a^{2 +} + S O_{4}^{2 -} white BaS O_{4} ↓

Practice Questions: Identify the Compounds

An aqueous solution of a gas X (i) turns acidified $K_{2} C r_{2} O_{7}$ green; (ii) after boiling with $H_{2} O_{2}$ , cooling and adding $BaC l_{2}$ , gives a precipitate insoluble in dil. $HCl$ ; (iii) gives turbidity with $H_{2} S$ . Identify X and give equations.Answer: X = $S O_{2}$ . (i) $3S O_{2} + C r_{2} O_{7}^{2 -} + 2 H^{+} \to 2C r^{3 +} + 3S O_{4}^{2 -} + H_{2} O$ ; (ii) $S O_{2} + H_{2} O_{2} \to H_{2} S O_{4}$ , then $BaS O_{4}$ ; (iii) $S O_{2} + 2 H_{2} S \to 3S + 2 H_{2} O$ .
A compound X used in the laboratory gives (i) with $CuS O_{4}$ a brown precipitate that turns white with excess $N a_{2} S_{2} O_{3}$ ; (ii) with $A g^{+}$ a yellow precipitate insoluble in $N H_{4} OH$ . Identify X.Answer: X = $KI$ . $2C u^{2 +} + 4 I^{-} \to 2CuI + I_{2}$ (hypo removes $I_{2}$ , leaving white $CuI$ ); $A g^{+} + I^{-} \to AgI$ (yellow).
A solid laboratory reagent A (i) gives a green flame; (ii) its solution gives no precipitate with $H_{2} S$ ; (iii) heated with $K_{2} C r_{2} O_{7}$ and conc. $H_{2} S O_{4}$ it gives a red gas that turns aqueous $NaOH$ yellow. Identify A.Answer: A = $BaC l_{2}$ : green flame (Ba), no group II sulphide, chromyl chloride test for $C l^{-}$ ( $Cr O_{2} C l_{2}$ gives yellow $N a_{2} Cr O_{4}$ ).
An unknown compound A (i) on heating gives a residue, oxygen and an oxide of nitrogen; (ii) its solution with tap water gives a turbidity insoluble in nitric acid; (iii) the turbidity dissolves in $N H_{4} OH$ . Identify A.Answer: A = $AgN O_{3}$ . $2AgN O_{3} \to 2Ag + 2N O_{2} + O_{2}$ ; chloride in tap water gives $AgCl$ , which dissolves in ammonia as $[Ag (N H_{3})_{2}]^{+}$ .
A black mineral A, heated in air, gives a gas B. A with dilute $H_{2} S O_{4}$ gives a gas C and a solution D. C passed into an aqueous solution of B gives a white turbidity. D gives a blue compound E with potassium ferricyanide. Identify A to E.Answer: A = $FeS$ , B = $S O_{2}$ , C = $H_{2} S$ , D = $FeS O_{4}$ , E = Turnbull's blue $F e_{3} [Fe (CN)_{6}]_{2}$ . $4FeS + 7 O_{2} \to 2F e_{2} O_{3} + 4S O_{2}$ ; $S O_{2} + 2 H_{2} S \to 3S + 2 H_{2} O$ .
2 M $HCl$ is added to a solution of $C a^{2 +}$ , $A g^{+}$ , $C u^{2 +}$ and $K^{+}$ : a white precipitate A forms. After filtering, $H_{2} S$ gives a black precipitate B. The filtrate, after removing B, gives a white precipitate C with conc. $N a_{2} C O_{3}$ . Identify A, B and C.Answer: A = $AgCl$ , B = $CuS$ , C = $CaC O_{3}$ ( $K^{+}$ stays in solution).
A solution of salt X gives a dirty white precipitate with $(N H_{4})_{2} S$ . After conc. $HN O_{3}$ and $Co (N O_{3})_{2}$ are added, a filter paper dipped in it and ignited gives a green ash. Heated with conc. $HN O_{3}$ and treated with ammonium molybdate, the salt gives a canary-yellow precipitate. Identify X.Answer: Zinc phosphate, $Z n_{3} (P O_{4})_{2}$ : $ZnS$ is dirty white, Rinmann's green is $CoZn O_{2}$ , and the canary-yellow precipitate is ammonium phosphomolybdate.
(i) An iodide A heated with $KOH$ solution gives a gas B and a solution of C. (ii) B burns in air to D and water. (iii) B gives a black precipitate with $CuS O_{4}$ . (iv) C gives a precipitate E with $CuS O_{4}$ . Identify A to E.Answer: A = $P H_{4} I$ , B = $P H_{3}$ , C = $KI$ , D = $P_{4} O_{10}$ , E = $CuI$ (with $I_{2}$ ). $P H_{4} I + KOH \to P H_{3} + KI + H_{2} O$ ; $4P H_{3} + 8 O_{2} \to P_{4} O_{10} + 6 H_{2} O$ ; $3CuS O_{4} + 2P H_{3} \to C u_{3} P_{2} + 3 H_{2} S O_{4}$ .
A compound X heated with excess $NaOH$ gives a gas Y that fumes with $HCl$ . After Y is expelled completely, the alkaline solution gives Y again when heated with zinc powder. X alone on heating does not give nitrogen. Identify X and Y.Answer: X = $N H_{4} N O_{3}$ , Y = $N H_{3}$ . Zn/NaOH reduces $N O_{3}^{-}$ to $N H_{3}$ ; $N H_{4} N O_{3}$ gives $N_{2} O$ on heating, not $N_{2}$ (unlike $N H_{4} N O_{2}$ ).
An orange solid A on heating gives a green residue B, a colourless gas C and water vapour. Dry C over heated Mg gives a white solid D, which with water gives a gas E that fumes with $HCl$ . Identify A to E.Answer: A = $(N H_{4})_{2} C r_{2} O_{7}$ , B = $C r_{2} O_{3}$ , C = $N_{2}$ , D = $M g_{3} N_{2}$ , E = $N H_{3}$ .
Calcium burns in nitrogen to a white powder that dissolves in water giving a gas A and an alkaline solution. On exposure to air the solution forms a thin solid layer B. Identify A and B.Answer: A = $N H_{3}$ (from $C a_{3} N_{2}$ + $H_{2} O$ ), B = $CaC O_{3}$ ( $Ca (OH)_{2}$ + $C O_{2}$ from air).
A colourless salt A decomposes completely at about $250^{\circ} C$ to only B and C, leaving no residue. C is a neutral liquid oxide; B is a neutral gaseous oxide. White phosphorus burns in excess B to give a strong white dehydrating agent. Write the equations.Answer: A = $N H_{4} N O_{3}$ , B = $N_{2} O$ , C = $H_{2} O$ . $N H_{4} N O_{3} \to N_{2} O + 2 H_{2} O$ ; $P_{4} + 10 N_{2} O \to P_{4} O_{10} + 10 N_{2}$ .
