30. QUALITATIVE ANALYSIS
1. INTRODUCTION
Salt → Cation (C+− )+ Anion (A ) from base from acid (basic radical) (acidic radical)
Determination of quality of any salt mixture is called Qualitative analysis or Salt analysis.
Flowchart 30.1: Procedure of salt analysis 30.2 | Qualitative Analysis
2. PHYSICAL APPEARANCE
Table 30.1: Physical properties
Experiment Observation Inference Color (due to d-d transition of e- in Blue or bluish green Cu2+ or Ni2+ d-block elements and f-f transition Green Ni+2 of e- in f-block elements) Reddish brown Fe+2 Pink violet Fe+3 Light pink Co+2 White Mn+2
Absence of above transition metals ions Ba+2 ,Sn + 2 ,Ca + 2 ,Mg ++ 23 ,Al ,Zn + 2 etc
Smell Rotten egg S2-
- Vinegar CH3COO + Ammonical NH4 Density Heavy Pb++22 ,Ba can beCO 2− Light powder 3
++23 + 2 Deliquescence: Colorless →Mg ,Al ,Zn ++23 Colored → Cu ,Fe Deliquescence→ Substance absorbs moisture from the atmosphere until it dissolves in the absorbed water and forms a solution Hygroscopic→ Substance absorbs and hold water molecules from the surroundings (Waterof crystallisation)
3. DRY HEATING TESTS
Table 30.2: Dry tests
Observation Inference 1. Gas evolved: 2−− Colorless + odorless CO33 ,HCO Colorless 2− Rotten egg (HS2 ↑) S − Vinegar like CH3 COOH CH3 COO Burning sulphur 2− 22 −− SO3 ,S 23 O ,S Ammonical + NH4 Cl− Colored (Pungent Smell) −− NO32 , NO Reddish brown (NO2 ↑) Cl− Yellowish green Br− Reddish brown (Br2 ↑) − Violet black (I2 ↑) I Chemistry | 30.3
Observation Inference
2. Sublimate: Decomposition of solid substance and deposition White NH4Cl,Hg(I),Hg(II) chlorides As(III),Sb(III) or Se(IV) in the upper portion of test-tube. Yellow S , As S ,Hg(I) / Hg(II)iodides ⇾ 23 free exception ⇾ sulphur toabove
3. Decripitation: Decomposition with crackling sound on halides of alkali metals, Pb( NO ) heating. Salts not having water of crystallization. { 3 2}
4. Fusion: Salts with water of crystallization will fuse generally. CuSO42 .5H O,FeSO 4 .7H 2 O
5. Swelling: 33−− PO43 , BO 6. Residue: Yellow when hot and white when cold ZnO Brown when hot and yellow when cold PbO ++22 + 2 + 2 White residue which glows on heating Ba,Sr,Ca,Mg 2+ Original salt is blue in color, becomes white on heating Cu or CuSO4 Colored salts become brown or black on heating Co+++232 ,Fe ,Fe ,Cr + 3 ,Cu ++ 22 ,Ni ,Mn + 2 ,etc
4. OTHER DRY TESTS
Figure 30.1: Different regions of flame Figure 30.2: Diagram showing non luminous and luminous flame
5. CHARCOAL CAVITY TEST
Charcoal piece (Carbon) ZnSO4+ NaCO 23 →+ ZnCO 3 NaSO 24 Bunsen burner charcoalwill absorbNa SO 24
ZnCO32→+ ZnO CO →→ CuSO4 CuO Cu reddish residue of metallic Cu
Metal carbonate decompose on heating to form metal oxides and Salt + Na23CO may further get reduced to metal especially if metals are less reactive Figure 30.3: Set up for charcoal cavity test (e.g Ag, Cu, gold). Charcoal absorbs the anion formed by Na. 30.4 | Qualitative Analysis
6. COBALT NITRATE TEST
Residue of charcoal cavity test + cobalt nitrate. If residue in charcoal cavity test is white, then this test is performed.
Table 30.3: Charcoal cavity test
Test Inference
Residue+ Co NO +∆ Distinct colour residue is obtained for different metal cations. ( 3 )2 They are mixed cobalt-metal oxides. E.g. 2Co NO→++ 2CoO 4NO O ( 3)2 22
ZnO+→ CoO ZnO.CoO CoZnO2 Rinman's green ZnO.CoO Green
Al2O3.CoO Blue MgO.CoO Pink SnO.CoO Bluish
7. FLAME TEST
Generally, alkali and alkaline earth metal salts impart characteristic color to the flame. For this metals only, the electronic de-excitation will be in visible range. To perform this test, the metal salts are converted into their corresponding chlorides [generally, these metal chlorides are volatile] (volatile vaporizes easily)
⇾
Figure 30.4: Steps to preform flame test Chemistry | 30.5
Table 30.4: Flame test for transition metals
Sr. No. Metals Color of flame 1 Sr , Li Crimson Red 2 Ca Brick Red 3 Ba Apple Green 4 K Violet (lilac) 5 Cu Blue Green 6 Na Golden Yellow 7 Ni Brown
8. BORAX BEAD TEST
→ Borax salt of boric acid [Na2B4O7.10H2O]. The free end of Pt wire is coiled into a small loop. This loop is heated in Bunsen flame until it is red hot and then is quickly dipped into powdered form of solid. The adhering solid is held in the hottest Red Hot part of the flame. At first, salt loses water of Pt. Wire crystallization and shrinks on the loop forming a Figure 30.5: Heating a Pt wire colorless, transparent glass like bead consisting of mixture of sodium meta borate and boric anhydride. Na B O .10H O →∆∆ Na B O → NaBO+ B O 24 7 2 24 7 2 2 3 colorless,tranparent glasslikebead.
The bead is moistened and dipped into the finely powdered subs. (salt) so that a minute of it sticks to the bead and this bead is heated in lower reducing flame, is allowed to cool and the color is observed. This is then heated in an oxidizing flame, allowed to cooled and color is observed again. Coloured beads are due to the formation of various coloured transition metal borates. The Bead “R” and Bead “O” is due to the variable oxidation states of the metal ions.
CuCO32∆+ CuO CO
CuO+→ B23 O Cu( BO2) 2 blue colored bead is oxdising flame
2Cu(BO23 )+ 2NaBO2 +→ C CuBO2 + Na 2 B 4 O 7 + CO Copper( I) metaborate→ Colorless
2Cu(BO23 )+ 4NaBO2 +→ 2C 2CuBO2 + 2Na 2 B 4 O 7 + 2CO Red colored bead in reducing flame
Table 30.5: Colour of flame of different ions during borax bead test
Color in oxidizing flame Color in reducing flame Metal When Hot When Cold When Hot When Cold Copper Green Blue Colorless Brown red 30.6 | Qualitative Analysis
Color in oxidizing flame Color in reducing flame Metal When Hot When Cold When Hot When Cold Iron Brown yellow Pale yellow Bottle green Bottle green Chromium Green Green Green Green Cobalt Blue Blue Blue Blue Manganese Violet Red Grey Grey Nickel Violet Brown Grey Grey
Non luminous flame = Oxidizing flame. Luminous flame = Reducing flame.
9. MICROCOSMIC SALT BEAD TEST
Na(NH ).HPO .4H O →∆ NaPO + H O ↑ NH ↑ 4 42 3 2 3 micro cosmic salt
(Ammonium sodium phosphate)
Na(NH4).HPO4 when heated first, colorless transparent bead of sodium metaphosphate is obtained. This combines with metallic oxide to form orthophosphate which are of characteristic colors. It will have color similar to borax bead test. NaPO+→ CoO NaCoPO orthophosphate 34 (blue color)
10. CONFIRMATORY TESTS (WET TESTS)
Table 30.6: Wet tests
Sr. No. Acid radical Characteristics
1 2− ++ + + + CO3 All carbonates are ppt. except Na ,K ,Rb ,Cs ,NH4
2 − All bicarbonates are soluble (i.e. are found in soln. phase only). HCO3 + Only alkali metals NH4 bicarbonate can be obtained in solid form or can be isolated.
3 2− + SO3 All sulphides are insoluble except group I-NH4
4 2− + S The acid , bisulphide, polysulphides of alkali metal & NH4 are soluble, rest are ppt. Generally S2- ppt are common.
Sn2+ are normally soluble − HS
5 −− −All are soluble except AgNO ,CH COOAg, CH COOHg NO2 ,NO 33 ,CH COO 2 3 3 Chemistry | 30.7
Sr. No. Acid radical Characteristics
6 − +22 ++ + + + Cl All soluble except Ag ,Hg2 ,Cu ,BiO ,SbO , Pb soluble in hot water
7 − + +22 ++ Br All except Ag , Pb ,Cu ,Hg2 soluble in hot water
8 − +2 + + 2 ++ 32 + I All soluble except Ag ,Hg2 ,Hg ,Cu ,Bi ,Pb +2 +2 9 2− All soluble except Pb ,Ba SO4 ++22 Sr ,Ca or Hg(II) slightly soluble
Basic sulphates of Hg, Bi and Cr are insoluble but these dissolves in dil.HCl or dil.HNO3 E.g. HgO. HgSO→ insoluble 4 (basic) (neutral salt)
10 3− All phosphates are ppt except PO4 + GrpI+ NH 4 slightly soluble
Primary phosphates of alkaline earth metals are also soluble (dihydrogen phosphate) Mg H PO E.g. ( 24)2
11. ANALYSIS OF ANIONIC OR ACIDIC RADICAL
Flowchart 30.2: Analysis of anionic or acidic radical 30.8 | Qualitative Analysis
Table 30.7: Analysis of group A and group B radicals
Group A (dil H2SO4) Sr. No. Observation Gases Inference
1 Effervescence of a colorless & odorless gas , which turns lime water milky CO 2− 2 2- HCO CO3 , 3
2- 2 Evolution of colorless, suffocating gas with burning sulphur smell which SO2 SO3
turns K2Cr2O7 paper green 2– → 3+ Cr2O7 + SO2 Cr (green)
3 Evolution of colorless gas with rotten egg smell which turns lead acetate H2S 2− paper black(PbS) S − 4 Evolution of pungent smelling reddish brown gas which turns starch iodide NO2 NO2 paper blue
- → I +NO2 I2 (I2+starch-blue complex)
- 5 Evolution of colorless gas having vinegar like smell CH3COOH CH3COO
Group A(conc. H2SO4) 1 Colorless, pungent smelling gas which gives white fumes with rod dipped in HCl Cl- − NH3 (conc. H2SO4 cant oxidise Cl toCl2 )
- 2 Reddish brown gas with pungent smell and intensity of these fumes/ Br2 Br
vapours increase on addition of a pinch of MnO2 & these also turn starch iodide paper orange red.
– 3 Evolution of violet vapours which turn starch paper blue I2 I
- 4 Evolution of brown fumes, intensity of which increase on addition of Cu NO2 NO3 pieces/turnings and turns starch iodide paper blue.
Group B Ppt.
2- 1 Water extract (WE)+sodium carbonate extract+soda extract(SE)+BaCl2(aq) White insoluble SO4 ppt.
3- 2 WE+SE+conc.HNO3(1N or 2N)+ammonium molybdate Canary yellow PO4 ppt.
