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### Topic: 9th problem IChO-analytical chemistry  (Read 9387 times)

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##### 9th problem IChO-analytical chemistry
« 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+ AcOPb+ + H+ , lgβAcOPb+=2.68;
2 AcOH + Pb2+ Ac(OPb)2 + 2H+
lgβ(AcO)2Pb=4.08
Pb2+ + H2O PbOH+ + H+
lgβ3=-7.8
CrO42- + H2O 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?

« Last Edit: February 10, 2014, 07:11:37 PM by Borek »

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##### Re: 9th problem IChO-analytical chemistry
« Reply #1 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?

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##### Re: 9th problem IChO-analytical chemistry
« Reply #2 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.

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##### Re: 9th problem IChO-analytical chemistry
« Reply #3 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.

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##### Re: 9th problem IChO-analytical chemistry
« Reply #4 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+.

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##### Re: 9th problem IChO-analytical chemistry
« Reply #5 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)+  Pb(AcO)2) rather than cumulative (Pb2+ + 2 AcO-  Pb(AcO)2)?

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##### Re: 9th problem IChO-analytical chemistry
« Reply #6 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.

#### Rutherford

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##### Re: 9th problem IChO-analytical chemistry
« Reply #7 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.

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##### Re: 9th problem IChO-analytical chemistry
« Reply #8 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.

#### Rutherford

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##### Re: 9th problem IChO-analytical chemistry
« Reply #9 on: February 28, 2014, 12:10:12 PM »
What about CH3COO-? Did you that [HCrO4-]=const=2.9·10-5M?

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##### Re: 9th problem IChO-analytical chemistry
« Reply #10 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.

#### Rutherford

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##### Re: 9th problem IChO-analytical chemistry
« Reply #11 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 .

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##### Re: 9th problem IChO-analytical chemistry
« Reply #12 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.

#### Rutherford

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##### Re: 9th problem IChO-analytical chemistry
« Reply #13 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- Pb(AcO)2 neglected?

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##### Re: 9th problem IChO-analytical chemistry
« Reply #14 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).