[Rutherford]

I skipped quantum mechanics and problem 5 because of Kp unit ambiguity (waiting for organizers response). If someone wants to check with me the results of problem 6th: http://icho2014.hus.edu.vn/document/01-24-2014%20IChO46-Preparatory.pdf I got:
1. ²⁰⁴Pb
2. 21
3. pH=1.86 2.63
4. 16.7%
5. I need help here . I thought first to do two extractions, and I set up a system of equations and got negative result for the volumes . Then I tried with three successive extractions, but I had more unknowns than equations. What else to try?

[CrazyAssasin]

1.204Pb
2. I don't get this one, how we can calculate without knowing time
3. I get pH=2.63
4.16.6%
5. Have no idea

[Rutherford]

2. Time needs to be calculated. You need to set up a system of equations:
-the ratio of Pb isotopes 206 and 204 in nature;
-the ratio of Pb isotopes 206 and 204 in the zircon;
-the rate law of ²⁰⁶Pb production (it is produced from ²³⁸U decay);
Keeping in mind that the mass of ²⁰⁴Pb didn't change, from these equations you can calculate the years that passed, I got 1.93·10¹⁰y.
Then setting up the rate law for Uranium 238 decay and using uranium isotope ratio after the years passed, you can calculate the ratio of the isotopes at the time zircon formed. Lots of calculations in this problem.

3. I found mine mistake. Got the same result as you now.

[curiouscat]

5. I need help here . I thought first to do two extractions, and I set up a system of equations and got negative result for the volumes . Then I tried with three successive extractions, but I had more unknowns than equations. What else to try?

You are on the right track I think. In four extractions I'm getting 96.01% extracted. So should be OK.

It's messy bookeeping but roughly the expression I get is:

% extracted = 0.5  / n × 10 × 0.01 ×{1/(10 × 0.5 / n +1)+1/(10 × 0.5 / n +1)²+ 1/(10 × 0.5 / n +1)³ + 1/(10 × 0.5 / n +1)⁴} / 0.01 × 100

n being number of extractions. In the summation use n terms. I've shown 4 here since n=4 worked.

Sorry, too lazy to use Latex. If you are stuck, just ask.

[Rutherford]

Wow, I gave up after three . I did my attempts through volumes, not number of extractions, so I am unfamiliar with the expression you wrote. Could you explain the logic behind it?

[curiouscat]

Wow, I gave up after three . I did my attempts through volumes, not number of extractions, so I am unfamiliar with the expression you wrote. Could you explain the logic behind it?

How much % extracted did you get for 1, 2 and 3 extractions?

[Rutherford]

For 1 83.3%, for 2 I assumed that it is possible to get 96% and I got negative volume, so it isn't possible, for 3 I got more equations than unknowns.

[curiouscat]

For 1 83.3%, for 2 I assumed that it is possible to get 96% and I got negative volume, so it isn't possible, for 3 I got more equations than unknowns.

Note, I could still be wrong. 

[Rutherford]

Hope you aren't. As you said, you got 96%, but I think that volumes of the organic solvent in each extraction have to be mentioned, because they asked us to propose a scheme. I don't think that only 4 extractions without the volume data would be enough.

[curiouscat]

For 1 83.3%, for 2 I assumed that it is possible to get 96% and I got negative volume, so it isn't possible, for 3 I got more equations than unknowns.

You are doing it the reverse way, I think. By putting % extraction exactly equal to 96% you overconstrain the problem. No wonder you get more equations than unknowns.

What you have is an inequality not a strict equality. Any extraction better than 96% is always ok.

Instead, for 2 extractions determine the % extracted. Then for 3. And so on till you exceed 96% at some point.

[curiouscat]

Hope you aren't. As you said, you got 96%, but I think that volumes of the organic solvent in each extraction have to be mentioned, because they asked us to propose a scheme. I don't think that only 4 extractions without the volume data would be enough.

Yeah, sure. I meant (implicitly) 4 extractions, equal volumed. i.e. 125 ml organic used for each extraction.

[Rutherford]

Okay, I got 96.2% using your way. Nice idea to use equal volumes in each extraction. Thank you very much for the help.

[curiouscat]

Okay, I got 96.2% using your way. Nice idea to use equal volumes in each extraction. Thank you very much for the help.

Sure. Glad to help.