Cations

11

CationsCations

22

CationsCations

Cations are the positive part of any saltCations are the positive part of any saltThe cations cannot be tested directly as anionsThe cations cannot be tested directly as anionsTherefore, we classified cations according to Therefore, we classified cations according to the solubility of their salt in Hthe solubility of their salt in H22OO

i.e we found that the chlorides of Pbi.e we found that the chlorides of Pb2+2+ , Ag , Ag++ , , HgHg22

2+2+ are insoluble in water, so we put them in are insoluble in water, so we put them in one group and say thatone group and say that “ “The group reagent is The group reagent is cold dil HClcold dil HCl””Also we found that the sulphides of PbAlso we found that the sulphides of Pb2+2+, Hg, Hg2+2+, , BiBi3+3+, Cu, Cu2+2+, Cd, Cd2+2+are insoluble in water, so we put are insoluble in water, so we put them in one group and say thatthem in one group and say that “The group “The group reagent is Hreagent is H22S”S”

33

GroupGroup

reagentIons

Distinguishing features

I (Silver group)

Dilute HClAg+, Pb++, Hg2+

+

Ppt as Chlorides

II(Acid sulphide

gp)

HCl + H2SHg++, Pb++, Bi+++, Cu++, Cd++,

Sn++, As+++, Sb+++, Sn++++

Ppt as Sulphides in acid medium

III(Iron group)

NH4OH + NH4ClAl+++, Cr+++, Fe+++

PPt as Hydroxides

IV(Alkaline sulphide

gp)

NH4OH + NH4Cl +H2S

Ni++, CO++, Mn++, Zn++

Ppt as Sulphides in alkaline medium

V(Alkaline earth

group)

NH4OH + NH4Cl + (NH4)2CO3

Ba++, Sr++, Ca+

+

Ppt as Carbonates.

VI(Alkali metal

group)

Nogroup reagent

Mg++, Na+, K+, NH4

+

Ions not pptd in previous groups.

Classification of cationsClassification of cations

44

Group IGroup I( Pb( Pb2+2+ , Ag , Ag++ , Hg , Hg22

2+2+ ) )

55

Group IGroup I( Pb( Pb2+2+ , Ag , Ag++ , Hg , Hg22

2+2+ ) )The chloride salt of these metals have low solubility so they are The chloride salt of these metals have low solubility so they are ppt as Clppt as Cl-- and called and called “ HCl group”“ HCl group”Group reagent : Group reagent : cold dil HCl slight excess with stirringcold dil HCl slight excess with stirring

Procedure for separation :-Procedure for separation :- 1 ml mix + 1 ml cold dil HCl centrifuge1 ml mix + 1 ml cold dil HCl centrifuge

Residue CentrifigateResidue CentrifigateGroup I Rest of groupsGroup I Rest of groups

Pb ClPb Cl22 white ppt of lead chloride white ppt of lead chloride

Ag Cl white ppt of silver chlorideAg Cl white ppt of silver chloride

HgHg22ClCl22 silky white ppt of mercurus chloride silky white ppt of mercurus chloride

Ques.1Ques.1 : Why cold HCl ? : Why cold HCl ?

Ans:Ans: Because Pb Cl Because Pb Cl22 is soluble on hot so escape to group II is soluble on hot so escape to group II

66

N.B.N.B.

Pb ClPb Cl22 have the highest Ksp, therefore it is the last to ppt. have the highest Ksp, therefore it is the last to ppt.

Pb ClPb Cl22 Ksp = 1.6 x 10 Ksp = 1.6 x 10-6-6

Ag Cl Ksp = 1.0 x 10Ag Cl Ksp = 1.0 x 10-10-10

HgHg22 Cl Cl22 Ksp = 1.0 x 10 Ksp = 1.0 x 10-18-18

Therefore the order of ppt HgTherefore the order of ppt Hg22+2+2 then Ag then Ag++ then Pb then Pb+2+2

Ques.2Ques.2 : Why dil. HCl ? : Why dil. HCl ?

Ans:Ans: Because Pb Cl Because Pb Cl22 tends to form soluble complex and tends to form soluble complex and escape to group II in conc. HCl (i.e PbClescape to group II in conc. HCl (i.e PbCl22 is soluble in is soluble in excess Clexcess Cl--))

PbPb+2+2 + Cl + Cl-- PbCl PbCl22

PbClPbCl22 + Cl + Cl-- [PbCl [PbCl33]]-- + Cl + Cl-- [PbCl [PbCl44]]2-2-

Soluble complexSoluble complex

77

Ques.3Ques.3 : Why slight excess with stirring ? : Why slight excess with stirring ?

Ans :Ans : We add slight excess HCl to ensure complete pptn. By We add slight excess HCl to ensure complete pptn. By the common ion effect.the common ion effect.

(Strong acid) HCl H(Strong acid) HCl H++ + Cl + Cl-- (strongly ionize) (1) (strongly ionize) (1)

(ppt ) AgCl Ag(ppt ) AgCl Ag++ + Cl + Cl- - (partially ionize)(2)(2)

common ion effectcommon ion effect

By common ion effect reaction 2 moves to the left By common ion effect reaction 2 moves to the left preventing the dissolution of Ag Clpreventing the dissolution of Ag Cl

With stirring because Pb ClWith stirring because Pb Cl22 tends to form tends to form

supersaturated soln. i.e. go to soln. therefore we have supersaturated soln. i.e. go to soln. therefore we have to break this supersaturation by stirring.to break this supersaturation by stirring.

88

Procedure for separationProcedure for separationResidue (group I) + 2 ml hot water Residue (group I) + 2 ml hot water

centrifugecentrifugeResidue centrifugate (1) Residue centrifugate (1) AgCl, Hg AgCl, Hg22ClCl2 2 PbClPbCl22

+ dil NH+ dil NH33

centrifugecentrifuge

Residue centrifugate (2) Residue centrifugate (2) Hg Hg AgAgHgHg22ClCl22 + NH + NH33 Hg(NH Hg(NH22)Cl + Hg)Cl + Hg00 AgCl + NH AgCl + NH33 Ag(NH Ag(NH33))22ClCl

white blackwhite blackIt is called “ disproportionation reaction” because (Hg2

2+) tends to form (Hg0) mercury and (Hg2+) mercuric.

99

Confirmation of lead(PbConfirmation of lead(Pb2+2+):-):-1- By cooling centrifugate(1) 1- By cooling centrifugate(1) white ppt of lead chloride white ppt of lead chloride2- Centrifugate(1)+ KI yellow ppt of Pb I2- Centrifugate(1)+ KI yellow ppt of Pb I22 ( (soluble in excess KI)soluble in excess KI)

PbIPbI22 + excess I + excess I-- [PbI [PbI44]]2-2- soluble complex soluble complex 3- Centrifugate + acetic acid + K3- Centrifugate + acetic acid + K22CrOCrO44 (pot.chromate) (pot.chromate) yellow ppt of lead chromateyellow ppt of lead chromate

PbClPbCl22 + K + K22CrOCrO44 PbCrO PbCrO44 + 2 KCl + 2 KClQues.:Ques.: yellow pptyellow ppt

Why acidification with acetic acid and not strong mineral acid ?Why acidification with acetic acid and not strong mineral acid ?Ans :Ans : Because in strong acid Because in strong acid 2CrO2CrO44

2-2- +2 H +2 H++ Cr Cr22OO772- 2- + H+ H22O O

i.e. dichromate is formed and form soluble lead dichromate Pb Cri.e. dichromate is formed and form soluble lead dichromate Pb Cr22OO77

In presence of alkalinity the yellow ppt turns redIn presence of alkalinity the yellow ppt turns red 2 PbCrO2 PbCrO44 + H + H22O PbO Pb22CrOCrO55 + H + H22CrOCrO44

red ppt (basic lead chromate)red ppt (basic lead chromate)

1010

In presence of high alkalinity the ppt will dissolve because Pb In presence of high alkalinity the ppt will dissolve because Pb is amphotericis amphoteric

PbCrOPbCrO44 + 2 OH + 2 OH- - Pb(OH) Pb(OH)22 + CrO + CrO442-2-

white pptwhite ppt

Pb(OH)Pb(OH)22 + OH + OH-- [HPbO [HPbO22]]-- + H + H22OO

soluble plumbite ionsoluble plumbite ion

N.B.N.B.