16.8 g of a solid X on heating gives 4.4 g of a gas A (turns lime water milky) and 1.8 g of a gas B (condenses to a colourless liquid). The residue Y dissolves in water to an alkaline solution, which gives a white precipitate Z with excess $BaC l_{2}$ ; Z effervesces with acid. Identify A, B, X and Y.Answer: X = $NaHC O_{3}$ (0.2 mol), A = $C O_{2}$ (0.1 mol), B = $H_{2} O$ (0.1 mol), Y = $N a_{2} C O_{3}$ , Z = $BaC O_{3}$ . $2NaHC O_{3} \to N a_{2} C O_{3} + C O_{2} + H_{2} O$ .
A salt X (i) gives an alkaline solution; (ii) swells to a glassy mass on strong heating; (iii) with conc. $H_{2} S O_{4}$ added to its hot concentrated solution, gives white crystals of a weak acid. Identify X with equations.Answer: X = borax, $N a_{2} B_{4} O_{7} \cdot 10 H_{2} O$ . (i) $N a_{2} B_{4} O_{7} + 7 H_{2} O \to 2NaOH + 4 H_{3} B O_{3}$ ; (ii) $N a_{2} B_{4} O_{7} \to 2NaB O_{2} + B_{2} O_{3}$ ; (iii) $N a_{2} B_{4} O_{7} + H_{2} S O_{4} + 5 H_{2} O \to N a_{2} S O_{4} + 4 H_{3} B O_{3}$ .
A red solid turns blue on heating. Its solution gives a black precipitate with $N H_{4} Cl$ , $N H_{4} OH$ and $H_{2} S$ , and a white precipitate insoluble in acids with $BaC l_{2}$ . Identify the solid.Answer: Hydrated cobalt(II) sulphate, $CoS O_{4} \cdot 7 H_{2} O$ : it turns blue as it loses water; $CoS$ is black; $BaS O_{4}$ is white.
A white amorphous powder A on strong heating gives a colourless, non-combustible gas B and a solid C. B turns lime water milky and the turbidity clears with excess gas. C in dil. $HCl$ gives a white precipitate with $K_{4} [Fe (CN)_{6}]$ . A in dil. $HCl$ gives a white precipitate D with $H_{2} S$ in excess $N H_{4} OH$ . Identify A to D.Answer: A = $ZnC O_{3}$ , B = $C O_{2}$ , C = $ZnO$ , D = $ZnS$ (same chemistry as Solved Example 20).
Y is a pale pink solid. Its solution gives a white precipitate with $NaOH$ that turns black in air; boiled with conc. $HN O_{3}$ and $Pb O_{2}$ it gives a violet solution; with lead acetate it gives a white precipitate soluble in $NaOH$ . Identify Y.Answer: Y = $MnS O_{4} \cdot 4 H_{2} O$ . $Mn (OH)_{2}$ is oxidised in air; $2M n^{2 +} + 5Pb O_{2} + 4 H^{+} \to 2Mn O_{4}^{-} + 5P b^{2 +} + 2 H_{2} O$ ; $PbS O_{4}$ dissolves in $NaOH$ as plumbite.
A solution of X (i) decolourises acidified $KMn O_{4}$ with evolution of $O_{2}$ ; (ii) liberates $I_{2}$ from acidified $KI$ ; (iii) gives a brown precipitate with alkaline $KMn O_{4}$ with evolution of $O_{2}$ ; (iv) removes black stains from old oil paintings. Identify X.Answer: X = $H_{2} O_{2}$ . (iv) $PbS + 4 H_{2} O_{2} \to PbS O_{4} + 4 H_{2} O$ (black to white); (iii) brown $Mn O_{2}$ forms.
A compound X (i) with $KI$ in acetic acid liberates iodine; (ii) its turbid aqueous suspension gives a precipitate with $C O_{2}$ ; (iii) its paste heated with ethanol gives an anaesthetic. Identify X.Answer: X = bleaching powder, $CaOC l_{2}$ . (ii) $CaC O_{3}$ forms; (iii) ethanol gives chloroform, $CHC l_{3}$ .
A Lewis acid X (i) fumes in moist air; (ii) fumes more near a rod dipped in $N H_{4} OH$ ; (iii) its acidic solution gives, with $N H_{4} Cl$ and $N H_{4} OH$ , a white precipitate soluble in $NaOH$ ; (iv) its acidic solution gives no precipitate with $H_{2} S$ . Identify X.Answer: X = anhydrous $AlC l_{3}$ . $AlC l_{3} + 3 H_{2} O \to Al (OH)_{3} + 3HCl$ ; $HCl$ + $N H_{3}$ gives $N H_{4} Cl$ fumes; $Al (OH)_{3}$ dissolves in $NaOH$ as $NaAl O_{2}$ .
(i) A black mineral A with dilute $NaCN$ in air gives a solution of B and C. (ii) B with zinc gives metal D. (iii) D in dil. $HN O_{3}$ gives a white precipitate E with dil. $HCl$ . (iv) E fused with $N a_{2} C O_{3}$ gives D. (v) E dissolves in ammonia to give F. Identify A to F.Answer: A = $A g_{2} S$ , B = $Na [Ag (CN)_{2}]$ , C = $N a_{2} S$ (oxidised by air), D = $Ag$ , E = $AgCl$ , F = $[Ag (N H_{3})_{2}] Cl$ .
A mixture of gases X, Y and Z is passed through acidified $K_{2} C r_{2} O_{7}$ (X absorbed, solution turns green), then lime water (Y absorbed, milky), then alkaline pyrogallol (Z absorbed, solution turns black). Identify X, Y and Z.Answer: X = $S O_{2}$ , Y = $C O_{2}$ , Z = $O_{2}$ . $S O_{2}$ is removed first because it would also turn lime water milky.
A yellow solid A, unaffected by acids and bases and insoluble in water, dissolves slowly in hot conc. $HN O_{3}$ with a brown gas B. A dissolves in boiling sodium sulphite to give a clear solution C, which on acidification gives a colourless gas D and a milky precipitate E. Identify A to E.Answer: A = S, B = $N O_{2}$ , C = $N a_{2} S_{2} O_{3}$ , D = $S O_{2}$ , E = S. $S + 6HN O_{3} \to H_{2} S O_{4} + 6N O_{2} + 2 H_{2} O$ .