12. SODIUM CARBONATE EXTRACT
Filter paper
2g salt + Funnel 3 times Na23CO Wire gauze + 20 ml distilled Tripod stand water Burner Beaker Sodium carbonate extract
Figure 30.6: Set up for preparation of sodium carbonate extract
Sodium carbonate reacts with other inorganic salts to form water soluble salt of acid radical and the cation of salt 2- will be ppt. in the form of CO3 . SE is used when given salt is partially soluble or insoluble in water. Cation of the salt interfere with the test of acid radical. Chemistry | 30.9
2- Dil.H2SO4 grp (CO3 ,CO2) Table 30.8: Sodium carbonate extract test
Sr. No. Reagent Observation Remarks
1 Dil.H2SO4 Colorless, Some carbonates like FeCO3,MgCO3 22−− and CaCO should be powdered for CO+ H SO → SO ++ H O CO Odourless, brisk 3 3 24 4 2 2 appreciable reaction. Effervescence gas H23 CO
2 Lime water Ca(OH)2 White turbidity White turbidity dissolves in dil. acid liberating CO gas. +→ ↓ 2 Ca(OH)22 CO CaCO 3 (For CaCO3, prolonged passage of CO CaCO↓+ HCl → CaCl + CO ↑+ H O Ba(OH)+→ CO BaCO ↓ 2 3 222 22 3 dissolves the turbidity) dil. → CaCO3+CO2
Ca(HCO3)2
3 MgSO4/BaCl2 soln (colorless) White 2+ 2- → Mg +CO3 MgCO3 2+ → Mg +2HCO3 Mg(HCO3)2 ∆ Mg(HCO32 ) → MgCO3 ↓ + CO 2 + H 2 O
4 AgNO3 soln. White ppt(dissolves in Carbonates are more easily polarized NH or HNO ) + 2- → 3 3 2Ag +CO3 AgCO3 Metal carbonates are generally constant AgCO →∆ or Ag O + CO Yellow/brown ppt. and not stable to heat. 3standing 22 1 →∆ 2Ag + O 2 2
5 HgCl2/Hg(NO3)2 Reddish brown ppt 2+ →- Hg +HCO3 No reaction 2+ 2- → Hg +3CO3 +3H2O 3HgO.Hg(CO3)2.Hg(CO3)2 (basic mercuric carbonate)
2- 6 Phenolphthalein (HPh) CO3 +HPh = Pink
HCO3+HPh = Colo rless
ACTION OF HEAT Bicarbonates →∆ Carbonates
∆ 2NaHCO3 → Na 23 CO + CO 2 ↑ + H 2 O Carbonates →∆ metal oxide+ CO 2 ++ + + exception Na , K , Rb , Cs
Li23 CO→+ Li 2 O CO 2; PbCO32→+ PbO CO
Carbonates of less reactive metals
∆ Ag,Au,Hg,Cu → metal + CO22 ↑ + O ↑ 1 Ag CO →∆ 2Ag + CO ↑ + O ↑ 23 22 2 30.10 | Qualitative Analysis
Table 30.9: Confirmatory test for sulphite radical
S. No. Reagent Observation Remarks
2− SO3
1 Dil.H2SO4 Colorless, suffocating SO22−−+ H SO( aq) → SO ↑+ H O + SO burning sulphur smelling 3 24 2 2 4 gas.
2- 2- Above gas+lime/baryta water Milkiness If both SO3 ,CO3 are present,then SO 2- will Ca(OH)+ SO → CaSO ↓+ H O 3 22 32 2- be oxidized to SO4 by
K2Cr2O7
On prolonged passage, milkiness disappears No milkiness (a) and (b) not given by CO 2- → 3 CaSO3+H2O+SO2 Ca(HSO3)2
Green + →32+− + a) acidifiedK2 Cr 27 O SO 2 Cr SO4 green moist filterpaper
−−2 Blue b) SO+ IO →+ I SO 2 3 24 starchiodate (blue) filter paper
2 BaCl2 Soln. White ppt. BaSO3 dissolves in- 22+−+→ ↓ → Ba SO33 BaSO (i) Dil.HCl BaSO3 + HCl
On exposure to air, changes to BaSO4. SO2+H2O + BaCl2
Oxidising agents show such changes. (ii) Dil.H2SO4 − H O→→ H O;Br Br → 2- 2222 BaSO3 + H2SO4 SO3 −+2 HNO34→→ NO; MnO Mn
2- 2- Sodium nitroprusside Rosy-red coloration If both SO3 and S are present, this test is not Na[Fe(CN) NO]+ SO 2- → Na [Fe(CN) NO]SO 5 3 4 5 3 used.
3 AgNO3 Soln. No Visible change Ag+−+ SO2 → [Ag(SO )]− 33 soluble sulphitoargentate
On adding more AgNO3 to the above complex White ppt.
- + → [Ag(SO3)] +Ag Ag2SO3 Dissolves in NH3 and dil.
HNO3 2- Dissolves in excess of SO3
2- → - Ag2SO3+SO3 2[Ag(SO3)] (aq) − →∆ ↓ +2− + 2Ag( SO3 ) 2Ag SO4 S O2 Chemistry | 30.11
S. No. Reagent Observation Remarks
4 Pb(NO3)2/ Pb(CH3COO)2 White +−22 O2 Pb+ SO33 → PbSO → PbSO 4 5 Zn+ H24 SO (aq) 22−+ + Zn+ SO3 + H → Zn + H22 S ↑+ H O
Illustration 1: M2X.7H2O(A) has water and M2X (M is monovalent alkaline cation and X is divalent anion) in 1:1 ratio + by wt. (A). On reaction with dil. H2SO4 gives a gas that turns K2Cr2O7/H soln green. Identify (A) and explain.
→ SO2− Sol: Anion 3 M2SO3 2x+32+48 = 18x7 = 126 2x = 46 x = 23
A Na2SO⇒3
⇾ Table 30.10: Confirmatory tests for sulphide and nitrite ion
S2- Reagent Observation Reaction
1 Dil H2SO4 Colorless, rotten egg smell.
2- 2- S +H2SO4 → H2S+SO4
Black (i) Above gas+ Pb( NO33) / Pb( CH COO) →↓ PbS 22 moist filterpaper
+2 Yellow (ii) Cd+→ H2 S CdS ↓
− All sulphide +−2 2OH (iii) Pb →Pb( OH) ↓ 2OH 2 One ppt except Al S (aq.) 2− 2 3 − HS2 PbS+ 2H2 O + 2OH → Pb( (OH) 4 H2S+Al(OH)3
2 Oxidising agent Yellow/white ppt. −+HS2 2 (i) MnO4 → S ↓ + Mn 23−+HS2 (ii) Cr27 O → S ↓ + Cr − HS2 (iii) I3 → 3I + S + 3 Sodium Nitroprusside Purple colour
S2− +→ Na Fe CN NO Na Fe CN NO 24( )55( )
4 Methylene blue test Methylene blue 30.12 | Qualitative Analysis
HC 3 CH3 N (CH )N + N(CH ) 32 S 32 3+ 2+ ++ + Fe (soln.) + HS2 + Fe +NH+4 H N
NH2 N, N-Dimethyl phenyl diamine
- NO2 Reagent Observation Reaction
1 Dil.H2SO4 Pale blue liquid (contains
HNO2+N2O3) gives reddish Dil.H2SO4 + solid nitrites brown NO2 fumes-Pungent smell)
2 BaCl2 soln. 2+− Ba+→ NO3 No Reaction
3 Conc.AgNO3 White ppt +− Ag+ NO22 → AgNO 4 KI Soln (excess) Yellow brown vapours
−− + − 2NO2+ 3I + 4H ⇒ I 32 + 2NO ↑+ H O
5 MnO− (acidic) Decolourised by a soln. of 4 nitrite − − −+2 NO2+ MnO 43 →+ NO Mn (Nogas)
6 Solid urea NO− + urea → N ↑+ CO ↑ 2 22 acidified
7 Solid thiourea SCN– ions can be −+ − confirmed by reaction (NH22) CS+ NO → H + N 2 ↑+ SCN + H 2 O 2 with FeCl + HCl acidified 3
Fe3+−+ 3SCN → ↓ Fe SCN Blood red colour ( )3 (aq.)
13. BROWN RING TEST
When nitrite solution is added carefully along the sides of test tube to a saturated solution of FeSO4, acidified with acetic acid or dil. H2SO4, a brown ring is formed due to formation of complex of variable composition best represented as Chemistry | 30.13
Brown Solution of Ring +2 nitrite [Fe(H25O) NO] - (NO)2 Saturated
FeSO4 Solution +
dil HS24O - + Reaction : NO2 + H HNO2
3HNO2 HNO3+ 2NO + HO2 2+ 2- NO +Fe++SO4 +5HO2 [Fe(H25O) NO]SO4
Figure 30.7: Brown ring test for estimation of unsaturation
NO− − NO− − 3 will also give similar test if NO2 and 3 both are present, then NO2 can be selectively decomposed by using
Sulphamic acid (NH23− SO H) O 2−+ HO− S − NH2 + 2HNO 2 → N 24 ↑ + SO + 2H + H 2 O O
Sodium oxide (NaN3) −−+ 2− + NO23+ N + H → N 2 ↑+ SO 4 + 2H + H 2 O
Boil the mixture with NH4Cl NO− 2 is decomposed to N2 2−+ NaNO24+ NH Cl → N 2 ↑+ SO 4 + 2H + H 2 O
Table 30.11: Confirmatory test for acetate ion
‒ CH3COO Test Observation Remarks 1 Vinegar like smell All acetate are soluble Dil.H SO 2 4 except. −+ CH33 COO+→ H CH COOH CH3COOAg and
CH3COOHg 2 CH COO−++→ Ag CH COOAg White ppt. AgNO3 Soln: 33
– → 3 BaCl2 soln: BaCl2 + CH3COO No reaction 4 Fruity smell of an C25 H OH+ conc.H2 SO 4 ester (2− 3ml) ( 1ml)
+1gm acetate salt+heat
n 5 Neutral FeCl3 Red blood sol 6CH COO−+++→ 3Fe3 2H O Fe H O CH COO 3 2 32( )26( 3 )
∆ Reddish brown Above redblood soln. 4H2O 30.14 | Qualitative Analysis
Illustration 2: An aqueous solution of salt containing an anion Xn− gives the following reactions: It gives the purple or violet coloration with sodium nitroprusside solution. It liberates a colorless, unpleasant smelling gas with dilute Xn− H2SO4 which turns lead acetate paper black. Identify the anion ( ) and write the chemical reactions involved.
n− Sol: X is S2- because 24−− Fe CN NO +→ S2− Fe CN NOS ( )55 ( ) (purple or violet coloration) 2− SO2− ++ H S Pb CH COO → PbS ↓ (black) + 2CH COOH S+ HSO24 →↑ HS 2 (colorless, unpleasant smelling) 42( 3)2 3
Illustration 3: Sulphite on treatment with dil H2SO4 liberates a gas which: Turns lead acetate paper black Turns with blue flame Smells like vinegar Turns acidified K solution green
22−− Sol: SO3+ H 24 SO →+ SO 2 SO 4 + H 2 O
SO2 turns acidified K2 Cr 27 O solution green K Cr O+ H SO +→ 3SO Cr SO (Green)+ K SO + H O 22724 2 2( 4)3 24 2 Therefore , (iv) option is correct.
