Question

1) Identify cations by selective precipitation and specific reaction of these mixture:

Cu(2+); Bi (3+), Cd (2+), Sb(5+); Sn (2+), Fe(3+), Co(2+), Mg(2+), Li+

2) Calculate the solubility of each of the following compounds in moles per liter. Ignore
any acid–base properties.

PbI2, K(sp) = 1.4 * 10^-8

CdCO3, K(sp)= 5.2 * 10^-12

Sr3(PO4)2, K(sp)= 1*10^-31

Answer 1

1)

• Metal cations are classified in to Five groups for qualitative analysis
• This classification is based on: the behaviour of the cations with some reagents ande solubilities of their chlorides, carbonates and sulphides.

Group2:

Devided in to 2 groups 2A and 2B ; Based on the solubility of the cations in ammonium polysuphide (NH₄)₂S_x

Cu²⁺; Bi³⁺, Cd²⁺; these are insoluble in ammonium polysulphide. So they belong to group 2A

Sb(5+), Sn (2+); Soluble in ammonium polysulphide. So they belong to group 2B

Group 3:

Fe(3+), Co(2+); the group reagent is ammonium sulphide solution or hydrogen sulphide gas in the presence of ammonia and ammonium chloride. These cations won't react with 2^nd group reactants.

Group 5:

Mg²⁺ belongs to this group 5.
It has similar characteristics to alkali metals. Its general properties are similar to that of potassium as the sizes of both ions are identical.

2)

PbI₂

K_sp = [Pb²⁺] [I^-]²

Solving for x will give :

CdCo₃K_sp = [Cd²⁺] [Co^2-]³

Sr₃(PO₄)₂ [Sr²⁺]³ [Po^3-]⁴

from this x can be calculated