Amphoteric metals (i.e. soluble in both acid and alkaline Amphoteric metals (i.e. soluble in both acid and alkaline medium)medium)

PbPb2+2+ , , SnSn2+,4+2+,4+, As, As3+,5+3+,5+, Sb, Sb3+,5+3+,5+, , AlAl3+3+,Cr,Cr3+3+, , Zn Zn2+2+

gp I gp IIb gp III gp IVgp I gp IIb gp III gp IV

4- centrifugate + H4- centrifugate + H22SOSO44 white ppt of lead sulphate white ppt of lead sulphate

PbClPbCl22 + H + H22SOSO44 PbSO PbSO44

1111

Confirmation of silver (AgConfirmation of silver (Ag++):):

1- acidify centrifugate(2) with dil HNO1- acidify centrifugate(2) with dil HNO33 white ppt of white ppt of silver chloride soluble in dil. NHsilver chloride soluble in dil. NH33..[Ag(NH[Ag(NH33))22]Cl + 2 HNO]Cl + 2 HNO3 3 AgCl + NH AgCl + NH44NONO33 white pptwhite ppt

Silver amine complex have high KSilver amine complex have high Kstabstab,so it ionizes to give a quantity ,so it ionizes to give a quantity of Ag insufficient to exceed Ksp of Ag Cl (10of Ag insufficient to exceed Ksp of Ag Cl (10-10-10) therefore AgCl is ) therefore AgCl is soluble in dil NHsoluble in dil NH33..

The addition of acid increase the dissociation of silver amine The addition of acid increase the dissociation of silver amine complex to give more silver which exceeds Ksp of AgCl and a complex to give more silver which exceeds Ksp of AgCl and a white ppt of AgCl is formed.white ppt of AgCl is formed. [Ag(NH[Ag(NH33))22]]++ Ag Ag+++ 2 NH+ 2 NH33

NHNH33 + H + H++ NH NH44++

push the reaction forward forming more Ag+push the reaction forward forming more Ag+

The addition of HThe addition of H++ convert NH convert NH33 to NH to NH44++ so increase the dissociation so increase the dissociation

of the silver amine complex.of the silver amine complex.

1212

2- Centrifugate 2 + KI 2- Centrifugate 2 + KI yellow ppt of silver yellow ppt of silver

iodideiodide

[Ag(NH[Ag(NH33))22]Cl Ag]Cl Ag++

+ KI+ KI

AgI (AgI (insoluble in NHinsoluble in NH33 but soluble in KCN) but soluble in KCN)

Ag amine complex ionizes to give AgAg amine complex ionizes to give Ag++ which exceeds Ksp of which exceeds Ksp of AgI (10AgI (10-16-16)(ver low Ksp) so AgI ppt)(ver low Ksp) so AgI ppt

N.B.N.B.

KCN dissolves all Ag salts except AgKCN dissolves all Ag salts except Ag22S which have low Ksp S which have low Ksp

(10(10-50-50))

AgAg++ + 2CN + 2CN-- [Ag(CN) [Ag(CN)22]]--

Ag cyanide complex (very stable)Ag cyanide complex (very stable)

Ag cyanide complex has very high KAg cyanide complex has very high Kstabstab, so it yields a small , so it yields a small

amount of Agamount of Ag++ insufficient to exceed Ksp of AgI insufficient to exceed Ksp of AgI

1313

Confirmation of mercurus (HgConfirmation of mercurus (Hg222+2+):):

The ppt (HgThe ppt (Hg00 + Hg (NH + Hg (NH22)Cl) is dissolved in aqua regia [HNO)Cl) is dissolved in aqua regia [HNO33 1: 3 HCl] then heat to evaporate excess aqua regia.1: 3 HCl] then heat to evaporate excess aqua regia.Hg(NHHg(NH22)Cl + HNO)Cl + HNO33 + HCl HgCl + HCl HgCl22 + NO + N + NO + N22 + H + H22OO

HgHg00 + HNO + HNO33 + HCl HgCl + HCl HgCl22 + NO + H + NO + H22OO

i.e. the ppt is transformed to soluble mercuric chloridei.e. the ppt is transformed to soluble mercuric chloride..1- Add stannous chloride1- Add stannous chloride

2 HgCl2 HgCl22 + SnCl + SnCl22 Hg Hg22ClCl22 + SnCl + SnCl44

whitewhite

HgHg22ClCl22 + excess SnCl + excess SnCl2 2 2 Hg 2 Hg0 0 + SnCl + SnCl44 blackblack

2- Add KI 2- Add KI scarlet red ppt of mercuric iodide soluble in scarlet red ppt of mercuric iodide soluble in excess KI.excess KI.

HgHg+2+2 + KI HgI + KI HgI22 + 2K + 2K++

HgIHgI22 + excess I + excess I-- [HgI [HgI44]]-2-2 + 2K + 2K++

soluble Nesselar reagentsoluble Nesselar reagent

1414

Group IIA,BGroup IIA,B

1515

Group IIA,BGroup IIA,BIIA PbIIA Pb2+2+, Hg, Hg2+2+, Bi, Bi3+3+, Cu, Cu2+2+, Cd, Cd2+2+

IIB AsIIB As3+,5+3+,5+, Sb, Sb3+,5+3+,5+, Sn, Sn2+,4+2+,4+

It is called the It is called the “ Acid Sulphide group”“ Acid Sulphide group”Group reagent : Group reagent : dil HCl + Hdil HCl + H22SSBoth group II and group IV have their sulphides insoluble in Both group II and group IV have their sulphides insoluble in water. But group IV have higher Ksp (needs high Swater. But group IV have higher Ksp (needs high S-2-2 to ppt) to ppt)So to ppt gp II only we use dil HCl with HSo to ppt gp II only we use dil HCl with H22SS(strong acid) HCl H(strong acid) HCl H++ + Cl + Cl-- (completely ionize) (completely ionize) ( weak acid) H( weak acid) H22S 2HS 2H++ + S + S-2-2 (partialy ionize) (partialy ionize)


By common ion effect the ionization of HBy common ion effect the ionization of H22S is decreased S is decreased therefore decrease sulphide ion concentration which become therefore decrease sulphide ion concentration which become sufficient to ppt group II (have low Ksp easily exceeded) but sufficient to ppt group II (have low Ksp easily exceeded) but insufficient to ppt group IV.insufficient to ppt group IV.