A is a binary compound of a univalent metal. 1.422 g of A reacts completely with 0.321 g of sulphur in a sealed tube to give 1.743 g of a white solid B, which forms a hydrated double salt C with $A l_{2} (S O_{4})_{3}$ . Identify A, B and C.Answer: B = $K_{2} S O_{4}$ (0.01 mol, 174.3 g/mol); A = $K O_{2}$ ( $2 \times 71.1 = 142.2$ g per 0.01 mol of S); $2K O_{2} + S \to K_{2} S O_{4}$ ; C = potash alum $K_{2} S O_{4} \cdot A l_{2} (S O_{4})_{3} \cdot 24 H_{2} O$ .
A colourless crystalline solid A turns skin black. At $950^{\circ} C$ it gives gases B and C and a residue D. B dissolves in water giving E, in which D dissolves. A gives a precipitate with $NaOH$ that dissolves in excess $N H_{4} OH$ and gives a silver mirror with glucose. A gives a white precipitate with $KCN$ , soluble in excess. Identify the compounds.Answer: A = $AgN O_{3}$ , B = $N O_{2}$ , C = $O_{2}$ , D = $Ag$ , E = $HN O_{3}$ . The $NaOH$ precipitate is $A g_{2} O$ (brown), which gives Tollens' reagent; $AgCN$ dissolves in excess $KCN$ as $K [Ag (CN)_{2}]$ .
Identify A, B, C and D: (i) A + $NaOH$ gives $NaCl$ + $N H_{3}$ + $H_{2} O$ ; (ii) $N H_{3}$ + $C O_{2}$ + $H_{2} O$ gives B; (iii) B + $NaCl$ gives C + $N H_{4} Cl$ ; (iv) C on heating gives $N a_{2} C O_{3}$ + $H_{2} O$ + D.Answer: A = $N H_{4} Cl$ , B = $N H_{4} HC O_{3}$ , C = $NaHC O_{3}$ , D = $C O_{2}$ (Solvay process).
A metal A boiled with dilute nitric acid gives a salt B and an oxide of nitrogen C. B gives a precipitate D with brine, soluble in $N H_{4} OH$ . B added to hypo gives a white precipitate E that turns into a black compound F. Identify A to F.Answer: A = $Ag$ , B = $AgN O_{3}$ , C = $NO$ , D = $AgCl$ , E = $A g_{2} S_{2} O_{3}$ , F = $A g_{2} S$ . $3Ag + 4HN O_{3} \to 3AgN O_{3} + NO + 2 H_{2} O$ .
A compound A loses water of crystallisation on heating, then gives a blackish-brown powder B and two oxides of sulphur C and D. B boiled with $HCl$ gives a yellow solution E. $H_{2} S$ with E gives a white turbidity F and a pale green solution G. E gives a blood-red compound H with thiocyanate. Identify A to H.Answer: A = $FeS O_{4} \cdot 7 H_{2} O$ , B = $F e_{2} O_{3}$ , C = $S O_{2}$ , D = $S O_{3}$ , E = $FeC l_{3}$ , F = S, G = $FeC l_{2}$ , H = $[Fe (SCN)]^{2 +}$ . $2FeS O_{4} \to F e_{2} O_{3} + S O_{2} + S O_{3}$ .
A compound X gives a golden-yellow flame and (i) dissolves Zn powder on boiling with evolution of $H_{2}$ ; (ii) with $SnC l_{2}$ gives a white precipitate that dissolves in excess X; (iii) makes soap with fats; (iv) is standardised against oxalic acid with phenolphthalein; (v) precipitates $A l^{3 +}$ and $C r^{3 +}$ hydroxides that dissolve in excess. Identify X.Answer: X = $NaOH$ . $Zn + 2NaOH \to N a_{2} Zn O_{2} + H_{2}$ ; $Sn (OH)_{2}$ dissolves as $N a_{2} Sn O_{2}$ ; with $B r_{2}$ water, chromite gives yellow chromate.
A white solid A decrepitates on heating and gives brown vapour B. With conc. $H_{2} S O_{4}$ and copper turnings it gives dense brown fumes of B and a blue solution C. Its solution gives a yellow precipitate D with $K_{2} Cr O_{4}$ and a white precipitate E with dil. $H_{2} S O_{4}$ . Identify A to E.Answer: A = $Pb (N O_{3})_{2}$ , B = $N O_{2}$ , C = $C u^{2 +}$ solution ( $CuS O_{4}$ ), D = $PbCr O_{4}$ , E = $PbS O_{4}$ .
(i) A compound A forms when a gas B is passed through a concentrated liquor of sodium sulphide and sodium sulphite. (ii) A with dilute $AgN O_{3}$ gives a white precipitate that quickly turns black (C). (iii) A few drops of $FeC l_{3}$ in excess A give a violet colour D that fades quickly. (iv) A with $CuC l_{2}$ gives a white precipitate that dissolves in excess A to give E. Identify A to E.Answer: A = $N a_{2} S_{2} O_{3}$ , B = $S O_{2}$ ( $2N a_{2} S + N a_{2} S O_{3} + 3S O_{2} \to 3N a_{2} S_{2} O_{3}$ ), C = $A g_{2} S$ , D = $[Fe (S_{2} O_{3})_{2}]^{-}$ (reduced to $F e^{2 +}$ ), E = $N a_{4} [C u_{6} (S_{2} O_{3})_{5}]$ (a copper(I) thiosulphate complex).
A black compound A with dil. $H_{2} S O_{4}$ gives a gas B, which gives a white turbidity D with a solution of an acid C. B in an acidified solution of E gives a precipitate F soluble in dil. $HN O_{3}$ ; after boiling, excess $N H_{4} OH$ gives a blue compound G, and acetic acid with $K_{4} [Fe (CN)_{6}]$ gives a chocolate-brown precipitate H. $BaC l_{2}$ with E gives a white precipitate insoluble in $HN O_{3}$ . Identify A to H.Answer: A = $FeS$ , B = $H_{2} S$ , C = an oxidising acid such as $HN O_{3}$ , D = S, E = $CuS O_{4}$ , F = $CuS$ , G = $[Cu (N H_{3})_{4}] S O_{4}$ , H = $C u_{2} [Fe (CN)_{6}]$ .
(i) $H_{2} S$ through a solution of A in $N H_{4} OH$ gives a black compound B. (ii) B with $HCl$ and $KCl O_{3}$ gives A. (iii) A with $KCN$ gives a buff precipitate that dissolves in excess to give C. (iv) C changes to D when its solution is boiled. (v) A with excess $NaHC O_{3}$ and bromine water gives a green compound E; heating changes nothing. Identify A to E.Answer: A = $CoC l_{2}$ , B = $CoS$ , C = $K_{4} [Co (CN)_{6}]$ , D = $K_{3} [Co (CN)_{6}]$ (air oxidation on boiling), E = $N a_{3} [Co (C O_{3})_{3}]$ (green). Nickel would give a black precipitate on heating instead.