Illustration 4: A colorless pungent smelling gas (x) is obtained when a salt is reacted with dil. H2SO4. The gas (X) responds to the following properties. It turns lime water milky It turns acidified potassium dichromate solution green
It gives white turbidity when H2S gas is passed through it aqueous solution. Its aqueous solution in NaOH gives a white ppt. with barium chloride which dissolves in dil HCl liberating (X). Identify (X) and write the chemical equations involved.
Sol: As gas X turns lime water milky it may be CO2 or SO2. But CO2 is colorless and odorless , so ‘X’ may be SO2 . This is further , confirmed by the following reactions:
22−− SO3+ H 24 SO → SO 4 ++ SO 2 H 2 O
Ca(OH)22+→ SO CaSO 3 (Milky) + H 2 O
K Cr O+ H SO +→ 3SO K SO + Cr SO (Green)+ H O 22724 2 24 2( 4)3 2
SO22+ 2H S →↓ 3S (White) + 2H2 O
SO2+→ 2NaOH Na23 SO + H 2 O
Na23 SO+→ BaCl 2 BaSO 3 ↓ (White) + 2NaCl
BaSO3+ 2HCl → BaCl2 (Soluble) ++ SO22 H O Chemistry | 30.15
Conc: H2SO4 group Table 30.12: Confirmatory test for chloride ion
[Cl‒] Reagent Observation Remarks
1 Dipped rod of Being a stronger oxidizing agent, Cl is −−→H24 SO ↑ + 2 Cl HCl HSO4 NH3+HCl=white fumes not produced.
of NH4Cl + H34 PO ↓ − HCl↑+ H24 PO
2 Yellowish green; turns Permanently bleaches dyes by oxidation MnO( OH) + 2H SO +→ 2Cl− 2 24 iodide paper blue +−22 Mn+ Cl2 ↑+ 2SO 42 + 3H O
3 Ag+−+→ Cl AgCl ↓ White ppt soluble in NH3 (AgNO3 )
AgCl+ AsO3− arsenite →↓ Ag ASO Yellow (distinct from 3 33 AgBr/AgI)
4 Pb2+−+→ 2Cl PbCl White ppt-soluble in hot 2 water) conc.HCl/KCl Pb CH COO ( 3 )2
5 Chromyl chloride test Red vapours Cl2 & K2Cr2O7.have equal oxidizing power. Br- & I-don’t give similar test since - → K2Cr2O7+conc.H2SO4 + 4Cl CrO2Cl2 2- - - Cr2O7 oxidises Br , KI to Br2 & I2
22−+ Bright yellow ppt. CrO44+→ Pb PbCrO ↓
Illustration 5: Comp X imparts imparts a golden yellow flame and shows following reaction
n (1) Zn (powder) when boiled with conc. Ag. Sol of X dissolves and H2 gas is evolved . n n (2) When Ag. Sol of X is added to Ag Sol of SnCl2, a white ppt is obtain which dissolves in excess of X
When dissolved in Zn which is amphoteric, H2O gas is evolved which signified Basic nature
NaOH+ Zn → Na22 ZnO
NaOH+ ZnCl2 → Sn(OH)2 → Na 22 SnO
n Illustration 6: A gaseous mixture of X, Y, Z when passed into acidified K2Cr2O7, gas X was absorbed and sol turned green. The remainder mixture was passed through lime which turned milky by absorbing gas Y. The residue gas when passed through alc. pyrogallol solution turned black.
Sol: K2Cr2O7 and lime water to differentiate between SO2 and CO2 .pyrogallol is used to absorb O2 gas Q.
Illustration 7: Colourless salt (A) gives apple green flame with conc. HCl,(A) or reaction with dil. H2SO4 gives brown fumes (D) turing KI starch paper blue (A) + CH COOH+ K CrO → Yellow yellow ppt ppt (A) + H SO → White ppt (C) +(D) 3 24 24 → (D) + CH32 NH → CH 3 OH + H 2 O + gas (E) (E) + Mg (F) 30.16 | Qualitative Analysis
(F)+ HO2 → NH2 Identify A to F
Sol: Apple green → Ba2+ − Brown fumes with dil H24 SO → NO 2
Ba(NO22 )+ CH 3 COOH+ K2 CrO 4 → BaCrO4(yellow ppt)
Ba(NO )+ H SO → BaSO + HNO 22 2 4 4 2 whiteppt
HNO2+ CH 32 NH → CH 3 OH + H 2 O + N 2 ↑
N2+ Mg → Mg32 N
Mg32 N+ H 2 O → NH3
Table 30.13: Confirmatory test for bromide ion
– Br Reagent Observation Remarks
1 − −+ Reddish brown Br vapours and pungent i) 2Br+ SO2 + 4H →+ Br SO + 2H O 2 4+ 2 22 smell −− ii) Br+ H34 PO →+ HBr H 24 PO
2 Pinch of MnO2 added to above test-tube. Intensity of coloured fumes increases. Fluorescence Cl2,
−+2 Irritating pungent smell. Br2, I2, Br+ MnO22 →+ Br Mn + H 2 O Bleaches litmus paper. Colourless red violet Filter paper dipped in fluorescein dye(yellow) turns red. Turns starch iodide paper blue. 3 Ag+−+→ Br AgBr ↓ Pale yellow ppt. (AgNO ) 3 Dissolves in conc. NH3, KCN hypo soln.
+2 - 4 Pb +2Br →PbBr2 (Pb-acetate) White ppt.Soluble in conc.HBr or conc, KBr
5 - → - 8HNO3+6Br 3Br2↑+NO↑+6NO3 +4H2O 6 −+3 K2 Cr 27 O+ H 2 SO 4 + Br → Br2 + Cr
n 7 Layer test Br2 vapours colours the sol orange red.
With Cl2 water. − i)Cl22 water+ Br →↑ Br Dropwise
ii) Above soln. +CHCl3+shake Upper layer: colourless Lower layer: Red brown
Greater solubility of Br2 in organic solvent
iii) Excess of Cl2water Yellow soln. −− Br+→+ Cl22 Br 2Cl
Br22+→ Cl 2BrCl Chemistry | 30.17
Table 30.14: Confirmatory test for Iodide ion
I‒ Reagent Observation Remarks 1 Violet coloured vapours All are soluble except HI+ HSO− I ↑+ SO ↑ 42 2 Ag+,Hg+2, Cu+, Pb+2, Pb+2 = ↑↑ soluble in hot water
I+ H24 SO ↓↓
I2↑+ S ↓ I 22 ↑+ HS ↑
– – Feasible only with I , not with Br HI is a good reducing agent
2 −+2 Intensity of violet vapours MnO2+ H 24 SO +→+ I I 2 Mn increase 3 Yellow. Ag+−+→ I AgI ↓ Soluble in KCN, AgNO3
Partially soluble in conc. NH3.
4 +−2 +→ ↓ Yellow Pb 2I PbI2 Pb− acetate
5 −+3 I+ K2 Cr 27 O + H 2 SO 4 →+ I 2 Cr
6 −− KNO2+ I + H 24 SO → I 3 + NO ↑+ H2 O
7 −− Scarlet Red + HgCl22+ 2I → HgI ↓+ 2Cl K24 HgI KOH 2− − Nessler 'sreagent HgI24+→ 2I HgI + Used for detection of NH4
and NH3 8 Iodometric Titrations +−2 + i) 2Cu+→ 2I 2Cu + I2 22−− − ii) I2+ 2S 23 O →+ 2I S46 O
9 − +−2 2FeCl33+→ 3I 2Fe + I
10 Layer Test Brown soln. – → Cl2 water +I I2
Above solution. + CHCl3 Upper layer: Colourless lower layer: violet layer
Excess of Cl2 water Colourless iodic acid formed −− i) Cl22+→ I Cl + I
−− II23+→ I ii) − I3++ 8Cl 22 9H O → − −+ 3IO3 ++ 16Cl 18H 30.18 | Qualitative Analysis
Remark for HgCl2 + K2[HgI4] + KOH as alkaline solution of K2HgI4 . Is called Nessler’s reagent used for detection of NH4 or NH3
Table 30.15: Confirmatory test for nitrate ion
‒ NO3 Reagent Observation Remarks
1 − All nitrate are soluble i) NO3+→ H 24 SO HNO 3 − +HSO4
1 Reddish Brown ii) 2HNO→ 2NO ↑+ O ↑ H O 3 22 22 2 Cu turnings+Above soln.→ Intensity of above colour increases +−22 Cu+ 2NO ↑+ 4SO42 + 4H O
3 Diphenyl /blue ring test Blue complex NO− +↓ C H NH 3( 65)2
Ph22−−− N N Ph
4 Brown Ring Test
Brown ring Test
- 3 – 4 ml conc. H2SO4 + 2ml of NO 3 solution +mix+cool the mixture under a stream of cold water+ saturated solution of FeSO4 from the side of the test tube to form a layer on the top of the liquid. A brown ring will formed at the junction of two liquids. This brown ring complex is unstable and decomposes on shaking or on heating into NO and a yellow colour solution of Fe+3 ion. − − + Reduction of NO3 or NO2 is alkaline medium into NH4 or NH3 can be done by Zn, Al , Devarda’s Alloy(50% Cu, 45% Al, 5% Zn) −− − NO3+ 4Zn + 7OH + 6H23 O → NH ↑+ 4[Zn(OH) 4 ] −− − 3NO3+ 8Al + 5OH + 18H23 O → 3NH ↑+ 8[Zn(OH) 4 ]
Action of heat on nitrate →∆ Alkali metal nitrate Metal nitrate + O2 1 NaNO →∆ NaNO + O 3 222 (E xcept Li)
∆ Metal nitrate → Metal oxide + NO22 + O (Except Na++ ,K ,Rb+ ,Cs +) 1 3LiNO →∆ Li O + 2NO + O 3 2 222 ∆ 2Cu(NO32 ) → 2CuO + 4NO2 + O 2
∆ 2Pb(NO32 ) → 2PbO + 4NO2 + O 2
∆ 2Hg(NO32 ) → 2Hg + 4NO2 + O 2 ∆ 2AgNO3 → 2Ag + 4NO22 + O Chemistry | 30.19
∆ NH43 NO → N 2 O + 2H 2 O ∆ NH42 NO → N 2 + 2H 2 O ∆ NHNO25 2 → NH 2 + 2HO 2
Nitrogen oxides are supporter of combustion as these fumes in to N2 and O2.
