Qualitative Analysis Theory_E

Qualitative Analysis

Introduction :

Qualitative analysis involves the detection of cation(s) and anion(s) of a salt or a mixture of salts. The systematic procedure for qualitative analysis of an inorganic salt involves the following steps : (a) Preliminary tests

 Physical appearance (colour and smell).

 Dry heating test.

 Charcoal cavity test.

 Charcoal cavity and cobalt nitrate test.

 Flame test.

 Borax bead test.

 Dilute sulphuric acid test.

 Potassium permanganate test.

 Concentrated sulphuric acid test.

 Tests for sulphate, phosphate and borate. (b) Wet tests for acid radicals.

(c) Wet tests (group analysis) for basic radicals.

1.

Physical Examination Of the Mixture :

The physical examination of the unknown mixture involves the study of colour, smell and density. Table : 1

Physical Examination

Ex pe rim e nt Obse rva tions Infe re nce

(a ) Colour Blue or Bluish green Cu2+or Ni2+

Greenish Ni2+

Light green Fe2+

Dark brown Fe3+

Pink, violet Co2+

Light pink, flesh colour or dull

earthy colour M n2+

W hite

Shows the absence of Cu2+,Ni2+,Fe2+,Fe3+

Mn2+, Co2+ (b) Sm e ll

Ammonical smell NH4+

Vinegar like smell CH3COO–

Sm ell like that of rotten eggs S2–

(i) Heavy Salt of Pb2+or Ba2+– (ii) Light fluffy powder Carbonate salts

(d) De lique sce nce

Salt absorbs m oisture and becomes

paste like

(i) If coloured, may be Cu(NO3)2,

FeCl3

(ii) If colourless, may be Zn(NO3)2, chlorides of

Zn2+, Mg2+etc. Take a pinch of the salt between

your fingers and rub with a drop of water

2.

Dry Heating Test :

This test is performed by heating a small amount of mixture in a dry test tube. Quite valuable information can be generated by carefully performing and noting the observations here. On heating some salts undergo decomposition thus evolving the gases or may undergo characteristic changes in the colour of residue. These observations are tabulated below along with the inferences that you can draw.

Table : 2

Observation

Inference

1. Gas evolved

(a) Colourless and odourless gas

CO₂gas – turns lime water milky CO₃2–

(b) Colourless gas with odour

(i) H₂S gas–Smells like rotten eggs, turns Hydrated S2– lead acetate paper black.

(ii) SO₂gas–Characteristic suffocating SO₃2– smell, turns acidified potassium dichromate

solution or paper green.

(iii) HCl gas – Pungent smell, white fumes with Cl– ammonia, white precipitate with silver nitrate solution.

(iv) Acetic acid vapours–Characteristic vinegar CH₃COO– like smell.

(v) NH₃gas– Characteristic smell, turns NH₄+ Nessler's solution brown.

(c) Coloured gases – Pungent smell

(i) NO₂gas – Reddish brown, turns ferrous NO₂–_{or NO} 3

– sulphate solution black.

(ii) Cl₂gas – Greenish yellow, turns starch Cl– iodide paper blue.

(iii) Br₂vapours – Reddish brown, turns starch Br– paper orange red.

(iv) I₂vapours – Dark violet, turns starch paper – blue.

2. Sublimate formed

(a) White sublimate NH₄+

(b) Black sublimate accompanied by violet – vapours.

3. Fusion

The mixture fuses. Alkali metal salts or salt containing water of crystallisation.

4. Swelling

The mixture swells up into voluminous mass. PO₄3–_{, BO} 3

3–_indicated

5. Residue

(i) Yellow when hot, white when cold. Zn2+ (ii) Brown when hot and yellow when cold Pb2+

(iii) Original salt blue becomes white on heating Hydrated CuSO₄indicated

(iv) Coloured salt becomes brown or black on Co2+_{, Fe}2+_{, Fe}3+_{, Cr}3+_{, Cu}2+_{, Ni}2+_{, Mn}2+

heating. indicated.

Note :



_{Use a perfectly dry test–tube for performing this test. While drying a test–tube, keeps it in slanting}

position with its mouth slightly downwards so that the drops of water which condense on the upper cooler parts, do not fall back on the hot bottom, as this may break the tube.



_{For testing a gas, a filter paper strip dipped in the appropriate reagent is brought near the mouth of}

the test tube or alternatively the reagent is taken in a gas–detector and the gas is passed through it.

Figure : Detection of gas evolved.



_{Do not heat the tube strongly at one point as it may break.}

3.

Charcoal Cavity Test :

This test is based on the fact that metallic carbonates when heated in a charcoal cavity decompose to give corresponding oxides. The oxides appear as coloured incrustation or residue in the cavity . In certain cases, the oxides formed partially undergo reduction to the metallic state producing metallic beads or scales. Example :

(a) ZnSO₄+ Na₂CO₃  ZnCO₃+ Na₂SO₄

ZnCO₃  ZnO (Yellow when hot, white when cold) + CO₂ (b) CuSO₄+ Na₂CO₃  CuCO₃+ Na₂SO₄

CuCO₃  CuO + CO₂

CuO + C  Cu (Reddish scales) + CO Table : 3

Inference

Incrusta tion or Re sidue M e ta llic be a d

Yellow when hot, white when cold None Zn2+

Brown when hot, yellow when cold Grey bead which

marks the paper Pb

2+

No characteristic residue Red beads or scales Cu2+ W hite residue which glows on heating None Ba2+,Ca2+, Mg2+

Black None Nothing definite–generally

coloured salt

Observation

4.

Cobalt Nitrate Test :

In case the residue is white in colour after charcoal cavity test, add a drop of cobalt nitrate in the charcoal cavity. A drop of water is then added and the mass is heated in an oxidising flame using blow pipe. It is cooled and one or two drops of cobalt nitrate solution is added and then again heated in the oxidising flame. Different metal salts give different coloured mass as given in the table. To illustrate :

ZnSO₄+ Na₂CO₃  ZnCO₃+ Na₂SO₄ ; ZnCO₃  ZnO + CO₂

2Co (NO₃)₂  2CoO + 4 NO₂+ O₂; ZnO + CoO  ZnO. CoO (or CoZnO₂) (Rinmann's green)

Table : 4

S.No. Metal Colour of the mass 1. Zinc Green 2. Aluminium Blue 3. Magnesium _Pink Tin 4. Bluish - green

5.

Flame test :

The chlorides of the metals are more volatile as compared to other salts and these are prepared in situ by mixing the compounds with a little concentrated hydrochloric acid. On heating in a non-luminous Bunsen flame they are volatilized and impart a characteristic colour to the flame as these absorb energy from the flame and transmit the same as light as characteristic colour .

Table : 5

Colour of Flame Inference

Crimson Red / Carmine Red Lithium

Golden yellow Sodium

Violet/Lilac Potassium

Brick red Calcium

Crimson Strontium

Apple Green/Yellowish Green Barium Green with a Blue centre/Greenish Blue Copper

Figure : Flame test

6.

Borax Bead test :

On Heating borax forms a colourless glassy bead of NaBO₂ and B₂O_{3 .} Na₂B₄O₇.10H₂O  Na₂B₄O₇  2NaBO₂+ B₂O₃

On heating with a coloured salt , the glassy bead forms a coloured metaborate in oxidising flame. For example, in oxidising flame copper salts give blue bead.

CuSO₄  CuO + SO₃ ; CuO + B₂O₃  Cu(BO₂)₂ (blue bead) However, in reducing flame the colours may be different due to different reactions. 2Cu(BO₂)₂ + C  2CuBO₂+ B₂O₃+ CO

Table : 6

When Hot When Cold When Hot When Cold

Copper Green Blue Colourless Brown red Iron Brown yellow Pale yellow/Yellow Bottle green Bottle green

Chromium Yellow Green Green Green

Cobalt Blue Blue Blue Blue

Manganese Violet/Amethyst Red/Amethyst Grey/Colourless Grey/Colourless Nickel Violet Brown/Reddish brown Grey Grey

Colour in oxidising flame Colour in reducing flame Metal



Non luminous flame is called oxidising flame.



