When you mix two liquids 50/50, and one has a freezing point of 20 Celcius and the other has a freezing point of 0 Celcius, does the new liquid have a freezing point of 10 Celcius?
In: 6
**ELI5:** No. Boiling points and freezing points are based on the amount of stuff in the more prominent liquid. The more stuff dissolved, the lower the freezing point will be. In your case of two liquids, the liquid that was present in lesser amount would be considered to be “dissolved” in the greater amount liquid. (assuming they’re miscible.)
**ELI’m Older:** Boiling and Freezing points (along with the oft forgotten about vapor pressure) are colligative properties. The equation that determines these things is pretty much all the same.
DT = K * m * i or, in words “The change in temperature of the freezing point (compared to the freezing point of the pure liquid) is the proportionality constant for that liquid (which you can look up in a table) times the concentration of whatever you’re dissolving in that liquid (in molality) times the van’t hoff factor for that thing(or how much the thing you’re dissolving splits apart.)
Basically, ignoring the constant because…. it’s a constant, the freezing point will decrease if you A: dissolve MORE of something or B: dissolve something that physically splits apart more.
For example. 1 molal CaCl2 will have a LOWER freezing point than 1 m NaCl because CaCl2 splits apart into 3 things (one Ca^(2+) ion and 2 Cl^- ions). But 2 m NaCl will have a LOWER freezing point than 1 m CaCl2 because there is literally 4 molal of “stuff” in solution (2 molal Na^+ and 2 Molal Cl^-) where as for 1 m CaCl2 there is only 3 molal of “stuff” in solution (1 molal Ca^(2+) and 2 molal Cl^-.)
Basically, colligative properties only depend on the amount of “stuff” in solution. They do not particularly care what that stuff is.
The van’t hoff factor is the way we get away from “ideal” materials. IDEALLY, 1 M (yes I’m switching to molarity) NaCl would produce 2 molar of “stuff” in solution. 1 molar Na and 1 Molar Cl. But in practice that’s not what we see. The more “stuff” that’s in solution (so the higher the concentration) the less more “stuff” wants to go into solution. So the van’t hoff factor for NaCl is like 1.9 instead of the idealized 2 because, on average, you’ll get 1.9 “things” when you dissolve NaCl in water (instead of the 2 like we’d expect.)
Worth noting that yes, van’t hoff factors DO change with concentration, and the 1.9 above is measured at 0.05 Molar (which is a pretty low concentration.)
**As for WHY?** Well, colligative properties are determined by intermolecular forces. (Forces that occur BETWEEN molecules.) The stronger the forces, the higher the boiling point/lower the freezing will be (because the molecules LITERALLY “hold onto” each other harder therefore making boiling more difficult and making rearranging into a solid more difficult.) When you dissolve something into something else, you often incur STRONGER intermolecular forces such as ion-dipole (which are the strongest intermolecular forces). That’s why it’s not just the pure liquids’ temps averaged. You literally have stronger forces now. The sum of the parts is greater than that of the whole is one way to put it. With the pure liquid you have no way to get ion-dipole forces, but when you dissolve something in it, you do!
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