Cations

1. Cations

2. Cations • Cations are the positive part of any salt • The cations cannot be tested directly as anions • Therefore, we classified cations according to the solubility of their salt in H2O • i.e we found that the chlorides of Pb2+ , Ag+ , Hg22+ are insoluble in water, so we put them in one group and say that “The group reagent is cold dil HCl” • Also we found that the sulphides of Pb2+, Hg2+, Bi3+, Cu2+, Cd2+are insoluble in water, so we put them in one group and say that“The group reagent is H2S”

3. Classification of cations

4. Group I ( Pb2+ , Ag+ , Hg22+ )

5. Group I( Pb2+ , Ag+ , Hg22+ ) • The chloride salt of these metals have low solubility so they are ppt as Cl- and called “ HCl group” • Group reagent : cold dil HCl slight excess with stirring • Procedure for separation :- • 1 ml mix + 1 ml cold dil HCl centrifuge • Residue Centrifigate • Group I Rest of groups • Pb Cl2 white ppt of lead chloride • Ag Cl white ppt of silver chloride • Hg2Cl2 silky white ppt of mercurus chloride • Ques.1 : Why cold HCl ? • Ans: Because Pb Cl2 is soluble on hot so escape to group II

6. N.B. • Pb Cl2 have the highest Ksp, therefore it is the last to ppt. • Pb Cl2 Ksp = 1.6 x 10-6 • Ag Cl Ksp = 1.0 x 10-10 • Hg2 Cl2 Ksp = 1.0 x 10-18 • Therefore the order of ppt Hg2+2 then Ag+ then Pb+2 • Ques.2 : Why dil. HCl ? • Ans: Because Pb Cl2 tends to form soluble complex and escape to group II in conc. HCl (i.e PbCl2 is soluble in excess Cl-) • Pb+2 + Cl- PbCl2 • PbCl2 + Cl- [PbCl3]- + Cl- [PbCl4]2- • Soluble complex

7. Ques.3 : Why slight excess with stirring ? • Ans : We add slight excess HCl to ensure complete pptn. By the common ion effect. • (Strong acid) HCl H+ + Cl- (strongly ionize) (1) • (ppt ) AgCl Ag+ + Cl- (partially ionize)(2) • common ion effect • By common ion effect reaction 2 moves to the left preventing the dissolution of Ag Cl • With stirring because Pb Cl2 tends to form supersaturated soln. i.e. go to soln. therefore we have to break this supersaturation by stirring.

8. Procedure for separationResidue (group I) + 2 ml hot water centrifuge • Residue centrifugate (1) AgCl, Hg2Cl2 PbCl2 • + dil NH3 • centrifuge • Residue centrifugate (2) Hg Ag • Hg2Cl2 + NH3 Hg(NH2)Cl + Hg0 AgCl + NH3 Ag(NH3)2Cl • white black It is called “ disproportionation reaction” because (Hg22+) tends to form (Hg0) mercury and (Hg2+) mercuric.

9. Confirmation of lead(Pb2+):- • 1- By cooling centrifugate(1) white ppt of lead chloride • 2- Centrifugate(1)+ KI yellow ppt of Pb I2 (soluble in excess KI) • PbI2 + excess I- [PbI4]2- soluble complex • 3- Centrifugate + acetic acid + K2CrO4 (pot.chromate) • yellow ppt of lead chromate • PbCl2 + K2CrO4 PbCrO4 + 2 KCl • Ques.:yellow ppt • Why acidification with acetic acid and not strong mineral acid ? • Ans : Because in strong acid • 2CrO42- +2 H+ Cr2O72- + H2O • i.e. dichromate is formed and form soluble lead dichromate Pb Cr2O7 • In presence of alkalinity the yellow ppt turns red • 2 PbCrO4 + H2O Pb2CrO5 + H2CrO4 • red ppt (basic lead chromate)

10. In presence of high alkalinity the ppt will dissolve because Pb is amphoteric • PbCrO4 + 2 OH- Pb(OH)2 + CrO42- • white ppt • Pb(OH)2 + OH- [HPbO2]- + H2O • soluble plumbite ion • N.B. • Amphoteric metals (i.e. soluble in both acid and alkaline medium) • Pb2+, Sn2+,4+, As3+,5+, Sb3+,5+, Al3+,Cr3+, Zn2+ • gp I gp IIb gp III gp IV • 4- centrifugate + H2SO4 white ppt of lead sulphate • PbCl2 + H2SO4 PbSO4