N2 O → Supporter of combustion (better than air) 2N O → 2N + O 2 22 66.67% 33.33%
Air → O 2 20%
Table 30.16: Confirmatory test for sulphate ion
2‒ SO4 Reagent Observation Remarks
1 Ba+−22+→ SO BaSO ↓ White ppt. All sulphates are 44 soluble except BaCl2 Insoluble in dil.HCl Ba+2,Pb+2,Sn+2,Ag+,Ca+2
Soluble in hot, conc. H2SO4 which are partially soluble. 2 2Ag+−+→ SO2 Ag SO ↓ White ppt. 4 24 AgNO3
3 +−22+→ ↓ White ppt Pb SO44 PbSO Pb NO ( 3 )2
4 3Hg+−22+ SO +↓ 2H O 2HgO.HgSO Yellow ppt. 42 4 Hg NO ( 3 )2
5 Match Stick Test White ppt +−22 Ba+→ SO44 BaSO ↓
Above ppt+ Na23 CO Purple coloration (s) paste
Apply paste on wooden piece of match stick+Burn in reducing flame+Dip in sodium nitroprusside soln.
i) Paste →+BaCO3 Na 24 SO
ii) Na24 SO+→ 4C Na2 S + 4CO ↑
iii) Na S+ Na Fe( CN) NO 225 2− S Na Fe CN NOS 4 ( )5 30.20 | Qualitative Analysis
Table 30.17: Confirmatory test for phosphate ion
3‒ PO4 Reagent Observation Remarks 1 White ppt Ba+−23+ PO → Ba PO ↓ 4 34( )2 BaCl2
2 White ppt 3Ag+−+ PO3 → Ag PO ↓ 4 34 Soluble in NH , KCN, hypo, etc. AgNO3 3
3 Magnesia i) PO32−+++ Mg NH + 44 White ppt. Mg( NO3 ) 2 Mg NH PO↓+ Mg+ → Mg PO ↓ ( 44) 34
ii) AsO32−+++ Mg NH + White ppt 44 Brownish red ppt Mg NH AsO↓+ Mg+ → Mg AsO ↓ ( 44) 34
4 3− + Canary yellow ppt. PO4 conc.HNO3+ammonium molybdate (excess) NH MoO↓↓ NH PO .12MoO ( 4)23 4 ( 44) 3
++H2 O NH 43 NO + NaNO 3
Table 30.18: Confirmatory test for borate ion
3‒ BO3 Reagent Observation Remarks
0.2g salt+Conc. H2SO4(1ml)+C2H5OH(4- Green edged flame 5ml)+.Ignite on a Bunsen flame. 3H24 SO 2Na33 BO →2H33 BO −3Na24 SO 3C25 H OH 2H3 BO 3→(C2 H 53 ) BO 3 −3H2 O Group O
1 + - Ammoniacal smell NH4 + dil.NaOH - White fumes (NH Cl) with HCl ↓ 4
+ - HgNO3 on filter paper becomes black (Hg) NH32↑+ H O + Na - Red litmus turns blue
- (MnCl2+H2O2)on filter paper becomes brown
black (MnO2[MnO(OH)2]) 2 + NH42+ K HgI 4 − +OH Nessler'sreagent ↓ HgO.Hg NH I ( 2 ) Chemistry | 30.21
3 Yellow ppt 3NH+ + Na Co NO 43( 2)6 ↓ NH Co NO↓+ 3Na+ ( 42)36( )
4 + Yellow ppt + 2NH4+ Na 26 PtCl NH4 salts ↓ are insoluble NH PtCl↓+ 2Na+ ( 46)2
Illustration 8: 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 ppt. with AgNO3 solution and (C) gives white ppt. with BaCl2 solution. Identify (A), (B) & (C).
2- - Sol: As mixture gives white ppt. with BaCl2 and AgNO3, it should be SO4 and Cl ions. As SO2Cl2 when dissolves in water gives, a mixture of H2SO4 & HCl which then reacts with KOH to form KCl and K2SO4. Therefore (A) is SO2Cl2 and (B) & (C) are K2SO4 and KCl respectively.
Vapour density of SO2Cl2 = molecular weight/2.
Vapour density of SO2Cl2 = 135/2 = 67.2.
→ Illustration 9: Na2S2O3 + I2 NaI + ………[X], [X] is:
(A) Na2S4O6 (B) Na2SO4 (C) Na2S (D) Na3ISO4
→ Sol: 2Na2S2O3 + I2 2NaI + Na2S4O6 Therefore A is correct option.
Illustration 10: Column I and 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
2- (B) White ppt. on addition of AgNO3 (q) S (C) Ppt. with solution containing Pb+2 ions − (r) NO2 (D) Its acidified salt solution decolourises pink KMnO solution. 2− 4 (s) SO3
Sol: A → p, q, s; B → p, r, s; C → p, q, s; D → p, q, r, s
−−2−+ (A) Cl+ H24 SO →↑ HCl (Colourless) + HSO4; S+→↑ 2H H2 S (Colourless)
NO−++→ 2H NO ↑ (Redish brown) + H O 2−+ 22 2; SO32+→↑ 2H SO (Colourless) + H 2 O
Ag+−+→ Cl AgCl ↓ (White) +−2 (B) ; Ag+→ S Ag2 S ↓ (Black) +− +2 − Ag+→ NO2 Ag NO2 ↓ (White); 2Ag + SO3 → Ag 23 SO ↓ (White)
2+ − 22+− (C) Pb+ 2Cl → PbCl2 ↓ (White); Pb +→ S PbS ↓ (Black) Pb2+−+→ NO PbNO (Soluble); Pb22++ SO − → PbSO ↓ (White) 22 3 3 30.22 | Qualitative Analysis
− 2+− (D) 2MnO42+ 16HCl →+ 5Cl 2Mn ++ 6Cl 8H 2 O − ++2 2MnO42+ 5H S + 6H → Mn + 5S ↓+ 8H2 O − −+2 + − 2MnO4++→++ 5NO 2 6H Mn 5NO32 3H O − − 22+ −+ 2MnO4++ 5SO 22 2H O → 2Mn + 5SO4 + 4H
Classification of Cation Radicals: For the purpose of systematic qualitative analysis, the cations are classified into various groups and the classification is based on whether the cation with the given group reagent and forms a ppt or not
Table 30.19: Classification of cation radicals
Group Cation Reagent Observation
+ Zero NH4 NaOH or Ca(OH)2, heat if Ammonium gas is evolved. required AgCl + +2 +2 Hg Cl I. Ag Hg2 Pb dil. HCl 22 White PbCl2
+2 +2 +2 +3 +2 II (A) Cu , Hg , Pb , Bi , Cd H S gas in dil. HCl insoluble +2 2 Cu CuS in YAS (Yellow Ammonium Hg+2 HgS black Sulphide) (NH4)2Sn Pb+2 PbS +3 Bi S Bi 23
CdS→ yellow CdS → II (B) Sn++++++243535 ,Sn ,As ,As ,Sb ,Sb SnS Brown ppt. → SnS2, As2S3 Yellow → Sb2S3 Orange → As2S5 Yellow solution → Sb2S5 Orange solution +3 +3 +3 → III Fe , Al , Cr NH4OH in presence of NH4Cl Fe(OH)3 Reddish brown → Al(OH)3 Gelatinous white → Cr(OH)3 Dirty/ grey green +2 +2 +2 +2 → IV Mn , Co , Zn , Ni H2S in presence of NH4OH and NiS, CoS Black
NH4Cl ZnS → Dirty black MnS → Buff colored (light pink)
+2 +2 +2 (V) Ba , Sr , Ca (NH4)2CO3 in presence of NH4OH BaCO3 and NH Cl Na HPO in presence 4 2 4 SrCO of NH OH and NH Cl 3 White ppt 4 4 CaCO3
Mg(NH4)PO4 ↓ (White)
+ Preparation of Original Solution (O.S): Original solution is used for the analysis if basic radicals except NH4 . It is prepared by dissolving given salt or mixture in a suitable solvent as follows:
1. H2O 2. Dil. HCl 3. 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 Chemistry | 30.23
In this way after selecting suitable solvent, given salt or mixture is dissolved in small quantity in the solvent and filtered. Obtained filtered is called as original solution (O.S.) and that is used for the detection of basic radicals + except NH4 .
Remarks:
+2 +2 1. Pb is placed in both group I and group II because PbCl2 is soluble is not water and all of Pb is not separated by addition of HCl.
2. In group II, only those sulphides are ppt. which have very low values of Ksp for this. H2S gas is added in acidic
medium. Dil. HNO3 cant be added to prevent oxidation of H2S to sulphur. (It is yellow and may be confused
with CdS). Dil. H2SO4 cant be added to prevent the formation of ppt. of sulphates.
3. Before proceeding with test of group III, the solution is boiled to remove dissolved H2S gas to prevent ppt. +2 of sulphides in group III. Then dil. HNO3 is added as we don’t have to perform tests with Fe .HNO3 oxidises Fe+2 to Fe+3
4. NH4OH is added in presence of NH4Cl to decrease the degree of dissociation of NH4OH by common ion effect.
So, only those salts are ppt. which have low values of Ksp. (NH4)2SO4 and (NH4)2CO3 cannot be used in place
of NH4Cl to prevent the ppt of carbonates and sulphates.
5. For the test of group IV, H2S is added in presence of NH4OH to increase the degree of dissociation of H2S resulting in increase of S2- conc. So that sulphides of higher values of Ksp can be separated. CO2− 6. We cant use NaCO3 in test of group V as Na2CO3 is highly soluble resulting in high conc. of 3 which may lead to ppt of group VI, Mg+2 ions.
Table 30.20: Classification of cation radicals
Group 0: Reagent Observation Remarks 1. Dil.NaOH + → ↑ Ammoniacal smell forms white fumes NH4 +dil. NaOH NH3 +H2O+Na in presence of HCl. NH3(g)+HCl → NH4Cl(g) (dense White fumes)
Filter paper dipped in HgNO3 (aq.) becomes black NH3 HgNO3 →
Hg+↓ Hg(NH23 )NO Black wht ie
Filter paper dipped in red litmus become blue
Filter paper dipped in MnCl2+H2O2 (aq.) Becomes brown black due to
formation of MnO2 [MnO(OH)2]
2. Test with K HgI +− ⊕Yellow / brown 2 4 NH+ K [HgI ] + OH → Hg(NH )I + 2K 42 4 2 Nesslare's reagent basic mercuric amido iodide
3. Test with ++Yellow ppt. 3NH4+→ Na 3 [Co(NO 26 ) ] (NH 43 ) [Co(NO 26 ) ] + 3Na Na3[Co(NO2)6] Yellow
4. Na [PtCl ] or H Yellow ppt 2 6 2 + 2NH → ↓ + [PtCl6] {sodiumchloro 4 +Na2[PtCl6] (NH4)2[PtCl6] +2Na + Generally, NH4 salts are insoluble. palatinate} or {chloropalatinate acid} 30.24 | Qualitative Analysis
Group I
Flowchart 30.3: Systematic path for the analysis of group I radicals Chemistry | 30.25
Table 30.21: Analysis of group I and group II radical
Pb2+ Test/Reagents Observation
Dilute HCl solution: White ppt is formed in cold solution. 2+ + Pb +HCI →↓PbCI2 (white) + 2H White ppt is soluble in hot water. White ppt is also 2− − soluble in concentrated HCl or concentrated KCI. PbCl24↓+ 2Cl → PbCl
Sodium hydroxide solution White ppt is formed which is soluble in excess of Pb2+−+→ 2OH Pb OH ↓ ; the reagent. ( )2 2− Black/brownish black Pb OH↓+ 2OH− → Pb OH ( )24( ) 2− Pb OH+ H O → PbO ↓+ 2H O + 2OH− ( )4 22 2 2 2− Pb OH+ S O22−− → PbO ↓+ 2H O + 2SO ( )4 28 2 2 4
Potassium iodide solution A yellow ppt is formed which is soluble in excess more conc.(6M) soln of the reagent. PbCl22+ 2Kl → Pbl ↓+ 2KCl;
Yellow ppt of Pbl2 is moderately soluble in boiling Pbl2+ Kl → K 24 Pbl ↓ water to give a colourless solution.