A colourless solid A on heating gives a white solid B and a colourless gas C; B gives reddish-brown fumes with dilute acids. B heated with $N H_{4} Cl$ gives a colourless gas D and residue E. A heated with $(N H_{4})_{2} S O_{4}$ gives a colourless gas F and a white residue G. E and G give a bright yellow flame. C with hot Mg gives a white powder that forms $Mg (OH)_{2}$ with water; D is absorbed by hot calcium, and the product gives ammonia with water. Identify A to G.Answer: A = $NaN O_{3}$ , B = $NaN O_{2}$ , C = $O_{2}$ , D = $N_{2}$ , E = $NaCl$ , F = $N_{2} O$ , G = $N a_{2} S O_{4}$ . $NaN O_{2} + N H_{4} Cl \to NaCl + N_{2} + 2 H_{2} O$ ; $2NaN O_{3} + (N H_{4})_{2} S O_{4} \to N a_{2} S O_{4} + 2 N_{2} O + 4 H_{2} O$ .
A solution of salt X gives a yellow precipitate with $H_{2} S$ . $N H_{4} OH$ gives a white precipitate soluble in excess. Shaken with $CC l_{4}$ and chlorine water, the solution gives a violet layer. Identify X.Answer: X = $Cd I_{2}$ : yellow $CdS$ , $Cd (OH)_{2}$ dissolving as $[Cd (N H_{3})_{4}]^{2 +}$ , and $I_{2}$ in $CC l_{4}$ .
A mixture of two salts (i) with $Mn O_{2}$ and conc. $H_{2} S O_{4}$ gives a yellowish-green gas; (ii) with $NaOH$ gives a gas that turns red litmus blue; (iii) in water gives a blue precipitate with $K_{4} [Fe (CN)_{6}]$ and a red colour with ammonium thiocyanate; (iv) boiled with $KOH$ , gives a gas that forms a brown precipitate with alkaline $K_{2} Hg I_{4}$ . Identify the salts.Answer: $N H_{4} Cl$ and $FeC l_{3}$ .
A gas A passed over dry $KOH$ at low temperature gives a deep red compound B and a gas C. A with but-2-ene, then Zn/ $H_{2} O$ , gives acetaldehyde. Identify A, B and C.Answer: A = $O_{3}$ (ozonolysis), B = potassium ozonide $K O_{3}$ , C = $O_{2}$ . $4KOH + 4 O_{3} \to 4K O_{3} + O_{2} + 2 H_{2} O$ .
A solution of salt A gives a white crystalline precipitate B with $NaCl$ ; the filtrate gives a black precipitate C with $H_{2} S$ . B dissolved in hot water gives a yellow precipitate D with $NaI$ on cooling. A gives no gas with dil. $HCl$ but gives a reddish-brown gas on heating. Identify A to D.Answer: A = $Pb (N O_{3})_{2}$ , B = $PbC l_{2}$ , C = $PbS$ (lead left in solution), D = $Pb I_{2}$ .
A pale green amorphous solid (i) gives a brown precipitate with potassium ferrocyanide; (ii) gives a blue colour with $N H_{4} OH$ ; (iii) effervesces with dil. $HCl$ . Identify the ions and the formula.Answer: $C u^{2 +}$ , $C O_{3}^{2 -}$ and $O H^{-}$ : basic copper carbonate, $CuC O_{3} \cdot Cu (OH)_{2}$ .
A compound X (i) gives a brown precipitate with $H_{2} S$ in acid; (ii) the precipitate dissolves in yellow ammonium sulphide; (iii) $NaOH$ gives a white precipitate soluble in excess; (iv) its solution reduces $FeC l_{3}$ . Identify the cation and give equations.Answer: $S n^{2 +}$ . $S n^{2 +} + H_{2} S \to SnS + 2 H^{+}$ ; $SnS + (N H_{4})_{2} S_{2} \to (N H_{4})_{2} Sn S_{3}$ ; $Sn (OH)_{2} + 2O H^{-} \to Sn O_{2}^{2 -} + 2 H_{2} O$ ; $S n^{2 +} + 2F e^{3 +} \to S n^{4 +} + 2F e^{2 +}$ .
Salt A is a white deliquescent solid that gives brown fumes on heating. $NaOH$ gives a white gelatinous precipitate, which forms a blue lake with blue litmus. Diphenylamine gives a deep blue colour. Identify the salt.Answer: $Al (N O_{3})_{3}$ : blue lake shows $A l^{3 +}$ ; brown $N O_{2}$ and diphenylamine show $N O_{3}^{-}$ .
X is a white deliquescent solid used as a desiccant. Its solution is said to turn acidified $K_{2} C r_{2} O_{7}$ green, gives a curdy white precipitate with $AgN O_{3}$ , a brick-red flame, and a white precipitate with ammonium oxalate insoluble in acetic acid. Identify X.Answer: X = $CaC l_{2}$ (brick-red flame, $Ca C_{2} O_{4}$ , $AgCl$ ). Note: a chloride does not turn acidified dichromate green; that observation fits the chromyl chloride test on the solid instead.
A scarlet compound A with conc. $HN O_{3}$ gives a chocolate-brown precipitate B. The filtrate, neutralised and treated with $KI$ , gives a yellow precipitate C. B warmed with conc. $HN O_{3}$ and $Mn (N O_{3})_{2}$ gives a pink solution D. Identify A to D.Answer: A = $P b_{3} O_{4}$ , B = $Pb O_{2}$ , C = $Pb I_{2}$ , D = $HMn O_{4}$ . $P b_{3} O_{4} + 4HN O_{3} \to 2Pb (N O_{3})_{2} + Pb O_{2} + 2 H_{2} O$ ; $5Pb O_{2} + 2Mn (N O_{3})_{2} + 6HN O_{3} \to 2HMn O_{4} + 5Pb (N O_{3})_{2} + 2 H_{2} O$ .
A white solid X heated with conc. $HN O_{3}$ and then ammonium molybdate gives a canary-yellow precipitate. $NaOH$ gives a white precipitate insoluble in excess. The cobalt nitrate ash test gives a pink ash. Identify the salt.Answer: Magnesium phosphate, $M g_{3} (P O_{4})_{2}$ : phosphate from the molybdate test, $Mg (OH)_{2}$ and a pink ash for magnesium.

