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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Radu on February 04, 2014, 09:18:56 AM

Title: 9th problem IChO-analytical chemistry
Post by: Radu on February 04, 2014, 09:18:56 AM
    Hi,
 I have a problem(4, page 25): Calculate the product of solubilty of PbCrO4 , if its solubilty in 1M AcOH is 2,9*105.
    Here's how I dealt with it: Ks=S2αCrO42-αPb2+, where S is the aparent solubility, and α the coefficients, defined as the raport of the species it denote and the total concentration  (protonated, free, complexated etc) of the  species.
   I had to take into consideration the following equilibriums:
         AcOH + Pb2+ ::equil:: AcOPb+ + H+ , lgβAcOPb+=2.68;
        2 AcOH + Pb2+ ::equil:: Ac(OPb)2 + 2H+
 lgβ(AcO)2Pb=4.08
         Pb2+ + H2O ::equil:: PbOH+ + H+
 lgβ3=-7.8
       CrO42- + H2O ::equil:: HCrO4- + HO-.
           pKaHCrO4-=6.5,   we are also given pKaAcOH=4.76
           I did the matter balance:   S= [Pb2+]+[PbOH+] + [PbAcO+]+ [Pb(AcO)2] , and also S= [CrO42-]+[HCrO4-]. Assuming the ph stays constant(PH=2.38, 1M AcOH solution),
   I have obtained αCrO42-=7.5852*10-5.
                         αPb2+=1.44519*10-9.
    Ks≈10-23, when it actually is ≈10-14. Where am I wrong?

                         

 
         
Title: Re: 9th problem IChO-analytical chemistry
Post by: Big-Daddy on February 10, 2014, 04:45:31 PM
Out of interest, for part 3 did you get: PbC2O4 precipitates first, then Pb(IO3)2, then PbSO4, then PbI2, then PbCl2; the concentration of lead nitrate solution is around 2.44*10-5 M?
Title: Re: 9th problem IChO-analytical chemistry
Post by: Radu on February 10, 2014, 05:09:52 PM
   The order of precipitation is the same, but I have obtained a lead nitrate concentration of 0.02347M.
    νPb(NO3)2=0.02*20+0.001*10+0.005*20=0.51 mmoles.
           So, we have [Pb2+]=0.51/21.6=0.023M.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Big-Daddy on February 10, 2014, 05:23:16 PM
Regarding part 3: Why 0.001*10 instead of 0.001*20?

I have very little faith in this procedure. You're saying that firstly (1) all of each salt precipitating before PbI2, precipitates entirely before PbI2, which seems like an approximation; 2) after the previous three salts are entirely precipitated, precipitation of PbI2 begins instantly, rather than on addition of some other amount of Pb2+. It doesn't feel like this is very precise to me. I am interested to learn if there is an exact way to solve this question?

As to your original question, you can check your calculation by writing the 11 exact equations for the system and using an equation solver. But it does seem to me that it should be a good approximation that the pH of 1.0 M acid solution will not be affected too much by the presence of 2.9*10-5 M lead chromate(VI). If [H+] is correctly found from the acid alone then you can calculate chromate concentration as a function of the solubility pretty easily (it's just Ka*s/([H+]+Ka). Lead ion's concentration can be found similarly by substitution in terms of [Pb2+] into the mass balance for Pb. It is a long but doable problem.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Radu on February 10, 2014, 05:42:17 PM
   Firstly, because lead iodate precipitates using two iodate molecules per 1 Pb2+.
    Secondly, I have done some further calculations, and it is really accurate.
    Here's what I found:
         when PbI2 starts to precipitate, the concentrations of free oxalate and iodate are of 10-6 order, and that of sulphate is of 10-4 order, so basically all of them are precipitated.
       How I reached these numbers: KsPbC2O4 / KsPbI2 = [C2O42-] / [I-]2. PbI2 will start to precipitate when this raport of concentrations is obtained,  that is to say when [oxalate]=.; analogously for the others.
       PbCl2, moreover, doesn't co-precipitate with PbI2, through anologous calculations.

    I have arrived at the correct result lately, I had defined the stability constants wrongly, they include just AcO- and Pb2+.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Big-Daddy on February 10, 2014, 05:50:16 PM
   Firstly, because lead iodate precipitates using two iodate molecules per 1 Pb2+.
    Secondly, I have done some further calculations, and it is really accurate.
    Here's what I found:
         when PbI2 starts to precipitate, the concentrations of free oxalate and iodate are of 10-6 order, and that of sulphate is of 10-4 order, so basically all of them are precipitated.
       How I reached these numbers: KsPbC2O4 / KsPbI2 = [C2O42-] / [I-]2. PbI2 will start to precipitate when this raport of concentrations is obtained,  that is to say when [oxalate]=.; analogously for the others.
       PbCl2, moreover, doesn't co-precipitate with PbI2, through anologous calculations.

    I have arrived at the correct result lately, I had defined the stability constants wrongly, they include just AcO- and Pb2+.

So you observed that I- concentration must be much greater than the other anions' concentrations who have already mostly precipitated, by the point when PbI2 starts precipitating (since at that point Ksp of PbI2 as well as Ksp of all formerly precipitating anions are all met, since it is known that all of these have started precipitating). Looks good.

