Question

A solution is known to contain Bi3+ ion or Cd2+ ion or both. Describe the test(s) that could be used to separate and identify these ions. Describe the results you would expect to see for each ion, if that ion was present.

Answer 1

Apply

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

Cd2+ + H2S --> CdS(s)(yellow) + 2 H +

then

add HNO3

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

3 CdS(s) + 8 HNO3 --> 3 Cd(NO3)2 + 2 NO + 3 S + 4 H2O

then add NH3

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

Cd(NO3)2 + 2 NH3*H2O--> Cd(OH)2(s) + 2 NH4NO3

NOW, the trick is to add plenty of NH3

then

Cd(OH)2 + 4 NH3*H2O --> Cd(NH3)4(OH)2 + 4H2O

the Bi(OH)3 is ready to take out as a precipitate

2 Bi(OH)3(s) + 3 Na2SnO2 --> 2 Bi(s) + 3 Na2SnO3 + 3 H2O

whereas for Cd

Cd(NH3)4(OH)2 + 4 KCN --> K2[Cd(CN)4] + 4 NH3 + 2 KOH

for separatoin

Cd(NH3)4(OH)2 + H2Saq --> CdS(s)+ 4NH3 + 2H2O