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

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 HSO_{4}^{} and SO_{4}^{2} are identical?

Hopefully I am not too rusty at this...
I used the reactions:
IO_{3}^{} + 5I^{} + 6H^{+} :rarrow: 3I_{2} + 3H_{2}O
I_{2} + 2S_{2}O_{3}^{2} :rarrow: 2I^{} + S_{4}O_{6}^{2}
I calculated n(S_{2}O_{3}^{2}) = 0.000328984 mol
Therefore n(I_{2}) = 0.000164492 mol
I assumed the KI and KIO_{3} were both in excess and it was the acid that was limiting.
So n(H^{+}) = 0.000328984 mol
That means n(H_{2}SO_{4}) = 0.000164492 mol
Hows that so far?

I assumed the KI and KIO_{3} 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.

What was pH of the initial sulfuric acid solution, if at pH 2.00 concentrations of HSO_{4}^{} and SO_{4}^{2} are identical?
I don't see the significance of this statement :\
Is it used for finding the value of K_{a2}?

Correct line of thinking.

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?

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, HSO_{4}^{} is a relatively weak acid.
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!

Assume that the first proton completely dissociates, then 0.000164492 mol of H^{+} and HSO_{4}^{} forms. The calculated value of K_{a2} is 10^{2}. HSO_{4}^{} further dissociates into SO_{4}^{2} and H^{+}. Set up a variable x which is the amount of SO_{4}^{2} formed. Substitute into K_{a2} 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

(0.000164492 + x)(x)/(0.000164492  x)
These are moles, not concentrations?

Whoops, sloppy work. How about pH = 1.78 :X

Whoops, sloppy work. How about pH = 1.78 :X
Even sloppier.

:[ Ok, I think I finally have the correct answer, pH = 2.27.

:[ Ok, I think I finally have the correct answer, pH = 2.27.
2.27 it is ;D
CU on Monday.

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?

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.

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.
It seems that I had made a computation error before. :3

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 HSO_{4}^{}.
At pH 2, Ka2 = [10^2][SO4 2]/[HSO4], and the sulfate and hydrogen sulfate concentrations are the same, leaving Ka2 = 10^2.
Right.
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.