11. Confirmation of silver (Ag+): • 1- acidify centrifugate(2) with dil HNO3 white ppt of silver chloride soluble in dil. NH3. • [Ag(NH3)2]Cl + 2 HNO3 AgCl + NH4NO3 • white ppt • Silver amine complex have high Kstab,so it ionizes to give a quantity of Ag insufficient to exceed Ksp of Ag Cl (10-10) therefore AgCl is soluble in dil NH3. • The addition of acid increase the dissociation of silver amine complex to give more silver which exceeds Ksp of AgCl and a white ppt of AgCl is formed. • [Ag(NH3)2]+ Ag++ 2 NH3 • NH3 + H+ NH4+ • push the reaction forward forming more Ag+ • The addition of H+ convert NH3 to NH4+ so increase the dissociation of the silver amine complex.

12. 2- Centrifugate 2 + KI yellow ppt of silver • iodide • [Ag(NH3)2]Cl Ag+ • + KI • AgI (insoluble in NH3 but soluble in KCN) • Ag amine complex ionizes to give Ag+ which exceeds Ksp of AgI (10-16)(ver low Ksp) so AgI ppt • N.B. • KCN dissolves all Ag salts except Ag2S which have low Ksp (10-50) • Ag+ + 2CN- [Ag(CN)2]- • Ag cyanide complex (very stable) • Ag cyanide complex has very high Kstab, so it yields a small amount of Ag+ insufficient to exceed Ksp of AgI

13. Confirmation of mercurus (Hg22+): • The ppt (Hg0 + Hg (NH2)Cl) is dissolved in aqua regia [HNO3 1: 3 HCl] then heat to evaporate excess aqua regia. • Hg(NH2)Cl + HNO3 + HCl HgCl2 + NO + N2 + H2O • Hg0 + HNO3 + HCl HgCl2 + NO + H2O • i.e. the ppt is transformed to soluble mercuric chloride. • 1- Add stannous chloride • 2 HgCl2 + SnCl2 Hg2Cl2 + SnCl4 • white • Hg2Cl2 + excess SnCl2 2 Hg0  + SnCl4 • black • 2- Add KI scarlet red ppt of mercuric iodide soluble in • excess KI. • Hg+2 + KI HgI2 + 2K+ • HgI2 + excess I- [HgI4]-2 + 2K+ • soluble Nesselar reagent

14. Group IIA,B

15. Group IIA,BIIA Pb2+, Hg2+, Bi3+, Cu2+, Cd2+IIB As3+,5+, Sb3+,5+, Sn2+,4+ • It is called the “ Acid Sulphide group” • Group reagent : dil HCl + H2S • Both group II and group IV have their sulphides insoluble in water. But group IV have higher Ksp (needs high S-2 to ppt) • So to ppt gp II only we use dil HCl with H2S • (strong acid) HCl H+ + Cl- (completely ionize) • ( weak acid) H2S 2H+ + S-2 (partialy ionize) • common ion effect • By common ion effect the ionization of H2S is decreased therefore decrease sulphide ion concentration which become sufficient to ppt group II (have low Ksp easily exceeded) but insufficient to ppt group IV.

16. Ques.1: • Why not use conc HCl to ensure that group IV will not interfere? • Ans: Because if we use conc HCl the concentration of S2- ion will decrease dramatically that it becomes insufficient to ppt some cations of group II which has slightly high Ksp eg: Cd, Pb, Sn. • 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 (external indicator) to adjust acidity. • blue add HCl 0.25 N HCl add H2O green or yellow • i.e low acidity bluish green i.e high acidity • gp IV will ppt gp II will • with gp II incompletely ppt