Yellow ppt reappears on dilution with water. Yellow
ppt of Pbl2 does not dissolve in excess of dilute solution of Kl.
Potassium chromate solution (in neutral, acetic or A yellow ppt is formed. ammonia solution) Yellow ppt is soluble in sodium hydroxide and +→ ↓ PbCl22 K CrO 4 PbCrO 4 2KCl HNO3 (nitric acid).Both reversible reactions on +−+−2222 buffering the solution with ammonia or acetic acid 2PbCrO2PbCrO++ 2H 2H+−+− 2Pb 2Pb2222 ++ ++ Cr Cr O O H H O O 2PbCrO2PbCrO44++ 2H 2H 2Pb 2Pb ++ ++ Cr Cr2727 O O H H 2 2 O O 4 27 2 respectively, PbCrO4 reprecipitates. 22−− ++++22−− PbCrOPbCrO4444++++ 4OH 4OH Pb Pb(( OH OH)) CrO CrO 4444 (( ))44
Ammonia solution With ammonia solution, Pb2+ gives a white ppt of Pb2+++ 2NH OH → Pb OH ↓+ 2NH lead hydroxide 44( )2
Dilute H2SO4: White ppt is formed which is soluble in more conc. ammonium acetate (6M) solution or ammonium + → ↓+ PbCl2 H 24 SO PbSO 4 2HCl tartrate in the presence of ammonia. 2− −−2 PbSO43↓+ 4CH COO → Pb( CH3 COO) + SO 4 4 Hot,conc.H2SO4 dissolves the ppt due to the 2− formation of PbHSO . PbSO↓+ 2C H O22−− → Pb C H O+ SO 4 4 44 6 ( 44 6)2 4 2 − PbSO4↓+ H 24 SO → Pb + 2HSO4 30.26 | Qualitative Analysis
2+ Hg2 Test/Reagents Observation
Dilute HCl solution White ppt is formed in cold solution. 2++ Hg2 +→ 2HCl Hg22 Cl ↓( white) + 2H
Ammonia solution A mixture of mercury metal (black ppt) and basic mercury (II) amido chloride (white ppt) is formed. 2Hg22 Cl+ 4NH 4 OH → HgO.Hg( NH2) Cl↓+ Hg ↓
++3NH42 Cl 3H O
Dissolution of white ppt. Hg2Cl2 in aquaregia (a) Stannous chloride test: White ppt is formed which finally turns to black. 3Hg22 Cl+ 2HNO 3 +→ 6HCl 6HgCl ++ 2NO 4H2 O (b) Potassium iodide test: Scarlet/red ppt is 2HgCl+ SnCl → Hg Cl ↓+ SnCl ; 2 2 22 4 formed which is soluble in excess of the reagent. Hg Cl+ SnCl →↓ 2Hg( black) + 2SnCl 22 2 4 (c) Copper chips test: Shining grey deposition of HgCl22+ KI → HgI ↓+ 2KCl; mercury on copper chips is formed. HgI2+→ KI(excess) K24 HgI( soluble)
HgCl22+ Cu → Hg ↓+ CuCl
Potassium iodide solution: A green ppt is formed. Green ppt in excess of reagent undergoes 2+− disproportionation reaction and a soluble Hg2 +→ 2l Hg22 l ↓ 2− 2− Hgl − 4 ions and black mercury are formed. Hg22 l+→ 2l Hgl4 +↓ Hg( finely divided) Hg l→ Hgl ↓+ Hg ↓ Boiling the mercury (l) iodide ppt with water, 22 2 disproportionation takes place and a mixture of red mercury (ll) iodide ppt and black mercury is formed.
Potassium chromate solution A red crystalline ppt is formed which turns black 22+− when solution of sodium hydroxide is added. Hg2+→ CrO 4 Hg 24 CrO ↓ ; −−2 Hg24 CrO+ 2OH → Hg2 O ↓+ CrO 4 + H 2 O
Potassium cyanide solution A black ppt of mercury is obtained Hg2+−+ 2CN → Hg ↓+ Hg CN soluble . 2 ( )2 ( )
(Ag+) Dilute hydrochloric acid/soluble chlorides White ppt Ag+++ HCl → AgCl ↓+ H Soluble in conc.HCl − On dilution with water, the equilibrium shifts back AgCl+ Cl− → AgCl 2 to the left and the ppt reappears. Dil.ammonia + − solution dissolves the ppt forming a soluble AgCl+ 2NH33 → Ag( NH) + Cl 2 complex. Ag NH Cl+ 2HNO → AgCl ↓+ 2NH NO Dil.HNO or HCl neutralizes the excess ammonia ( 3)2 3 43 3 and the ppt reappears because the equilibrium is shifted backwards.
Potassium iodide solution A bright yellow ppt. is formed which is insoluble in Ag+−+→ l Agl ↓ dilute ammonia but partially soluble in concentrated − ammonia. The yellow ppt. is soluble in KCN and in −− Agl+→ 2CN Ag CN + l ; Na S O . ( )2 2 2 3 3− Agl+→ 2S O2 Ag S O + l− 23 ( 23)2 Chemistry | 30.27
Potassium chromate solution Red ppt. is formed which is soluble in dilute HNO3 2Ag+−+−+→ CrO22 Ag CrO ↓ and in ammonia solution. 2Ag2Ag+→+→ CrO CrO44 Ag Ag 24 24 CrO CrO ↓ ↓ 2Ag CrO+ 2H++++ 4Ag ++ Cr O22 − − H O 2Ag2Ag22 CrO CrO 4 4++ 2H 2H 4Ag 4Ag ++ ++ Cr Cr2727 O O H H 2 2 O O ++ 2− 2Ag2Ag CrO CrO+→+→ 4NH 4NH 2 2 Ag Ag( NH NH) + + CrO CrO 22−− 2424 3 3 (( 33))22 4 4
Disodium hydrogen phosphate solution In neutral solution a yellow ppt. is formed with the 3Ag+−+ HPO2 → Ag PO ↓+ H + reagent. The yellow ppt. is soluble in nitric acid and 4 34 ammonia solution.
Hydrazine sulphate (saturated) With diammineargentate (l) reagent forms finely + divided silver which adheres to the cleaned glass 4 Ag NH+− H N NH .H SO → ( 3)2 2 22 4 walls of the test tube forming an attractive mirror. +−2 4Ag↓+ N2 ↑+ 6NH 4 + 2NH 34 + SO
Ammonia solution ++ 2Ag+ 2NH32 + H O → Ag 2 O ↓+ 2NH 4 Brown ppt. is formed + Ag O↓+ 4NH + H O → 2 Ag NH+ 2OH− 2 32 ( 3)2 Ppt. dissolves in ammonia.
Hg2+ Test/Reagents Observation
Precipitation with H2S in acidic medium Black ppt. is formed. + 2++H Insoluble in water, hot dilHNO , alkali hydroxides, or Hg+ H2 S → HgS ↓ + 2H 3 2− colourless ammonium sulphide. 2− HgS+→ S HgS2 Na2S(2M) dissolves the ppt. forming soluble 3HgS++ 6HCl 2HNO3 → complex. 3HgCl+ 3S ↓+ 2NO ↑+ 4H O 22 Aqua regia dissolves the ppt. 2HNO+→ S SO2−+ + 2H + 2NO ↑ 34 HgCl2 is undissociated. On heating, white ppt. of
sulphur dissolves forming H2SO4.
Stannous chloride solution When added in moderate amounts silky white ppt. is formed. 2HgCl2+→+ SnCl 2 SnCl 4 Hg 22 Cl ↓ HgCl+ SnCl → SnCl +↓ 2Hg If more reagent is added, Hg (I) chloride is reduced 2 2 42 to black ppt. of mercury.
Potassium iodide solution On slow addition red ppt. is formed. Ppt. dissolves 2+− in excess of Kl forming colourless soluble complex. Hg+→ 2I HgI2 ↓ 2− KCN does not have any effect. − HgI24+→ 2I HgI
Copper chips, sheet or coin A black ppt. of mercury is formed. Hg22+++ Cu → Hg ↓+ Cu
Sodium hydroxide solution When added in small quantity brownish-red ppt. of Hg2+−+ 2OH → HgO ↓+ H O varying composition is formed and in stoichiometric 2 amounts ppt. turns to yellow when Hg (II) oxide is formed. Ppt. is insoluble in excess reagent but dissolves readily in acids and this can be used to differentiate Hg (I) from Hg (II). 30.28 | Qualitative Analysis
Ammonia solution White ppt. of mixed composition (Mercury (II) 2+− oxide+Mercury (II) amido nitrate) is formed with 2Hg+ NO3 + 4NH 32 +→ H O + metal nitrate. HgO.Hg( NH23) NO↓+ 3NH 4
Cobalt (II) thiocyanate test When reagent is added to an aqueous solution of 2 2+ Hg22++++ CO 4SCN − → CO +2 Hg( SCN) ↓ Hg ions and the walls of the test tube is stirred with 4 a glass rod, deep-blue crystalline ppt. is formed. In
place of Cobalt (II) thiocyanate, Co(CH3COO) and
NH4SCN can be added to the aqueous solution of Hg2+ ions.
Cu2+ Test/Reagents Observation
Precipitation with H2S in acidic medium Black ppt. is formed. + Cu2+++ H S H → CuS ↓ + 2H 2 Ppt. is insoluble in boiling dilute (M) H2SO4 (distinction from cadmium), in NaOH, 3CuS+→ 8HNO3 Na Sand NH S. Ppt. dissolves in hot conc. 3Cu NO blue++ 2NO 4H O + 3S 24( )2 ( 32)2 ( ) HNO3 3− 2CuS↓+ 8CN−− → 2 Cu CN+ S 2 ( )4 2 When boiled for longer, S is oxidized to H2SO4 and
a clear solution of Cu(NO3)2 is obtained. KCN dissolves the ppt. forming a clear solution. (disulphide ion)
Ammonia Solution When added sparingly a blue ppt. of basic salt 22− (basic copper sulphate) is formed with CuSO . 2 Cu+ SO4 ++→ 2NH 32 2H O 4 Cu OH .CuSO↓+ 2NH + It is soluble in excess of reagent forming a deep ( )2 44 blue colouration. Cu OH .CuSO+→ 8NH ( )2 43 2+ 2 Cu NH++ SO2−− 2OH ( 34)4
Sodium hydroxide in cold solution A blue ppt. is formed. Cu2 +→ 2 OH− Cu OH ↓ ( )2 Cu OH Heat → CuO ↓ black + H O ( )2 ( ) 2
Potassium iodide It gives a white ppt. of Cu (l) iodide but the solution 2+− − is intensely brown because of the formation of tri- 2Cu+ 5l → Cu22 l ↓+ l 3 iodide ions (or iodine). The soln becomes colourless −22 −− − l3+ 2S 23 O →+ 3l S46 O and a white ppt. is visible when excess of sodium thiosulphate solution is added. These reactions are used in quantitative analysis for the iodometric determination of copper.
