# Can compounds dissolved in different miscible solvents react?

For example, can crystal iodine dissolved in ethanol react with sodium thiosulfate dissolved in water when mixed together?

• Yes, used widely in chemistry – Waylander Nov 17 '19 at 15:11
• As ethanol and water are miscible in any ratio, it is not the best example. Additionally, for nonmiscible solvents, compounds have distribution constants, so they are present in both solvents, even if in some cases, they prefer one solvent almost exclusively. – Poutnik Nov 17 '19 at 17:17
• @Poutnik, the question concerns miscible solvents, so the ethanol/water example fits. – DrMoishe Pippik Nov 17 '19 at 20:10
• I see, but I do not see a sense of the question, regarding miscible solvents. They mix themselves and compounds react. – Poutnik Nov 17 '19 at 20:13
• @sekharm, Why wouldn't they react, assuming at least some of both solutes stay dissolved after mixing? Try it. – DrMoishe Pippik Nov 17 '19 at 20:24

There are two ways to add two miscible solvents to each other. One is the bartender skill, whereby you attempt to create two different layers of solvents and then try to keep the flask motionless to prevent mixing. The other way is to mix them all along, whether by shaking, careless pouring, or a stirring bar. I will concentrate on the second as that is, in my opinion, the far more common use in chemistry.

Mixing to miscible solvents, even with compounds dissolved in them, results in a mixed solvent that for all intents and purposes of this question can be considered a new solvent. The question is no longer whether the individual solutes dissolve in their respective solvents but whether they dissolve in the mixed solvent. In many reactions, pouring the crude reaction mixture into a different solvent (miscible with the reaction solvent) serves as an easy way for the product to precipitate ready for isolation. For example, when xanthones are synthesised by dehydration in concentrated sulphuric acid, the crude mixture is then poured into water. Xanthones are not soluble in water but also not in water/sulphuric acid mixtures if the acidity is not sufficiently high so the xanthones precipitate and can be filtered off.

Obviously, there are three cases here: both compounds remain soluble, neither compound remains soluble and one of the compounds remains soluble. The first two cases are pretty trivial: if both compounds are soluble they can react as they would in a single solvent. Likewise, if neither compound is, they will both precipitate and not react (disregarding very slow, very minor solid state reactions if there are any relevant interfaces between the precipitates—but I digress).

The interesting case is the third case: what happens when one compound remains soluble but the other does not. Interestingly, with sufficient stirring there is typically enough contact surface exposed to molecules in solution that these two compounds will still react. Furthermore, precipitation doesn’t always mean entirely insoluble; the solubility might be low but sufficient to sustain a reaction. Thus, in most cases if one reactant is not sufficiently soluble, a reaction will still occur.

I will spare a few words on the bartender skill as there is an example reaction undergrad students may well perform as part of their lab courses: a reaction to test for the presence of nitrate. At first, nitrate is reduced by iron(II) salts in diluted sulphuric acid giving nitrogen oxide:

$$\ce{3Fe^2+ + NO3- + 4H+ -> 3Fe^3+ + NO + 2 H2O}\tag{1}$$

Then, the solution is layered with concentrated sulphuric acid whose greater density means it will drop to the bottom of the test tube. At the interphase, water is drawn into the sulphuric acid leading to the formation of pentaaquanitrosyliron(II) ions $$\ce{[Fe(H2O)5(NO)]^2+}$$ from unreacted iron(II). This complex has a dark-red, violet or brown colour and forms in a ring-like manner at the bottom of the water phase as shown in the image from Wikipedia. In German, this is known as the ring test.

$$\ce{[Fe(H2O)6]^2+ + NO <=>> [Fe(H2O)5(NO)]^2+ + H2O}\tag{2}$$

This shows how the compounds may still be able to react.