17. Procedure for separation:- • 1- On the filterate from group I , adjust acidity to 0.25 N HCl using C.V. + excess H2S centrifuge • Residue Centrifugate • gpII Test for complete pptn • Pb S black ppt Add excess H2S,centrifuge • Hg S black ppt • Cu S black ppt • Bi2S3 brown ppt Residue Centrifugate • Cd S canary yellow ppt added to gp II Rest of gps • As2S5, As2S3 yellow ppt • Sb2S5, Sb2S3 orange ppt • SnS2, SnS yellow ppt

18. Group IIA “ Copper Subgroup” • Pb2+, Hg2+, Bi3+, Cu2+, Cd2+ • Procedure for separation :- • Residue of group IIA + dil HNO3 + dil H2SO4 (for Pb)  centrifuge • Residue Centrifugate • HgS, PbSO4 Bi+3, Cu+2, Cd+2 • PbS+HNO3+H+ Pb2++NO+S+H2O CuS+HNO3+H+ Cu2++NO+S+H2O • soluble Bi2S3+HNO3+H+ Bi3++NO+S+H2O • Pb2++H2SO4 PbSO4 (white ppt) CdS+HNO3+H+ Cd2++NO+S+H2O • Residue (Pb and Hg) + saturated ammonium acetate centrifuge • Residue Centrifugate • Hg Pb • Confirm as in gp I Confirm as in gp I

19. Centrifugate ( Bi, Cu, Cd) + conc NH4OH centrifuge • Residue Centrifugate • Bi3+ Cu2+, Cd2+ • Bi3++NH4OH Cu2++NH4OH [Cu(NH3)4]2++H2O soluble Cu amine complex (blue) • Bi(OH)3 + NH4+ Cd2++NH4OH [Cd(NH3)4]2++H2O • White ppt soluble Cd amine complex (colorless)

20. Confirmation of Bismuth (Bi+3) :- • 1- Bismuth hydroxide (white ppt) + alkali stannite ( SnCl2 + excess NaOH) black ppt of metallic bismuth. • SnCl2 + NaOH Sn(OH)2 + NaCl • white ppt • Sn(OH)2 + excess NaOH NaHSnO2 alkali stannite • 2 Bi(OH)3 + 3 [HSnO2]- 2Bi0+ 3 [HSnO3]- + 3 H2O Black ppt

21. Confirmation of copper (Cu2+) and cadmium (Cd2+) :- • On the centrifugate from conc NH3 [ Cu amine complex (blue col.) and Cd amine complex (colorless)] • 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 them • Cyanide procedure :- • Divide the solution into two unequal portions : • The smaller portion :- Test for Cu2+ • 1- Acidify with acetic acid + potassium ferrocyanide • Cu2++ [Fe(CN)6]4- Cu2[Fe(CN)6] reddish brown ppt • N.B. Acetic acid is used to decompose Cu amine complex • [Cu(NH3)4]2+ + CH3COOH Cu(CH3COO)2 + NH3 • 2- Soln + KI white ppt of cuprous iodide in brown soln of iodine • 2Cu2+ + 4 I- Cu2I2 + I2 white ppt brown soln

22. 3- Soln + KCNS black ppt • Cu2+ + CNS- Cu(SCN)2 • The larger portion :- Test for Cd2+ • First we should mask Cu by adding KCN till blue color disappear • (Cu(NH3)4]2+ + 2 CN- Cu(CN)2(greenish yellow ppt) + 4 NH3 • 2 Cu(CN)2 Cu2(CN)2(white ppt) + (CN)2(cyanogen) Cu2(CN)2 + 4 CN- 2 [Cu(CN)3]2- cuprocyanide complex (very stable) • [Cd(NH)3)4]2+ + 2CN- Cd(CN)2(buff ppt) + 4 NH3 • Cd(CN)2 + 2 CN- (Cd(CN)4]2-Cadmicyanide complex(unstable) • Then add H2S canary yellow ppt of Cd S is formed. • Since Cd cyanide complex is unstable. Therefore it yields sufficient Cd+2 to react with H2S and exceed Ksp of CdS • [Cd(CN)4]2- + H2S CdS + HCN + CN- • N.B. • If Cu is absent in the centrifugate from conc. NH4OH (no blue color) • Therefore you can test for Cd directly by adding H2S canary yellow • ppt.

