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Iodine Clock Reaction Question 
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#1
Sep1010, 12:16 AM

P: 93

1. The problem statement, all variables and given/known data
Find the concentration of I^{} involved in the Iodine Clock Reaction 2. Relevant equations I think this will involve m_{1}v_{1}=m_{2}v_{2}  but I am not sure. 3. The attempt at a solution We did nine 'iodine clock' reactions to demonstrate chemical kinetics. I'll use the second one as an example. We added 7.50mL of 0.200 M KI, 2.50ml of 0,200 KCl, and 5.00mL of 0.0050 S_{2}O_{3}^{2}, and one drop of starch to one flask. We then added 5.00mL of S_{2}O_{8}^{2} to another flask. We then mixed them together in a beaker. The total amount of solution is 20mL. It turns dark blue/black in about a minute. I need to figure out the concentration of I^{}. I understand that the reaction I need to investigate is: 2I^{}(aq)+S_{2}O_{8}^{2}(aq) [tex]\rightarrow[/tex]I_{2}+2SO_{4}^{2} I'm not sure how I can get the concentration from this. I know how to get the concentration of KI, but I don't know how to get only the I^{}. Again, I think it has to do with the equation listed above, but I'm not sure. 


#2
Sep1010, 01:09 AM

P: 93

I've read up more on this and think I have my answer.
KI is ionic, so in a solution the salt will turn into K^{+} and I^{}. Since KI is one to one, if 0.200 M of KI, that means I have 0.200 M of I^{}. Is this correct? 


#3
Sep1010, 04:05 AM

Admin
P: 23,393

You have 0.200M I^{} in the solution that you mixed with others  which is (from the pov of I^{}) just a dilution.
And yes, dilution can be calculated with C_{1}V_{1} = C_{2}V_{2}. See http://www.chembuddy.com/?left=conce...ilutionmixing. You will just need final volume, which can be easily calculated assuming volumes are additive. In general they are not, but differences are negligible in this case. 


#4
Sep1010, 02:00 PM

P: 93

Iodine Clock Reaction Question
Yes, that makes sense. Yesterday made for a nice refresher course in dilutions and ionic solutions.
I have another question concerning the Iodine Clock; I don't think the lab instructor was entirely clear on getting the rate from the timing agent. He said that in order to find the rate for I^{}, use the following equation [tex]\Delta[/tex][S_{2}O_{3}^{2}] / [tex]\Delta[/tex]t I understand that our initial concentration for thiosulfate is always the same at 1.25x10^{3} and that the final concentration will always be zero since it's used up. However, don't the coefficients need to be considered? The equation in question is: I_{2}(aq)+2S_{2}O_{3}^{2}(aq)[tex]\rightarrow[/tex]2I^{}+S_{4}O_{6}^{2}(aq) That tells me that the rate for thiosulfate should be multiplied by 1/2 and the iodine should be multiplied by 1. For example, my first iodine clock reaction took 50 seconds for the starchiodine complex to appear. I started with 1.25x10^{3} concentration of thiosulfate. Using the rate equation, I end up 1.25x10^{3} and divide that by 50 since that's how many seconds the reaction took. That gives me an answer of 2.5x10^{5}, but then I must multiply this by 1/2 because the coefficient of the thiosulfate in the chemical reaction is 2. This gives me 1.25x10^{5}. I then multiply this by 50 (since this equals the change of concentration of iodine divided by the change in time which is 50 seconds) and that will give me the rate for iodine. Am I on the right track? I want to be sure, because I have eight other reactions to calculate and graph. 


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