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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: dgolverk on October 01, 2008, 02:43:30 PM

Title: Titration
Post by: dgolverk on October 01, 2008, 02:43:30 PM
Hi,
I'm struggling a bit with the next question. I am not sure what steps I should in order to solve it.
Calculate the molar concentration of a thiosulfate solution given that 25.00mL of 0.0195mol/L KIO3 solution in a flask containing 2.00g of KI and 10 mL of 0.500mol/L H2SO4 requird 34.81mL of thiosulfate solution to reach the starch endpoint.

So first of all I was thinking of writing a balanced equation, but when I tried to do that I couldn't find where the H2SO4 should be placed.
KIO3 + 5 KI + 3 H2SO4 = 3 K2SO4 + 3 H2O + 3 I2

I know that my colume for thiosulfate should be 34.81mL, and that I will eventually need to use C=nV. But I can't figure out how to put it all together.

I will be very thankful if someone can explain what I should do to solve it.
Thanks,
Title: Re: Titration
Post by: Borek on October 01, 2008, 03:43:22 PM
Google or search forums for iodometry.
Title: Re: Titration
Post by: dgolverk on October 01, 2008, 05:13:57 PM
Thanks for the input.
Unfortunately, I did not manage to find anything very useful.
I would be very glad if you can explain a little bit about the steps to solve this problem.
Thank you for your time.
Title: Re: Titration
Post by: Borek on October 01, 2008, 05:51:02 PM
So you have not tried hard enough. Iodometry is based on the reaction between iodine and thiosulfate. One of the ways of making iodine is in the reaction between iodate and iodide, you need strong acid to capture oxygen from iodate to form water. These reactions have been listed many times and in many places, both on forums and on the web.
Title: Re: Titration
Post by: dgolverk on October 01, 2008, 06:11:10 PM
Thanks for replying.
I did try to find equations, but this is not the problem.
I understand what you are sying about Iodometry, however according to the question H2SO4 should be present in the reaction, but I couldn't find anything that involves it. For example:
IO3- + 5I- + 6H+ -> 3I2 + 3H2O or
6S2O32- + 3I2 ->  6I- + 3S4O62-


Title: Re: Titration
Post by: Borek on October 01, 2008, 06:34:30 PM
Quote from: dgolverk on October 01, 2008, 06:11:10 PM
IO3- + 5I- + 6H+ -> 3I2 + 3H2O

Where do you take H+ from? SO42- is just a spectator.
Title: Re: Titration
Post by: dgolverk on October 01, 2008, 07:45:35 PM
Alright, so now I have
IO3- + 5I- + 6H+ -> 3I2 + 3H2O
2S2O3 2- + I2 -> S4O6 2- + 2I-

I read in few sites that the ratio between iodate and thiosulfate should be 1:6. But I don't understand where is the 6 coming from. Can you please explain it?
Thank you for your time.
Title: Re: Titration
Post by: Borek on October 02, 2008, 02:58:19 AM
Quote from: dgolverk on October 01, 2008, 07:45:35 PM
I read in few sites that the ratio between iodate and thiosulfate should be 1:6. But I don't understand where is the 6 coming from. Can you please explain it?

Just look at these reactions and stoichiometry coefficients. Assume you started with 1 mole of iodate.
Title: Re: Titration
Post by: dgolverk on October 02, 2008, 07:15:54 AM
So I know that for 1 mole of iodate ion I get 3 moles of I2. Looking at reaction 1. For 1 moles of I2 I get 2 moles of thiosulfate.
Therefore, when calculating it (2X3=6). I think I got it.
However, only one question: Should I write the 6 thoisulfate in my final equation? or just leave it as part of my calculations?
Thanks,