23. Group III

24. Group IIIFe3+, Al3+, Cr3+ • It is called “ Ammonium hydroxide” group as they are ppt as hydroxide • Group reagent : NH4Cl + NH4OH • Procedure for separation :- • 1- We have to boil the filtrate from group II to expel H2S.Gr. • Ans.: • Because group III ppt in alkaline medium and group IV ppt as sulphide in alkaline medium so if H2S is not expelled group IV will ppt with group III • 2- Test for iron • Drops of the filtrate + conc HNO3 cool + NH4SCN • blood red color , therefore iron is present • If iron is present, add 2 drops of conc. HNO3 to the filtrate to oxidise Fe2+ to Fe3+

25. Ques.1. Why do we oxidize Fe2+ to Fe3+ ? • Ans. : • Because H2S used in pptn of group II is a reducing agent which reduces Fe+3 (if present) to Fe+2 and Fe(OH)2 has high Ksp so not ppt in conditions such that of group III but Fe(OH)3 has a lower Ksp which can ppt with group III and iron does not escape • 3-Soln + conc HNO3 cool + NH4Cl + NH4OH  cool, • centrifuge • Residue Centrifugate • Gp III + Mn3+ (from gp IV) other gps • Fe3+ + OH- Fe(OH)3 Ferric hydroxide (reddish brown ppt) • Al3+ + OH- Al(OH)3 Aluminium hydroxide (white gelatinous ppt) • Cr3+ + OH- Cr(OH)3 Chromium hydroxide (greyish green ppt) • Mn3+ + OH- Mn(OH)3 Manganic hydroxide ( brown ppt)

26. Ques.2. Why do we add NH4Cl ? • Ans.: • 1- Group IV and Mg from group VI ppt as OH- but it has higher Ksp i.e needs excess OH- ions, therefore by adding NH4Cl the ionization of NH4OH decrease due to common ion effect so it gives OH‑ ions sufficient to ppt group III but insufficient to ppt group IV • Salt (completely ionize) NH4Cl NH4+ + Cl- • Weak base (partially ionize) NH4OH NH4+ + OH- • common ion effect • 2- NH4Cl is an electrolyte which aids in coagulation of the gelatinous ppt of group III • 3- It reduces OH- ion conc so prevent solubility of amphoteric cations Al3++ OH- Al(OH)3 + excess OH- AlO2- soluble aluminate • Cr3++ OH- Cr(OH)3 + excess OH- CrO2- soluble chromite • 4- NH4Cl forms soluble complex with group IV cations.

27. Ques.3. Why do we boil after NH4Cl ? • Ans.: • 1- To coagulate the ppt • 2- To break Cr Amine complex • [Cr(NH3)6] (OH)3 Cr(OH)3 + 6NH3 • soluble Cr amine complex • 3- To expel O2 and prevent oxidation of manganus (Mn2+) which has higer Ksp i.e. [not ppt as OH- ] to manganic (Mn3+) which has low Ksp and can be ppt as Mn(OH)3 so it escapes from group IV to group III

28. Separation of group III :- • ppt of group III + NaOH + H2O2 centrifuge • Residue Centrifugate • Fe3+, Mn3+(from gp IV) AlO2-, CrO42- • It depends on the amphoteric character of Al3+ and Cr3+ so dissolve in NaOH, they dissolve as stable AlO2- and unstable chromite CrO2‑ which is oxidised by H2O2 to the stable chromate CrO42- (has yellow color) • Al(OH)3 + OH- AlO2- + 2H2O • Cr(OH)3 + OH- CrO2- + 2H2O • unstable chromite • CrO2- + H2O2 CrO42- + 4 H2O • stable yellow chromate

29. Separation of ferric and manganic :- • Residue + HNO3 + H2O2 dissolve, divide to 2 portions • Fe(OH)3 + 3HNO3 Fe(NO3)3 + 3H2O • MnO(OH)2 + 3HNO3 + H2O2 Mn(NO3)3 + 3H2O + O2 • First portionTest for iron :- • 1- soln + NH4SCN blood red color • Fe3+ + SCN- [Fe(SCN)]2+ • 2- soln + potassium ferrocyanide prussian • blue ppt • Fe3+ + K4[Fe(CN)6] K Fe [Fe(CN)6]