Luminous flame is called reducing flame.

Figure : Borax bead test



All acid radicals which are in JEE syllabus are colourless and diamagnetic. Hence the colour of the salts is only due to the basic radicals.

Table : 7 SOLUBILITY CHART S.No. Anion Solubility / Exception

1. CO₃2– _{Except carbonates of alkali metals and of ammonium, all other normal carbonates} are insoluble.

2. SO₃2– _{Only the sulphites of the alkali metals and of ammonium are water soluble. The} sulphite of other metals are either sparingly soluble or insoluble.

3. S2– _{The acid, normal and polysulphide of alkali metals are soluble in water. The normal} sulphides of most other metals are insoluble; those of the alkaline earths are sparingly soluble, but are graduallychanged by contact with water into soluble hydrogen sulphides. 4. NO₂–_{, NO}

3

– _{Almost all nitrites and nitrates are soluble in water. AgNO}

2sparingly soluble. Nitrates of mercury and bismuth give basic salts on treatment with water. These are soluble in dilute nitric acid.

5. CH₃COO– _{Acetates are water soluble except Ag(I) and Hg(II) acetates which are sparingly} soluble.

6. Cl– _{Most chlorides are soluble in water. PbCl}

2(sparingly soluble in cold but readily soluble in boiling water), Hg₂Cl₂, AgCl, CuCl, BiOCl, SbOCl and Hg₂OCl₂are insoluble in water.

7. Br– _{Silver, mercury(I) and copper(I), bromides are insoluble. Lead bromide is sparinglysoluble} in cold but more soluble in boiling water. All other bromides are soluble in water. 8. I– _{Silver, mercury(I), mercury(II), copper(I), lead and bismuth(III) iodides are the least}

soluble salts. All other iodides are water soluble.

9. SO₄2– _{The sulphates of barium, strontium and lead are insoluble in water, those of calcium} and mercury(II) are slightly soluble. Some basic sulphates of mercury, bismuth and chromium are also insoluble, but these dissolves in dilute hydrochloric or nitric acid. 10 PO₄3– _{The phosphate of the alkali metals, with the exception of lithium and ammonium,} are soluble in water ; the primary phosphate of the alkaline earth metals are soluble. All the phosphates of the other metals and also the secondary and tertiary phosphate of the alkaline earth metals are sparingly soluble or insoluble in water.

Analysis of ANIONS (Acidic Radicals) :

Analysis of anions (acidic radicals) can be broadly divided in to two groups.

(A) GROUP 'A' RADICALS : It involves those anions which are characterised by volatile products by reaction with HCl/ H₂SO₄It is further subdivided in to two groups as given below.

(a) Dilute Sulphuric acid/Dilute Hydrochloric acid : The anions of this group liberate gases or acid vapours with dilute sulphuric acid/hydrochloric acid.

Table : 8

Observation Inference

Gas Radical

Effervescence with the evolution of a colourless CO₂ CO₃2– and odourless gas which turns lime water milky.

Evolution of colourless gas having smell of rotten

egg which turns lead acetate paper black. H₂S S2– Colourless gas having suffocating odour (like burning SO₂ SO₃2– sulphur) which turns acidified K₂Cr₂O₇paper green.

Evolution of reddish brown pungent smelling gas which turns

(i) FeSO₄solution brownish-black and NO₂ NO₂– (ii) wet starch –iodide paper blue.

Colourless gas having smell of vinegar. HAC(g) CH₃COO–

No peculiar gas is evolved. – All above are absent

(b) Concentrated Sulphuric acid group : The anions of this group liberate acid vapours or gases with conc. H₂SO₄.

Table : 9

Observation Inference

Gas Radical

Colourless gas with pungent smell which gives dense HCl Cl– white fumes with a glass rod dipped in NH₄OH.

Reddish brown gas with pungent smell, intensity of Br₂ Br– reddish brown fumes increases on addition of a pinch

of solid MnO₂. Also it turns starch paper orange red.

Evolution of violet vapours which turns starch paper blue.

I

₂

I

– Evolution of reddish brown fumes which intensifies on NO₂ NO₃– addition of copper turnings or bits of filter paper.

Starch iodide paper develops a blue–black spot due to the formation of a I₂–starch complex. (NO₂liberated acts as oxidising agent).

(B) GROUP 'B' RADICALS : Anions of this group do not give acid vapours or gases with dilute as well as concentrated H₂SO₄but are characterised by their specific reactions in solutions. This group is further sub divided into two groups based on the type of the reactions.

(a) Oxidation and reduction in solutions : CrO₄2–_{, Cr} 2O7

2– _etc.

(b) Precipitation reactions : These are given by SO₄2–_{, PO} 4

3–_etc.

Table : 10

Observation Inference

W.E. or S.E. + BaCl₂(aq)  White precipitate insoluble SO₄2– in dil. HCl and HNO_3.

W.E or S.E + conc HNO₃(1–2 mL) + PO₄3– ammonium molybdate and boil  Canary yellow precipitate



W.E. = Water extract. (Salt is dissolved in distilled water)



S.E. = Sodium carbonate extract

Preparation of sodium carbonate extract :

Take 1-2 g of salt/salts mixture and three times the amount of pure solid sodium carbonate in a borosil conical flask. Add 20 mL of distilled water and boil the contents for 10 minutes. Cool the solution and then filter. The Filtrate is termed as "Sodium carbonate extract".

Sodium carbonate reacts with the inorganic salt to form water soluble sodium salt of the acid radical. BaCl₂+ Na₂CO₃  BaCO₃ (white) + 2NaCl (aq)

Cd₃(PO₄)₂+ 3Na₂CO₃  3CdCO₃ + 2Na₃PO₄(aq) Sodium carbonate extract is used when

(a) salt is only partially soluble in water or insoluble

(b) cations interfere with the tests for acid radicals or the coloured salt solutions may be too intense in colour that the test results are not too clear.



As sodium carbonate extract contains excess of sodium carbonate, it should be neutralised with a suitable acid before proceeding for analysis of an anion.

Figure : Preparation of sodium carbonate extract

Individual tests :

(A) GROUP 'A' RADICALS :

(a) DILUTE SULPHURIC ACID/DILUTE HYDROCHLORIC ACID GROUP :

1. CARBONATE ION (CO₃2–_{) :}

 Dilute H₂SO₄test : A colourless odourless gas is evolved with brisk effervescence. CaCO₃+ H₂SO₄  CaSO₄+ H₂O + CO₂

 Lime water/Baryta water (Ba(OH)₂) test : The liberated gas can be identified by its property of rendering lime water (or baryta water) turbid.

CO₂+ Ca(OH)₂  CaCO_3milky+ H₂O On prolonged passage of CO₂the milkiness disappears.

CaCO₃+ CO₂+ H₂O  Ca(HCO₃)₂(soluble)  CaCO₃+ H₂O + CO₂

 Magnesium sulphate test (for soluble carbonates) : CO₃2–_{(aq) + MgSO}

4(aq)  MgCO3 (white) + SO42–(aq)

 Silver nitrate solution : White precipitate is formed CO₃2–_{+ Ag}+  _Ag

2CO3

White precipitate is soluble in HNO₃and ammonia. The precipitate becomes yellow or brown upon addition of excess reagent owing to the formation of silver oxide ; the same happens if the mixture is boiled.

Ag₂CO₃  Ag₂O + CO₂



Mercury(II) chloride does not form precipitate with hydrogen carbonate ions, while in a solution of normal carbonates a reddish–brown precipitate of basic mercury(II) carbonate (3HgO. HgCO₃= Hg₄O₃CO₃) is formed.