Ah. You mean the formation constants are stepwise (the second one is AcO- + Pb(AcO)+  ::equil:: Pb(AcO)2) rather than cumulative (Pb2+ + 2 AcO-  ::equil:: Pb(AcO)2)?
Title: Re: 9th problem IChO-analytical chemistry
Post by: Radu on February 10, 2014, 06:18:31 PM
  No, they are cumulative, if you'll look at what I initially wrote you'll see I had considered the equilibriums involving AcOH and liberation of protons and included them all in β-s. Actually β-s are for the equilibriums with AcO-, without any protons involved.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Rutherford on February 28, 2014, 11:44:08 AM
What did you get at the end? I got using the approximation that [H+]=const. Ksp=1.83·10-13.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Radu on February 28, 2014, 12:05:45 PM
  I actually got 1.99*10-14.
       I made the approximation that pH=cst=2.38 and [AcOH]≈cst≈1M.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Rutherford on February 28, 2014, 12:10:12 PM
What about CH3COO-? Did you that [HCrO4-]=const=2.9·10-5M?
Title: Re: 9th problem IChO-analytical chemistry
Post by: Radu on February 28, 2014, 02:27:01 PM
No, 2.9*10-5 is the total concentration of chromate in solution, that is to say [CrO42-] + [HCrO4-]. You don't actually need to calculate [HCrO42-] , you need to calculate αCrO42-, look at my initial work, you will see that we need only the fractions of free Pb2+ and CrO42- in solution.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Rutherford on March 01, 2014, 03:12:50 AM
I just wanted to say that at this pH the chromate is almost fully protonated to HCrO4-.

The expression for the constant of formation of Pb(OAc)2 is problematic for me.
β(AcO)2Pb=1.20·104=1/([Pb2+][CH3COO-]2), as [CH3COO-] is known from the dissociation of AcOH, and it is equal to [H+], then the concentration of [Pb2+] can be calculated, but it is unreasonably high. So I think that the concentration of AcO- isn't constant, therefore the pH isn't constant, too ???.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Big-Daddy on March 01, 2014, 07:43:41 AM
The expression for the constant of formation of Pb(OAc)2 is problematic for me.
β(AcO)2Pb=1.20·104=1/([Pb2+][CH3COO-]2), as [CH3COO-] is known from the dissociation of AcOH, and it is equal to [H+], then the concentration of [Pb2+] can be calculated, but it is unreasonably high. So I think that the concentration of AcO- isn't constant, therefore the pH isn't constant, too ???.

Why did you write it as 1? [Pb(OAc)2] should not have unity activity, because it will still be dissolved so then β(AcO)2Pb=1.20·104=[Pb(OAc)2]/([Pb2+][CH3COO-]2). Otherwise, how can you find [Pb2+] from this, as [Pb(OAc)2] is not known?

Lead acetate is not completely insoluble, and my guess is you assume it is always soluble for this problem. I'm sure the Ksp condition would be nowhere near to met for this case.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Rutherford on March 01, 2014, 08:23:58 AM
Yeah, that would give the correct answer. Thanks for the correction.
At what concentration we could assume activity 1?
How can this reaction: Pb(AcO)++AcO- :rarrow: Pb(AcO)2 neglected?
Title: Re: 9th problem IChO-analytical chemistry
Post by: Big-Daddy on March 01, 2014, 08:30:30 AM
Yeah, that would give the correct answer. Thanks for the correction.


Correct answer, meaning 1.99*10-14 for Ksp?

At what concentration we could assume activity 1?

We can't, unless concentration itself is 1 or the species is solid etc. (I don't know exactly what the conditions are yet - but it's the same thing as what you can leave out of the expression for the equilibrium constant or reaction quotient, e.g. [H2O] in aqueous solution-based equilibria, is left out from the expression, i.e. taken to be 1).
Title: Re: 9th problem IChO-analytical chemistry
Post by: Rutherford on March 01, 2014, 08:33:28 AM
Yes.

How can this reaction: Pb(AcO)++AcO- :rarrow: Pb(AcO)2  be neglected?
Title: Re: 9th problem IChO-analytical chemistry
Post by: Radu on March 01, 2014, 11:23:18 AM
  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.
 
Title: Re: 9th problem IChO-analytical chemistry
Post by: Radu on March 01, 2014, 11:29:41 AM
  Raderford, β=[Pb(AcO)2]/ {[Pb2+][AcO-]2 = [Pb(AcO)2]*[H3O+]2/{[Pb2+]kaAcOH[AcOH]2.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Rutherford on March 01, 2014, 11:57:28 AM
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?
Title: Re: 9th problem IChO-analytical chemistry
Post by: Radu on March 01, 2014, 12:16:15 PM
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.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Rutherford on March 01, 2014, 12:41:51 PM
If it was step-wise there would be a little more calculation. I think that this is clear to me now. Thanks.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Big-Daddy on March 02, 2014, 03:10:14 PM
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.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Rutherford on March 03, 2014, 05:55:53 AM
I asked the organizers. They said that it will be revised soon.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Radu on March 03, 2014, 10:13:58 AM
 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.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Big-Daddy on March 03, 2014, 01:39:03 PM
   - 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) 'Kf' 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 Kf was step-wise formation/stability - I have seen that somewhere ...
Title: Re: 9th problem IChO-analytical chemistry
Post by: Rutherford on March 03, 2014, 02:01:53 PM
I think that I read this from a IUPAC document (maybe the symbols were different, except β):
Mx++yL- :rarrow: [MLy]x-y βy=? 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.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Big-Daddy on March 03, 2014, 02:07:58 PM
I think that I read this from a IUPAC document (maybe the symbols were different, except β):
Mx++yL- :rarrow: [MLy]x-y βy=? 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-  ::equil:: 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-  ::equil:: Pb(OAc)2. Thanks. I find Kf notation more common in IChO - Kf is definitely stepwise, e.g. Kf3 refers to binding from the form with 2 ligands to the form with 3 ligands.
Title: Re: 9th problem IChO-analytical chemistry
Post by: Rutherford on March 03, 2014, 03:12:14 PM
But, after all, the organizers promised a revision soon, so we will see how they will formulate it then.