30. Second portion • Test for manganus :- • 1-With red lead Pb3O4 • soln + dil HNO3 + H2O2 till no O2 bubbles • + PbO2 pink color • Pb3O4 + Mn2+ + H+ MnO4- + Pb2+ + H2O • pink color • H2O2 must be completely removed because it reduces MnO4- so bleach color formed • MnO4- + H2O2 + H+ Mn2+ + O2 + H2O • colorless • N.B • H2O2 is oxidising agent but with MnO4-, it is reducing agent because it has lower oxidation potential • [ MnO4- has the highest oxidation potential]

31. Separation of aluminium and chromium :- • Divide the soln. containing AlO2- and CrO4- to 2 portions: • First portion • Test For CrO42- :- • 1- With pb acetate yellow ppt of lead chromate • Pb(CH3COO)2 + (CrO4)2- PbCrO4 • 2- Perchromic acid test • soln + dil H2SO4 + ether + 1 drop H2O2 • blue color in ether layer • CrO42- + 2H+ Cr2O72- + H2O+H2O2 [CrO8]3- • dichromate perchromate

32. Second portion • Test for AlO2- :- • 1- Soln + NH4Cl solid  cool white gelatinous ppt • Al(OH)3 + OH- AlO2- + H2O • white ppt • NH4Cl reduces OH- concentration and reverse the reaction i.e move backward and ppt Al(OH)3 • NH4+ + OH- NH4OH NH3 + H2O

33. Group IV

34. Group IVMn2+, Zn2+, Co2+, Ni2+ • This group is called “ Ammonium sulphide” group • Group reagent is : NH4Cl + NH4OH + H2S • It is ppt as sulphide in alkaline medium (weak acid, partially ionize) H2S 2H+ + S2- (weak base, partially ionize) NH4OH OH- + NH4+ • H2O • Due to formation of water which is weakly ionized. The ionization of H2S is pushed forward and increase S-2 ion conc so ppt group IV as sulphide (has higer Ksp)

35. Procedure for separation :- • 1- Centrifugate from group III + NH4Cl + NH4OH  • + H2S centrifuge • Residue Centrifugate • gp IV rest of groups • MnS  buff ppt Manganous sulphide • ZnS  white ppt Zinc sulphide • CoS  black ppt Cobalt sulphide • NiS  black ppt Nickel sulphide • Ques.1. Why do we add NH4Cl ? • Ans. : 1- By common ion effect it decreases OH- conc so prevent pptn of group IV as hydroxide • Salt (completely ionize) NH4Cl NH4+ + Cl- • Weak base (partially ionize) NH4OH NH4+ + OH- • common ion effect • 2- NH4Cl is an electrolyte to coagulate the colloidal ppt of group IV eg ZnS, MnS so it become easily filtered

36. Ques.2. Why do we boil after add NH4Cl ? • Ans.: • 1- To decompose Ni amine complex and ppt NiS • Ni(NH3)6S  NiS  • soluble Ni amine complex black ppt • 2- By boiling CoS and NiS undergo condensation and change in physical characters so become insoluble in very dil HCl • 3- To coagulate the ppt • N.B. • If iron is not completely oxidised in group III it will ppt as FeS (black ppt) in group IV

37. Separation of group IV :- • ppt of group IV + very dil HCl centrifuge • Residue Centrifugate • NiS , CoS Zn2+, Mn2+ • black ppt • It depends on the amphoteric character of Zn, therefore soluble in dil HCl and NaOH • Centrifugate  + NaOH + Br2 water  centrifuge • Residue Centrifugate • MnO2  ZnO22- • black ppt • Zn2+ + OH- Zn(OH)2 zinc hydroxide • Zn(OH)2 + OH- (excess) ZnO22- soluble zincate • While Mn(OH)2 is oxidized by hypobromite to MnO2 which is black ppt • OH- + Br2 OBr- + Br- + H2O • Mn2+ + Br2 + OH- MnO2 + 2Br- + 2H2O