CO₃2–_{+ 4 Hg}2+_{+ 3 H}

2O  Hg4O3CO3 + 6H+



Lime water milky test is also shown by SO₂but CO₂does not turn the filter paper soaked in acidified K₂Cr₂O₇green.



Soluble bicarbonates give white precipitate with MgSO₄(aq) / MgCl₂(aq) only on heating. Mg2+_{+ 2HCO} 3 –  _Mg(HCO 3)2   MgCO₃ + H₂O + CO₂



Action of heat : Bicarbonates : 2NaHCO₃  Na₂CO₃+ H₂O + CO₂

Carbonates : Except carbonates of Na, K, Rb, Cs ; the Li₂CO₃ and all alkaline earth metals decompose as given below :

Li₂CO₃  Li₂O + CO₂; MgCO₃  MgO + CO₂; AgAg₂CO₃  2Ag + CO₂



CaCO₃  CaO + CO₂ 2. SULPHITE ION (SO₃2–_{) :}

 Dilute H₂SO₄test : Decomposition of salt is more rapidly on warming, with the evolution of sulphur dioxide.

CaSO₃+ H₂SO₄  CaSO₄+ H₂O + SO₂ SO₂has suffocating odour of burning sulphur.

 Acidified potassium dichromate test : The filter paper dipped in acidified K₂Cr₂O₇turns green. Cr₂O₇2–_{+ 2H}+_{+ 3SO}

2  2Cr3+(green) + 3SO42– + H2O.

 Barium chloride/Strontium chloride solution : White precipitate of barium (or strontium) sulphite is obtained.

SO₃2–_{+ Ba}2+_/Sr2+  _BaSO

3/SrSO3 (white).



White precipitate dissolves in dilute HCl, when sulphur dioxide is evolved. BaSO₃ + 2H+  _Ba2+_{+ SO}

2 + H2O.

White precipitate (BaSO₃) on standing is slowly oxidised to sulphate which is insoluble in dilute mineral acids. This change is rapidly effected by warming with bromine water, a little concentrated nitric acid or with hydrogen peroxide.

2 BaSO₃ + O₂  2 BaSO₄

BaSO₃ + Br₂+ H₂O  2 BaSO₄+ 2 Br–_{+ 2H}+

Hence, reddish brown colour of bromine water is decolourised. 3BaSO₃ + 2 HNO₃  3 BaSO₄ + 2NO  + H₂O BaSO₃ + H₂O₂  BaSO₄ + H₂O



These reactions are not given by carbonates (distinction from carbonates).

 Zinc and sulphuric acid test : Hydrogen sulphide gas is evolved. SO₃2–_{+ 3Zn}2+_{+ 8H}+  _H

2S + 3Zn2++ 3H2O

 Lime water test : A white precipitate is formed. The precipitate dissolves on prolonged passage of the gas, due to the formation of hydrogen sulphite ions.

Ca(OH)₂+ SO₂  CaSO₃(milky) + H₂O CaSO₃ + SO₂+ H₂O  Ca(HSO₃)₂(soluble)



A turbidity is also produced by carbonates ; sulphur dioxide must therefore be first removed when testing for the latter. This may be affected by adding potassium dichromate solution to the test–tube before acidifying. The dichromate oxidizes and destroys the sulphur dioxide without affecting the carbon dioxide.

 Lead acetate or lead nitrate solution : White precipitate of PbSO₃is obtained. SO₃2–_{+ Pb}2+  _PbSO

3

White precipitate gets soluble in dil. HNO₃on boiling. The precipitate is oxidized by atmospheric oxygen and PbSO₄is formed.

3. SULPHIDE ION (S2–_{) :}

 Dilute H₂SO₄test : Pungent smelling gas like that of rotten egg is obtained. S2–_{+ 2H}+  _H

2S

 Lead acetate test : Filter paper moistened with lead acetate solution turns black. (CH₃COO)₂Pb + H₂S  PbS (black) + 2CH₃COOH.

 Sodium nitroprusside test : Purple coloration is obtained. S2–_{+ [Fe(CN)}

5(NO)]2–  [Fe(CN)5NOS]4–(violet).

It is a ligand exchange reaction not a redox.

No reaction occurs with solution of H₂S or free gas. If however, filter paper moistened with a solution of the reagent is made alkaline with NaOH or NH₃solution, a purple colouration is produced with free H₂S also.



H₂S does not provide sufficient concentration of S2–_{ions so that it does not give sodium nitroprusside}

test. Solubility is low 0.1 M and K₁is just 10–7_.

 Cadmium carbonate suspension/ Cadmium acetate solution : Yellow precipitate is formed. Na₂S + CdCO₃  CdS+ Na₂CO₃



Filter paper moistened with cadmium acetate when brought in contact with evolving gas it turns yellow.

S2–_{+ 2H}+  _H

2S ; H2S + Cd2+  CdS + 2H+.

 Silver nitrate solution : Black precipitate is formed which is insoluble in cold, but soluble in hot, dilute nitric acid.

Ag+_{+ S}2– _{ Ag} 2S

 Methylene blue test : NN–Dimethyl–p–phenylenediamine is converted by iron(III) chloride and hydrogen sulphide in strongly acid solution into the water–soluble dyestuff, methylene blue. This is a sensitive test for soluble sulphides and hydrogen sulphide.

4. NITRITE ION (NO₂¯ ) :

 Dilute H₂SO₄test : Solid nitrite in cold produces a transient pale blue liquid (due to the presence of free nitrous acid, HNO₂or its anhydride, N₂O₃) first and then

evolution of pungent smelling reddish brown vapours of NO₂takes place. NO₂–_{+ H}+ _HNO

2; (2HNO2  H2O + N2O3);

3HNO₂  HNO₃+ 2NO + H₂O ; 2NO + O₂  2NO₂ 

 Starch iodide test : The addition of a nitrite solution to a solution of potassium iodide, followed by acidification with acetic acid or with dilute sulphuric acid, results in the liberation of iodine, which may be identified by the blue colour produced with starch paste. A similar result is obtained by dipping potassium iodide–starch paper moistened with a little dilute acid into the solution.

2NO₂–_{+ 3I}–_{+ 4CH}

3COOH  I3–+ 2NO+ 4CH3COO–+ 2H2O

Starch + I₃–  _{Blue (starch iodine adsorption complex)}

 Ferrous sulphate test (Brown ring test) : When 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 appears due to the formation of [Fe(H₂O)₅NO]SO₄at the junction of the two liquids. If the addition has not been made slowly and caustiously, a brown colouration results.

NO₂–_{+ CH}

3COOH  HNO2+ CH3COO–

3HNO₂  H₂O + HNO₃+ 2NO Fe2+_{+ SO}

42–+ NO  [Fe, NO]SO4

 Thiourea test : When a dilute acetic acid solution of a nitrite is treated with a little solid thiourea, nitrogen is evolved and thiocyanic acid is produced. The latter may be identified by the red colour produced with dilute HCl and FeCl₃solution.

NaNO₂+ CH₃COOH  HNO₂+ CH₃COONa

HNO₂+ H₂NCSNH₂(s) (thiourea)  N₂+ HSCN + 2H₂O FeCl₃+ 3HSCN dilHClFe(SCN)₃(blood red colouration) + 3HCl

 Acidified potassium permanganate solution : Pink colour of KMnO₄is decolourised by a solution of a nitrite, but no gas is evolved.

5 NO₂– _{+ 2 MnO}

4–+ 6 H+  5 NO3–+ 2 Mn2++ 3 H2O

 Silver nitrate solution : White crystalline precipitate of silver nitrite from concentrated solutions. NO₂–_{+ Ag}+  _AgNO

2 5. ACETATE ION (CH₃COO¯)

 With dilute H₂SO₄a vinegar like smell is obtained.

(CH₃COO)₂Ca + H₂SO₄  2CH₃COOH + CaSO₄

 Neutral ferric chloride test : A deep red/ blood red colouration (no precipitate) indicates the presence of acetate.

