I investigated two mixtures with different solvents, one with water and one with n-heptane. Both contained iodine $\ce{I2}$ as a solute. To both solutions I added a bit of starch.

As I remember this resulted in a colourchange. So the solution turned from yellowish to dark blue (if I remember correctly!).

Now according to wikipedia starch and iodine indeed form a structure which has a dark blue colour. But it only forms in the presence of $\ce{I^-}$.

This leaves me wondering, why do I remeber the solution to be dark blue, eventhough I think there was no $\ce{I^-}$ present? Could it be the solution turned dark blue only after I added some sodium thiosulfate? Because in the next step I did a titration with $\ce{Na2S2O3}$.

In this case I don't see which reaction could have produced the $\ce{I^-}$ though. I thought only $\ce{NaI}$ is produced after adding the sodium thiosulfate.

$$\ce{I_2 + 2Na_2S_2O_3 -> 2NaI + Na_2S_4O_6} \tag{1}$$

So at which point did the solution turn dark blue and where did the $\ce{I^-}$ come from, that was needed for the formation of the starch-iodine-compound? Could it be there is an intermediate step to (1) in which $\ce{I^-}$ is formed and this $\ce{I^-}$ was used to produce the dark blue starch-iodine compound?


I don't think your memory is serving you right. That is why we write everything in the notebook, especially color changes.

I think you are doing distribution experiments where iodine is distributed between aqueous layer and an organic layer. When we add indicator for titration, it is not a solid starch but starch which is boiled in water. So when you added starch $solution$ to heptane which contained iodine, I would not be surprised if the starch solution turned blue.

Remember that iodine is strong oxidizing agent as well. A very small fraction of it can easily convert into iodide. You really really need a trace of the triiodide ion to form a dark blue iodine complex.

  • $\begingroup$ Your assumptions are correct. This is my first chemistry lab. And yes I should've wrote everything down more carefully. One question for clarification: You think the Iodine interacted with the sodium thiosulphate, forming some $\ce{I^-}$ which then lead to the reaction $\ce{I^-}+\ce{I_2}+\textrm{starch}\leftrightarrow\textrm{dark blue starch}$? And when adding more and more thiosulphate all of the $I_2$ and consequently all of the dark blue starch reacted to the colourless $\ce{I^-}$? $\endgroup$ Apr 18 '20 at 7:23
  • 1
    $\begingroup$ Right, this is what I think happened in your case. $\endgroup$
    – M. Farooq
    Apr 18 '20 at 13:04

The molecular iodine $\ce{I_2}$ is poorly soluble in water : maximum $0.0011$ M. If starch is added to this solution, the iodine will react with starch and the solution is dark blue. In the lab, this experiment is rarely done with simple $\ce{I_2}$ solutions, because the solutions to be titrated are usually more concentrated than $0.001$ M. Usually $\ce{I_2}$ is dissolved in $\ce{KI}$ solutions, producing $\ce{KI_3}$ or $\ce{I_3^-}$ ions.$$\ce{KI + I_2 <=> KI_3}$$ The "solubility" of $\ce{I_2}$ as combined in $\ce{KI_3}$ is at least $1000$ times higher than $\ce{I_2}$ in water. So in the presence of $\ce{KI}$ in solution, more $\ce{I_2}$ can stay in solution. And if some starch is added to a $\ce{KI_3}$ solution, it will produce a dark blue-black color, due to the small amount of free $\ce{I_2}$ in the $\ce{KI_3}$ solution.

When titrating either $\ce{I_2}$ or $\ce{KI_3}$ by adding thiosulfate ions $\ce{S_2O_3^{2-}}$, the free $\ce{I_2 }$ is consumed. $$\ce{I_2 + 2 S_2O_3^{2-}-> S_4O_6^{2-} + 2 I^-}$$

$\ce{I_2}$ is consumed by adding $\ce{S_2O_3^{2-}}$. But as the equilibrium $\ce{KI + I_2 <=> KI_3}$ is rapid, new $\ce{I_2}$molecules are continuously regenerated from $\ce{KI_3}$, so that the starch solution stays dark blue up to the end of the titration. Apparently, the titration proceeds as if the solution of $\ce{KI_3}$ is a solution of $\ce{I_2}$.


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