38. N.B. • The centrifugate must be heated to expel H2S because if H2S is still present, so on adding NaOH, ZnS will ppt as it is ppt as sulphide in alkaline medium • Confirmation of zinc:- • 1- By adding excess H2S white ppt of zinc sulphide • ZnO22- + S2- + H+ Zn S  + OH- • 2- Add acetic acid + pot. ferrocyanide sky blue • ppt • Zn2+ + [Fe(CN)6]4- Zn2[Fe(CN)6]  • Zinc ferrocyanide • Confirmation of Manganus :- • As under group III

39. Confirmation of cobalt and nickel :- • ppt is dissolved in aqua regia [ HNO3 1 : 3 HCl ] • 3CoS + 2HNO3 + 6HCl 3CoCl2 + 2NO + 3S + 4H2O • 3NiS + 2HNO3 + 6HCl 3NiCl2 + 2NO + 3S + 4H2O • Divide the soln to 2 portions : • First portionTest for Co2+ :- • 1-soln + NH4Cl + NH4OH + pot ferricyanide red ppt • Co2+ + [Fe(CN)6]3- Co3[Fe(CN)6]2 red ppt • cobaltus ferricyanide insol in NH3 • N.B. Nickelous ferricyanide is yellow and soluble in NH3 • (that is why we add NH4OH) • 2- Vogel test • soln + dil H2SO4 + SnCl2 + ether + NH4SCN blue color in etherial layer • Co2+ + SCN- [Co(SCN)4]2- blue color • N.B. • H2SO4 contains traces of Fe+3 which interfere with the test by giving blood red color with SCN- • Therefore we add reducing agent eg SnCl2 to reduce Fe+3 to Fe+2 or we add F- or PO4- to make complex with Fe+3

40. Second portionTest for Ni2+ :- • 1- Dimethyl glyoxime test • Soln + NH4Cl + NH4OH + D.M.G. red color • 2- Test with KCN • soln + KCN + NaOH + Br2 water black ppt of NiO2 • OH- + Br2 OBr- + Br- + H2O • Ni2+ + 2CN- Ni(CN)2(excess CN-) [(Ni(CN4)]2- OBr- NiO2 . • unstable Ni cyanide complex • Cobalt will not interfere • Co2++ 2CN- Co(CN)2(excess CN-) [Co(CN)6]4- atm. O2 [Co(CN)6]3- • stable cobalticyanide complex • Cobalticyanide complex does not interfere as it is an oxidising agent which does not react with OBr-

41. Group V

42. Group VBa2+, Sr2+, Ca2+ • It is called “Alkaline earth group” or “ Ammonium carbonate group” • Group reagent : NH4Cl + NH4OH + (NH4)2CO3 • The cations ppt as carbonate in alkaline medium • The three cations belong to the same group in the periodic table, therefore have similar characters and their separation is somewhat difficult • Procedure for separation :- • Centrifugate from group IV is evaporated nearly to dryness + NH4Cl + NH4OH  + (NH4)2 CO3 soln  to 60oC wait, • centrifuge • Residue Centrifugate • Gp V gp VI • BaCO3 white ppt • SrCO3 white ppt • CaCO3 white ppt

43. Ques.1. Why we evaporate the centrifugate from group IV till nearly dryness ? • Ans. : • To expel H2S, to prevent oxidation of S2- to SO42- and prevent pptn of Ba2+ and Sr2+ as SO42- [ CaSO4 has higher Ksp so need more SO42- ion conc. to ppt] • Ques.2. We must add a moderate amount of NH4+ ? • Ans. : • Because excess NH4+ will decrease carbonate ion concentration. • NH4+ + CO32- NH3 + HCO3- • N.B. • The soln must be a little basic. If acidic the carbonate will change to HCO3- which are soluble • CO3- + H+ HCO3- • Ques.3. Why we use NH4Cl ? • Ans. : • 1- By common ion effect it decreases OH- ion conc. and CO32- ion conc. so prevent pptn of Mg as Mg(OH)2 or as MgCO3 • 2- To coagulate the ppt

