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.

$$\ce{I_2 + 2Na_2S_2O_3 -&gt; 2NaI + Na_2S_4O_6} \tag{1}$$

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.