Qualitative Analysis : Analysis of group II

Analysis of group II

Radicals –   IIA – Hg2+, Pb2+, Bi2+, Cu2+ and Cd2+

                    IIB – As3+, Sn2+, Sn4+, Sb3+

Group reagent – H2S in presence of dil. HCl.

Principle: Sulphides of cations mentioned above are insoluble in dilute HCl. Among these sulphides, some are of acidic nature and some are of basic nature. Basic sulphides are insoluble in yellow ammonium sulphides and are grouped in subgroup-II A. Sulphides of acidic nature are soluble in yellow ammonium sulphide due to formation of complex. They are grouped in sub-group II-B.

Separation and analysis of II_A and II-B radicals is based on following sequence of reactions:

Procedure: Add dilute HCl in filtrate of gp. I and pass H2S in warm solution. Formation of coloured ppt. (black, yellow, orange etc.) indicates the presence of gp-II. Pass H2S repeated till ppt. are formed. Collect ppt. and wash. Residue will be used to analyse gp-II while filtrate will be used for analysis of gp-III. Residue obtained is washed with hot water and is treated with warm yellow ammonium sulphide. Residue will contain cations of II-A gp. While filtrate contains soluble complexes of II-B cations.

(i)     Analysis of II-A cations: The residue obtained after extraction of yellow ammonium sulphides is washed with hot water till yellow colour disappears. After that the ppt. is heated with 1 : 1 HNO3 and filtered.            

(ii)    Analysis of II-B cations: Cations of II-B group (As3+, Sb3+ and Sn2+) form soluble this complexes with yellow ammonium sulphide. The filtrate obtained after treating gp-II sulphides with yellow ammonium sulphide is acidified with dil. HCl. Formation of coloured ppt. indicates the presence of gp. II-B wash these coloured ppt. and boil with conc. Filter the solution.

 

Residue: Black ppt. (HgS) black is washed with water and is heated with aqua-regia, till evolution of NO continues. The solution thus obtained is divided into three parts.

Part – I: White ppt. is obtained on adding SnCl2 which turns grey on standing.

Part –II: Yellow ppt. is obtained on adding KI, which dissolves in excess of KI

Part – III: Surface of Cu turning turns silvery Hg2+ confirm

Filtrate: May contain nitrates of Pb2+, Cu2+, Bi3+ or Cd2+. Add dil. H2SO4 and C2H5OH in hot solution. If white ppt. is formed, filtrate it.
Residue: It may be PbSO4. Add hot ammonium acetate solution and divide the solution into two parts:

Part – I: Yellow ppt. is obtained on adding K2CrO4 solution and ppt. is soluble in NaOH.

Part – II: Yellow ppt. is obtained on adding KI solution which is soluble in hot water Pb2+ confirms

Filtrate: May be nitrate or sulphate of Bi3+, Cu2+ or Cd2+. Add NH4OH solution and filter. White ppt. is analysis for Bi3+ and filtrate for Cu2+ or Cd2+
Residue: White ppt. may be of Bi(OH)3. It is dissolved in dil. HCl and is divided into three parts.

Part – I: On adding excess of water in solution, the solution turns turbid.

Part – II: Black ppt. is obtained on adding sodium stannite.

Part – III: On adding thiourea in presence of conc. HNO3 yellow ppt. is obtained.Bi3+ confirm.

Filtrate: It may contain complex of Cu2+ or Cd2+. If solution is blue, it will contain Cu2+, otherwise Cu2+ will be absent.
Coloured solution: it may contain Cu2+ or Cu2+/Cd2+ both (i) in Part –I, add CH3COOH and then K4[Fe(CN)6]. If chocolate brown ppt. is formed then Cu2+ confirm.

(ii) When both Cu2+ and Cd2+ are present.

(a) If blue colour of solution disappears on adding KCN and yellow ppt. is obtained on passing H2S.

Cd2+ confirm.

(b) In part – II, and dil. H2SO4 and Zn or Fe turning. Heat and neutralize with NH4OH add 1-2 drop HCl and pass H2S. yellow ppt. Cd2+ confirm.

Colourless solution: In colourless solution H2S is passed, yellow ppt. indicates Cd2+.

 

 

Residue: Yellow ppt. may be of As3+. Divide the yellow ppt. in three parts:

Part – I: Heat the yellow ppt. with conc. HNO3 and add ammonium molybdate solution, a canary yellow ppt. is obtained.

