Question

You have been given a solution that contains a Group I, a Group II, and a Group III cation. · Develop a qualitative analysis scheme for these cations by selecting reagents and conditions for reactions. · Write the balanced net reaction for each cation reaction.

Answer 1

First, identify cations and groups:

Group 1 = (Ag+ , Pb2+, Hg2 2+)

Group 2 = ( Bi3+ , Cu2+ , Cd2+)

Group 3 = (Co2+, Ni2+, Fe3+, Mn2+, Cr3+, Al3+, Zn2+)

Now..

Group 1 -> Add HCl concentrated, Ag+ , Pb2+, Hg2 2+ precipitate as Chlorides:

Ag+ + Cl- = AgCl(s)

Pb+2 + 2Cl- = PbCl2(s)

Hg2+2 + 2Cl- = Hg2Cl2(s)

Then, separate filter (Groups 2 +3 ions)

Treat preicpitates with acid,

Pb+2 gets ionic, then add SO4-2 so

Pb+2 + SO4-2 = PObSO4(s) separated

For AgCl, add plenty in basic media

Hg(s) due to basicit

Hg2+2 + 2 NH3 --> Hg(s) + HgNH2Cl(s) + NH4Cl

filter Hg, then add acidic media once again, Recover AgCl

This is now group 1 separated.

Group 2:

Add H2S in adicid media

Bi3+ , Cu2+ , Cd2 ions precipitate in aidic media as Sulfides

Bi3+ ,+ H2S --> Bi2S3(s)

Cu2+ + S-2 = CuS(s)

Cd2+ + S- 2= CdS(s)

Add plenty of base, NH4(OH)

so

Bi(NO3)3 + 3 NH3 -- >Bi(OH)3(s) + 3 NH4NO3 --> filter

For copper recovery: add KCN agent, so Cu forms as:

2 Cu(NH3)4*(NO3)2 + 10KCN + H2O --> 2K3(Cu(CN)4 + 4KNO3 + NH4CN + NH4CNO + 6NH3 (all complex

Add H2S again, so Cd+2 form CdS again but Copper remains as ionic complex

now..

Cd can be filtered, Copper remains in solution... which is no seprated

Now, Group 3.

Group 3 = (Co2+, Ni2+, Fe3+, Mn2+, Cr3+, Al3+, Zn2+)

For nickel, add HDMG

Ni(NO3)2 + 2 HDMG --> Ni(DMG)2 + 2 HNO; which can be filtered alone

Then

Fe+3 ions: add KsCN for complex formation (acidic)

Fe(NO3)3 + 6 KSCN --> 3K+ + [Fe(SCN)6] + 3 KNO3

For Co2+ add NH4SCN, which will form (basic complex)

Co(NO3)2 + 4 NH4SCN  (NH4)2[Co(SCN)4] + 2 NH4NO3

Recover Mn+2:

2 Mn(NO3)2 + 5NaBiO3 + 16HNO3 --> 2 HMnO4 + 5Bi(NO3)3 + 5Na+ + 5NO3-

Add NaOH high... in excess, omplex Zn(OH)4-2, Al(OH)4- and CrO4-2 forms

Zn(OH)4-2, Al(OH)4- and CrO4-2

--> add BaCl2 in acidic media, BaCrO4 forms... solid, filter it

Now,

Only Zinc copmelx is present, addH2S and recover:

[Zn(OH)4]2- + H2S --> ZnS(s) + 2 NaOH + 2H2