6CH₃COO–_{+ 3Fe}3+_{+ 2H}

2O  [Fe3(OH)2(CH3COO)6]++ 2H+



When solution is diluted with water and boiled, brownish red precipitate of basic iron (III) acetate is obtained.

[Fe₃(OH)₂(CH₃COO)₆]+_{+ 4H}

2O  Boil 3Fe(OH)2CH3COO ¯ + 3CH3COOH + H+

 Silver nitrate solution test : A white crystalline precipitate is produced in concentrated solution in the cold.

CH₃COO–_{+ Ag}+ _CH

3COOAg

Precipitate is more soluble in boiling water and readily soluble in dilute ammonia solution.

Example-1 An aqueous solution of salt containing an anion Xn–_{gives the following reactions :}

(i) It gives the purple or violet colouration with sodium nitroprusside solution.

(ii) It liberates a colourless unpleasant smelling gas with dilute H₂SO₄which turns lead acetate paper black. Identify the anion (Xn–_{) and write the chemical reactions involved.}

Solution Xn–_{is S}2–_because

(i) [Fe(CN)₅NO]2–_{+ S}2–  _[Fe(CN)

5NOS]4–(purple or violet colouration)

(ii) S2–_{+ H}

2SO4  H2S (colourless unpleasant smelling) + SO42–

H₂S + Pb(CH₃COO)₂  PbS (black) + 2CH₃COOH Example-2 Sulphite on treatment with dil. H₂SO₄liberates a gas which :

(A) turns lead acetate paper black (B) burns with blue flame

(C) smells like vinegar (D) turns acidified K₂Cr₂O₇ solution green Solution SO₃2–_{+ H}

2SO4  SO2+ SO42–+ H2O

SO₂turns acidified K₂Cr₂O₇solution green.

K₂Cr₂O₇+ H₂SO₄+ 3SO₂  Cr₂(SO₄)₃(Green) + K₂SO₄+ H₂O Therefore, (D) option is correct.

Example-3 A colourless pungent smelling gas (X) is obtained when a salt is reacted with dil. H₂SO₄. The gas (X) responds to the following properties.

(A) It turns lime water milky

(B) It turns acidified potassium dichromate solution green

(C) It gives white turbidity when H₂S gas is passed through its aqueous solution.

(D) Its aqueous solution in NaOH gives a white precipitate with barium chloride which dissolves in dil. HCl liberating (X).

Identify (X) and write the chemical equations involved.

Solution As gas X turns lime water milky it may be CO₂or SO₂. But CO₂is colourless and odourless, so 'X' may be SO₂. This is further, confirmed by the following reactions :

SO₃2–_{+ H}

2SO4  SO42–+ SO2+ H2O ; Ca(OH)2+ SO2  CaSO3(milky) + H2O

K₂Cr₂O₇+ H₂SO₄+ 3SO₂  K₂SO₄+ Cr₂(SO₄)₃(green) + H₂O

SO₂+ 2H₂S  3S (white) + 2H₂O ; SO₂+ 2NaOH  Na₂SO₃+ H₂O

(b) CONC . H

₂

SO

₄

GROUP :

1. CHLORIDE ION (Cl¯) :

 Concentrated H₂SO₄test : Colourless pungent smelling gas is evolved which gives fumes of NH₄Cl when a glass rod dipped in dil. HCl is brought in contact with evolving gas.

Cl–_{+ H}

2SO4  HCl + HSO4–

 NH₄OH + HCl  NH₄Cl (white fumes) + H₂O.



2NaCl + MnO₂+ 2H₂SO₄(conc.)  Na₂SO₄+ MnSO₄+ 2H₂O + Cl₂

 Silver nitrate test :

Cl–_{+ Ag}+  _AgCl (white)

With sodium arsenite it is converted into yellow precipitate (distinction from AgBr and AgI) but insoluble in dilute nitric acid.

3AgCl + AsO₃3–  _Ag

3AsO3 + 3Cl–.



White precipitate is soluble in aqueous ammonia and precipitate reappears with HNO₃. AgCl + 2NH₄OH  [ Ag(NH₃)₂]Cl (Soluble) + 2H₂O

[Ag(NH₃)₂]Cl + 2H+  _AgCl + 2NH 4+.

 Chromyl chloride test : 4Cl–_{+ Cr}

2O72–+ 6H+(conc.)  2CrO2Cl2(deep red vapours) + 3H2O

When deep red vapours are passed into sodium hydroxide solution, a yellow solution of sodium chromate is formed, which when treated with lead acetate gives yellow precipitate of lead chromate.

CrO₂Cl₂+ 4OH–  _CrO

42–+ 2Cl–+ 2H2O

CrO₄2–_{+ Pb}+2  _PbCrO

4(yellow)



Heavy metal chlorides such as Hg₂Cl₂, HgCl₂, SnCl₂, AgCl, PbCl₂and SbCl₃do not respond to this test as they are partially dissociated. This test is given generally by ionic chlorides.



Test should be carried out in a dry test tube otherwise chromic acid will be formed. CrO₂Cl₂+ 2H₂O H₂CrO₄+ 2HCl

2. BROMIDE ION (Br¯) :

 Concentrated H₂SO₄test : First a reddish-brown solution is formed, then reddish-brown bromine vapour accompanies the hydrogen bromide (fuming in moist air) is evolved.

2NaBr + H₂SO₄  Na₂SO₄+ 2HBr 2HBr + H₂SO₄  Br₂+ 2H₂O + SO₂



2KBr + MnO₂+ 2H₂SO₄  Br₂+ K₂SO₄+ MnSO₄+ 2H₂O

 Silver nitrate test : Pale yellow precipitate is formed NaBr + AgNO₃  AgBr + NaNO₃



Yellow precipitate is partially soluble in dilute aqueous ammonia but readily dissolves in concentrated ammonia solution.

AgBr + 2NH₄OH  [Ag(NH₃)₂] Br + H₂O

 Lead acetate test : Bromides on treatment with lead acetate solution, gives a white crystalline precipitate of lead bromide, which is soluble in boiling water giving colourless solution.

2Br–_{+ Pb}+2  _PbBr 2

 Chlorine water test (organic layer test) : When to a sodium carbonate extract of metal bromide containing CCl₄, CHCl₃or CS₂, chlorine water is added and the content is shaken and then allow to settle down reddish brown colour is obtained in organic layer.

2Br–_{+ Cl}

2  2Cl–+ Br2 .

Br₂+ CHCl₃/ CCl₄ Br₂dissolve to give reddish brown colour in organic layer.. With excess of chlorine water, the bromine is converted into yellow bromine monochloride and a pale yellow solution results.

Br₂+ Cl₂  2BrCl

 Starch paper test : When starch paper is brought in contact with evolving bromine gas orange red spots are produced.

Br₂ + starch  starch bromine adsorption complex (orange red)

 Potassium dichromate and concentrated H₂SO₄: When a mixture of solid bromide, K₂Cr₂O₇ and concentrated H₂SO₄is heated and evolved vapours are passed through water, a orange red solution is obtained.

3. IODIDE ION¯) :

 Concentrated H₂SO₄test : Pungent smelling violet vapours are evolved. 2Na + H₂SO₄  Na₂SO₄+ 2HI

2HI + H₂SO₄  I₂ (dark violet) + 2H₂O + SO₂



Evolution of dark violet fumes intensifies on adding a pinch of MnO₂. 3I–_{+ MnO}

2+ 2H2SO4  I3– + Mn2++ 2SO42–+ 2H2S

 Starch paper test : Iodides are readily oxidised in acid solution to free iodine; the free iodine may than be identified by deep blue colouration produced with starch solution.

3I–_{+ 2NO}

2–+ 4H+  I3– + 2NO + 2H2O.

 Silver nitrate test : Bright yellow precipitate is formed.

