21
$\begingroup$

Some weeks ago one of my friend needed to use turpentine. As a good but ignorant chemist on that point I did some research about it. I found (at the beginning) in the French wikipedia page of turpentine :

"L'essence de térébenthine est un mélange de molécules variées, comprenant en particulier des terpéniques, des acides et des alcools. Ce liquide est insoluble dans l'eau, dans l'alcool absolu et dans l'éther, cependant il est légèrement soluble dans un mélange eau-alcool."

Which translated in English gives,

"Turpentine is a mixture of different molecules, particularly terpenes, acids, and alcohols. This liquid is insoluble in water, absolute alcohol, and ether, but it is slightly soluble in a mixture of water-alcohol."

I found no reference about that. Is it really possible? It sounds particularly strange to me. And obviously if that is possible, how?

$\endgroup$
  • 9
    $\begingroup$ An analogous situation might be the dissolution of metals in aqua regia. $\endgroup$ – a-cyclohexane-molecule May 24 '17 at 22:03
  • 2
    $\begingroup$ Remember that solubility and insolubility are nebulous terms. Even if a compound is insoluble a "tiny" amount will actually dissolve. So it would better to think of the compound being more soluble in a mixture of solvents than either solvent alone. $\endgroup$ – MaxW May 24 '17 at 22:27
  • 1
    $\begingroup$ @MaxW I know that. But I was not asking a question as a nitpick ^^ I mean at least measurable :) $\endgroup$ – ParaH2 May 24 '17 at 22:59
  • 2
    $\begingroup$ @a-cyclohexane-molecule Aqua regia isn't really an example of this. In that case, two different chemical reactions are taking place: nitric acid oxidizes a small amount of gold while HCl solubilizes the oxidized gold atoms. Chloroauric acid (the result of these two acting together) is soluble in water, which is the only solvent per se in the aqua regia. No chemical changes are occurring in the asker's question. $\endgroup$ – levineds May 24 '17 at 23:36
  • $\begingroup$ @Hexacoordinate-C - I'm an analytical chemist. You can "measure" extremely small amounts in solution. Much too small to directly weigh on a balance. $\endgroup$ – MaxW May 25 '17 at 16:07
16
$\begingroup$

The short answer is: yes, this is possible. Unfortunately, solubility is a fairly complex phenomenon to explain simply. Let's start with some examples where solubility is higher in a binary mixture than either solvent alone.

For a solid-liquid-liquid example: phenanthrene-cyclohexane-diiodomethane.[1]
For a liquid-liquid-liquid example like the one you were asking about: trifluoromethane-tetrafluoromethane-ethane at low temperatures and high pressures.[2]

In the Hildebrand Theory of solutions, there is a quantity called the "internal pressure", loosely how much these compounds want to break apart. In this model, this is usually something like the molar energy of vaporization divided by the molar volume. If the internal pressure of the solid is between the internal pressures of the liquids (as in this case), the solid is more soluble in the mixture than in either one individually. Another way to asking your question is about phase separation in ternary mixtures. You can find many examples of curved lines in ternary phase diagrams which are due to the ternary mixture not being the sum of its parts.

Qualitatively, you could imagine that some solvent solvates some parts of the molecule well and other poorly and vice versa for the other solvent. Hence, together they solvate better than each individually.

  1. Gordon, L. J.; Scott, R. L. J. Am. Chem. Soc. 1952, 74 (16), 4138–4140. DOI: 10.1021/ja01136a054
  2. Paas, R.; Peter, K. H.; Schneider, G. M. J. Chem. Thermodyn. 1976, 8 (8), 741-747. DOI: 10.1016/0021-9614(76)90053-7
$\endgroup$
  • $\begingroup$ Looking up Hansen parameters may be more relevant than Hildebrand for this question $\endgroup$ – Beerhunter May 24 '17 at 23:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.