What explains the stark white colour of my iodometric titration analyte solution? During my Advanced Higher (secondary school) investigation, I reacted pure copper with concentrated nitric acid to acquire a copper nitrate solution. This was neutralised with excess sodium carbonate and brought back to a clear solution with ethanoic acid. 

I then reacted the solution with potassium iodide in a conical flask before titrating it with sodium thiosulfate.:

$\ce{2Cu^2+(aq) + 4I^-(aq) → 2CuI(s) + I_2(aq)}$

$\ce{I_2(aq) + I^-(aq) ⇌ I_3^-(aq)}$

$\ce{I_3^-(aq) + 2S_2O_3^2-(aq) → S_4O_6^2-(aq) + 3I^-(aq)}$

This analyte solution was at first brown, but when titrated it turned a pale coffee colour. When starch was added just before the end point, it turned a dark grey-brown colour, as expected - the iodine forming a complex with starch. https://en.wikipedia.org/wiki/Iodine_test

But when all of the iodine had reacted with the sodium thiosulfate, I was left with a stark white solution. I tested to make sure no iodine was left over by adding in another drop of starch - no reaction. So, this seems to be the endpoint. 

Copper iodide seems the likely explanation, but the wikipedia page shows that it is mostly insoluble in water. It is soluble in iodine solutions but all the iodine is gone, presumably. 

Does it take time for the copper iodide to precipitate? I quickly rinsed out the conical flask. There was some white staining on the sides which I think is copper iodide.
You’re correct that this is just a suspension of copper(I) iodide. If CuI were soluble in water, the solution would be clear; the white colour is a sign of (very small) particles of the solid suspended in solution.