I

–_{+ Ag}+  _Ag

_I





Bright yellow precipitate is insoluble in dilute aqueous ammonia but is partially soluble in concentrated ammonia solution.

 Chlorine water test (organic layer test) : When chlorine water is added to a solution of iodide, free iodine is liberated which colours the solution brown and on shaking with CS₂, CHCl₃or CCl₄, it dissolves in organic layer forming a violet solution, which settles below the aqueous layer.

2NaI + Cl₂  2NaCl + I₂

I₂+ CHCl₃  I₂dissolves to give violet colour in organic layer.. If excess of chlorine water is added, I₂is oxidised to iodic acid (colourless).

I₃–+ 8Cl₂+ 9H₂O  3 IO₃+ 16Cl–_{+ 18 H}+

 Lead acetate solution : A yellow precipitate is formed which is soluble in hot water forming a colourless solution and yielding golden yellow plates ('spangles') on cooling.

2I–_{+ Pb}+2  _PbI 2

 Potassium dichromate and concentrated sulphuric acid : Violet vapours are liberated, and no chromate is present in distillate.

6I–_{+ Cr}

2O72+ 2H2SO4  3I2 +Cr3++7SO42–+ 7H2O



Action of heat :

Most of halides are stable but few decompose as

2FeCl₃  2FeCl₂+ Cl₂; MgCl₂. 6H₂O  MgO + 2HCl + 5H₂O Hg₂Cl₂  HgCl₂+ Hg ; NH₄Cl  NH₃+ HCl

2CuI₂  Cu₂I₂+ I₂(without heating) 4. NITRATE ION (NO₃¯) :

 Concentrated H₂SO₄test : Pungent smelling reddish brown vapours are evolved. 4NO₃–_{+ 2H}

2SO4  4NO2 + O2+ 2SO42–+ 2H2O



Addition of bright copper turnings or paper pellets intensifies the evolution of reddish brown gas. 2NO₃–_{+ 4H}

2SO4+ 3Cu  3Cu2++ 2NO + 4SO42–+ 4H2O ; 2NO + O2 2NO2

4 C (paper pellet) + 4HNO₃  2H₂O + 4NO₂+ 4CO₂.

 Brown ring test : When a freshly prepared saturated solution of iron (II) sulphate is added to nitrate solution and then concentrated H₂SO₄is added slowly from the side of the test tube, a brown ring is obtained at the junction of two layers.

NaNO₃+ H₂SO₄  NaHSO₄+ HNO₃

6FeSO₄+ 2HNO₃+ 3H₂SO₄  3Fe₂(SO₄)₃+ 2NO + 4H₂O or 2NO₃–_{+ 4H}

2SO4+ 6Fe2+  6Fe3++ 2NO + 4SO42–+ 4H2O.

Fe2+_{+ NO} + 5H

2O  [Fe(H2O)5NO+]2+(brown ring).



On shaking and warming the mixture, NO escapes and a yellow solution of iron(iii) ions is obtained.



Bromides and iodides interfere in brown ring test as liberated halogens obscure the brown ring. Nitrites also interfere the brown ring test and can be removed by adding a little sulphamic acid, or urea. H₂NHSO₃+ NO₂–  _N 2 + SO42–+ H++ H2O NO₂–_{+ H}+  HCl _HNO 2 CO(NH₂)₂+ 2HNO₂  2N₂ + CO₂ + 3H₂O

 Diphenyl amine test : Blue ring is formed at the junction of two liquids (reagent and nitrate salt solutions).

NaNO₃+ H₂SO₄  NaHSO₄+ HNO₃ 2HNO₃  H₂O + 2NO₂+ [O]

2C₆H₅NHC₆H₅+ [O]  (C₆H₅)₂N – N (C₆H₅)₂(blue ring) + H₂O.



This test is also given by various oxidising agents like CrO₄2–_{, Cr}

2O72–, ClO3–, BrO3–, IO3–, NO2–

etc.

TOdistinguish Br₂with NO₂(both are reddish brown gases)

(a) Br₂+ starch–iodide paper Blue black colour spots do not develop immediately as Br₂is a weaker oxidising agent whereas NO₂being strong oxidising agent develops the blue black colour immediately.

(b) Bromine develops orange–red colour spots on starch paper.

(B) GROUP 'B' RADICALS :

Group of anions which do not give any gas with dilute as well as concentrated H₂SO₄in cold but give precipitate with certain reagents :

These acid radicals are identified in inorganic salts by their individual tests as given below 1. SULPHATE ION (SO₄2–_{) :}

 Barium chloride test :

W.E. or S.E. + Barium chloride (aq)  White precipitate Na₂SO₄+ BaCl₂  BaSO₄white+ 2NaCl.



White precipitate is insoluble in warm dil. HNO₃as well as HCl but moderately soluble in boiling concentrated hydrochloric acid.

 Lead acetate test :

W.E. or S.E. + Lead acetate  white precipitate

Na₂SO₄+ (CH₃COO)₂Pb  PbSO₄White + 2CH₃COONa



White precipitate soluble in excess of hot ammonium acetate.

PbSO₄+ 2CH₃COONH₄  (CH₃COO)₂Pb (soluble) + (NH₄)₂SO₄

 Match stick test :

(a) W.E. or S.E. + Barium chloride  white precipitate Na₂SO₄+ BaCl₂  2NaCl + BaSO₄ (white)

(b) White precipitate + Na₂CO₃(s) mix and apply the paste on the end of the carbonized match stick or a wooden splinter. Put it in the reducing flame.

BaSO₄(s) + Na₂CO₃(s)  Na₂SO₄+ BaCO₃ (white) Na₂SO₄+ 4C  Na₂S + 4CO

(c) Now dip the match stick in sodium nitroprusside solution, purple colour near the fused mass is developed.

Na₂S + Na₂[Fe(CN)₅NO]  Na₄ [Fe(CN)₅NOS] (purple)

 Mercury II chloride test : Yellow precipitate is formed SO₄2–_{+ 3Hg}2+_{+ 2H}

2O  HgSO4. 2H2O (basic mercury II sulphate) + 4H+

 Silver nitrate test : White precipitate is obtained. SO₄2–_{+ 2Ag}+  _Ag

2. PHOSPHATE ION (PO₄3–_{) :}

 Ammonium molybdate test :

Na₂HPO₄(aq) + 12(NH₄)₂MoO₄+ 23HNO₃ (NH₄)₃PMo₁₂O₄₀ (canary yellow) + 2NaNO₃+ 21NH₄NO₃+ 12H₂O



Some times ammonium phosphomolybdate is also represented by the formula (NH₄)₃PO₄. 12MoO₃

 Magnesium nitrate or magnesia mixture test : W.E. or S.E + Magnesium nitrate reagent (3-4 mL) and allows to stand for 4-5 minutes, white crystalline precipitate is formed.

Na₂HPO₄(aq) + Mg(NO₃)₂(aq) + NH₄OH(aq)  Mg(NH₄) PO₄ (white) + 2NaNO₃+ H₂O Magnesia mixture is a solution containing MgCl₂, NH₄Cl and a little aqueous NH₃.



PO₄3–_{also gives BaCl}

2test due to the formation of white precipitate of Ba3(PO4)2. So phosphate

test should be carried out first and then conclude if PO₄3– _{is present or absent before proceeding}

with the test for SO₄2–_.

 Silver nitrate solution : Yellow precipitate is formed which is soluble in dilute ammonia and in dilute nitric acid.

PO₄3–_{+ 3Ag}+  _Ag 3PO4

Ag₃PO₄+ 6NH₃  3[Ag(NH₃)₂]+_{+ PO}

43–; Ag3PO4+ 2H+  H2PO4–+ 3Ag+

 Iron (III) chloride solution : Yellowish-white precipitate of FePO₄is obtained HPO₄2–_{+ Fe}3+  _FePO

4 3. BORATE ION (BO₃3–_{) :}

Salt (0.2 g) + conc. H₂SO₄(1 mL) + Ethyl alcohol (4-5 mL) mix in a test tube and then heat. Ignite the evolved vapours with the help of Bunsen flame, green edged flame is obtained.