1616

Ques.1:Ques.1: Why not use conc HCl to ensure that group IV will not interfere?Why not use conc HCl to ensure that group IV will not interfere?Ans:Ans: Because if we use conc HCl the concentration of S Because if we use conc HCl the concentration of S2-2- ion ion will decrease dramatically that it becomes insufficient to ppt will decrease dramatically that it becomes insufficient to ppt some cations of group II which has slightly high Ksp eg: Cd, Pb, some cations of group II which has slightly high Ksp eg: Cd, Pb, Sn.Sn.Therefore we have to adjust acidity. The suitable acidity was Therefore we have to adjust acidity. The suitable acidity was found to be 0.2 - 0.3 N HCl (0.25N). We use crystal violet found to be 0.2 - 0.3 N HCl (0.25N). We use crystal violet (external indicator) to adjust acidity.(external indicator) to adjust acidity.

blue blue add HCladd HCl 0.25 N HCl 0.25 N HCl add H2Oadd H2O green or yellow green or yellowi.e low acidity bluish green i.e high acidityi.e low acidity bluish green i.e high acidity

gp IV will ppt gp IV will ppt gp II willgp II willwith gp II incompletely pptwith gp II incompletely ppt

1717

Procedure for separation:-Procedure for separation:-

1- On the filterate from group I , adjust acidity to 0.25 N HCl 1- On the filterate from group I , adjust acidity to 0.25 N HCl using C.V. + excess Husing C.V. + excess H22S centrifugeS centrifuge

Residue CentrifugateResidue Centrifugate gpII Test for complete pptngpII Test for complete pptnPb S black ppt Add excess HPb S black ppt Add excess H22S,centrifugeS,centrifugeHg S black pptHg S black pptCu S black ppt Cu S black ppt BiBi22SS33 brown ppt Residue Centrifugate brown ppt Residue CentrifugateCd S canary yellow ppt added to gp II Cd S canary yellow ppt added to gp II Rest of gpsRest of gpsAsAs22SS55, As, As22SS33 yellow ppt yellow pptSbSb22SS55, Sb, Sb22SS33 orange ppt orange pptSnSSnS22, SnS yellow ppt, SnS yellow ppt

1818

Group IIA “ Copper Subgroup”Group IIA “ Copper Subgroup”

PbPb2+2+, Hg, Hg2+2+, Bi, Bi3+3+, Cu, Cu2+2+, Cd, Cd2+2+

Procedure for separation :-Procedure for separation :-Residue of group IIA + dil HNOResidue of group IIA + dil HNO33 + dil H + dil H22SOSO44 (for Pb) (for Pb) centrifuge centrifuge

Residue CentrifugateResidue Centrifugate

HgS, PbSOHgS, PbSO44 Bi Bi +3+3, Cu, Cu+2+2, Cd, Cd+2+2

PbS+HNOPbS+HNO33+H+H++ Pb Pb2+2++NO+S+H+NO+S+H22O CuS+HNOO CuS+HNO33+H+H++ Cu Cu2+2++NO+S+H+NO+S+H22OO

soluble Bisoluble Bi22SS33+HNO+HNO33+H+H++ Bi Bi3+3++NO+S+H+NO+S+H22OO

PbPb2+2++H+H22SOSO44 PbSO PbSO44 (white ppt) CdS+HNO (white ppt) CdS+HNO33+H+H++ Cd Cd2+2++NO+S+H+NO+S+H22OO

Residue (Pb and Hg) + saturated ammonium acetate Residue (Pb and Hg) + saturated ammonium acetate centrifuge centrifuge


Hg PbHg Pb

Confirm as in gp I Confirm as in gp IConfirm as in gp I Confirm as in gp I

1919

Centrifugate ( Bi, Cu, Cd) + conc NHCentrifugate ( Bi, Cu, Cd) + conc NH44OHOH


Residue CentrifugateResidue Centrifugate BiBi3+3+ Cu Cu2+2+, Cd, Cd2+2+

BiBi3+3++NH+NH44OH CuOH Cu2+2++NH+NH44OH [Cu(NHOH [Cu(NH33))44]]2+2++H+H22OO

soluble Cu amine complex (blue)soluble Cu amine complex (blue)

Bi(OH)Bi(OH)33 + NH + NH44+ + CdCd2+2++NH+NH44OH [Cd(NHOH [Cd(NH33))44]]2+2++H+H22OO

White ppt soluble Cd amine complex (colorless)White ppt soluble Cd amine complex (colorless)

2020

Confirmation of Bismuth (BiConfirmation of Bismuth (Bi+3+3) :-) :-

1- Bismuth hydroxide (white ppt) + alkali stannite ( SnCl1- Bismuth hydroxide (white ppt) + alkali stannite ( SnCl22 + +

excess NaOH) black ppt of metallic bismuth.excess NaOH) black ppt of metallic bismuth.

SnClSnCl22 + NaOH Sn(OH) + NaOH Sn(OH)22 + NaCl + NaCl

white pptwhite ppt

Sn(OH)Sn(OH)22 + excess NaOH NaHSnO + excess NaOH NaHSnO22 alkali stannite alkali stannite

2 Bi(OH)2 Bi(OH)33 + 3 [HSnO + 3 [HSnO22]]-- 2Bi 2Bi00+ 3 [HSnO+ 3 [HSnO33]]-- + 3 H + 3 H22OO

Black pptBlack ppt

2121

Confirmation of copper (CuConfirmation of copper (Cu2+2+) and cadmium (Cd) and cadmium (Cd2+2+) :-) :-On the centrifugate from conc NHOn the centrifugate from conc NH33 [ Cu amine complex (blue col.) and Cd [ Cu amine complex (blue col.) and Cd

amine complex (colorless)]amine complex (colorless)]

Cu can be tested in presence of Cd but the opposite is not true because Cu Cu can be tested in presence of Cd but the opposite is not true because Cu gives all tests of Cd. Therefore we have to separate themgives all tests of Cd. Therefore we have to separate them

Cyanide procedure :-Cyanide procedure :-

Divide the solution into two unequal portions :Divide the solution into two unequal portions :

The smaller portionThe smaller portion :- :- Test for CuTest for Cu2+2+

1- Acidify with acetic acid + potassium ferrocyanide1- Acidify with acetic acid + potassium ferrocyanide

CuCu2+2++ [Fe(CN)+ [Fe(CN)66]]4-4- Cu Cu22[Fe(CN)[Fe(CN)66] reddish brown ppt ] reddish brown ppt

N.B. Acetic acid is used to decompose Cu amine complexN.B. Acetic acid is used to decompose Cu amine complex

[Cu(NH[Cu(NH33))44]]22+ + CH+ + CH33COOH Cu(CHCOOH Cu(CH33COO)COO)22 + NH + NH33

2- Soln + KI 2- Soln + KI white ppt of cuprous iodide in brown soln of iodine white ppt of cuprous iodide in brown soln of iodine

2Cu2Cu2+2+ + 4 I + 4 I-- Cu Cu22II22 + I + I22

white ppt brown soln white ppt brown soln

2222

3- Soln + KCNS 3- Soln + KCNS black pptblack ppt

CuCu2+2+ + CNS + CNS-- Cu(SCN) Cu(SCN)22

The larger portionThe larger portion :- :- Test for CdTest for Cd2+2+

First we should mask Cu by adding KCN till blue color disappearFirst we should mask Cu by adding KCN till blue color disappear

