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Topic: Problem of the week - 06/02/2012  (Read 13960 times)

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Offline Borek

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Problem of the week - 06/02/2012
« on: February 06, 2012, 06:17:45 AM »
Perhaps a simpler one than the last week.

Around 0.1 g of potassium iodate and 0.1 g of potassium iodide were added to  50.0 mL solution of sulfuric acid. The resulting yellowish solution was titrated with a 0.04012M sodium thiosulfate solution. The endpoint was reached after addition of 8.20 mL of the titrant. What was pH of the initial sulfuric acid solution, if at pH 2.00 concentrations of HSO4- and SO42- are identical?
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Offline UG

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Re: Problem of the week - 06/02/2012
« Reply #1 on: February 08, 2012, 04:09:46 PM »
Hopefully I am not too rusty at this...
I used the reactions:
IO3- + 5I- + 6H+  :rarrow: 3I2 + 3H2O
I2 + 2S2O32- :rarrow: 2I- + S4O62-
I calculated n(S2O32-) = 0.000328984 mol
Therefore n(I2) = 0.000164492 mol
I assumed the KI and KIO3 were both in excess and it was the acid that was limiting.
So n(H+) = 0.000328984 mol
That means n(H2SO4) = 0.000164492 mol
Hows that so far?

Offline Borek

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Re: Problem of the week - 06/02/2012
« Reply #2 on: February 08, 2012, 04:52:20 PM »
I assumed the KI and KIO3 were both in excess and it was the acid that was limiting.

Yes. This is a less known fact, but iodometry can be used to determine amount of acid.
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Offline UG

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Re: Problem of the week - 06/02/2012
« Reply #3 on: February 08, 2012, 05:01:38 PM »
What was pH of the initial sulfuric acid solution, if at pH 2.00 concentrations of HSO4- and SO42- are identical?
I don't see the significance of this statement :-\
Is it used for finding the value of Ka2?

Offline Borek

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Re: Problem of the week - 06/02/2012
« Reply #4 on: February 09, 2012, 06:56:10 AM »
Correct line of thinking.
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Offline XGen

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Re: Problem of the week - 06/02/2012
« Reply #5 on: February 09, 2012, 09:31:43 PM »
I can not figure out why Ka and Ka2 are needed for sulfuric acid. Since you mention that it will be required I am most probably incorrect in my assumption that sulfuric acid completely and perfectly dissociates.

However, if we know the number of mols of H+ to be 0.00328984 mols, and we know the solution to be essentially 50 mL, why is the answer not -log(0.00328984/50), or about 4.18? This is definitely not correct because it does not involve the piece of information concerning [HSO4-] and [SO42-], but why?

Offline Borek

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Re: Problem of the week - 06/02/2012
« Reply #6 on: February 10, 2012, 03:00:45 AM »
I am most probably incorrect in my assumption that sulfuric acid completely and perfectly dissociates.

Good point. While assumption that first proton is completely dissociated holds even in highly concentrated solutions, HSO4- is a relatively weak acid.

Quote
However, if we know the number of mols of H+ to be 0.00328984 mols, and we know the solution to be essentially 50 mL, why is the answer not -log(0.00328984/50), or about 4.18?

Even assuming complete dissociation of sulfuric acid molarity of H+ would be not 0.00328984/50 - for two reasons. First, amount of H+ is not 0.00328984 moles, second, volume is not 50 L.

Come on folks, it is Friday - not much time till the next problem - and we are not done with this one yet!
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Offline UG

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Re: Problem of the week - 06/02/2012
« Reply #7 on: February 10, 2012, 03:23:50 AM »
Assume that the first proton completely dissociates, then 0.000164492 mol of H+ and HSO4- forms. The calculated value of Ka2 is 10-2. HSO4- further dissociates into SO42- and H+. Set up a variable x which is the amount of SO42- formed. Substitute into Ka2 expression.
10-2 = (0.000164492 + x)(x)/(0.000164492 - x)
Solve the quadratic equation to get x = 1.59332 x 10-4
So then n(H+) = x + 0.000164492 = 3.238244 x 10-4 mol
Concentration of H+ = 6.4765 x 10-3 mol L-1
pH comes out at 2.19 ;D

Offline Borek

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Re: Problem of the week - 06/02/2012
« Reply #8 on: February 10, 2012, 04:10:10 AM »
(0.000164492 + x)(x)/(0.000164492 - x)

These are moles, not concentrations?
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Offline UG

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Re: Problem of the week - 06/02/2012
« Reply #9 on: February 10, 2012, 04:15:56 AM »
Whoops, sloppy work. How about pH = 1.78 :-X

Offline Borek

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Re: Problem of the week - 06/02/2012
« Reply #10 on: February 10, 2012, 04:20:28 AM »
Whoops, sloppy work. How about pH = 1.78 :-X

Even sloppier.
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Offline UG

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Re: Problem of the week - 06/02/2012
« Reply #11 on: February 10, 2012, 04:26:26 AM »
:-[ Ok, I think I finally have the correct answer, pH = 2.27.

Offline Borek

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Re: Problem of the week - 06/02/2012
« Reply #12 on: February 10, 2012, 05:47:57 AM »
:-[ Ok, I think I finally have the correct answer, pH = 2.27.

2.27 it is  ;D

CU on Monday.
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Offline XGen

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Re: Problem of the week - 06/02/2012
« Reply #13 on: February 10, 2012, 03:35:50 PM »
Sorry if this is a trivial question, but I am not understanding what the volume would be to get the correct answer. I realize that the volume is indeed not 50 mL, as the result is a value around pH 2.17. However, this is a tenth of a pH too small. Could someone explain this to me?

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Re: Problem of the week - 06/02/2012
« Reply #14 on: February 10, 2012, 03:51:38 PM »
Please elaborate. Initial volume of the acid was 50 mL. Question asks about pH in this initial sample of the acid.

2.17 is a pH calculated with assumption that acid was 100% dissociated - it was not, that's why the pH was slightly higher.
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