2Na₃BO₃+ 3H₂SO₄  3Na₂SO₄+ 2H₃BO₃ 3C₂H₅OH + H₃BO₃  (C₂H₅)₃BO₃+ 3H₂O

Example-4 A compound (A) of S, Cl and O has vapour density of 67.5 (approx.). It reacts with water to form two acids and reacts with KOH to form two salts (B) and (C) while (B) gives white precipitate with AgNO₃ solution and (C) gives white precipitate with BaCl₂solution. Identify (A), (B) & (C).

Solution As mixture give white precipitate with BaCl₂and AgNO₃, it should contain SO₄2–_{and Cl}–_{ions. As}

SO₂Cl₂when dissolved in water gives, a mixture of H₂SO₄& HCl which then react with KOH to form KCl and K₂SO₄. Therefore, (A) is SO₂Cl₂and (B) & (C) are K₂SO₄and KCl respectively.

Vapour density of SO₂Cl₂= molecular weight / 2. Vapour density of SO₂Cl₂= 135 / 2 = 67.2. Example-5 Bromine vapours turn moist starch iodide paper :

(A) brown (B) red (C) blue (D) colourless

Solution 2I–_{+ Br}

2  I2+ 2Br–; I2+ starch  blue starch iodine adsorption complex.

Therefore, (C) option is correct.

Example-6 Na₂S₂O₃+₂  NaI + ... [X], [X] is :

(A) Na₂S₄O₆ (B) Na₂SO₄ (C) Na₂S (D) Na₃SO₄ Solution 2Na₂S₂O₃+₂  2NaI + Na₂S₄O_6.

Therefore, (A) option is correct.

Example-7 Column I and column II contains four entries each. Entries of column I are to be matched with some entries of column II. Each entry of column I may have the matching with one or more than one entries of column II.

Column I Column II

(A) Colourless gas evolved on addition of dil. H₂SO₄ (p) Cl–

(B) White precipitate on addition of AgNO₃ (q) S2–

(C) Precipitate with solution containing Pb+2_{ions. (r) NO} 2–

Solution (A - p, q, s) ; (B - p, r, s) ; (C - p, q, s) ; (D - p, q, r, s) (A) Cl–_{+ H}

2SO4  HCl (colourless) + HSO4–; S2–+ 2H+  H2S (colourless)

NO₂–_{+ 2H}+  _NO

2 (reddish brown) + H2O ; SO32–+ 2H+  SO2 (colourless) + H2O

(B) Ag+_{+ Cl}–  _AgCl (white) ; _Ag+_{+ S}2–  _Ag

2S (black)

Ag+_{+ NO}

2–  AgNO2 (white) ; 2Ag++ SO32–  Ag2SO3 (white)

(C) Pb2+_{+ 2Cl}–  _PbCl

2 (white) ; Pb2++ S2–  PbS (black)

Pb2+_{+ NO}

2–  PbNO2(soluble) ; Pb2++ SO32–  PbSO3 (white)

(D) 2MnO₄–_{+ 16HCl}  _5Cl 2+ 2Mn2++ 6Cl–+ 8H2O 2MnO₄–_{+ 5H} 2S + 6H+  Mn2++ 5S + 8H2O 2MnO₄–_{+ 5NO} 2–+ 6H+  Mn2++ 5NO3–+ 3H2O 2MnO₄–_{+ 5SO} 2+ 2H2O  2Mn2++ 5SO42–+ 4H+

Analysis of CATIONS (Basic Radicals) :

Table : 11

Group Group reagent Basic radical Composition and colour of precipitate

Zero NaOH or Ca(OH)₂, heat if required NH₄+ _{Ammonia gas is evolved.}

1. Dil HCl Ag+ _{AgCl ; White}

Hg₂2+ _Hg

2Cl2 ; White

Pb2+ _PbCl

2 ; White

2.(A) H₂S in presence of dil HCl Hg2+ _{HgS ; Black}

(Insoluble in YAS) Pb2+ _PbS; _Black

Bi3+ _Bi 2S3; Black Cu2+ _{CuS ; Black} Cd2+ _{CdS ; Yellow} 2.(B) H₂S in presence of dil HCl As3+ _As 2S3; Yellow (Soluble in YAS) Sb3+ _Sb 2S3; Orange Sn2+ _{SnS ; Brown} Sn4+ _SnS 2; Yellow 3. NH₄OH in presence of NH₄Cl Fe3+ _Fe(OH) 3; Reddish brown Cr3+ _Cr(OH) 3; Green Al3+ _Al(OH) 3; Gelatinous white 4. H₂S in presence of NH₄OH Zn2+ _{ZnS ; White}

and NH₄Cl Mn2+ _{MnS ; Buff (or Pink)}

Co2+ _{CoS ; Black} Ni2+ _{NiS ; Black} 5. (NH₄)₂CO₃in presence of NH₄OH Ba2+ _BaCO 3; White Sr2+ _SrCO 3; White Ca2+ _CaCO 3; White 6. Na₂HPO₄in presence of NH₄OH Mg2+ _Mg(NH 4)PO4; White



[YAS = Yellow ammonium sulphide. (NH₄)₂S_x].

There are some important points which should be kept in mind while doing the analysis of cations. 1. Group 1st_{radicals (Ag}+

,Pb2+,Hg22+) are precipitated as chloride because the solubility product of these

chlorides (AgCl, PbCl₂, HgCl₂) is less than the solubility products of chlorides of all other metal ions, which remain in solution. Lead chloride is slightly soluble in water and therefore, lead is never completely precipitated by adding dilute hydrochloric acid to a sample ; the rest of the lead ions are precipitated with H₂S in acidic medium together with the cations of the second group.

2. Group 2nd_{radicals are precipitated as sulphides because of their low solubility products whereas sulphides}

of other metals remain in solution because of their high solubility products. HCl acts as a source of H+_which

decreases the concentration of S2–_{due to common ion effect. Hence, the concentration of S}2–_{ion is too low}

that it exceeds only the solubility products of the metal sulphides of IInd_group.

We can not use H₂SO₄inplace of HCl because some cations of higher groups i.e. vth_{group will also precipitate}

as their sulphates like BaSO₄, SrSO₄, CaSO₄etc.

HNO₃can't be used in place of HCl. HNO₃is a powerful oxidising agent. HNO₃will oxidize H₂S forming sulphur (yellow precipitate) or colloidal solution causing confusion with CdS, As₂S₃even though Cd2+_{, As}3+

will be absent. The colloidal solution is white-yellow and that cannot be filtered causing unnecessary trouble. 3. Group 3rd_{radicals are precipitated as hydroxides and the addition of NH}

4Cl suppresses the ionisation of

NH₄OH so that only the group 3 cations are precipitated as hydroxides because of their low solubility products.

(i) Excess of NH₄Cl should not be added, as manganese will precipitate as MnO₂.H₂O

(ii) (NH₄)₂SO₄cannot be used in place of NH₄Cl because the SO₄2–_{will also give the precipitate of BaSO} 4,

SrSO₄etc.

(iii) While proceeding for 3rd_{group from 2}nd_{group, the filtrate of 2}nd_{group is boiled off to remove the dissolved}

H₂S and then one drop of concentrated HNO₃is added and again boil so that if Fe2+_{is present is oxidised to}

Fe3+_{. The K}

spof Fe2+is higher than Fe3+, therefore, it is partially precipitated and will thus interfere in the

analysis of 4th_{group radicals. In our scheme Fe}2+_{is not there even if it is present, we shall report only Fe}3+

(Fe2+_{needs other special tests).}

(iv) If the medium remains acidic the hydroxides do not precipitate and we would think that Fe3+_{, Al}3+_{, Cr}3+_are

absent even though they may be present.

