Inorganic answer 1
| The evolution of carbon dioxide on heating A suggests that it is a
carbonate, lead(II) carbonate; the residue B is therefore an oxide, lead(II) oxide: A à B: PbCO3 à PbO + CO2
B à C: 6PbO + O2 à 2 Pb3O4
C à D + E: Pb3O4 + 4HNO3 à 2Pb(NO3)2 + PbO2 + 2H2O.
D à F: Pb(NO3)2 + 2NaOH à Pb(OH)2 + 2NaNO3 or, ionically, Pb2+ + 2OH- à Pb(OH)2. F à G: Pb(OH)2 + 2NaOH à Na2Pb(OH)4 Or, ionically, Pb(OH)2 + 2OH - à
[Pb(OH)4]2-. Lead(IV) oxide is a strong oxidising agent and oxidises concentrated HCl to chlorine, J, and lead(II) chloride I. Lead(II) chloride complexes with chloride ions to give various species such as PbCl3- and PbCl42-, which are yellow and water-soluble: E à I + J: PbO2 + 4HCl à PbCl2 + 2H2O + Cl2. J à K: PbCl2 + Cl- à PbCl3- .
J à L: Cl2 + 2I- à I2 + 2Cl- I2 + I- à I3- .
D à M: Pb2+ + 2I- à PbI2 . |
© JRG Beavon 1999
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Dr Rod Beavon 17 Dean's Yard London SW1P 3PB
e-mail: rod.beavon@westminster.org.uk