Part – II: Heat the yellow ppt. with saturated solution of (NH4)2CO3. Add dil. HCl and pass H2S, a yellow ppt. is obtained.

Part – III: Dissolve the yellow ppt. in conc. HNO3 and make the solution ammonical. Addition of magnesia mixture gives white ppt.

As3+ confirm.

Filtrate: It may contains chlorides of Sb3+ or Sn4+. In this solution add NH4OH and solid oxalic acid, then pass H2S. Formation of orange ppt. indicates Sb3+ and if ppt. is white then Sb3+ is absent, Filter the ppt.
Residue: Orange ppt. may be of Sb2S3. Dissolve it in minimum amount of conc. HCl. Divide the solution in three parts.

Part – I: Black ppt. is obtained if Fe or Zn is added.

Part – II: Add NH4OH and excess of water, a while ppt. is obtained.

Part – III: Add NH4OH and pass H2S, an orange ppt. is obtained. Sb3+ confirm.

Filtrate: May contain Sn2+/Sn4+ Boil it with Zn dust and dil HCl and divide the solution in three parts:

Part – I: ON addition pf HgCl2 a white ppt, is obtained which turns black on standing.

Part – II: A deep blue solution or ppt. is obtained on adding ammonium molybdate solution.

Part – III: Pass H2S and a dark brown ppt. is obtained.

Sn2+/Sn4+ confirm.

 

NOTE:

  1. H2S must be passed in hot solution slowly to get more granular and easily filterable precipitates.
  2. To ensure complete precipitation. pH must be controlled. If solution is less acidic, then sulphides of gp-IV (e.g., ZnS) may be precipitated. If solution is much more acidic than PbS, CdS & SnS are not completely precipitated. For complete precipitation, the H+ conc. must be 0.2 – 0.3 N.
  3. To check the above concentration of H+, methyl violet indicator may be used. Add 1 – 2 drop indicator in solution and check the colour. If colour of solution is blue-violet, add NH4OH, till the colour of solution is changed to yellow-green. If colour is yellow, add dil. HCl to adjust the colour yellow-green.
  4. Before passing H2S into original solution; It must be free from any oxidizing agents like NO3, SO32– etc. They oxidize H2S to sulphur which appears as yellow ppt.
  5. If pH of original solution is too low, then red ppt, of lead sulpho chloride (PbSCl2) is obtained which gives black ppt. of PbS on dilution.
  6. To separate IIA and II B, colourless ammonium sulphide can’t be used because SnS is insoluble in it.
  7. If yellow or white ppt. is obtained on adding HCl in analysis of filtrate for II-B, then gp. II-B is absent and yellow ppt. is of sulphur.
  8. Sometimes on passing H2S, the colour of ppt, changes gradually with time e.g., in case of Hg at first a dirty white, then brown then orange and finally black ppt. is obtained.
  9. To dissolve sulphides of gp. II-A,, too much concentrated HNO3 should not be used. It may oxidize PbS to PbSO4 which will remain behind with HgS. Also the residue may not be black but yellow due to formation of Hg(NO3)2 HgS.
  10. When As2S3 is dissolved in conc. HNO3 a black mass floats in the test tube. It may be sulphur and must be removed with a glass rod.
  11. Sodium stannite is prepared by adding NaOH in SnCl2 solution till the precipitate formed dissolves.

SnCl2 + 2NaOH → Sn(OH)2 + 2NaCl

                                           ppt.

Sn(OH)2 + 2NaOH → Na2SnO2 + 2H2O

                                      Soluble

Reactions and Explanations:

Following ppt, are obtained in this group:

                   Black – HgS, PbS, Bi2S3, Cus

                   Yellow – CdS, SnS2 and As2S3

                   Brown – SnS

                   Orange – Sb2S3

On adding yellow ammonium sulphide, II-A cations remain insoluble while II-B cation sulphides form soluble thio complexes.

(i)      As2S3 + 2(NH4)2S2 → As2S5 + 2(NH4)2S

          As2S5 + 3(NH4)2S → 2(NH4)3AsS4

Ammonium tetrathioarsenate

                             (soluble)

 

(ii)     Sb2S3 + 2(NH4)2S2 → Sb2S5 + 2(NH4)2S

          Sb2S5 + 3(NH4)2S  → 2(NH4)3SbS4

          Ammonium tetrathioantimonate

                             (soluble)

 

(iii)    SnS + (NH4)2S2 → SnS2 + (NH4)2S

          SnS2 + (NH4)2S → (NH4)2.SnS3

          Ammonium tetrathiostannate.

