Question

Draw a flow diagram that summarizes the separation of the cations in these unknown: Pb2+, Mn2+, Bi3+, Ba2+, Cu2+, Mg2+.

At each step, give the correct chemical form of the cation(s) present. Alsdo indicate observed colors of precipitates and supernatants.

Answer 1

First, identify the type of group:

Pb+2 = group 1

Bi+3 and Cu+2 = Group 2

Mn+2 = Group 3

Ba+2 = Group 4

Mg+2 = Group 5

a)

For Pb+2, simply add HCl in order to precipitate PbCl2(s)

Pb+2(aq) + 2Cl-(aq) <-> PbCl2(s)

b)

Add H2S so

2Bi+3 + 3H2S <-> Bi2S3(s)

then add HNO3 so

Bi2S3 + HNO3 --> 2Bi(NO3)2

Add NH4OH

Bi+3 + 3NH4OH = --> Bi(OH)3(s)

then add:

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

For Cu+2 (which is already in Cu(OH)2 state, similar ot Bi(OH)3)

2 Cu(NH3)4(NO3)2 + 10KCN + H2O --> 2K3[Cu(CN)4] + 4KNO3 + NH4CN + NH4CNO + 6NH3

2K3[Cu(CN)4] color is blue

For

Ba+2

Add Na2CO3, so

Ba+2 + CO3-2 = BaCO3(s)

BaCO3(s) + 2 CH3COOH --> (CH3COO)2Ba + H2O + CO2(g)

(CH3COO)2Ba + K2CrO4 --> BaCrO4(s) + 2 CH3COOK

For Mg+2

Mg(NO3)2 + NH3 + Na2HPO4 + 5 H2O -->MgNH4PO4(s) + 2 NaNO3

Mg(NO3)2 + 2 NaOH --> Mg(OH)2(s) + 2 NaNO3