(Cu(NH(Cu(NH33))44]]2+2+ + 2 CN + 2 CN-- Cu(CN) Cu(CN)22(greenish yellow ppt)(greenish yellow ppt) + 4 NH + 4 NH33

2 Cu(CN)2 Cu(CN)22 Cu Cu22(CN)(CN)22(white ppt)(white ppt) + (CN) + (CN)22(cyanogen)(cyanogen)

CuCu22(CN)(CN)22 + 4 CN + 4 CN- - 2 [Cu(CN) 2 [Cu(CN)33]]2- 2- cuprocyanide complex (very stable)cuprocyanide complex (very stable)

[Cd(NH)[Cd(NH)33))44]]2+2+ + 2CN + 2CN-- Cd(CN) Cd(CN)22(buff ppt)(buff ppt) + 4 NH + 4 NH33

Cd(CN)Cd(CN)22 + 2 CN + 2 CN-- (Cd(CN) (Cd(CN)44]]2-2-Cadmicyanide complex(unstable)Cadmicyanide complex(unstable)

Then add HThen add H22S S canary yellow ppt of Cd S is formed. canary yellow ppt of Cd S is formed.

Since Cd cyanide complex is unstable. Therefore it yields sufficient CdSince Cd cyanide complex is unstable. Therefore it yields sufficient Cd+2+2 to react to react with Hwith H22S and exceed Ksp of CdSS and exceed Ksp of CdS

[Cd(CN)[Cd(CN)44]]2- + H + H22S CdS + HCN + CNS CdS + HCN + CN --

N.B.N.B.

If Cu is absent in the centrifugate from conc. NHIf Cu is absent in the centrifugate from conc. NH44OH (no blue color)OH (no blue color)

Therefore you can test for Cd directly by adding HTherefore you can test for Cd directly by adding H22S S canary yellow canary yellow

ppt.ppt.

2323

Group IIIGroup III

2424

Group IIIGroup IIIFeFe3+3+, Al, Al3+3+, Cr, Cr33++

It is called It is called “ Ammonium hydroxide”“ Ammonium hydroxide” group as they are ppt as hydroxide group as they are ppt as hydroxideGroup reagent : Group reagent : NHNH44Cl + NHCl + NH44OHOHProcedure for separation :-Procedure for separation :-1- We have to boil the filtrate from group II to expel H1- We have to boil the filtrate from group II to expel H22SS ..Gr.Gr.Ans.:Ans.: Because group III ppt in alkaline medium and group IV ppt as sulphide in Because group III ppt in alkaline medium and group IV ppt as sulphide in alkaline medium so if Halkaline medium so if H22S is not expelled group IV will ppt with group IIIS is not expelled group IV will ppt with group III

2- Test for iron 2- Test for iron Drops of the filtrate + conc HNODrops of the filtrate + conc HNO33 cool + NH cool + NH44SCN SCN blood red color , therefore iron is presentblood red color , therefore iron is present

If iron is present, add 2 drops of conc. HNOIf iron is present, add 2 drops of conc. HNO33 to the filtrate to oxidise Fe to the filtrate to oxidise Fe2+2+ to to FeFe3+3+

2525

Ques.1.Ques.1. Why do we oxidize Fe Why do we oxidize Fe2+2+ to Fe to Fe3+3+ ? ?Ans. :Ans. : Because HBecause H22S used in pptn of group II is a reducing agent which S used in pptn of group II is a reducing agent which reduces Fereduces Fe+3+3 (if present) to Fe (if present) to Fe+2+2 and Fe(OH) and Fe(OH)22 has high Ksp so has high Ksp so not ppt in conditions such that of group III but Fe(OH)not ppt in conditions such that of group III but Fe(OH)33 has a has a lower Ksp which can ppt with group III and iron does not lower Ksp which can ppt with group III and iron does not escapeescape

3-Soln + conc HNO3-Soln + conc HNO33 cool + NH cool + NH44Cl + NHCl + NH44OH OH cool, cool,


Residue CentrifugateResidue Centrifugate Gp III + MnGp III + Mn3+3+ (from gp IV) (from gp IV) other gpsother gps FeFe3+3+ + OH + OH-- Fe(OH) Fe(OH)33 Ferric hydroxide (reddish brown ppt) Ferric hydroxide (reddish brown ppt) AlAl3+3+ + OH + OH-- Al(OH) Al(OH)33 Aluminium hydroxide (white gelatinous ppt) Aluminium hydroxide (white gelatinous ppt) CrCr3+3+ + OH + OH-- Cr(OH) Cr(OH)33 Chromium hydroxide (greyish green ppt) Chromium hydroxide (greyish green ppt) MnMn3+3+ + OH + OH-- Mn(OH) Mn(OH)33 Manganic hydroxide ( brown ppt) Manganic hydroxide ( brown ppt)

2626

Ques.2.Ques.2. Why do we add NH Why do we add NH44Cl ?Cl ?

Ans.:Ans.: 1- Group IV and Mg from group VI ppt as OH1- Group IV and Mg from group VI ppt as OH-- but it has higher Ksp i.e needs but it has higher Ksp i.e needs excess OHexcess OH-- ions, therefore by adding NH ions, therefore by adding NH44Cl the ionization of NHCl the ionization of NH44OH OH

decrease due to common ion effect so it gives OHdecrease due to common ion effect so it gives OH‑‑ ions sufficient to ppt ions sufficient to ppt group III but insufficient to ppt group IVgroup III but insufficient to ppt group IV

Salt (completely ionize) NHSalt (completely ionize) NH44Cl NHCl NH44++ + Cl + Cl--

Weak base (partially ionize) NHWeak base (partially ionize) NH44OH NHOH NH44++ + OH + OH--


2- NH2- NH44Cl is an electrolyte which aids in coagulation of the gelatinous ppt of Cl is an electrolyte which aids in coagulation of the gelatinous ppt of

group IIIgroup III

3- It reduces OH3- It reduces OH-- ion conc so prevent solubility of amphoteric cations ion conc so prevent solubility of amphoteric cations

AlAl3+3++ OH+ OH-- Al(OH) Al(OH)3 3 + excess OH+ excess OH-- AlO AlO22-- soluble aluminate soluble aluminate

CrCr3+3++ OH+ OH-- Cr(OH) Cr(OH)3 3 + excess OH+ excess OH-- CrO CrO22-- soluble chromite soluble chromite

4- NH4- NH44Cl forms soluble complex with group IV cations.Cl forms soluble complex with group IV cations.

2727

Ques.3.Ques.3. Why do we boil after NH Why do we boil after NH44Cl ?Cl ?