Potassium ferrocyanide (Potassium hexacyanidoferrate 2+ (II) solution Cu ions gives brown/chocolate brown ppt. 2Cu2+++ K Fe CN → Cu Fe CN↓+ 4K 42( )66( )
3− 2+ 2 Fe( CN) +→ 3Cu Cu3 Fe( CN) ↓ 66 2 Green Chemistry | 30.29
Potassium cyanide When added sparingly forms first a yellow ppt. Cu2+−+→ 2CN Cu CN ↓ which decomposes into CuCN and cyanogen. ( )2 Excess reagent dissolves the ppt. forming a 2Cu( CN) →↓+ 2CuCN( white) 2 colourless soluble complex. CN↑ highlypoisonous ( )2 ( ) Complex is so stable that H2S cannot ppt. Cu(l) 3− CuCN+→ 3CN− Cu CN sulphide (distinction from cadmium). ( )4
Potassium thiocyanate solution +2 The Cu ions solution initially gives a black ppt. Cu+−2 +→ 2SCN Cu SCN ↓ ( )2 which then slowly decomposes to give white ppt. of 2Cu SCN→ 2CuSCN ↓+ SCN ↑ Cu (l) thiocyanate. ( )22( ) Cu (ll) thiocyanate can be immediately converted 2Cu( SCN) ++ SO 2H O → 2 22 into Cu (l) thiocyanate by adding a suitable reducing −2 −+ 2CuSCN+ 2SCN ++ SO4 4H agent like saturated solution of SO2.
Bi3+ Test/Reagents Observation
Precipitation with H2S in acidic medium Black ppt. is formed which is insoluble in cold dilute + HNO and yellow ammonium sulphide. 3++H 3 2Bi+ 3H2 S → Bi23 S ↓( black) + 6H Bi S+ 8HNO → 2Bi NO + 2NO ++ 3S 4H O 23 3 ( 3)3 2 3 − Bi23 S+ 6HCl(boiling,conc.)→++ 2Bi 6Cl 3H2 S
Sodium hydroxide White ppt. is formed with the reagent,slightly Bi3+−+→ 3OH Bi( OH) ↓ soluble in excess reagent in cold solution but 3 soluble in acids. Ppt. on boiling loses water and ++3 Bi OH+→+ 3H Bi H O – ( )3 2 turns yellowish white which is oxidised to BiO3 by H O . Bi OH→ BiO.OH ↓+ H O 2 2 ( )3 2 −+ BiO.OH+ H22 O → BiO 3 ++ H H 2 O
Ammonia solution White basic salt of variable composition is formed. Bi3+−+ NO + 2NH + 2H O → Bi OH NO↓+ 2NH + 3 32 ( )2 3 4
Alkaline sodium stannite (Sodium A black ppt. of metallic bismuth is obtained. tetrahydroxidostannate (ll)) The reagent must be freshly prepared and test must Bi3+−+→ 3OH Bi OH ↓ be carried out in cold solution. ( )3 2− 2− 2Bi OH+ Sn OH→ 2Bi ↓+ 3 Sn OH ( )34( ) ( )6
Dilution with water Solution of bismuth salts gives white ppt. when water is added in larger quantity. Soluble in Bi3+−+ NO + H O → BiO NO ↓+ 2H+ 32 ( 3) mineral acids (dilute) but insoluble in tartaric acid Bi3+−++ Cl H O → BiOCl ↓ + 2H+ (distinction from antimony) and in alkali hydroxide 2 bismuthoxychloride (distinction from tin). orbismuthylchloriode 30.30 | Qualitative Analysis
Potassium iodide When the reagent is added dropwise to a solution 3+ 3+− containing Bi ions, a black ppt. is formed. The Bi+→ 3I BiI3 ↓ − ppt. dissolves in excess Kl forming orange coloured BiI+ I− → BiI 34 soluble complex. On dilution the reaction is +− BiI32↓+ H O → BiOl ↓+ 2H + 2l reversed and black BiI3 turns orange.
(Cd2+) Test/Reagents Observation
Yellow ppt. is formed which dissolves in hot dil. Precipitation with HS2 in acidic medium + HNO . 2++H 3 Cd+ H2 S → CdS ↓ + 2H Ppt. does not dissolve in KCN. CdS+ 8HNO → 3Cd NO + 4H O ++ 2NO 3S 3( 32)2
Ammonia solution (Dropwise addition) Ammonium hydroxide first gives white ppt. of ++ 2 Cd(OH)2 which gets dissolve in excess of reagent Cd+ 2NH32 + 2H O → Cd( OH) ↓ + 2NH4 2 forming a soluble complex. 2+ Cd OH+→ 4NH Cd NH + 2OH− ( )2433( )
Potassium cyanide White ppt. of Cd(CN)2 is formed which in excess of Cd2+−+→ 2CN Cd CN ↓ reagent dissolves forming a soluble complex. ( )2 2− The colourless soluble complex is unstable,therefore, Cd CN+→ 2CN− Cd CN ( )24( ) reacts with H2S gas forming a yellow ppt. of CdS. 2− Cd CN+ H S → CdS ↓+ 2H+− + 4CN Kl forms no ppt. (distinction from Copper) ( )4 2
Sodium hydroxide White ppt. is obtained which insoluble in excess of Cd2++ aq+ 2NaOH aq→ Cd OH ↓+ 2Na NaOH ( ) ( ) ( )2
Illustration 11: When NaOH solution is mixed with aqueous solution of a salt ‘A’, and warmed, a black ppt. is formed.
Black ppt. is filtered and dissolved in concentrated HNO3 by boiling. The resulting solution gives a chocolate brown coloured ppt. with potassium ferrocyanide solution. The filtrate obtained after filtering off the black ppt., upon warming with Zn and NaOH evolves an alkaline pungent smelling gas. The resulting solution also responds to the brown ring test. The filtrate does not evolve 2N gas when it is boiled with urea in the presence of H2SO4. Identify the cation and anion present in the salt ‘A’.
Sol: Cu(NO ) + NaOH → Cu(OH) ↓ (Blue) + 2NaNO 3 2 ∆ 2 3 ↓ ↓ Cu(OH)2 CuO (Black) + H2O → ↑ 4Zn + NaNO3 + 7NaOH 4Na2ZnO2 + 2H2O + NH3 (Pungent smelling alkaline gas) ∆
CuO + 2HNO3 Cu(NO3)2 + H2O → ↓ 2Cu(NO3)2 + K4 [Fe(CN)6] Cu2 [Fe(CN)6] (Chocolate brown) + 4KNO3
Illustration 12: A compound on heating with an excess of caustic soda solution liberates a gas (B) which gives while 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.
Sol: As NH3 gives white fumes with HCl, therefore, (B) should be NH3 and (A) should be the salt of ammonium.
Further we know that nitrite of ammonium gives a NH3 with Zn and alkali and when heated alone gives neutral oxide (N2O) not N2. Hence the salt should be ammonium nitrate not ammonium nitrite. Chemistry | 30.31
NH43 NO (A)+ NaOH → NaNO3 + H 2 O + NH 3 ↑( B) ;NH 3 +→ HCl NH4 Cl( White fumes) Zn/NaOH NaNO3+ 8 H →NaOH + 2H2 O + NH 3 ;NH 43 NO→ N 2 O( Neutral) + 2H2 O
Illustration 13: A certain metal (A) is boiled with dilute HNO3 to give a salt (B) and an neutral oxide of nitrogen (C). An aqueous solution of (B) gives a white ppt. (D) with brine which is soluble in ammonium hydroxide. An aqueous solution of (B) also gives red/brick red ppt., (E) with potassium chromate solution. Identify (A) to (E) and write the chemical reactions involved.
Sol: As solution of (B) gives white ppt. with NaCl (aq) ppt. is soluble in ammonium hydroxide, it may be of silver salt. Further it gives brick red ppt. with K2CrO4, therefore, metal (A) may be silver.
3Ag(A)+ 4HNO33 → 3AgNO (B) ++ NO( C) 2H 2 O;
AgNO33+→↓ NaCl AgCl( White) (D) + NaNO AgCl+→ 2NH OH Ag NH Cl Soluble+ 2H O 43( )2 ( ) 2
2AgNO324+→ K CrO Ag 24 CrO ↓( Red / Brickred) (E) + 2KNO3
2+ Illustration 14: Which of the following salt will give white ppt. with the solution containing Pb ions?
(A) Na2CO3 (B) NaCl (C) Na2SO3 (D) All of these
22+− Sol: Pb+→ CO33 PbCO ↓( White) 2+− Pb+→ 2Cl PbCl2 ↓( White) 22+− Pb+→ SO33 PbSO ↓( White)
Therefore, (D) option is correct.
Table 30.22: Some important reactions of group III radicals
Fe+3 Test/Reagents Observation Potassium ferrocyanide (Potassium hexacyanidoferrate (ll)) Intense blue ppt. (Prussian blue) of
4− iron (lll) hexacyanidoferrate (ll) is +3 4Fe+→ 3 Fe( CN) Fe4 Fe( CN) ↓ formed. 68 3 4− This is insoluble in dilute acids but − Fe4 Fe( CN) + 12OH → Fe( OH) ↓+ 3 Fe( CN) 83 36 decomposes in concentrated HCl. A large excess of the reagent dissolves it partly or entirely, when an intense blue solution is obtained. Sodium hydroxide turns the ppt. red.
Oxalic acid also dissolves Prussian blue forming a blue-solution.
Important:
If iron (lll) chloride is added to an excess of potassium hexacyanidoferrate (ll), a product with the composition of KFe Fe CN is formed. This tends to form colloidal solutions (‘soluble Prussian blue’) and can not be filtered. ( )6 30.32 | Qualitative Analysis
Potassium ferricyanide (Potassium hexacyanidoferrate (lll)) A brown colouration is formed. 3− Upon adding hydrogen peroxide or Fe3+ +→ Fe CN Fe Fe CN ( )66( ) some tin (ll) chloride solution, the 34−− hexacyanidoferrate (lll) part of the Fe2+ + Fe CN →+ Fe Fe CN ( )66 ( ) compound is reduced and Prussian blue is ppt.. 4− 3+ 2+ 4Fe+ 3 Fe( CN) →+ Fe4 Fe( CN) 66 3 Fe gives dark blue ppt. with potassium ferricyanide. First hexacyanidoferrate (lll) ions oxidise iron (ll) to iron (lll), when hexacyanidoferrate (ll) is formed.
And these ions combine to form a ppt. called Turnbull’s blue.
Note: Composition of this ppt. is identical to that of Prussian blue. Earlier the composition suggested was Fe3 Fe( CN) , hence different name. 6 2 2+ + 3Fe+→ 2K33 Fe( CN) Fe Fe( CN) (Ferrous ferric cyanide)+ 6K 662
Turnbull’s blue
Fe (ll) in ammonical solution gives red solution with DMG-colouration fades on standing due to the oxidation of iron (ll) complex. Fe (lll) does not give such complex.
In complete absence of air, Fe (ll) ions produces white ppt. with potassium hexacyanidoferrate (ll). 4− Fe2 ++ 2K+ Fe CN → K Fe Fe CN ↓ ( )662 ( )
Under ordinary atmospheric conditions a pale-blue ppt. is formed.
Cr+3 Test/Reagents Observation
Acidified H2O2 test
Na2 CrO 4+→ H 24 SO Na 24 SO + H 2 CrO 4 ; Blue colouration Amylalchohol HCrO2 4+ 2HO 22 →CrO5 + HO 2 On acidifying the yellow solution
with dil.H2SO4+few drops of 4CrO52424+ 6H SO → 2Cr( SO) ++ 7O 22 6H O 3 ether/amyl alcohol +H2O2= Blue colouration ,can be extracted into the organic layer by gently shaking.
