Your reaction is not balanced. Could be a typo.
I don't know why sodium thiosulfate only use one sodium to react with iodine.
Sodium? Sodium is just a spectator.
What is normality of 0.1M I2 solution?
Ya it was a typo
2 Na2S2O3 + I2 2 NaI + 2 Na2S4O6
the normality of Iodine solution is 0.1 N
What I was trying to do is finding the % Sodium Lauryl Sulfate experiement.
In the wet analytical chemistry book instructed me to first run a blank experiment. In a 50 ml water, add in 10 ml acetic acid, then add 10 ml of 0.1 N iodine solution. (I think the acetic acid just making the water acidic, so the iodine would not react with water). Then titrate the solution with 0.1 N of Na
2S
2SO
3.
Then in the second run. Add 2.0 g of sodium lauryl sulfate (SLS) solution in the water/iodine/acetic acid solution. Suppose the SLS's sodium will react with the iodine in the solution before the titration. Therefore, we should actually use less of 0.1 N Na
2S
2SO
3 to titrate the iodine solution.
I was going to run a couple standards of 5%, 10%, 15%, 20%, and 30% of SLS solution, then make the calibration curve to compare my sample.
But now i am running into some quesitons.
1. How do I get the 0.1 n of Na
2S
2SO
3 solution? The F.W. is 158.11 g/mol. I want to make into 500mL.
So I was using 0.5 * 0.1 = 0.05 mol. Then I don't really know why in this case, everyone said 0.1 N of Na
2S
2SO
3 equals to 0.1 M Na
2S
2SO
3.
2. Since I really am just making a calibration curve, I suppose that even if I miscalculated the concertration of Na
2S
2SO
3, I would still be okay because I am just using the amount of my Na
2S
2SO
3 solution need to titrate the left over iodine in the solution after it react with SLS. However, I still can't get the curve right...