Ans.:Ans.:

1- To coagulate the ppt1- To coagulate the ppt

2- To break Cr Amine complex2- To break Cr Amine complex

[Cr(NH[Cr(NH33))66] (OH)] (OH)33 Cr(OH)Cr(OH)33 + 6NH + 6NH33 soluble Cr amine complexsoluble Cr amine complex

3- To expel O3- To expel O22 and prevent oxidation of manganus (Mn and prevent oxidation of manganus (Mn2+2+) )

which has higer Ksp i.e. [not ppt as OHwhich has higer Ksp i.e. [not ppt as OH-- ] to manganic (Mn ] to manganic (Mn3+3+) ) which has low Ksp and can be ppt as Mn(OH)which has low Ksp and can be ppt as Mn(OH)33 so it escapes so it escapes

from group IV to group IIIfrom group IV to group III

2828

Separation of group III :-Separation of group III :-

ppt of group III + NaOH + Hppt of group III + NaOH + H22OO22 centrifuge centrifuge


FeFe3+3+, Mn, Mn3+3+(from gp IV) AlO(from gp IV) AlO22--, CrO, CrO44

2-2-

It depends on the amphoteric character of AlIt depends on the amphoteric character of Al3+3+ and Cr and Cr3+3+ so so dissolve in NaOH, they dissolve as stable AlOdissolve in NaOH, they dissolve as stable AlO22

-- and unstable and unstable

chromite CrOchromite CrO22‑‑ which is oxidised by H which is oxidised by H22OO22 to the stable to the stable

chromate CrOchromate CrO442-2- (has yellow color) (has yellow color)

Al(OH)Al(OH)33 + OH + OH- - AlO AlO22-- + 2H + 2H22OO

Cr(OH)Cr(OH)33 + OH + OH-- CrO CrO22-- + 2H + 2H22OO

unstable chromiteunstable chromite

CrOCrO22-- + H + H22OO22 CrO CrO44

2-2- + 4 H + 4 H22OO

stable yellow chromatestable yellow chromate

2929

Separation of ferric and manganicSeparation of ferric and manganic :- :-

Residue + HNOResidue + HNO33 + H + H22OO22 dissolve, divide to 2 dissolve, divide to 2 portionsportions

Fe(OH)Fe(OH)33 + 3HNO + 3HNO33 Fe(NO Fe(NO33))33 + 3H + 3H22OO

MnO(OH)MnO(OH)22 + 3HNO + 3HNO33 + H + H22OO2 2 Mn(NO Mn(NO33))33 + 3H + 3H22O + OO + O22

First portionFirst portion Test for iron :-Test for iron :-1- soln + NH1- soln + NH44SCN blood red colorSCN blood red color

FeFe3+3+ + SCN + SCN-- [Fe(SCN)] [Fe(SCN)]2+2+

2- soln + potassium ferrocyanide 2- soln + potassium ferrocyanide prussian prussian blue pptblue ppt

FeFe3+3+ + K + K44[Fe(CN)[Fe(CN)66] K Fe [Fe(CN)] K Fe [Fe(CN)66]]

3030

Second portionSecond portion

Test for manganus :-Test for manganus :-

1-With red lead Pb1-With red lead Pb33OO44

soln + dil HNOsoln + dil HNO33 + H + H22OO22 till no O till no O22 bubbles bubbles

+ PbO+ PbO22 pink color pink color

PbPb33OO44 + Mn + Mn2+2+ + H + H++ MnO MnO44-- + Pb + Pb2+2+ + H + H22OO

pink colorpink color

HH22OO22 must be completely removed because it reduces MnO must be completely removed because it reduces MnO44--

so bleach color formedso bleach color formed MnOMnO44

-- + H + H22OO22 + H + H++ Mn Mn2+2+ + O + O22 + H + H22OO

colorlesscolorless

N.BN.B

HH22OO22 is oxidising agent but with MnO is oxidising agent but with MnO44--, it is reducing agent , it is reducing agent

because it has lower oxidation potentialbecause it has lower oxidation potential

[ MnO[ MnO44-- has the highest oxidation potential] has the highest oxidation potential]

3131

Separation of aluminium and chromium :-Separation of aluminium and chromium :-

Divide the soln. containing AlODivide the soln. containing AlO22-- and CrO and CrO44

-- to 2 portions: to 2 portions:

First portion First portion

Test For CrOTest For CrO442-2- :- :-

1- With pb acetate yellow ppt of lead chromate1- With pb acetate yellow ppt of lead chromate

Pb(CHPb(CH33COO)COO)22 + (CrO + (CrO44))2-2- PbCrO PbCrO44

2- Perchromic acid test 2- Perchromic acid test

soln + dil Hsoln + dil H22SOSO44 + ether + 1 drop H + ether + 1 drop H22OO22

blue color in ether layer blue color in ether layer

CrOCrO442-2- + 2H + 2H++ Cr Cr22OO77

2-2- + H + H22OO +H2O2+H2O2 [CrO [CrO88]]3-3-

dichromate perchromate dichromate perchromate

3232

Second portionSecond portion

Test for AlOTest for AlO22-- :- :-

1- Soln + NH1- Soln + NH44Cl solid Cl solid cool white gelatinous ppt cool white gelatinous ppt

Al(OH)Al(OH)33 + OH + OH-- AlO AlO22-- + H + H22OO

white pptwhite ppt

NHNH44Cl reduces OHCl reduces OH-- concentration and reverse concentration and reverse

the reaction i.e move backward and ppt Al(OH)the reaction i.e move backward and ppt Al(OH)33

NHNH44++ + OH + OH-- NH NH44OH NHOH NH33 + H + H22OO

3333

Group IVGroup IV

3434

Group IVGroup IVMnMn2+2+, Zn, Zn2+2+, Co, Co2+2+, Ni, Ni2+2+

This group is called This group is called “ “ Ammonium sulphideAmmonium sulphide”” group group

Group reagent is Group reagent is : : NHNH44Cl + NHCl + NH44OH + HOH + H22SS

It is ppt as sulphide in alkaline mediumIt is ppt as sulphide in alkaline medium

(weak acid, partially ionize) (weak acid, partially ionize) HH22S 2HS 2H++ + S + S2-2-

(weak base, partially ionize) (weak base, partially ionize) NHNH44OH OHOH OH-- + NH + NH44++

HH22OO

Due to formation of water which is weakly ionized. The Due to formation of water which is weakly ionized. The ionization of Hionization of H22S is pushed forward and increase SS is pushed forward and increase S-2-2 ion conc ion conc so ppt group IV as sulphide (has higer Ksp)so ppt group IV as sulphide (has higer Ksp)

3535

Procedure for separationProcedure for separation :- :-

1-1- Centrifugate from group III + NH Centrifugate from group III + NH44Cl + NHCl + NH44OH OH

+ H+ H22S centrifugeS centrifuge

Residue CentrifugateResidue Centrifugate gp IV rest of groupsgp IV rest of groups MnS MnS buff ppt Manganous sulphide buff ppt Manganous sulphide ZnS ZnS white ppt Zinc sulphide white ppt Zinc sulphide CoS CoS black ppt Cobalt sulphide black ppt Cobalt sulphide NiS NiS black ppt Nickel sulphide black ppt Nickel sulphide

Ques.1.Ques.1. Why do we add NH Why do we add NH44Cl ?Cl ?Ans. :Ans. : 1- By common ion effect it decreases OH 1- By common ion effect it decreases OH-- conc so prevent pptn of conc so prevent pptn of group IV as hydroxidegroup IV as hydroxide

Salt (completely ionize) NHSalt (completely ionize) NH44Cl NHCl NH44++ + Cl + Cl--

Weak base (partially ionize) NHWeak base (partially ionize) NH44OH NHOH NH44++ + OH + OH--


2- NH2- NH44Cl is an electrolyte to coagulate the colloidal ppt of group IV eg ZnS, Cl is an electrolyte to coagulate the colloidal ppt of group IV eg ZnS, MnS so it become easily filteredMnS so it become easily filtered

3636

Ques.2.Ques.2. Why do we boil after add NH Why do we boil after add NH44Cl ?Cl ?

