# 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 probably have the solution of potassium iodide in excess for the oxidation. $\ce{CuI}$ is somewhat soluble in in iodide solutions (not in iodine solution), due to common ion effect (see Wikipedia for solubility). – Mathew Mahindaratne Jan 8 at 17:14
• @MathewMahindaratne Yes, that makes sense! I can't believe I forgot about the excess potassium iodide. It's the small things that always end up being the key. Can you post your comment as an answer so I can accept it? – Lightrogen Jan 8 at 20:57

I agree with one of answers provided that what you obtained at the end point is a colloidal solution of copper(I) iodide. Since $$\ce{CuI}$$ is insoluble in water (0.0042% w/v), the stark white color (probably milky) solution is nano particles of the solid suspended in $$\ce{KI}$$ solution, assuming you have used significantly excess $$\ce{KI}$$ for the reaction. Thus your solution should be in following equilibrium: $$\ce{CuI + I- <=> [CuI2]-}$$

Wikipedia states that:

$$\ce{CuI}$$ is poorly soluble in water (0.00042 g/L at 25 °C), but it dissolves in the presence of $$\ce{NaI}$$ or $$\ce{KI}$$ to give the linear anion [$$\ce{[CuI2]-}$$. Dilution of such solutions with water reprecipitates $$\ce{CuI}$$. This dissolution–precipitation process is employed to purify $$\ce{CuI}$$, affording colorless samples.

Thus, your intention is to isolate solid $$\ce{CuI}$$, you must add extra $$\ce{KI}$$ while moderate heating your solution until you receive a clear solution. Let the solution to cool to room temperature, and add adequate amount of pure water (about 4 times by volume) to cause the precipitation. Upon standing for about $$\pu{1 h}$$, $$\ce{CuI}$$ will precipitate.

A typical procedure is given below in literature (Ref. 1):

Commercial copper(I) iodide or a sample that has become discolored ($$\pu{3.80g}$$, $$\pu{0.02 mole}$$) placed in a 250-mL Erlenmeyer flask is dissolved with magnetic stirring and moderate heating by the addition of a minimum volume (about $$\pu{125 mL}$$) of $$\pu{ 3.5 M}$$ potassium iodide (about $$\pu{3 hr}$$ is required). Activated charcoal (e.g., Matheson, Coleman, and Bell "Darco" (3-60) (about $$\pu{5 g}$$) is added to the resulting solution, which varies in color from light tan to dark brown, depending upon the extent of contamination. The mixture is stirred magnetically with moderate heating until the supernatant solution becomes colorless, at which time stirring is stopped and the charcoal is allowed to settle, indicating that the contaminating iodine has been completely adsorbed (about $$\pu{10 min}$$). The mixture is suction-filtered through a 150-mL (60-mm i.d.) sintered-glass funnel (medium porosity), and the filtrate is diluted with sufficient water (about $$\pu{500 mL}$$) to cause reprecipitation of the copper(I) iodide upon standing for about one hr.

Important: Vigorous heating in air causes $$\ce{CuI}$$ to decomposes to $$\ce{CuO}$$.

Ref. 1: Inorganic Syntheses, Volume 22; Smith L Holt, Jr., Ed.; Inorganic Syntheses, Inc.: New York, NY, 1983, Chapter 2: Transition Metal Complexes and Compounds (sub Title: 20. Purification of Copper (I) Iodide; submitted by G. B. Kauffman & L. Y. Fang; Checked by N. Vishwanathan & G. Townsend), pp 102–103 (https://doi.org/10.1002/9780470132531.ch20).

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.