 

(A)    Reaction of II-A radicals:

II-A sulphides form soluble nitrates (except HgS) with HNO3. HgS is insoluble in HNO3.

          3MS + 8HNO3 → 3M(NO3)2 + 2NO + 3S + 4H2O

                             (M = Cu, Cd or Pb)

          Bi2S3 + 8HNO3 → 2Bi(NO3)2 + 2NO + 3S + 4H2O

 

(a)    Test for Hg2+: HgS dissolves in aqua-regia to form HgCl2.

          3HgS + 2HNO3 + 6HCl → 3HgCl2 + 2NO + 3S + 4H2O

(i)      Reaction with SnCl2 gives white ppt. of Hg2Cl2, which decomposes to grey Hg.

          2HgCl2 + SnCl2 → Hg2Cl2↓ + SnCl4

                                White ppt.

          Hg2Cl2 + SnCl2 → 2Hg + SnCl4

                                Grey

 

(ii)     With KI, red ppt. of HgI2 is obtained.

          HgCl2 + 2KI → HgI2↓ + 2KCl

                             Red ppt.

          2KI + HgI2 → K2(HgI4)

                            Soluble

 

(iii)    With Cu turning; grey layer of Hg is deposited.

          HgCl2 + Cu → CuCl2 + Hg

 

(b)    Tests for Pb2+: When dil. H2SO4 is added in soluble nitrates of Pb2+, Cu2+, Bi3+ or Cd2+ insoluble PbSO4 is precipitated while other sulphate are soluble.

          Pb(NO3)2 + H2SO4 → PbSO4↓ + 2HNO3

                                    White ppt.

          This white ppt. is soluble in ammonium acetate solution.

          PbSO4 + 4CH3COONH4 → (NH4)2[Pb(CH3COO)4] + (NH4)2SO4.

Ammonium tetraacetato              Plumbate (II)

 

(i)     Reaction with K2CrO4:

(NH4)2[Pb(CH3COO)4] + K2CrO4 → PbCrO4↓ + 2CH3­COOK + 2CH3COONH4

          Lead chromate                                     (yellow ppt.)

PbCrO4 + 4NaOH → Na2[Pb(OH)4] + Na2CrO4

Sodium tetrahydroxoplumbate   (soluble)

 

(ii)    Reaction with KI:

(NH4)2[Pb(CH3COO)4] + 2KI → PbI2↓ + 2CH3COOK + 2CH3COONH4

          Lead iodide                            (yellow ppt.)

 

(C)    Test for Bi3+: When NH4OH is added in soluble nitrates/sulphates of Bi3+, Cu2+, Cd2+ and Bi3+ is precipitated as white ppt. of Bi(OH)3, while Cu2+ and Cd2+ remains as soluble complexes.

          Bi(NO3)2 + 3NH4OH → Bi(OH)3↓ + 3NH4NO3

                                      White ppt.

          It dissolves dil. HCl to form BiCl3.

          Bi(OH)3 + 3HCl → BiCl3 + 3H2O

          On adding excess water

          BiCl3 + H2O (hydrolysis) → BiOCl + 2HCl

          Bismuth oxychloride           (turbid solution)

Reaction with Na2SnO2:

2BiCl3 + 3Na2SnO2 + 6NaOH → 2Bi + 3Na2SnO3 + 6NaCl + 3H2O

                             Black ppt.

 

(d)    Tests for Cu2+: If solution obtained after testing of Bi3+ is blue, then Cu2+ will be necessarily present. With NH4OH, Cu2+ gives dark blue complex of copper tetrammine.

          CuSO4 + 4NH4OH → [Cu(NH3)4]SO4 + 4H2O

                             Dark blue colour.

(i)     Reaction with CH3COOH/K4[Fe(CN)]:

          [Cu(NH3)4]SO4 + 4CH3COOH → CuSO4 + 4CH3COONH4

          2CuSO4 + K4[Fe(CN)6] → Cu2[Fe(CN)6]↓ + 2K2SO4

                                        Copper ferrocyanide

                                      (Chocolate brown ppt.)

(ii)    When both Cu2+ and Cd2+ are present: On passing H2S in presence of KCN.

