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.


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  • $\begingroup$ $\rm I_3^-$ 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

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.


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