44. Ques.4. Why we use NH4OH ? • Ans • 1- It converts NH4HCO3 to (NH4)2 CO3 • NH4HCO3 + NH3 (NH4)2CO3 • 2- Converts ammonium carbamate into carbonate • NH4CO2NH2 + H2O (NH4)2CO3 • 3- To make the medium faintly alkaline to prevent HCO3- formation as all M(HCO3)2 are soluble • N.B. • Ammonium carbonate is prepared from equimolar concentration of ammonium bicarbonate NH4(HCO3)2 and ammonium carbamate NH4CO2NH2 • Ques.5. Why we heat to 60oc and not boil ? • Ans. : • 1- To convert HCO3- to CO32- • M(HCO3)2 MCO3 + CO2 + H2O • 2- Warming aid in the formation of more crystalline ppt • 3- Warming aid in the conversion of carbamate to carbonate • Excess heat cause loss of NH3 so convert CO32- to bicarbonate by pushing the reaction forward

45. Separation of group V :- • ppt + acetic acid The ppt will dissolve as acetate • BaCO3 + 2 CH3COOH Ba(CH3COO)2 + CO2 + H2O • SrCO3 + 2 CH3COOH Sr(CH3COO)2 + CO2 + H2O • CaCO3 + 2 CH3COOH Ca(CH3COO)2 + CO2 + H2O • Test for Barium :- • 1-2 drops of the soln + K2CrO4 (pot. chromate) If yellow ppt, • therefore Ba is present • If Ba is present add K2CrO4 soln till soln acquire orange tint, • centrifuge • Residue Centrifugate • BaCrO4  Sr2+, Ca2+ • Yellow ppt reppt [NH4OH + NH4Cl + • (NH4)2CO3 to 60oc • Centrifuge Residue Centrifugate SrCO3, CaCO3 reject white ppt

46. Test for Ba2+ :- • 1-Flame test • ppt of BaCrO4 + HCl soluble BaCl2 (volatile in flame ) and gives apple green color • If Ba is absent reject the portion used for test and the remainder of the soln is used to test for strontium and calcium • N.B. • If we add excess K2CrO4, SrCO4 and Ca CrO4 will also ppt as the chromate of this group are insoluble. But Ksp of BaCrO4 < SrCO4 < CaCrO4. So we add amount of K2CrO4 sufficient to ppt barium but not strontium and calcium • We use also acetic acid to convert part of K2CrO4 to K2Cr2O7 so decrease CrO42- conc • Ques: • Why we do not use HCl or HNO3 instead of acetic acid ? • Ans.: • Because they are strong acids which convert all CrO42- to Cr2O72- and no ppt is formed

47. Identification of Strontium and Calcium :- • Residue dissolve in acetic acid to dissolve • Soln + (NH4)2SO4  centrifuge • Residue Centrifugate • Sr2+ Ca2+ • Sr(CH3COO)2 + (NH4)2SO4 SrSO4 + 2CH3COONH4 • white ppt • The test depends on that Ca form double salt with ammonium sulphate which is soluble • Ca2+ + (NH4)2SO4 (NH4)2Ca(SO4)2 • double salt formation • N.B. • If we used dil H2SO4 instead of (NH4)2SO4, both Sr and Ca will ppt as SO4

48. Test for Calcium :- • 1- Soln + NH4OH + ammonium oxalate white ppt of calcium oxalate • C2O4(NH4)2 + Ca2+ C2O4Ca  + 2NH4Cl • white ppt • 2- Soln + NH4Cl + pot. ferrocyanide pale yellow ppt • Ca2+ + NH4+ + K+ + [Fe(CN)6]4- CaNH4K[Fe(CN)6] • triple ferrocyanide • Flame test :- • For Sr2+ • SrSO4 is heated in the reducing zone of the flame Sr S + HCl • SrCl2 volatile in flame with crimson red color • For Ca2+Ca C2O4 + HCl Ca Cl2 volatile in flame with brick red color.

49. Group VI

50. Group VIMg2+, Na+, K+, NH4+ • This group is called “ Alkali metal group” • It is also called the “ Soluble group “ because it has no group reagent (all Na, K and NH4 salts are soluble) • Removal of traces of alkaline earth group :- • Centrifugate of group V + ammonium oxalate [to ppt any Ca2+] + ammonium sulphate [to ppt any Ba2+ or Sr2+]  • centrifuge • Residue Centrifugate rejected gp VI • Divide the centrifugate into two unequal portions :