Ans.:Ans.:

1- To decompose Ni amine complex and ppt NiS1- To decompose Ni amine complex and ppt NiS

Ni(NHNi(NH33))66S S NiS NiS

soluble Ni amine complex black pptsoluble Ni amine complex black ppt

2- By boiling CoS and NiS undergo condensation and 2- By boiling CoS and NiS undergo condensation and change in physical characters so become insoluble in change in physical characters so become insoluble in very dil HCl very dil HCl


N.B.N.B.

If iron is not completely oxidised in group III it will ppt If iron is not completely oxidised in group III it will ppt as FeS (black ppt) in group IVas FeS (black ppt) in group IV

3737

Separation of group IVSeparation of group IV :- :-ppt of group IV + very dil HCl centrifugeppt of group IV + very dil HCl centrifuge


NiS , CoS ZnNiS , CoS Zn2+2+, Mn, Mn2+2+

black pptblack ppt

It depends on the amphoteric character of Zn, therefore soluble in dil HCl and It depends on the amphoteric character of Zn, therefore soluble in dil HCl and NaOHNaOH

Centrifugate Centrifugate + NaOH + Br + NaOH + Br22 water water centrifuge centrifuge


MnOMnO2 2 ZnO ZnO222-2-

black ppt black ppt

ZnZn2+2+ + OH + OH-- Zn(OH) Zn(OH)22 zinc hydroxide zinc hydroxide

Zn(OH)Zn(OH)22 + OH + OH-- (excess) ZnO (excess) ZnO222-2- soluble zincate soluble zincate

While Mn(OH)While Mn(OH)22 is oxidized by hypobromite to MnO is oxidized by hypobromite to MnO22 which is black ppt which is black ppt

OHOH-- + Br + Br22 OBr OBr-- + Br + Br-- + H + H22OO

MnMn2+2+ + Br + Br22 + OH + OH-- MnO MnO22 + 2Br + 2Br-- + 2H + 2H22OO

3838

N.B. N.B. The centrifugate must be heated to expel HThe centrifugate must be heated to expel H22S because if HS because if H22S is S is still present, so on adding NaOH, ZnS will ppt as it is ppt as still present, so on adding NaOH, ZnS will ppt as it is ppt as sulphide in alkaline mediumsulphide in alkaline medium

Confirmation of zincConfirmation of zinc:-:-1- 1- By adding excess HBy adding excess H22S white ppt of zinc S white ppt of zinc sulphidesulphide

ZnOZnO222-2- + S + S2-2- + H + H++ Zn S Zn S + OH + OH--

2- 2- Add acetic acid + pot. ferrocyanide Add acetic acid + pot. ferrocyanide sky blue sky blue

pptppt

ZnZn2+2+ + [Fe(CN) + [Fe(CN)66]]4-4- Zn Zn22[Fe(CN)[Fe(CN)66] ] Zinc ferrocyanideZinc ferrocyanide

Confirmation of Manganus :-Confirmation of Manganus :- As under group IIIAs under group III

3939

Confirmation of cobalt and nickelConfirmation of cobalt and nickel :- :-ppt is dissolved in aqua regia [ HNOppt is dissolved in aqua regia [ HNO33 1 : 3 HCl ] 1 : 3 HCl ] 3CoS + 2HNO3CoS + 2HNO33 + 6HCl 3CoCl + 6HCl 3CoCl22 + 2NO + 3S + 4H + 2NO + 3S + 4H22OO 3NiS + 2HNO3NiS + 2HNO33 + 6HCl 3NiCl + 6HCl 3NiCl22 + 2NO + 3S + 4H + 2NO + 3S + 4H22OO

Divide the soln to 2 portions :Divide the soln to 2 portions :First portionFirst portion Test for CoTest for Co2+2+ :- :-1-1-soln + NHsoln + NH44Cl + NHCl + NH44OH + pot ferricyanide OH + pot ferricyanide red ppt red ppt

CoCo2+2+ + [Fe(CN) + [Fe(CN)66]]3-3- Co Co33[Fe(CN)[Fe(CN)66]]22 red red ppt ppt

cobaltus ferricyanide insol in NHcobaltus ferricyanide insol in NH33

N.BN.B. Nickelous ferricyanide is yellow and soluble in NH. Nickelous ferricyanide is yellow and soluble in NH33 (that is why we add NH(that is why we add NH44OH)OH)

2- 2- Vogel testVogel test

soln + dil Hsoln + dil H22SOSO44 + SnCl + SnCl22 + ether + NH + ether + NH44SCN SCN blue color in etherial blue color in etherial layerlayer CoCo2+2+ + SCN + SCN-- [Co(SCN) [Co(SCN)44]]2-2- blue color blue color

N.B.N.B. HH22SOSO44 contains traces of Fe contains traces of Fe+3+3 which interfere with the test by giving blood which interfere with the test by giving blood red color with SCNred color with SCN--

Therefore we add reducing agent eg SnClTherefore we add reducing agent eg SnCl22 to reduce Fe to reduce Fe+3+3 to Fe to Fe+2+2 or we add or we add FF-- or PO or PO44

-- to make complex with Fe to make complex with Fe+3+3

4040

Second portionSecond portion Test for NiTest for Ni2+2+ :- :-

1- Dimethyl glyoxime test1- Dimethyl glyoxime test Soln + NHSoln + NH44Cl + NHCl + NH44OH + D.M.G. red colorOH + D.M.G. red color

2- Test with KCN2- Test with KCNsoln + KCN + NaOH + Brsoln + KCN + NaOH + Br22 water black ppt of NiO water black ppt of NiO22

OHOH-- + Br + Br22 OBr OBr-- + Br + Br-- + H + H22OO

NiNi2+2+ + 2CN + 2CN-- Ni(CN) Ni(CN)22 (excess CN-)(excess CN-) [(Ni(CN [(Ni(CN44)])]2- OBr- 2- OBr- NiONiO2 2 ..

unstable Ni cyanide complexunstable Ni cyanide complex

Cobalt will not interfereCobalt will not interfere

CoCo2+2++ 2CN+ 2CN-- Co(CN) Co(CN)22 (excess CN-) (excess CN-) [Co(CN) [Co(CN)66]]4- atm. O2 4- atm. O2 [Co(CN)[Co(CN)66]]3-3-

stable cobalticyanide complexstable cobalticyanide complex

Cobalticyanide complex does not interfere as it is an oxidising agent which Cobalticyanide complex does not interfere as it is an oxidising agent which does not react with OBr-does not react with OBr-

CH3 C NOH

CH3 C NOHNi2+

CH3 C NO

CH3 C NOH2

Ni2+ 2 H+

red ppt

4141

Group VGroup V

4242

Group VGroup VBaBa2+2+, Sr, Sr2+2+, Ca, Ca22++

It is called It is called ““Alkaline earth groupAlkaline earth group”” or or “ “ Ammonium carbonate groupAmmonium carbonate group””

Group reagent : NHGroup reagent : NH44Cl + NHCl + NH44OH + (NHOH + (NH44))22COCO33

The cations ppt as carbonate in alkaline mediumThe cations ppt as carbonate in alkaline medium

