0
$\begingroup$

I have a reaction equation, it's all balanced but I don't understand what the physical change is. It is as follows:

$$\ce{2Na2S2O3 + KI3 \to 2Na2S4O6 + 2NaI + KI}$$

It was used to prove if $\ce{S2O3^2-}$ was in the solution or not. I'm not sure but I think it changes color to violet and back to colorless due to $\ce{I-}$ but if someone could definitively say what you are supposed to observe it would be helpful.

$\endgroup$

closed as off-topic by Mithoron, andselisk, Melanie Shebel, user55119, Todd Minehardt May 4 at 17:47

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ $\rm I_3^-$ is colored pretty heavily (though I wouldn't call that color violet); $\rm I^-$ is colorless. What's unclear about that? $\endgroup$ – Ivan Neretin May 1 at 21:48
1
$\begingroup$

The keyword you should search is iodometric titration. Triiodide solutions are dark brown, just like the tincture of iodine. If you were to add thiosulfate ion (say from a buret) to the triiodide solution would become colorless. The change is not that sharp. Therefore this is not the way end point is detected. A small drop of starch is added. Starch makes a deep bluish purple complex with free iodine. The moment all the iodine (in the form of triiodie $\ce{I2.I-}$ ) is consumed by thiosulfate, there is a very sharp transition from bluish purple to a colorless solution.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.