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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 $\pu{3 h}$.

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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!

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  • $\begingroup$ I have one more question: why is the oil placed in a dark room? I understand that there will occur side reactions if not, but which side reactions exactly? Is it because of the formation of bromine radicals or will the iodide ions be oxidized to iodine? $\endgroup$ – Guest1 Dec 8 '19 at 8:12

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