Practice Questions: MCQs

In qualitative analysis Ca is precipitated in
(A) V group
(B) II group
(C) III group
(D) IV groupAnswer: (D) group IV (group V in the 0 to VI numbering), as $CaC O_{3}$ .
$A l^{3 +}$ , $F e^{3 +}$ and $C r^{3 +}$ are grouped together because their
(A) carbonates are insoluble in ammonia
(B) hydroxides are insoluble in ammonia
(C) sulphides are soluble in acids
(D) electronic charge is the sameAnswer: (B)
A colourless solid A dissolves in water. The solution gives a white precipitate B with $NaOH$ or $N H_{4} OH$ ; B dissolves in excess $NaOH$ but not in excess $N H_{4} OH$ . $BaC l_{2}$ gives a white precipitate C insoluble in dil. $HCl$ . A may be
(A) $A l_{2} (S O_{4})_{3}$
(B) $ZnS O_{4}$
(C) $SnS O_{4}$
(D) All the aboveAnswer: (A) $A l_{2} (S O_{4})_{3}$ ( $SnS O_{4}$ also fits). Not (D): $Zn (OH)_{2}$ dissolves in excess $N H_{4} OH$ .
Which ion liberates iodine when treated with $KI$ ?
(A) $C u^{2 +}$
(B) $F e^{2 +}$
(C) $P b^{2 +}$
(D) $S n^{2 +}$ Answer: (A)
A group I white precipitate is insoluble in boiling water but turns black with $N H_{4} OH$ . It is
(A) $PbC l_{2}$
(B) $AgCl$
(C) $HgC l_{2}$
(D) $H g_{2} C l_{2}$ Answer: (D)
$Cr O_{2} C l_{2}$ is passed into water and lead acetate is added. The precipitate is
(A) $PbCr O_{4}$
(B) $PbC l_{2}$
(C) $CrC l_{3}$
(D) $PbS O_{4}$ Answer: (A)
Which is insoluble in yellow ammonium sulphide?
(A) $SnS$
(B) $S b_{2} S_{3}$
(C) $B i_{2} S_{3}$
(D) $A s_{2} S_{3}$ Answer: (C)
Ammonia with Nessler's reagent (alkaline) gives a brown precipitate of
(A) $N H_{4} Hg I_{4}$
(B) $N H_{2} HgOHgI$
(C) $KI$
(D) $(N H_{4})_{2} Hg I_{4}$ Answer: (B)
A salt heated first with dilute and then with conc. $H_{2} S O_{4}$ shows no action. The salt must be a
(A) nitrite
(B) sulphide
(C) sulphite
(D) sulphateAnswer: (D)
Rinmann's green is
(A) cobalt nitrate
(B) sodium zincate
(C) zinc acetate
(D) cobalt zincateAnswer: (D)
A water-soluble solid gives a brown gas on heating, and its solution gives a white precipitate with dil. $HCl$ . The solid is
(A) $Ca (N O_{3})_{2}$
(B) $Zn (N O_{3})_{2}$
(C) $Pb (N O_{3})_{2}$
(D) $Ba (N O_{3})_{2}$ Answer: (C)
$K_{4} [Fe (CN)_{6}]$ is used to identify
(A) $C d^{2 +}$ ions
(B) $F e^{3 +}$ ions
(C) $C u^{2 +}$ ions
(D) All the aboveAnswer: (D)
The deep blue colour from $K_{3} [Fe (CN)_{6}]$ and ferrous sulphate is due to
(A) $Fe [Fe (CN)_{6}]$
(B) $F e_{3} [Fe (CN)_{6}]_{2}$
(C) $F e_{3} [Fe (CN)_{6}]_{4}$
(D) $F e_{4} [Fe (CN)_{6}]_{3}$ Answer: (B) Turnbull's blue.
The salt used for the bead test is
(A) $K_{2} S O_{4} \cdot A l_{2} (S O_{4})_{3} \cdot 24 H_{2} O$
(B) $FeS O_{4} \cdot (N H_{4})_{2} S O_{4} \cdot 6 H_{2} O$
(C) $Na (N H_{4}) HP O_{4} \cdot 4 H_{2} O$
(D) $CaS O_{4} \cdot 2 H_{2} O$ Answer: (C) microcosmic salt.
A nitrate boiled with aluminium turnings and conc. $NaOH$ gives
(A) $N O_{2}$
(B) $N H_{3}$
(C) $NO$
(D) $N_{2} O$ Answer: (B)
A mixture with dil. $H_{2} S O_{4}$ gives a gas that blackens lead acetate paper and turns dichromate paper green. The mixture contains
(A) sulphite
(B) sulphide
(C) nitrate
(D) nitriteAnswer: (B) $H_{2} S$ does both.
$N a_{2} HP O_{4}$ detects
(A) $N H_{4}^{+}$
(B) $N O_{3}^{-}$
(C) $K^{+}$
(D) $M g^{2 +}$ Answer: (D)
$H_{2} S$ through an ammoniacal solution X gives a buff precipitate. X contains
(A) $C o^{2 +}$
(B) $M n^{2 +}$
(C) $N i^{2 +}$
(D) $Z n^{2 +}$ Answer: (B) $MnS$ .
$N H_{4} Cl$ and aqueous $N H_{3}$ precipitate
(A) $B i^{3 +}$
(B) $A l^{3 +}$
(C) $Z n^{2 +}$
(D) $M g^{2 +}$ Answer: (B)
A bottle of an ammonium salt smells strongly of ammonia when opened. The salt is
(A) ammonium chloride
(B) ammonium sulphate
(C) ammonium carbonate
(D) ammonium nitrateAnswer: (C)