(v) In place of NH₄OH, NaOH solution can't be used for the precipitation as their hydroxides because in excess of it we get soluble complexes of Al3+_{and Cr}3+_.

4. In 4th_{group, ammonium hydroxide increases the ionisation of H}

2S by removing H+from H2S as unionised

water.

H₂S 2H+_{+ S}2–_{; H}+ _{+ OH}–  _H 2O

Now the excess of S2–_{ions is available and hence the ionic products of group 4}th_{group cations exceeds their}

solubility products and will be precipitated. In case H₂S is passed through a neutral solution, incomplete precipitation will take place due to the formation of HCl, which decreases the ionisation of H₂S. For example

MnCl₂+ H₂S  MnS + 2HCl

5. In 5th_{group the reagent ammonium carbonate should be added in alkaline or neutral medium. In the}

absence of ammonia or ammonium ions, magnesium will also be precipitated.

PREPARATION OF ORIGINAL SOLUTION (O.S) :

Original solution is used for the analysis of basic radicals except NH₄+_{. It is prepared by dissolving given salt} or mixture in a suitable solvent as follows :

H₂O

dil HCl

conc. HCl

Salt or Mixt. + H₂O  soluble (then H₂O is suitable solvent) If given salt or mixture is insoluble in H₂O then it is dissolved in dil HCl.

Salt or Mixt. + dil HCl  soluble (then dil HCl is taken as solvent) If given salt or mixture is insoluble in dilute HCl then it is dissolved in conc. HCl.

Salt or Mixt. + conc. HCl  soluble

In this way after selecting suitable solvent, given salt or mixture is dissolved in small quantity in the solvent and filtered. Obtained filtrate is called as original solution (O.S.) and that is used for the detection of basic radicals except NH₄+_.

ZERO GROUP :

1. AMMONIUM ION (NH₄+_{) :}

Sodium hydroxide solution : Ammonia gas is evolved on warming the solution containing ammonium salt and sodium hydroxide.

NH₄Cl + NaOH  NH₃+ H₂O + NaCl



The gas can be identified by the following characteristics / reactions. — Its characteristics smell.

— The evolution of the white fumes of ammonium chloride when a glass rod dipped in dilute HCl is held in the vapour.

NH₃+ HCl  NH₄Cl (white fumes)

— Its ability to turn filter paper moistened with Hg₂(NO₃)₂solution black. 2HgNO₃+ 2NH₃           lack b Hg NO ) NH ( Hg _{2 3} + NH₄NO₃ — Its ability to turns filter paper moistened with CuSO₄solution deep blue.

CuSO₄+ 4NH₃  [Cu(NH₃)₄]SO₄

— Filter paper moistened with a solution of manganese (II) chloride and hydrogen peroxide made alkaline with ammonia gives a brown colour due to the oxidation of manganese.

2NH₃+ Mn2+_{+ H}

2O2+ H2O MnO(OH)2+ 2NH4+

 Nessler's reagent (Alkaline solution of potassium tetraiodidomercurate(II) :

Brown precipitate or brown or yellow colouration is obtained according to the amount of ammonia or ammonium ions present. The precipitate is a basic mercury (II) amido–iodide.

NH₄+_{+ 2[HgI} 4]

2–_{+ 4OH}–  _{HgO Hg (NH}

2)I+ 7I–+ 3H2O

 Sodium hexanitrito–N–cobaltate (III) solution : NH₄+_{ions gives a yellow precipitate with the reagent.} 3NH₄+_{+ [Co(NO}

2)6]

3–  _(NH

4)[Co(NO2)6]

 Hexachloridoplatinate (IV) solution (i.e., hexachloroplatinic acid) : NH₄+_{ions gives a yellow precipitate with the reagent .}

2NH₄+_{+ [PtCl}

6]–  (NH4)2[PtCl6]yellow

 Saturated sodium hydrogen tartrate solution (NaHC₄H₄O₆) : NH₄+_{ions gives a white precipitate with the reagent .}

NH₄+_{+ HC}

4H4O6–  NH4HC4H4O6

 4-Nitrobenzene - diazonium chloride reagent : NH₄+_{gives red colouration with the reagent in}

presence of sodium hydroxide.

O₂N N=N—Cl + NH₄+_{+ OH}–  _O

I

st

_{GROUP (Pb}

2+

_{, Hg}

2

2+

_{, Ag}

+

_{) :}

1. LEAD ION (Pb

2+

_{) :}

 Dilute HCl solution : White precipitate is formed in cold solution. Pb2+_{+ HCl}  _PbCI

2 (white) + 2H +

White precipitate is soluble in hot water. White precipitate is also soluble in concentrated HCl or concentrated KCl.

PbCl₂+ 2Cl– _[PbCl 4]

2–_(colourless)

 Sodium hydroxide solution : White precipitate is formed which is soluble in excess of the reagent. Pb2+_{+ 2OH}–  _Pb(OH) 2 ; Pb(OH)2+ 2OH –  _[Pb(OH) 4] 2– [Pb(OH)₄]2–_{+ H}

2O2  PbO2 (black / brownish black) + 2H2O + 2OH – [Pb(OH)₄]2–_{+ S} 2O8 2–  _PbO 2 + 2H2O + 2SO4 2–

 Potassium iodide solution : A yellow precipitate is formed which is soluble in excess more concentrated (6M) solution of the reagent. Yellow precipitate of PbI₂is moderately soluble in boiling water to give a colourless solution.

PbCl₂+ 2KI  PbI₂ + 2KCl ; Pbl₂+ KI K₂[PbI₄]

Yellow precipitate reappears on dilution with water. Yellow precipitate of PbI₂does not dissolve in excess of dilute solution of KI.

 Potassium chromate solution (in neutral, acetic acid or ammonia solution) : A yellow precipitate is formed.

PbCl₂+ K₂CrO₄  PbCrO₄ + 2KCl

Yellow precipitate is soluble in sodium hydroxide and HNO₃(nitric acid). 2PbCrO₄+ 2H+ _2Pb2+_{+ Cr}

2O7 2–_{+ H}

2O PbCrO₄+ 4OH– _[Pb(OH)

4]

2–_{+ CrO} 4

2–

Both reversible reactions on buffering the solution with ammonia or acetic acid respectively, PbCrO₄ reprecipitates.

 Ammonia solution : With ammonia solution, Pb2+_{gives a white precipitate of lead hydroxide.} Pb2+_{+ 2NH}

4OH  Pb(OH)2 + 2NH4 +

 Dilute H₂SO₄: White precipitate is formed which is soluble in more concentrated ammonium acetate (6M) solution or ammonium tartrate in the presence of ammonia.

PbCl₂+ H₂SO₄  PbSO₄ + 2HCl PbSO₄ + 4CH₃COO–  _[Pb(CH 3COO)4] 2–_{+ SO} 4 2– PbSO₄ + 2C₄H₄O₆2–  _[Pb(C 4H4O6)2] 2–_{+ SO} 4 2–

Hot concentrated H₂SO₄dissolves the precipitate due to the formation of PbHSO₄. PbSO₄ + H₂SO₄  Pb2+_{+ 2HSO}

4 –

2. MERCURY(

I

) ION (Hg₂2+_{) :}

 Dilute HCl solution : White precipitate is formed in cold solution. Hg₂2+_{+ 2HCl  Hg}

2Cl2 (white) + 2H +

 Ammonia solution : A mixture of mercury metal (black precipitate) and basic mercury (II) amido chloride (white precipitate) is formed.

2Hg₂Cl₂+ 4NH₄OH               black 2)Cl Hg HgO.Hg(NH   + 3NH₄Cl + 3H₂O

 Dissolution of white precipitate (Hg₂Cl₂) in aquaregia : 3Hg₂Cl₂+ 2HNO₃+ 6HCl 6HgCl₂+ 2NO+ 4H₂O

(a) Stannous chloride test : White precipitate is formed which finally turns to black.

