Hi, I have this qn to ask:
The identity of B is PbCl2. How do I obtain that answer?
Thanks in advance! :)
There's a missing piece of info in the question (probably a typo
error by the question setter), which is the addition of aqueous
sodium chloride in between the stages "colourless solution
(nitrate)" and "yellow ppt B".
Pb2+(aq) ions were present from the beginning. With excess
OH-(aq) available, the white ppt of Pb(OH)2 dissolves to generate
the soluble complex ion tetrahydroxoplumbate(II) ions due to
competing equilibria.
Pb2+(aq) + 2OH-(aq) <---> Pb(OH)2 (s)
Pb2+(aq) + 4OH-(aq) <---> [Pb(OH)4]2-(aq)
With excess OH-(aq), the position of equilibrium for the
2nd equation shifts to the right, causing the position of
equilibrium for the 1st equation to shift to the left.
Alternatively, both equations can be combined into :
Pb(OH)2 (s) + 2OH-(aq) <---> [Pb(OH)4]2-(aq)