Question

Separate ions in a flow chart without sulfides

Ag bi sb ni

Answer 1

First, Ag, since it is cation group I,

use HCl --> Cl- ions precipitate Ag+ ions

Ag+ + Cl- --> AgCl(s)

then Bi+3

Add H2S to form the sulfide:

2 Bi3+ + 3 H2S --> Bi2S3(s)(brown) + 6 H +

Sb will precipitate as well:

Sb+3 + H2S --> Sb2S3(s)

separate Bi and Sb:

Bi2S3(s) + 8 HNO3 --> 2 Bi(NO3)3(aq) + 2 NO + 3 S + 4 H2O

HERE, remove Sb precipitate, since Bi+3 will be in solution

Then, after Sb is removed, go for Bi+3

Bi(NO3)3 + 3 NH3(g) + H2O --> Bi(OH)3(s) + 3 NH4NO3

2 Bi(OH)3(s) + 3 Na2SnO2  2 Bi(s) + 3 Na2SnO3 + 3 H2O

Finally, Ni2+ is only present in solution

Use OH- ions to preciptiate it as

Ni(OH)2(s); filter and you are done!