Determination of the iodine value of a fatty oil: Does the concentration of Na2SO3 effect the iodine value? I want to determine the iodine value of a fatty oil. For this I dissolve $\ce{1.0 g}$ of the oil in chloroform. Iodine bromide is then added and placed in a dark room for at least $\pu{30 min}$. After this, $\ce{KI}$ is added. Then, I am going to titrate with $\pu{0.1 M}$ $\ce{Na2SO3}$. Based on the added volume, I can calculate the iodine value of the unknown fatty oil.


Now I was wondering what the effect should be using $\pu{0.02 M}$ $\ce{Na2SO3}$ instead of $\pu{0.1 M}$? And is there a reason why the solution has to stay $\pu{30 min}$ in a dark room? I think there's no difference in outcome when it will be in a dark room, for example, say $\pu{3 h}$.
The reaction of $\ce{I2}$ or $\ce{IBr}$ with double bonds is slow. You have to wait half an hour for the reaction to be finished. But of course, you may wait a longer time, maybe $\pu{3 h}$ as you suggest. Or one night. Usually we are in a hurry, and we are looking for ways of avoiding to spend too much time for a given operation. That is why the procedure recommends half an hour. 

And of course you may carry out the titration with a solution 5 times more diluted than recommended. The only drawback is that you may have to refill many times your burette before going to the end of the titration. Suppose $\pu{42 mL}$ of the $\pu{0.1 M}$ solution is required for performing the titration. This could be done with a usual $50$-$\pu{mL}$ burette. With a $\pu{0.02 M}$ solution, you will use $5 \times \pu{42 mL} = \pu{210 mL}$. In this case you will have to refill four times your burette. Four times! Bad luck!