Question

Order the following species from the one which binds lead (II) ions least strongly to the one which (at this pH) binds them most strongly and leaves the smallest equilibrium concentration of lead ions in solution: chloride ion, chromate ion, EDTA ion, water molecule. Please explain the logic behind the order in which these were placed. I initially went by solubility but it does not seem to quite fit what the question is asking. Thanks in advance.

Answer 1

Reaction of lead ions with

chloride ion, chromate ion, EDTA ion, water molecule.

which binds lead (II) ions

EDTA would bind to lead ion most strongly as it will form a chelated compound its binding would be strongest

Now both chloride and chromate ions will react with lead Chromate is 2- charged and chloride is 1- charged.

Also the Ksp of Lead chromate is much smaller than lead chloride.

Lead(II) chloride PbCl2 Ksp =1.6×10–5

Lead(II) chromate PbCrO4 Ksp = 2.8×10–13 Since it is more negatively charged and also its solubility product Ksp is smaller tendency for the chromate to react with lead is higher as formation of a precipitate drives this reaction. Water does not react with lead ion as for example lead nitrate solution is formed in water. So the order would be from the one which binds lead (II) ions least strongly to the one which binds them most strongly water molecule, chloride ion, chromate ion, EDTA ion