The three cations belong to the same group in the periodic table, therefore The three cations belong to the same group in the periodic table, therefore have similar characters and their separation is somewhat difficulthave similar characters and their separation is somewhat difficult

Procedure for separation :-Procedure for separation :-Centrifugate from group IV is evaporated nearly to dryness + NHCentrifugate from group IV is evaporated nearly to dryness + NH44Cl + Cl +

NHNH44OH OH + (NH + (NH44))22 CO CO33 soln soln to 60 to 60ooC wait, C wait,



Gp V gp VIGp V gp VI

BaCOBaCO33 white ppt white ppt

SrCOSrCO33 white ppt white ppt

CaCOCaCO33 white ppt white ppt

4343

Ques.1Ques.1. Why we evaporate the centrifugate from group IV till . Why we evaporate the centrifugate from group IV till nearly dryness ?nearly dryness ?Ans. :Ans. :

To expel HTo expel H22S, to prevent oxidation of SS, to prevent oxidation of S2-2- to SO to SO442-2- and prevent pptn of Ba and prevent pptn of Ba2+2+

and Srand Sr2+2+ as SO as SO442-2- [ CaSO [ CaSO44 has higher Ksp so need more SO has higher Ksp so need more SO44

2-2- ion conc. to ion conc. to ppt] ppt]

Ques.2.Ques.2. We must add a moderate amount of NH We must add a moderate amount of NH44++ ? ?

Ans. :Ans. : Because excess NHBecause excess NH44

++ will decrease carbonate ion concentration. will decrease carbonate ion concentration.

NHNH44++ + CO + CO33

2-2- NH NH33 + HCO + HCO33--

N.B.N.B.

The soln must be a little basic. If acidic the carbonate will change to HCOThe soln must be a little basic. If acidic the carbonate will change to HCO33--

which are solublewhich are soluble

COCO33- - + H+ H++ HCO HCO33

--

Ques.3.Ques.3. Why we use NH Why we use NH44Cl ?Cl ?

Ans. :Ans. : 1- By common ion effect it decreases OH1- By common ion effect it decreases OH-- ion conc. and CO ion conc. and CO33

2-2- ion conc. so ion conc. so prevent pptn of Mg as Mg(OH)prevent pptn of Mg as Mg(OH)22 or as MgCO or as MgCO33


4444

Ques.4.Ques.4. Why we use NH Why we use NH44OH ?OH ?Ans Ans

1- It converts NH1- It converts NH44HCOHCO33 to (NH to (NH44))22 CO CO33

NHNH44HCOHCO33 + NH + NH33 (NH (NH44))22COCO33

2- Converts ammonium carbamate into carbonate2- Converts ammonium carbamate into carbonate

NHNH44COCO22NHNH22 + H + H22O (NHO (NH44))22COCO33

3- 3- To make the medium faintly alkaline to prevent HCOTo make the medium faintly alkaline to prevent HCO33-- formation as all formation as all

M(HCOM(HCO33))22 are soluble are soluble

N.B. N.B. Ammonium carbonate is prepared from equimolar concentration of Ammonium carbonate is prepared from equimolar concentration of ammonium bicarbonate NHammonium bicarbonate NH44(HCO(HCO33))22 and ammonium carbamate NH and ammonium carbamate NH44COCO22NHNH22

Ques.5.Ques.5. Why we heat to 60 Why we heat to 60ooc and not boil ?c and not boil ?Ans. :Ans. :

1- To convert HCO1- To convert HCO33-- to CO to CO33

2-2-

M(HCOM(HCO33))22 MCO MCO33 + CO + CO22 + H + H22OO

2- Warming aid in the formation of more crystalline ppt2- Warming aid in the formation of more crystalline ppt3- Warming aid in the conversion of carbamate to carbonate3- Warming aid in the conversion of carbamate to carbonate

Excess heat cause loss of NHExcess heat cause loss of NH33 so convert CO so convert CO332-2- to bicarbonate by pushing to bicarbonate by pushing

the reaction forwardthe reaction forward

4545

Separation of group VSeparation of group V :- :-ppt + acetic acid ppt + acetic acid The ppt will dissolve as acetate The ppt will dissolve as acetate

BaCOBaCO33 + 2 CH + 2 CH33COOH Ba(CHCOOH Ba(CH33COO)COO)22 + CO + CO22 + H + H22OO

SrCOSrCO33 + 2 CH + 2 CH33COOH Sr(CHCOOH Sr(CH33COO)COO)22 + CO + CO22 + H + H22OO

CaCOCaCO33 + 2 CH + 2 CH33COOH Ca(CHCOOH Ca(CH33COO)COO)22 + CO + CO22 + H + H22OO

Test for BariumTest for Barium :- :-1-2 drops of the soln + K1-2 drops of the soln + K22CrOCrO44 (pot. chromate) (pot. chromate) If yellow ppt, If yellow ppt,

therefore Ba is presenttherefore Ba is present

If Ba is present add KIf Ba is present add K22CrOCrO44 soln till soln acquire orange tint, soln till soln acquire orange tint,



BaCrOBaCrO4 4 Sr Sr 2+2+, Ca, Ca2+2+

Yellow ppt reppt [NHYellow ppt reppt [NH 44OH + NHOH + NH44Cl + Cl +

(NH(NH44))22COCO33 to 60 to 60oocc

CentrifugeCentrifuge


SrCOSrCO33, CaCO, CaCO33 reject reject

white pptwhite ppt

4646

Test for BaTest for Ba2+2+ :- :-

1-Flame test1-Flame testppt of BaCrOppt of BaCrO44 + HCl + HCl soluble BaCl soluble BaCl22 (volatile in flame ) and gives apple (volatile in flame ) and gives apple

green colorgreen color

If Ba is absent reject the portion used for test and the remainder of the soln If Ba is absent reject the portion used for test and the remainder of the soln is used to test for strontium and calciumis used to test for strontium and calcium

N.B.N.B. If we add excess KIf we add excess K22CrOCrO44, SrCO, SrCO44 and Ca CrO and Ca CrO44 will also ppt as the chromate will also ppt as the chromate

of this group are insoluble. But Ksp of BaCrOof this group are insoluble. But Ksp of BaCrO44 < SrCO < SrCO44 < CaCrO < CaCrO44. So we . So we

add amount of Kadd amount of K22CrOCrO44 sufficient to ppt barium but not strontium and calcium sufficient to ppt barium but not strontium and calcium

We use also acetic acid to convert part of KWe use also acetic acid to convert part of K22CrOCrO44 to K to K22CrCr22OO77 so decrease so decrease

CrOCrO442-2- conc conc

QuesQues::

Why we do not use HCl or HNOWhy we do not use HCl or HNO33 instead of acetic acid ? instead of acetic acid ?

Ans.:Ans.: Because they are strong acids which convert all CrOBecause they are strong acids which convert all CrO44

2-2- to Cr to Cr22OO772-2- and no ppt and no ppt

is formedis formed

4747

Identification of Strontium and Calcium :-Identification of Strontium and Calcium :-

Residue dissolve in acetic acid Residue dissolve in acetic acid to dissolve to dissolve

Soln + (NHSoln + (NH44))22SOSO4 4 centrifuge centrifuge

Residue CentrifugateResidue Centrifugate SrSr2+2+ Ca Ca2+2+

Sr(CHSr(CH33COO)COO)22 + (NH + (NH44))22SOSO44 SrSO SrSO44 + 2CH + 2CH33COONHCOONH44

white pptwhite pptThe test depends on that Ca form double salt with ammonium sulphate The test depends on that Ca form double salt with ammonium sulphate which is solublewhich is soluble

CaCa2+2+ + (NH + (NH44))22SOSO44 (NH (NH44))22Ca(SOCa(SO44))22

double salt formationdouble salt formationN.B.N.B.