2HgCl₂+ SnCl₂  Hg₂Cl₂ + SnCl₄; Hg₂Cl₂+ SnCl₂  2Hg (black) + 2SnCl₄ (b) Potassium iodide test : Scarlet/red precipitate is formed which is soluble in excess of the reagent. HgCl₂+ K

I

 Hg

I

₂+ 2KCl ; Hg₂+ K

I

(excess)  K₂[Hg₄] (soluble) (c) Copper chips test : Shining grey deposition of mercury on copper chips is formed.

HgCl₂+ Cu  Hg (grey) + CuCl₂

 Potassium iodide solution : A green precipitate is formed. Hg₂2+_{+ 2I}–  _Hg

2I2 

Green precipitate in excess of reagent undergoes disproportionation reaction and a soluble [HgI₄]2_ _{ions and} black mercury are formed.

Hg₂I₂ + 2I–  _[HgI 4]

2– + Hg (finely divided)

Boiling the mercury (I) iodide precipitate with water, disproportionation takes place and a mixture of red mercury (II) iodide precipitate and black mercury is formed.

Hg₂I₂  HgI₂ + Hg

 Potassium chromate solution : A red crystalline precipitate is formed which turns black when solution of sodium hydroxide is added.

Hg₂2+_{+ CrO} 4

2– _Hg

2CrO4 ; Hg2CrO4+ 2OH

–  _Hg

2O+ CrO4 2–_{+ H}

2O  Potassium cyanide solution : A black precipitate of mercury is obtained

Hg₂2+_{+ 2CN}– Hg + Hg(CN)

2(soluble).

3. SILVER ION (Ag+_{) :}

 Dilute hydrochloric acid/soluble chlorides : White precipitate is formed. Ag+_{+ HCl}  _AgCl + H+

The precipitate obtained after filtration is soluble in concentrated HCl. AgCl+ Cl– _[AgCl

2] –

On dilution with water, the equilibrium shifts back to the left and the precipitate reappears. Dilute ammonia solution dissolves the precipitate forming a soluble complex.

AgCl + 2NH₃ [Ag (NH₃)₂]+_{+ Cl}–

Dilute nitric acid or hydrochloric acid neutralizes the excess ammonia and the precipitate reappears because the equilibrium is shifted backwards.

[Ag(NH₃)₂]Cl + 2HNO₃  AgCl(white) + 2NH₄NO₃.

 Potassium iodide solution : A bright yellow precipitate is formed which is insoluble in dilute ammonia but partially soluble in concentrated ammonia.

Ag+ _{+ I}–  AgI

The yellow precipitate is soluble in KCN and in Na₂S₂O₃. AgI+ 2CN–  _[Ag(CN) 2] –_{+ I}–_; AgI+ 2S 2O3 2–  _[Ag(S 2O3)2] 3–_{+ I}–

 Potassium chromate solution : Red precipitate is formed which is soluble in dilute HNO₃and in ammonia solution. 2Ag+_{+ CrO}

4

2–  _Ag 2CrO4 2Ag₂CrO₄ + 2H+ _4Ag+_{+ Cr}

2O7 2–_{+ H}

2O 2Ag₂CrO₄ + 4NH₃  2[Ag(NH₃)₂]+_{+ CrO}

4 2–

 Disodium hydrogen phosphate solution : In neutral solution a yellow precipitate is formed with the reagent. 3Ag+_{+ HPO}

4

2–  _Ag

3PO4+ H +

The yellow precipitate is soluble in nitric acid and ammonia solution.

 Hydrazine sulphate (saturated) : With diammineargentate (I) reagent forms finely divided silver which adheres to the cleaned glass walls of the test tube forming an attractive mirror.

4[Ag(NH₃)₂]+_{+ H} 2N –_NH 2.H2SO4  4Ag+ N2+ 6NH4 +_{+ 2NH} 3+ SO4 2–  Ammonia solution : Brown precipitate is formed.

2Ag+_{+ 2NH}

3+ H2O  Ag2O+ 2NH4 + Precipitate dissolves in ammonia.

Ag₂O + 4NH₃+ H₂O  2[Ag(NH₃)₂]+_{+ 2OH}–

Example-8 A compound on heating with an excess of caustic soda solution liberates a gas (B) which gives white fumes on exposure of HCl. The resultant alkaline solution thus obtained after heating again liberates the same gas (B) when heated with zinc powder. Compound (A) on heating alone gives a neutral oxide of nitrogen not nitrogen gas. Identify (A) and (B) and give the relevant chemical reactions.

Solution As NH₃gives white fumes with HCl, therefore, (B) should be NH₃and (A) should be the salt of ammonium. Further we know that nitrite of ammonium gives NH₃with Zn and alkali and when heated alone gives neutral oxide (N₂O) not N₂. Hence the salt should be ammonium nitrate not ammonium nitrite. NH₄NO₃(A) + NaOH  NaNO₃ + H₂O + NH₃ (B) ; NH₃+ HCl  NH₄Cl (white fumes) NaNO₃+ 8[H] ZnNaOH NaOH + 2H₂O + NH₃; NH₄NO₃  N₂O (neutral) + 2H₂O Example-9 A certain metal (A) is boiled with dilute HNO₃to give a salt (B) and an neutral oxide of nitrogen (C).

An aqueous solution of (B) gives a white precipitate (D) with brine which is soluble in ammonium hydroxide. An aqueous solution of (B) also gives red / brick red precipitate, (E) with potassium chromate solution. Identify (A) to (E) and write the chemical reactions involved.

Solution As solution of (B) gives white precipitate with NaCl (aq) and precipitate is soluble in ammonium hydroxide, it may be of silver salt. Further it gives brick red precipitate with K₂CrO₄, therefore, metal (A) may be silver.

3Ag (A) + 4HNO₃  3AgNO₃(B) + NO(C) + 2H₂O; AgNO₃+ NaCl  AgCl (white) (D) + NaNO₃ AgCl + 2NH₄OH  [Ag(NH₃)]₂Cl (soluble) + 2H₂O

2AgNO₃+ K₂CrO₄  Ag₂CrO₄ (red / brick red) (E) + 2KNO₃

Example-10 Which of the following salt will give white precipitate with the solution containing Pb2+_{ions ?} (A) Na₂CO₃ (B) NaCl (C) Na₂SO₃ (D) All of these Solution Pb2+_{+ CO} 3 2–  _PbCO 3 (white) Pb2+_{+ 2Cl}–  _PbCl 2 (white) Pb2+_{+ SO} 3 2–  _PbSO 3 (white) Therefore, (D) option is correct.

II

nd

_{Group (Hg}

2+

_{, Pb}

2+

_{, Bi}

3+

_{, Cu}

2+

_{, Cd}

2+

_{, As}

3+

_{, Sb}

3+

_{, Sn}

2+

₎

On the basis of the solubility of the precipitates of the sulphides of II group cations in yellow ammonium sulphide, they have been classified into two subgroups as given below :

A : HgS, PbS, CuS, Bi₂S₃, all black but CdS is yellow. All insoluble in yellow ammonium sulphide.

B : SnS₂, As₂S₃are yellow, Sb₂S₃is orange & SnS is dark brown All soluble in yellow ammonium sulphide. IIA Group (Hg2+_{, Pb}2+_{, Bi}3+_{, Cu}2+_{, Cd}2+₎

1.

MERCURY (II) ION (Hg

2+

₎

_:

Precipitation with H₂S in acidic medium : Black precipitate is formed. Precipitate insoluble in water, hot dilute HNO₃,alkali hydroxides, or colourless ammonium sulphide.

Hg2+_{+ H}

2S 



H

HgS + 2H+



Na₂S (2M) dissolves the precipitate forming soluble complex. HgS + S2– _{ [HgS}

2] 2–



Aqua regia dissolves the precipitate.