# Titrating iodine starch solution with sodium thiosulphate - Colour change

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.

• 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 $\ce{I^-}$? Apr 18 '20 at 7:23
• Right, this is what I think happened in your case. 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 $$\ce{I_2}$$.