Here's the simplest picture of why solutes affect the melting temperature, T
m. First, we should explain that T
m is the temperature at which the net entropy (system plus environment) of the solid and liquid states is the same. (Entropy is a measure of how many different quantum states are available to the system.) The entropy lost as the atoms or molecules line up to form a solid is just balanced by the entropy gained as the heat released warms up the environment at T
m. Adding solutes to the liquid causes more entropy loss when the solid starts to form, because the solutes have less room to run around in the remaining liquid than they had before. So that favors keeping the liquid state, and therefore the solid doesn't form until T
m is reduced enough to compensate.
Notice that this effect depends only on how the available liquid space is reduced as the solid forms. To be more specific, entropy is the logarithm of the number of available states. If you take away 1% of the volume a particle had, its entropy will go down by ln(0.99) (times a conventional constant in some units). The type of particle doesn't matter at all.
This argument turns out to be exact so long as the solute particles are very dilute. Once the particles are concentrated enough to interact significantly with each other, some entropy and energy changes from the changing interactions can also have an effect, changing T
m one way or the other. For uncharged solutes, the interactions tend to be weak and the effect on the melting temperature has hardly any dependence on the type of solute, just the concentration of particles. For salts, I exaggerated a bit before, since the ions do interact significantly even at moderate concentrations. Thus you only get about the same effect for different salts at the same concentration. In particular, divalent ions (e.g. Mg
+2) interact more strongly than monovalent ones (e.g. Na
+), so say MgCl
2 and NaCl will have different effects for the same number of total ions.
(Now that you ask, I've gone back to soften the wording in the previous answer.)
Mike W.
(published on 04/15/13)