2[Cu(NH3)4]SO4 + 10KCN + 8H2O → 2K3[Cu(CN)4] + 2K2SO4 + 8NH4OH + (CN)2

          Potassium cuprocyanide                                       (colourless)        

          [Cd(NH3)4]SO4 + 4KCN + 4H2O → K2[Cd(CN)4] + K2SO4 + 4NH4

          Potassium                             Cadmium cyanide     

          On passing H2S, only CdS is precipitated.

          K2[Cd(CN4) + H2S   →  CdS↓ + 2KCN + 2HCN                                             

Cadmium sulphide

                                      (yellow ppt.)

In presence of H2SO4, Fe or Zn reduces Cu2+ to Cu and on passing H2S, yellow ppt, of CdS is obtained.

 

(e)     Tests for Cd2+: On passing H2S in colourless solution, yellow ppt. of CdS is obtained.

(B)    Analysis of II-B radicals:

Soluble this complexes of II-B radicals are decomposed into insoluble sulphides by dilute HCl.

          2(NH4)3.AsS4 + 6HCl → As2S5 + 6NH4Cl + 3H2S

          2(NH4)3.SbS4 + 6HCl → Sb2S5 + 6NH­4Cl + 3H2S

          (NH4)2SnS3 + 2HCl → SnS2 + 2NH4Cl + H2S

          (NH4)2S2 + 2HCl → 2NH4Cl + S + H2S

On boiling with conc. HCl, these sulphides are converted into soluble chlorides except arsenic sulphide.

                   As2S3 + HCl  \underrightarrow { \quad \Delta \quad } No reactions

                   Sb2S3 + 6HCl → 2SbCl­3 + 3H2S

                   Sb2S5 + 6HCl → 2SbCl3 + 3H2S + 2S

                   SnS2 + 4HCl → SnCl4 + 2H2S

 

(a)    Tests for As3+

(i)     Reaction with HNO3/ammonium molybdate:

          As­2S3 + 10HNO3 → 2H3AsO4 + 10NO2 + 2H2O + 3S

                                  Arsenic acid

In presence of HNO3. H3AsO4 gives yellow ppt. with ammonium molybdate.

H3AsO4 + 12(NH4)2MoO4 + 21HNO3 → (NH4)3[AsO4.12MoO3] + 21NH4NO3 + 12H2O

                   Ammonium-12-molybdo arsenate

(ii)    Reaction with (NH4)2CO3­ and then with HCl/H2S:  

          As2S3 + 3(NH4)2CO3 → (NH4)3.AsO3 + (NH4)3.AsS3 + 3CO2

                   Ammonium     Ammonium thio

                             arsenite     arsenite

          2(NH4)3.AsS3 + 6HCl → As2S3 + 3H2S + 6H2O

          2(NH4)3.AsO3 + 12HCl → 6NH4Cl + 2AsCl3 + 6H2O

          2AsCl3 + 3H2S → As2S3 ↓ + 6HCl

           Arsenic trisulphide      (yellow ppt.)

 

(b)    Tests for Sb3+:

(i)     Reaction with Fe:

          SbCl3 + Fe → Sb + FeCl3

                   Black ppt.

 

(ii)    Reaction with H2O:

          SbCl3 + H2O → SbOCl↓ + 2HCl

          Antimony oxychloride           (white ppt.)

 

(iii)   Reaction with H2S:

          2SbCl3 + 3H2S → Sb2S3↓ + 6HCl

                             Orange pp.t                                     

 

(c)     Tests for Sn2+/Sn4+: Filtrate may contain Sn2+/Sn4+ chlorides. On boiling with Zn/HCl, Sn4+ is reduced to Sn2+.

                   SnCl4 + Zn       ZnCl2 + SnCl2

 

(i)     Reaction with HgCl2:

          SnCl2 + 2HgCl2 → Hg2Cl2 + SnCl4

                                 White ppt.

          Hg2Cl2 + SnCl2 → 2Hg + SnCl4

                               Grey ppt.

 

(ii)    Reaction with ammonium molybdate:

SnCl2 + 8HCl + 3(NH4)2.MoO4 → Mo3.O8 + 6NH4Cl + 4H2O + SnCl4

                   Molybdenum oxide           (deep blue)

 

(iii)   Reaction with H2S:

                   SnCl2 + H2S → SnS↓ + 2HCl

                   Brown ppt.