The reagent used to detect phosphate is
(A) neutral $FeC l_{3}$
(B) magnesia mixture
(C) acidified $KMn O_{4}$
(D) conc. $H_{2} S O_{4}$ Answer: (B)
A crystalline solid gives a colourless solution that gives a yellow precipitate with $AgN O_{3}$ . The solid is most likely
(A) copper(II) oxide
(B) sodium chloride
(C) sodium iodide
(D) zinc sulphateAnswer: (C)
The correct order of $K_{s p}$ for $CuS$ , $ZnS$ and $CdS$ is
(A) $CuS$ $>$ $ZnS$ $>$ $CdS$
(B) $CdS$ $>$ $CuS$ $>$ $ZnS$
(C) $ZnS$ $>$ $CdS$ $>$ $CuS$
(D) $ZnS$ $>$ $CuS$ $>$ $CdS$ Answer: (C)
Magneson reagent detects
(A) $M n^{2 +}$
(B) $M g^{2 +}$
(C) $P O_{4}^{3 -}$
(D) $Mn O_{4}^{-}$ Answer: (B)
Which gives no precipitate with dilute $AgN O_{3}$ ?
(A) $C l^{-}$ (aq)
(B) $B r^{-}$ (aq)
(C) $I^{-}$ (aq)
(D) $F^{-}$ (aq)Answer: (D) $AgF$ is soluble.
$AgN O_{3}$ gives a red precipitate with
(A) $NaI$
(B) $KCl$
(C) $NaN O_{3}$
(D) $N a_{2} Cr O_{4}$ Answer: (D) $A g_{2} Cr O_{4}$ .
Which pair cannot be separated by $H_{2} S$ in ammoniacal medium?
(A) $Z n^{2 +}$ , $N i^{2 +}$
(B) $Z n^{2 +}$ , $M g^{2 +}$
(C) $M n^{2 +}$ , $C a^{2 +}$
(D) $C o^{2 +}$ , $C a^{2 +}$ Answer: (A)
Conc. $H_{2} S O_{4}$ on a salt gives a brown colour on the walls of the tube and violet vapour on heating. The brown colour is due to
(A) $I_{2}$ (s)
(B) $I^{+}$
(C) $I_{3}^{-}$
(D) $HI$ Answer: (C)
$MnS O_{4}$ boiled with sodium bismuthate and conc. $HN O_{3}$ gives
(A) $Mn O_{2}$
(B) $HMn O_{4}$
(C) $M n_{3} O_{4}$
(D) $PbMn O_{4}$ Answer: (B)
Conc. $H_{2} S O_{4}$ on dry $KN O_{3}$ gives brown fumes of
(A) $S O_{3}$
(B) $S O_{2}$
(C) $N O_{2}$
(D) $N_{2} O$ Answer: (C)
Which pair cannot be separated by $H_{2} S$ in dilute $HCl$ ?
(A) $B i^{3 +}$ , $S n^{2 +}$
(B) $A l^{3 +}$ , $H g^{2 +}$
(C) $Z n^{2 +}$ , $C u^{2 +}$
(D) $N i^{2 +}$ , $C u^{2 +}$ Answer: (A) both are group II.
$S r^{2 +}$ can be precipitated by
(A) $(N H_{4})_{2} S O_{4}$
(B) $N H_{4} Cl$
(C) $(N H_{4})_{2} C_{2} O_{4}$
(D) Both (A) and (C)Answer: (D)
Excess sodium thiosulphate with $AgBr$ gives
(A) $A g_{2} S_{2} O_{3}$
(B) $Na [Ag (S_{2} O_{3})]$
(C) $N a_{3} [AgB r_{4}]$
(D) $N a_{3} [Ag (S_{2} O_{3})_{2}]$ Answer: (D)
Nessler's reagent is
(A) $K_{2} Hg I_{4}$ in excess $KI$
(B) $K_{2} Hg I_{4}$ in excess $KOH$
(C) $K_{2} Hg I_{4}$ in excess $HCl$
(D) $Hg I_{2}$ in excess $KOH$ Answer: (B)
$H_{2} S$ through a dilute $HCl$ solution gives a black precipitate when the solution contains
(A) $C u^{2 +}$ and $P b^{2 +}$
(B) $Z n^{2 +}$ and $M g^{2 +}$
(C) $M g^{2 +}$ and $F e^{3 +}$
(D) $M g^{2 +}$ and $N i^{2 +}$ Answer: (A)
A salt solution gives, with dil. $HCl$ , a white precipitate soluble in hot water. The salt contains
(A) $F e^{2 +}$
(B) $H g^{2 +}$
(C) $A g^{+}$
(D) $P b^{2 +}$ Answer: (D)
$M g^{2 +}$ imparts no colour to the flame because it has
(A) high charge
(B) high excitation energy
(C) small size
(D) none of the aboveAnswer: (B) the energy needed to excite its tightly held electrons is too high (a result of its small size).
The solid formed with excess $SnC l_{2}$ and $HgC l_{2}$ solution is
(A) white
(B) black
(C) yellow
(D) brownAnswer: (B) finely divided Hg.
Which statement(s) about ferrous and ferric ions is/are correct?
(A) $F e^{3 +}$ gives a green colour with potassium thiocyanate
(B) $F e^{2 +}$ gives a blue precipitate with potassium ferricyanide
(C) $F e^{3 +}$ gives a red colour with potassium thiocyanate
(D) $F e^{2 +}$ gives a brown colour with ammonium thiocyanateAnswer: (B) and (C)
Aqua regia is
(A) conc. $HCl$ + conc. $HN O_{3}$ , 1 : 3
(B) dil. $HCl$ and dil. $HN O_{3}$ , 3 : 1
(C) dil. $HCl$ and conc. $HN O_{3}$ , 3 : 1
(D) conc. $HCl$ and conc. $HN O_{3}$ , 3 : 1Answer: (D)