If we used dil HIf we used dil H22SOSO44 instead of (NH instead of (NH44))22SOSO44, both Sr and Ca will ppt as SO, both Sr and Ca will ppt as SO44

4848

Test for CalciumTest for Calcium :- :-

1- Soln + NH1- Soln + NH44OH + ammonium oxalate OH + ammonium oxalate white ppt of calcium white ppt of calcium oxalateoxalate

CC22OO44(NH(NH44))22 + Ca + Ca2+2+ C C22OO44Ca Ca + 2NH + 2NH44ClCl

white pptwhite ppt

2- 2- Soln + NHSoln + NH44Cl + pot. ferrocyanide Cl + pot. ferrocyanide pale yellow ppt pale yellow ppt

CaCa2+2+ + NH + NH44++ + K + K++ + [Fe(CN) + [Fe(CN)66]]4-4- CaNH CaNH44K[Fe(CN)K[Fe(CN)66]]

triple ferrocyanidetriple ferrocyanide

Flame testFlame test :- :-For SrFor Sr2+2+

SrSOSrSO44 is heated in the reducing zone of the flame is heated in the reducing zone of the flame Sr S + HCl Sr S + HCl

SrClSrCl22 volatile in flame with crimson red color volatile in flame with crimson red color

For CaFor Ca2+2+ Ca CCa C22OO44 + HCl Ca Cl + HCl Ca Cl22 volatile in flame with brick red color. volatile in flame with brick red color.

4949

Group VIGroup VI

5050

Group VIGroup VIMgMg2+2+, Na, Na++, K, K++, NH, NH44

++

This group is called This group is called “ “ Alkali metal groupAlkali metal group””It is also called the It is also called the “ “ Soluble groupSoluble group “ “ because it has no group because it has no group reagent (all Na, K and NHreagent (all Na, K and NH44 salts are soluble) salts are soluble)

Removal of traces of alkaline earth group :-Removal of traces of alkaline earth group :-

Centrifugate of group V + ammonium oxalate [to ppt any CaCentrifugate of group V + ammonium oxalate [to ppt any Ca2+2+] ] + ammonium sulphate [to ppt any Ba+ ammonium sulphate [to ppt any Ba2+2+ or Sr or Sr2+2+] ] centrifugecentrifuge

Residue CentrifugateResidue Centrifugate rejected gp VIrejected gp VI

Divide the centrifugate into two unequal portions :Divide the centrifugate into two unequal portions :

5151

The smaller portion :The smaller portion :

Test for MgTest for Mg2+2+ :- :-

1- soln + conc HCl + Na1- soln + conc HCl + Na22HPOHPO44

render alkaline by excess NHrender alkaline by excess NH44OH OH

white gel ppt white gel ppt

MgMg2+2+ NH NH44++ + PO + PO44

3-3- MgNH MgNH44POPO44 white ppt white ppt

MgMg2+2+ + OH + OH--(excess NH(excess NH44OH) Mg(OH)OH) Mg(OH)22 white ppt white ppt

5252

The larger potion :The larger potion : Test for NaTest for Na++ and K and K++ :- :-

N.B.N.B.

Before we test for KBefore we test for K++ we must remove NH we must remove NH44++. . Gr.Gr.

Ans. :Ans. :

Because NHBecause NH44++ gives all tests for K gives all tests for K++

Removal of NHRemoval of NH44++ :- :-

Larger portion Larger portion + conc. + conc. HNOHNO33 evaporate evaporate

to dryness, to dryness,

Residue is dissolved in water and divided to 2 portionsResidue is dissolved in water and divided to 2 portions This treatment also remove MgThis treatment also remove Mg+2+2 as MgO as MgO

NHNH44++ + HNO + HNO33 NH NH44NONO3 3

NHNH44NONO3 3 N N22O + NHO + NH33

5353

Test for SodiumTest for Sodium :- :-

1- By pot. dihydrogen antimonate or pot. 1- By pot. dihydrogen antimonate or pot. pyroantimonatepyroantimonate

Soln + KHSoln + KH22SbOSbO44 or KH or KH22SbSb22OO77 white pptwhite ppt

KHKH22SbOSbO44 + Na + Na++ NaH NaH22SbOSbO44 + K + K++

KK22HH22SbSb22OO77 + 2Na + 2Na++ Na Na22HH22SbSb22OO77 + 2K + 2K++

2- By magnesium uranyl acetate 2- By magnesium uranyl acetate yellow ppt yellow pptNaNa+++ HMg(UO+ HMg(UO22))33(CH(CH33COO)COO)99NaMg(UONaMg(UO22))33(CH(CH33COO)COO)99+ H+ H++

3- Flame test :- golden yellow color3- Flame test :- golden yellow color

5454

Test for PotassiumTest for Potassium :- :-1- By sodium cobaltinitrite yellow ppt 1- By sodium cobaltinitrite yellow ppt

NaNa33[Co(NO[Co(NO22))66] + 3K] + 3K++ K K33[Co(NO[Co(NO22))66]] + 3 Na + 3 Na++

2- Flame test 2- Flame test violet color in violet color in flameflame

5555

Test for Ammonium :-Test for Ammonium :-It must be tested on the original solid because we It must be tested on the original solid because we add NHadd NH44Cl and NHCl and NH44OH during the course of analysisOH during the course of analysis

1- Solid + Na OH NH1- Solid + Na OH NH33 odour odour

NHNH44++ + Na OH NH + Na OH NH33 + H + H22OO

Other tests for NHOther tests for NH44++ :- :-

2- Turns red litmus paper blue2- Turns red litmus paper blue

3- With Nesselar reagent brown ppt3- With Nesselar reagent brown ppt

NHNH33 + 2[HgI + 2[HgI44]]2-2- + 3OH + 3OH NHg NHg22I. HI. H22O O + 7I + 7I-- + 2H + 2H22OO

brown pptbrown ppt

5656

The smaller portion :The smaller portion : Test for MgTest for Mg2+2+ :- :-

1- soln + conc HCl + Na1- soln + conc HCl + Na22HPOHPO44 render alkaline by render alkaline by

excess NHexcess NH44OH white gel pptOH white gel ppt

MgMg2+2+ NH NH4+4+ + PO + PO443-3- MgNH MgNH44POPO44 white ppt white ppt

MgMg2+2+ + OH + OH--(excess NH(excess NH44OH)OH) Mg(OH) Mg(OH)22 white ppt white ppt

2- 2- Test with Magneson reagent (blue lake test)Test with Magneson reagent (blue lake test)

Mg(OH)Mg(OH)22 + HCl soluble MgCl + HCl soluble MgCl22 + Magneson I reagent (p- + Magneson I reagent (p-

nitro benzene azo resorcinol) a blue dye + NaOH nitro benzene azo resorcinol) a blue dye + NaOH

Mg(OH)Mg(OH)22

The ppt adsorb the blue dyeThe ppt adsorb the blue dye

Cations

Documents

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