Consider the following reactions. (Hot solution)PbCl2 + K2CrO4 A + 2KCl A + NaOH B + Na2CrO4 PbSO4 + 4CH3COONH4 (NH4)2SO4 + X In the above reactions, A, B and X are respectively.
- Na2[Pb(OH)2], PbCrO4 and [Pb(NH3)4]SO4
- Na2[Pb(OH)2], PbCrO4 and (NH4)2[Pb(CH3COO)4]
- PbCrO4, Na2[Pb(OH)4] and (NH4)2[Pb(CH3COO)4]
- PbCrO4, Na2[Pb(OH)4] and [Pb(NH3)4]SO4
Answer
(3) PbCrO4, Na2[Pb(OH)4] and (NH4)2[Pb(CH3COO)4]
Solution
PbCl2 is soluble in hot water; with K2CrO4 it gives a yellow precipitate of lead chromate:
PbCl2 + K2CrO4 PbCrO4 + 2KCl
A = PbCrO4 (yellow)
PbCrO4 dissolves in NaOH because Pb2+ forms the soluble plumbite complex:
PbCrO4 + 4NaOH Na2[Pb(OH)4] + Na2CrO4
B = Na2[Pb(OH)4]
PbSO4 (white) dissolves in ammonium acetate forming a soluble acetato complex:
PbSO4 + 4CH3COONH4 (NH4)2[Pb(CH3COO)4] + (NH4)2SO4
X = (NH4)2[Pb(CH3COO)4]
Answer: (3)