[Rutherford]

Yes.

How can this reaction: Pb(AcO)⁺+AcO^- Pb(AcO)₂  be neglected?

[Radu]

  Yes, it is almost fully protonated, but, again, we don't care about the actual concentration of Pb²⁺. We know that the total concentration of lead in solution is s= 2.9*10^-5=[Pb²⁺]+[Pb(AcO)]⁺] + [Pb(AcO)₂] + [Pb(OH)⁺]. We explicit every concentration from β-s ,and make the approximations: [AcOH]≈1M and [H₃O⁺]=cst=10^-2.38.
 

[Radu]

  Raderford, β=[Pb(AcO)₂]/ {[Pb²⁺][AcO^-]² = [Pb(AcO)₂]*[H₃O⁺]²/{[Pb²⁺]ka_AcOH[AcOH]².

[Rutherford]

I did it through the actual concentrations, it's pretty much the same, as I got the same answer as you.

I just don't know why should β(Pb(AcO)₂) represent the cumulative constant and not the step-wise one. They should have marked this with β₂ or β₁, respectively. And how can the reaction of the PbAcO⁺ and AcO^- be neglected?

[Radu]

Humm, I think it's not neglected, it is included in β₂( the cumulative one), which is β₁*k₂, where k₂ is the step-wise constant.
      I had this problem, too, this is why I got a wrong answer at the beginning, I had considered them as step-wise.

[Rutherford]

If it was step-wise there would be a little more calculation. I think that this is clear to me now. Thanks.

[Big-Daddy]

So how was it decided that the constant is cumulative, rather than step-wise?  I can't tell from the question.

Inter-conversion is easy but we need to know whether the constant for which the value is given refers to the cumulative or step-wise equilibrium.

[Rutherford]

I asked the organizers. They said that it will be revised soon.

[Radu]

 I looked through and I found a book where they say there is a kind of convention:
   - k-s for step-wise stabilty constants, K-s for step-wise instability constants, β-s for cumulative stability, β^-1 for instability cumulative constants.
    I know this is kind of foolish, because the inorganicians use opposite conventions, so they should have simply explicited them.

[Big-Daddy]

   - k-s for step-wise stabilty constants, K-s for step-wise instability constants, β-s for cumulative stability, β^-1 for instability cumulative constants.

Sounds good, but I think that a) 'k' can't be used for any equilibrium constant per standard procedure, it refers to rate constants explicitly b) 'K_f' may be for step-wise stability constant but I don't think it would be for an instability constant.

Honestly I don't think instability constants are really symbolized in any way - it's just (stability constant)^-1. It would make some sense if β was cumulative formation/stability, and K_f was step-wise formation/stability - I have seen that somewhere ...

[Rutherford]

I think that I read this from a IUPAC document (maybe the symbols were different, except β):
M^x++yL^- [ML_y]^x-y β_y=? where y represents the number of the ligands participating in the reaction. In our case, if y=1, we could understand it as a step-wise constant, but if y=2 then it would be cumulative. So β₁(Pb(AcO)₂) represents the constant of the reaction of one ligand and another molecule.

[Big-Daddy]

I think that I read this from a IUPAC document (maybe the symbols were different, except β):
M^x++yL^- [ML_y]^x-y β_y=? where y represents the number of the ligands participating in the reaction. In our case, if y=1, we could I see. So β₁(Pb(AcO)₂) refers to Pb(OAc)⁺ + AcO^-  understand it as a step-wise constant, but if y=2 then it would be cumulative. So β₁(Pb(AcO)₂) represents the constant of the reaction of one ligand and another molecule.

Pb(OAc)₂ whereas β₂(Pb(AcO)₂) refers to Pb²⁺ + 2AcO^-  Pb(OAc)₂. Thanks. I find K_f notation more common in IChO - K_f is definitely stepwise, e.g. K_f3 refers to binding from the form with 2 ligands to the form with 3 ligands.

[Rutherford]

But, after all, the organizers promised a revision soon, so we will see how they will formulate it then.