Blue colouration fades slowly due to the decomposition of perchromic acid (or chromium peroxide) with the liberation of oxygen.
Tests for: In excess of NaOH. Sodium meta-aluminate (soluble). White Al OH+→ NaOH NaAlO + 2H O ( )3 22 gelatinous ppt. NaAlO++ H O NH Cl → Al OH + NaCl + NH 22 4 ( )3 3 Chemistry | 30.33
Fe3+ Test/Reagents Observation Tests for Fe OH Red brown ( )3 dissolves in dil. Fe( OH) +→ 3HCl FeCl32 + 3H O 3 HCl as 4FeCl34+→ 3K Fe( CN) Fe4 Fe( CN) + 12KCl Reaction withK Fe CN : 663 4 ( )6 Ferric ferrocyanide (Dark Blue) FeCl32+→ KSCN Fe( SCN) Cl + KCl Reaction with KCNS:
Ferrithiocyanate (dark red colour)
Cr3+ Test/Reagents Observation Tests for On boiling with CH COO Pb / CH COOH NaOH / Br ,Cr OH gives ( 33)2 2 ( )3 sodium chromate which gives 2NaOH+→ Br22 NaOBr + NaBr + H O yellow ppt. of PbCrO4 with NaOBr→+ NaBr O 2Cr OH+ 4NaOH +→ 3 O 2Na CrO + 5H O Yellow solution ( )3 24 2 Yellowsolution Yellow ppt. Na CrO+ CH COO Pb→ PbCrO ↓+ 2CH COONa 24( 3 )2 4 3 Yellowppt. Yellow solution 2Cr OH+ 2Na CO + 3KNO → 2Na CrO + 3KNO ++ 3H O 2CO ( )3 23 3 2 4 2 2 2 Yellowsolution Yellow ppt. Na CrO+ CH COO Pb→ PbCrO ↓+ 2CH COONa 24( 3 )2 4 3 Yellowppt.
Illustration 15: A black coloured compound (A) on reaction with dil. H2SO4 gives a gas (B) and a green colour solution. The gas (B) on passing in a solution of an acid (C) gives a white/yellow turbidity (D). Gas (B) when passed in acidified solution of (E) gives a ppt. (F) soluble in dil HNO3. After boiling this solution when excess of NH4OH is added, a blue coloured compound (G) is formed. To this solution on addition of acetic acid and aqueous potassium ferrocyanide a chocolate coloured ppt. (H) is obtained. On addition of an aqueous solution of BaCl2 to an aqueous solution of (E), a white ppt. insoluble in HNO3 is obtained. Green colour solution on reaction with ammonium hydroxide in presence of air gives reddish brown ppt.. Identify (A) to (H).
Sol: FeS( A) +→+ H24 SO FeSO 4 H 2 S( B)
HNO3( C) → NO 22 + H O + O;H 2 S + O → H 2 O +↓ S( D)
CuSO4( E) + HS 2 → CuS ↓+( F) HSO24 3CuS+ 8HNO → 3Cu NO+ 2NO ↑+ 4H O + 3S ↓ 33( )2 2 2+ Cu2+ +→ 4NH Cu NH G 33( )4 ( ) 2Cu2+++ K Fe CN → Cu Fe CN ↓+ H 4K 42( )66( ) ( ) 22− Ba+→ SO44 BaSO ↓( white) 23++ + Fe+ 2H +→ O Fe + H2 O Fe3++++→ 3NH 3H O Fe OH ↓ reddish brown + 3NH 32 ( )3 ( ) 4
Hence,(A)= FeS;(B) = H23 S;(C) = HNO ;(D) = S;(E) = CuSO 4 ;(F) = CuS; (G)= Cu NH NO ;(H)= Cu Fe CN ( 33)42( ) 2( )6 FeS( A) +→+ H24 SO FeSO 4 H 2 S( B)
HNO3( C) → NO 22 + H O + O;H 2 S + O → H 2 O +↓ S( D)
CuSO4( E) + HS 2 → CuS ↓+( F) HSO24 3CuS+ 8HNO → 3Cu NO+ 2NO ↑+ 4H O + 3S ↓ 33( )2 2 2+ Cu2+ +→ 4NH Cu NH G 30.34 | Qualitative33 Analysis( )4 ( ) 2Cu2+++ K Fe CN → Cu Fe CN ↓+ H 4K 42( )66( ) ( ) 22− Ba+→ SO44 BaSO ↓( white) 23++ + Fe+ 2H +→ O Fe + H2 O Fe3++++→ 3NH 3H O Fe OH ↓ reddish brown + 3NH 32 ( )3 ( ) 4
Hence,(A)= FeS;(B) = H23 S;(C) = HNO ;(D) = S;(E) = CuSO 4 ;(F) = CuS; (G)= Cu NH NO ;(H)= Cu Fe CN ( 33)42( ) 2( )6
Flowchart 30.4: Systematic path for the analysis of group IV radicals Chemistry | 30.35
14. REACTIONS OF Ni2+ (NICKEL) AND Co2+ (COBALT)
The black ppt. is dissolved in aqua-regia.
3NiS+ 6HCl + 2HNO32 → 3NiCl + 2NO ++ 3S 4H 2 O
3CoS+ 6HCl + 2HNO32 → 3CoCl + 2NO ++ 3S 4H 2 O
The solution is evaporated to dryness and residue extracted with dilute HCl. It is divided into three parts.
Table 30.23: Some important reaction of group IV
Part I Part II Part III
Excess CH3COOH (excess)+KNO2=yellow ppt. confirms Solution containing either nickel or cobalt is treated
NH4OH+ the presence of cobalt. with NaHCO3 and bromine water. Appearance of apple dimethyl KNO+→ CH COOH CH COOK + HNO green colour confirms the presence of cobalt. If no glyoxime = 2 3 3. 2 apple green colour is observed, the solution is heated CoCl+→ 2KNO Co NO + 2KCl rosy red ppt. 22( 2)2 when black ppt. is formed, which shows the presence if nickel is Co NO+ 2HNO → Co NO ++ NO H O of nickel. ( 22)23( 2) 2 present CoCl+→ 2NaHCO Co HCO + 2NaCl Co NO+→ 3KNO K Co NO 23( 3)2 ( 2)36 23( 2) Co HCO+→ 4NaHCO Na Co CO 3H O ( 3)233 4( 32)
+3CO22 Br +→ H 2 O 2HBr + O 2Na Co CO+ H O +→ O 2Na Co CO+ NaOH 4( 32)33 3( 3) Sod. cobalt carbonate (Green colouration)
NiCl2+ 2NaHCO 3 → NiCO 3 + 2NaCl ++ H 22 O CO
2NiCO3+→ O Ni 23 O + 2CO 2 Black
Zn2+ (zinc) The sulphide dissolves in HCl.
ZnS+→ 2HCl ZnCl22 + H S When the solution is treated with NaOH, first a white ppt. appears which dissolves in excess of NaOH
ZnCl+ 2NaOH → Zn OH ↓+ 2NaCl 2 ( )2 Whiteppt. Zn OH+→ 2NaOH Na ZnO + 2H O ( )2 22 2 Soluble
On passingHS2 , white ppt. of zinc sulphide is formed,
NaZnO2 22+→ HS ZnS + 2NaOH Whiteppt
Mn2+ (manganese) Manganese sulphide dissolves in HCl.
MnS+→ 2HCl MnCl22 + H S
On heating the solution with NaOH and Br2 -water, manganese dioxide gets ppt.. 30.36 | Qualitative Analysis
MnCl+→ 2NaOH Mn OH + 2NaCl 2 ( )2 Mn OH+→ O MnO + H O ( )2 22
The ppt. is treated with excess of HNO3 and PbO2 or Pb3O4 (red lead). The contents are heated. The formation of permanganic acid imparts pink colour to the supernatant liquid. 2MnO+ 4HNO → 2Mn NO ++ 2H O O 2 3( 3)2 22 2Mn NO++ 5Pb O 26HNO → 2HMnO + 15Pb NO + 12H O ( 3)22 34 3 4 ( 3) 2 Permanganicacid(pink)
BaCO+ 2CH COOH → CH COO Ba ++ CO H O 3 3 ( 3 )2 22 SrCO+ 2CH COOH → CH COO Sr ++ CO H O 3 3 ( 3 )2 22 CaCO+ 2CH COOH → CH COO Ca ++ CO H O 3 3 ( 3 )2 22
Flowchart 5: For the analysis of group V radical Chemistry | 30.37
GROUP V (Ba (II), Sr (II), Ca (II)) Ammonium carbonate ppt. V group radicals in the form of carbonates.These carbonates are soluble in acetic acid.
Table 30.24: Confirmatory test for group V radicals
Ba2+ Test/Reagents Observation
Ba CH COO+ K CrO → BaCrO ↓+ 2CH COOK White ppt. ( 3)2 24 4 3 Ba CH COO+ NH SO→ BaSO ↓+ 2CH COONH ( 3)22( 44) 4 3 4 Ba CH COO+ NH C O→ BaC O ↓+ 2CH COONH ( 3 )22( 4) 24 24 3 4 White ppt. Sr CH COO+ NH SO→ SrSO ↓+ 2CH COONH ( 3)22( 44) 4 3 4 Whiteppt.White ppt. Sr CH COO NH C O→ SrC O ↓+ 2CH COONH White ppt. ( 3 )22( 4) 24 24 3 4 Calcium chlorate and calcium Ca CH COO+ NH C O→ CaC O ↓+ 2CH COONH ( 3 )22( 4) 24 24 3 4 sulphate are soluble. WhiteWhiteppt. ppt.
VITH GROUP Table 30.25: Confirmatory test for group VI radical
Mg2+ Test/Reagents Observation Disodium hydrogen phosphate solution White crystalline ppt. is formed in
22− presence of NH4Cl prevent precipitation Mg++ NH3 HPO 4 → Mg( NH44) PO ↓ of Mg(OH)2 and NH3 soln Mg22+→ HPO− MgHPO ↓ 44 white flocculent ppt. Ammonia solution White gelatinous ppt is sparingly 2 + soluble in water but readily soluble in Mg+ 2NH44 OH → Mg( OH) ↓+ 2NH 2 ammonium salts. Mg OH → Mg+−2 + 2OH ( )2 + − NH4 ions ‘remove’ OH causing the +− NH44 Cl→+ NH Cl ; hydroxide to dissolve more. Not possible +− with NaCl. NH44+→ OH NH OH(weakbase)
Ammonium carbonate solution + Absence of NH4 salts. 22+− − 5Mg+ 6CO32 + 7H O → 2MgCO 3 .Mg( OH) .5H 2 O↓+ 2HCO 3 + 2 In the presence of NH4 salts no +−2 − precipitation occurs, because the NH43+ CO → NH 3 + HCO 3 equilibrium is shifted towards the − formation of HCO3 ions. Ksp of the
ppt. being high (Ksp of pure MgCO3 −5 is1× 10 ), the concentration of carbonate ions necessary to produce a ppt. is not attained. 4-(4-Nitrophenyl azo resorcinol) or Magneson I Ppt. is dissolved in dilute HCl (min.) +NaOH +0.5ml magneson-I reagent = MgCl2 +→ 2NaOH Mg( OH) + 2NaCl 2 Blue lake Magneson reagent = p-nitrobenzene-azo resorcinol, a dye stuff,
absorbed over Mg(OH)2 to give a blue coloured lake. 30.38 | Qualitative Analysis
Titan yellow (a water soluble yellow dyestuff) Deep red colour or ppt.