Common Mistakes to Avoid

Watch out

Testing a carbonate + sulphite mixture with lime water only: $S O_{2}$ also turns lime water milky, so pass the gas through acidified $K_{2} C r_{2} O_{7}$ first.
Writing mercury(I) as $H g^{+}$ : it is $H g_{2}^{2 +}$ , and $H g_{2} C l_{2}$ turns black with $N H_{4} OH$ instead of dissolving like $AgCl$ .
Replacing $N H_{4} Cl$ by $(N H_{4})_{2} S O_{4}$ or $N H_{4} N O_{3}$ in group IIIA: sulphate removes $B a^{2 +}$ early, and nitrate oxidises $M n^{2 +}$ so that it precipitates in the wrong group.
Skipping the boiling-off of $H_{2} S$ and the conc. $HN O_{3}$ step before group IIIA: $F e^{2 +}$ precipitates incompletely, and leftover $H_{2} S$ drags group IIIB sulphides down early.
Mixing up Prussian blue ( $F e^{3 +}$ + ferrocyanide) and Turnbull's blue ( $F e^{2 +}$ + ferricyanide).
Expecting chromyl bromide or chromyl iodide: only chloride forms $Cr O_{2} C l_{2}$ ; bromide and iodide give free $B r_{2}$ and $I_{2}$ , and mercury chlorides do not respond at all.
Using the brown ring for nitrate while a nitrite is present: nitrite gives the ring even with dilute acid, so destroy it first.
Mixing the two group numberings: $C a^{2 +}$ is group IV in the IIIA/IIIB scheme but group V in the 0 to VI numbering; $Z n^{2 +}$ is IIIB or IV. Answer from the reagent, not the number.

Frequently Asked Questions

What is qualitative analysis in chemistry?

Qualitative analysis identifies which ions a substance contains, without measuring amounts. For a salt, dry tests and acid tests find the anion, and group reagents separate the cations by solubility before each ion is confirmed by a specific colour or precipitate. It rests on the solubility product and the common ion effect.

Why is ammonium chloride added with ammonium hydroxide in group III analysis?

$N H_{4} Cl$ supplies $N H_{4}^{+}$ , which suppresses the ionisation of $N H_{4} OH$ by the common ion effect. The hydroxide ion concentration falls to about $1 0^{- 6}$ M: enough to precipitate $Fe (OH)_{3}$ , $Al (OH)_{3}$ and $Cr (OH)_{3}$ , which have tiny solubility products, but too low to precipitate $Mg (OH)_{2}$ or the hydroxides of later groups.

Why is hydrogen sulphide passed in acid for group II but in ammoniacal solution for zinc and manganese?

In dilute $HCl$ , $H^{+}$ suppresses the ionisation of $H_{2} S$ , so the sulphide ion concentration is very low and only sulphides with the smallest solubility products, such as $CuS$ , precipitate. Ammonia removes $H^{+}$ , raising the sulphide ion concentration enough to precipitate $ZnS$ , $MnS$ , $NiS$ and $CoS$ .

How can chloride, bromide and iodide ions be distinguished?

