# Chemical Forums

## Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on February 06, 2012, 06:17:45 AM

Title: Problem of the week - 06/02/2012
Post by: Borek 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?
Title: Re: Problem of the week - 06/02/2012
Post by: UG 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?
Title: Re: Problem of the week - 06/02/2012
Post by: Borek 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.
Title: Re: Problem of the week - 06/02/2012
Post by: UG 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?
Title: Re: Problem of the week - 06/02/2012
Post by: Borek on February 09, 2012, 06:56:10 AM
Correct line of thinking.
Title: Re: Problem of the week - 06/02/2012
Post by: XGen 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?
Title: Re: Problem of the week - 06/02/2012
Post by: Borek 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!
Title: Re: Problem of the week - 06/02/2012
Post by: UG 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
Title: Re: Problem of the week - 06/02/2012
Post by: Borek on February 10, 2012, 04:10:10 AM
(0.000164492 + x)(x)/(0.000164492 - x)

These are moles, not concentrations?
Title: Re: Problem of the week - 06/02/2012
Post by: UG on February 10, 2012, 04:15:56 AM
Whoops, sloppy work. How about pH = 1.78 :-X
Title: Re: Problem of the week - 06/02/2012
Post by: Borek on February 10, 2012, 04:20:28 AM
Whoops, sloppy work. How about pH = 1.78 :-X

Even sloppier.
Title: Re: Problem of the week - 06/02/2012
Post by: UG on February 10, 2012, 04:26:26 AM
:-[ Ok, I think I finally have the correct answer, pH = 2.27.
Title: Re: Problem of the week - 06/02/2012
Post by: Borek 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.
Title: Re: Problem of the week - 06/02/2012
Post by: XGen 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?
Title: Re: Problem of the week - 06/02/2012
Post by: Borek 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.
Title: Re: Problem of the week - 06/02/2012
Post by: XGen on February 10, 2012, 08:51:50 PM
Let me see if I understand it correctly.

Taking the 0.000164492 mol of H+ and dividing it by 0.05 L gives 0.00328984 M of [H+] and 0.00328984 M for [HSO4-].
At pH 2, Ka2 = [10^-2][SO4 2-]/[HSO4-], and the sulfate and hydrogen sulfate concentrations are the same, leaving Ka2 = 10^-2.
Then, letting x = the amount of hydrogen sulfate dissociated, 10^-2 = [0.00328984 + x][ x]/[0.00328984 - x]. Solving, x = 0.00213308 M. Then, 0.00328984 + 0.00213308 = 0.00542292 M, or 2.266, which rounds to 2.27.

Title: Re: Problem of the week - 06/02/2012
Post by: Borek on February 11, 2012, 05:04:13 AM
Taking the 0.000164492 mol of H+ and dividing it by 0.05 L gives 0.00328984 M of [H+] and 0.00328984 M for [HSO4-].

Not exactly. You don't know these concentrations. All you can say at this moment is that analytical concentration of sulfuric acid is 0.000164492/2/0.05L. It doesn't say anything about equilibrium concentrations of H+ and HSO4-.

Quote
At pH 2, Ka2 = [10^-2][SO4 2-]/[HSO4-], and the sulfate and hydrogen sulfate concentrations are the same, leaving Ka2 = 10^-2.

Right.

Quote
Then, letting x = the amount of hydrogen sulfate dissociated, 10^-2 = [0.00328984 + x][ x]/[0.00328984 - x]. Solving, x = 0.00213308 M. Then, 0.00328984 + 0.00213308 = 0.00542292 M, or 2.266, which rounds to 2.27.

Right.