It is adsorbed by Mg(OH)2 Dissolve ppt. in dil.HCl(min.)+1 drop of NaOH soln (2M) + 1 drop titan Deep red colour solution or ppt. is yellow soln obtained. 22++ Ba andCa do not react but intensify the colour.
Table 30.26: Action of heat on different compounds
(a) Some oxides liberate O : 2 (e) Some sulphates liberate SO2 : 2HgO Heat → 2Hg + O ↑ Heat 2 2MgSO4 → 2MgO + 2SO22 + O (Red) (Silvery deposit) Heat (hightemp) 2ZnSO4→2ZnO + 2SO22 + O Heat Heat 2Pb34 O → 6PbO + O2 ↑ 2BeSO4 → 2BeO + 2SO22 + O (Red) (Yellow) (f) Some sulphates lose water of crystallization: 2PbO Heat → 2PbO + O ↑ 22 Heat Heat 2CaSO.2HO( ) → 2CaSO.HO( ) + 2HO 2Ag O → 4Ag + O ↑ 4 2 42 2 22 70°° C 100 C ZnSO42 .7H O→ ZnSO42 .6H O → (b) Some carbonates liberate CO: −−HO225HO 450° C Heat ZnSO.HO → ZnSO 42 −HO 4 CuCO32 → CuO + CO ↑ 2 (Green) (black) (g) Some nitrates liberate NO2 and O2: ZnCO Heat → ZnO + CO ↑ 32 2Zn( NO ) Heat →2ZnO + 4NO + O (white) Yellow(hot) 32 22 white Brown white(cold) Heat Heat 2Cu( NO3) →2CuO + 4NO22 + O 2Ag23 CO → 4Ag + 2CO2 ↑ + O 2 ↑ 2 Heat 2Pb( NO ) Heat →2PbO + 4NO + O CaCO32 → CaO + CO ↑ 32 22 Heat 2Mg NO Heat →2MgO + 4NO + O MgCO32 → MgO + CO ↑ ( 3)2 22 Li CO Heat → Li O + CO ↑ 2Ca NO Heat →2CaO + 4NO + O 23 2 2 ( 3)2 22 Heat (c) Some bicarbonates liberate CO2 : 2LiNO3 → Li 222 O + 2NO + 1 / 2O Hg NO Heat →Hg + 2NO + O 2NaHCO3→ Na 23 CO ++ CO 2 H 2 O ( 3)2 22 NH HCO→++ NH CO H O Heat 4 3 3 22 2AgNO3 → 2Ag + 2NO22 + O 2Co NO Heat →2CoO + 4NO + O ( 3)2 22 (d) Some sulphates liberate SO3 : Heat (h) Some nitrates liberate O2 : CuSO.5HO42→ CuSO;4 −5H2 O Heat Heat 2NaNO3 → 2NaNO22 + O →CuO + SO3 Heat Heat 2AgNO3 → 2AgNO22 + O 2FeSO4 → .Fe23 O + SO 2 + SO 3 Al SO →RedHot Al O + 3SO 2( 4)3 23 3 Chemistry | 30.39
(i) Some nitrates liberate N2O: (k) Some chlorides decompose as: Heat → + Heat NH43 NO N 2 O 2H 2 O 2FeCl3 → 2FeCl22 + Cl Heat 2CuCl2 → Cu22 Cl + Cl 2 (j) Hydrated chlorides liberate HCl: Heat NH43 Cl → NH + HCl Heat 2 AlCl3 .6H 2 O → Al23 O + 6HCl + 9H2 O Heat Hg22 Cl → HgCl2 + Hg Heat MgCl22 .6H O → MgO + 2HCl + 5H2 O Heat (l) Some other salts decompose as: ZnCl22 .2H O → Zn( OH) Cl+ HCl + H2 O Heat Heat (NH4) Cr 27 O → N2 + Cr 23 O + 4H 2 O 2( ZnCl22 .H O) → Zn2 OCl 2 + 2HCl + H2 O 2 orange Green Heat 2 FeCl3 .6H 2 O → Fe23 O + 6HCl + 9H2 O 4K Cr O Heat → 4K CrO + 2Cr O + 3O SnCl .2H O Heat → Sn OH Cl + HCl+ HO 2 27 2 4 23 2 22 ( ) 2 Heat NH42 NO → N 2 + 2H 2 O Heat NH43 NO → N 2 + 2H 2 O Heat 2Mg( NH4) PO 4 → Mg22 P O 7 + H 2 O + 2NH 3 Heat 2Zn( NH4) PO 4 → Zn22 P O 7 + H 2 O + 2NH 3 CH COO Pb Heat → PbCO+ CH COCH ( 3 )2 33 3 Heat FeC24 O → FeO + CO2 + CO Heat SnCO24 → SnOCO +2 + CO Heat CaC24 O → CaCO3 + CO K Fe CN Heat →4KCN + Fe + 2C + N 42( )6 100°° C 160 C H3 BO 3 → HBO2 → H24 B O 7 ; Red Hot →BO23 Heat 2KCiO32 → 2KCl + 3O Heat 2KMnO 4 →K2 MnO 4 + MnO 22 + O Heat Na24 B O 7 .10H 2 O→ Na24 B O 7 ; −10H2 O Heat →2NaBO + B O 2 23 (Glassy bead) ∆ Na NH HPO → NH + NaPO + H O Na(NH( 44) )HPO4 4 NH 33 + NaPO 323 + H2O
Table 30.27: Different colored inorganic compounds
Black Coloured Blue Coloured Compounds Green Coloured Compounds Yellow Coloured Compounds Compounds Light Blue
1. PbS 1. Cu(OH)2 1. Ni(OH)2 (Green ppt.) 1. As2S3
2. Ag2S 2. Cu(NO3)2 2. Hg2I2 (Green ppt.) 2. As2S5
3. CuS 3. CuCl2 3. Cr2O3 (Green solid) 3. CdS
4. Cu2S 4. CuSO4.5H2O (Blue vitrol) 4. Cr2(SO4)3 4. SnS2 (Artificial gold) 30.40 | Qualitative Analysis
Black Coloured Blue Coloured Compounds Green Coloured Compounds Yellow Coloured Compounds Compounds
5. NiS 5. Zn2[Fe(CN)6] (Bluwish white 5. CrCl3 5. FeS2 (Fool’s gold) ppt) 6. CoS 6. FeSO4. 7H2O 6. (NH4)2Sx (where X=2 to 5)
6. Co(OH)2 7. HgS 7. FeCl2 7. PbCrO4
8. FeS 8. FeSO4.(NH4)2SO4.6H2O 8. BaCrO4 Deep Blue (Mohr’s salt) 9. NiO 9. SrCrO4
10. MnO 1. [Cu(NH3)4]SO4 (Swizzer’s 9. Na2MnO4 10. AgBr (light yellow) reagent) 11. FeO 10. K2MnO4 11 AgI (Dark yellow)
2. [Cu(NH3)4](NO3)2 11. B(OC2H5)3 (Burns with green 12. CuO 12. PbI2 3. Fe4[Fe(CN)6]3 (Prussian’s blue) edge flame) 13. PbO2 13. PbO (in Cold) 4. Fe3[Fe(CN)6]2 (Turnbull’s blue) 12. CoO.ZnO (Riemann’s green) 14. MnO2 14. ZnO (in Hot) 5. Na4[Fe(CN)5(NOS)] (Violet) 15. Mn3O4 15. HgO(Yellow ppt.)
16. Fe3O4 16. Na2O2(Pale yellow)
17. Co3O4 17. Ag3PO4
18. Ni(OH)3 18. Ag2CO3
19. Cu3P2 19. Ag3AsO4
20. BiI3 20. Cu(CN)2
21. Hg+Hg(NH2)Cl 21. K3[Co(NO2)6]
22. (NH4)3PO4.12MoO3
23. (NH4)3AsO4.12MoO3
24. Na2CrO4 2- 25. CrO4 (Yellow in solution)
Red Coloured Brown Coloured Compounds Orange Coloured Pink Coloured Compounds Compounds Compounds
1. Ag2CrO4 (Brick red) 1. SnS 1. Sb2S3 Mn(OH)2
2. Hg2CrO4 (Brick red) 2. Bi2S3 2. Sb2S5 MnS
- 3. HgI3 (Scarlet red) 3. CdO 3. KO3 MnO4
4. Pb3O4 (2PbO + PbO2) 4. PbO2 4. CsO2 (pink or purple in aq.soln)
2- 5. CrO2Cl2 (Reddish Brown) 5. Fe(OH)3 (Reddish Brown) 5. Cr2O7 (orange in aq. soln) Co(CN)2
6. Fe(CH3COO)3 6. Fe2O3 (Reddish Brown solid) (NH4)2SnCl6 (Blood red) CoCl2.6H2O
7. Fe(SCN)3 (Blood Red) 7. Fe2(CO3)3
8. AsI3 8. Cu2O (Reddish Brown)
9. SbI3 9. Ag3AsO4 (Reddish Brown) − 10. SnI2 10. Cu2I2 + I3 (Brown ppt.)
11 CuBr2 11 Cu2[Fe(CN)6] (Chocolate Brown)
12. [Ni(DMG)2] (Rosy red) 12. NO2 (Brown gas)
13. [Fe(H2O)5(NO)]SO4 (Brown ring) Chemistry | 30.41
POINTS TO REMEMBER
An aqueous solution containing :
+ +2 +2 +2 +3 +2 +2 , Mn+2 Ag , Hg2 , Pb , Hg , Bi , Cu , Ni , +2 +2 + +2 Fe+3 , Al+3 ,Ba+2, Sr , Ca , K , Na+ , Mg
Add dilute HCl Precipitate forms Ions remain in soln. Group II-V: Group I: +2 +2 +3 +2 +2 +2 +3 +3 AgCl(s) Pb ,Hg ,Bi ,Cu , Ni ,Mn ,Fe ,Al
Hg22Cl (s) +2 +2 +2 + + + +2 Ba ,Sr ,Ca ,NH4 , K ,Na ,Mg PbCl2(s)
Add HS2 in 0.3 M HCl
Precipitate forms Ions remain in soln. Group II: Group III-V:
+2 +3 +3 +3 +2 +2 PbS(s) Ni ,Mn ,Fe ,Al , Ba ,Sr , HgS(s) +2 + + + +2 Ca ,NH4 ,K ,Na ,Mg Bi22S (s)
CuS(s) Add HS2 in basic soln.
(NH34/NH Cl buffer).
Precipitate forms Ions remain in soln. Group III: Group IV-V:
NiS(s) Ba+2 , Sr+2 , Ca+2 MnS(s) + + +2 NH +, K ,Na , Mg FeS(s) 4
Al(OH)3(s) Add (NH42)CO3 in basic
soln. (NH34/NH Cl buffer).
Precipitate forms Ions remain in soln. Group IV: Group V: + + NH4 , K , BaCO3(s) Na+ , Mg+2 SrCO3(s)
CaCO3(s)