For a), this qn is a matter of memorising the facts of cation
test. When NaOH is added to a suspension containing Lead(II)
Hydroxide, a white ppt is formed which is soluble in excess NaOH,
giving a colourless solution.
Chemical Eqn is as follows: Pb2+ + 2OH - reversible reaction sign
Pb(OH)2
When added in excess,
Pb(OH)2 + 2OH- reversible reaction sign [Pb(OH)4]2-
For b), a green ppt soluble in excess aqueous NH3 to give a blue
solution.
Chemical Eqn: Ni2+ + 2OH- reversible reaction sign Ni(OH)2
When added in excess,
Ni(OH)2 + 6NH3 
[Ni(NH36)]2+ + 2OH-
For c), a blue ppt soluble in excess aqueous NH3 to give a deep
blue solution.
Chemical Eqn: Cu2+ + 2OH- reversible reaction sign Cu(OH)2
When added in excess,
Cu(OH)2 + 4NH3 [Cu(NH34)]2+ + 2OH-
For d), I look at my QA book then I manage to find out the ans.
lol
Blue solution turning to yellow solution as more concentratd HCl is
added.
Chemical Eqn: Cu(OH)2 + 2H+ Cu2+ + 2H2O
Cu2+ + 6H2O reversible reaction sign [Cu(H2O)6]2+
Cu(H2O)6 + 4Cl- CuCl42- + 6H2O
These 4 qns are about ligand exchange. When solution such as NaOH
is added into Ni2+ solution, the basic property of NaOH will react
with nickel solution causing ppt to be formed. Then, when excess
NaOH is added, ligand exchange occurs. As OH- ligand has more lone
pairs of electrons than H2O ligand, the pd-pd attraction of OH- is
stronger than H2O.
For the last qn, if you can give me the chemical formula of the
compound, it will really help me in finding out the structure of
the compound.