With silver nitrate, chloride gives white $AgCl$ soluble in dilute ammonia, bromide gives pale yellow $AgBr$ soluble only in concentrated ammonia, and iodide gives yellow $AgI$ insoluble in ammonia. Chlorine water with carbon tetrachloride gives an orange-brown layer for bromide and a violet layer for iodide. Only chloride gives the chromyl chloride test.

What causes the brown ring in the brown ring test?

Nitrate is reduced to nitric oxide by iron(II) in concentrated sulphuric acid. Nitric oxide then replaces a water molecule in the hexaaquairon(II) ion, giving the brown complex $[Fe (H_{2} O)_{5} (NO)]^{2 +}$ . It forms only at the junction of the acid layer and the solution, and it decomposes on shaking or warming.

How do you distinguish a carbonate from a bicarbonate?

Both give carbon dioxide with dilute acid. A carbonate solution gives a white precipitate with magnesium sulphate or barium chloride in the cold and turns phenolphthalein pink. A bicarbonate precipitates only after boiling, stays nearly colourless with phenolphthalein, and its dry solid gives carbon dioxide and water on heating.

Which salt analysis topics are in the NEET syllabus?

NEET (NMC 2026 syllabus, Unit 20) asks for the chemical principles of salt analysis for the cations $P b^{2 +}$ , $C u^{2 +}$ , $A l^{3 +}$ , $F e^{3 +}$ , $Z n^{2 +}$ , $N i^{2 +}$ , $C a^{2 +}$ , $B a^{2 +}$ , $M g^{2 +}$ and $N H_{4}^{+}$ , and the anions carbonate, sulphide, sulphate, nitrate, nitrite, chloride, bromide and iodide. Questions test group reagents, colours and the reasons behind them.

How is qualitative analysis tested in JEE Main and JEE Advanced?

JEE Main shares the NEET list of ions and asks mostly single-step reagent and colour questions. JEE Advanced adds $A g^{+}$ , $H g^{2 +}$ , $C r^{3 +}$ and $M n^{2 +}$ , and bicarbonate, and favours multi-step identification problems and solubility-product reasoning, such as why $ZnS$ does not precipitate in acid.

Previous year questions on Principles Of Qualitative Analysis

24 questions from past papers, each with a step-by-step solution.

Show all 24 questions

Ready to master Principles Related to Practical Chemistry?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test

1. What Qualitative Analysis Means

1.1 Ions covered

1.2 Preparing the solutions

2. Preliminary (Dry) Tests

2.1 Physical examination

2.2 Effect of heating (dry heating test)

2.3 Flame test

2.4 Borax bead test

2.5 Charcoal cavity and cobalt nitrate tests

3. Classification of Anions

4. Class A(i): Anions Detected with Dilute Acid

4.1 Carbonate

4.2 Bicarbonate (hydrogen carbonate)

4.3 Sulphite

4.4 Sulphide

4.5 Nitrite

4.6 Acetate

4.7 Thiosulphate

5. Class A(ii): Anions Detected with Conc. H2​SO4​

5.1 Chloride

5.2 Bromide

5.3 Iodide

5.4 Nitrate

6. Class B Anions: Tests in Solution

6.1 Sulphate

6.2 Chromate and dichromate

6.3 Permanganate

7. Classification of Cations into Groups

7.1 The same groups numbered 0 to VI

8. Why the Group Reagents Work: Solubility Product and Common Ion Effect

8.1 Group I: chlorides with dilute HCl

8.2 Group II and group IIIB: sulphides

8.3 Group IIIA: hydroxides with NH4​Cl and NH4​OH

8.4 Group IV: carbonates

8.5 Points to remember

9. Confirmatory Tests for Cations

9.1 Group I: Pb2+, Ag+, Hg22+​

9.2 Group IIA: Hg2+, Cu2+, Bi3+, Cd2+

9.3 Group IIB: As3+, Sb3+, Sn2+

9.4 Group IIIA: Al3+, Fe3+, Cr3+

9.5 Group IIIB: Ni2+, Co2+, Mn2+, Zn2+

9.6 Group IV: Ba2+, Sr2+, Ca2+

9.7 Group V: NH4+​, Na+, K+, Mg2+

10. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What is qualitative analysis in chemistry?

Why is ammonium chloride added with ammonium hydroxide in group III analysis?

Why is hydrogen sulphide passed in acid for group II but in ammoniacal solution for zinc and manganese?

How can chloride, bromide and iodide ions be distinguished?

What causes the brown ring in the brown ring test?

How do you distinguish a carbonate from a bicarbonate?

Which salt analysis topics are in the NEET syllabus?

How is qualitative analysis tested in JEE Main and JEE Advanced?

Previous year questions on Principles Of Qualitative Analysis

Ready to master Principles Related to Practical Chemistry?

5. Class A(ii): Anions Detected with Conc. $H_{2} S O_{4}$

8.1 Group I: chlorides with dilute $HCl$

8.3 Group IIIA: hydroxides with $N H_{4} Cl$ and $N H_{4} OH$

9.1 Group I: $P b^{2 +}$ , $A g^{+}$ , $H g_{2}^{2 +}$

9.2 Group IIA: $H g^{2 +}$ , $C u^{2 +}$ , $B i^{3 +}$ , $C d^{2 +}$

9.3 Group IIB: $A s^{3 +}$ , $S b^{3 +}$ , $S n^{2 +}$

9.4 Group IIIA: $A l^{3 +}$ , $F e^{3 +}$ , $C r^{3 +}$

9.5 Group IIIB: $N i^{2 +}$ , $C o^{2 +}$ , $M n^{2 +}$ , $Z n^{2 +}$

9.6 Group IV: $B a^{2 +}$ , $S r^{2 +}$ , $C a^{2 +}$

9.7 Group V: $N H_{4}^{+}$ , $N a^{+}$ , $K^{+}$ , $M g^{2 +}$