Does exothermic solvation mean solute is more soluble at low temp?

Question

(I'm surprised this question or a very similar one hasn't been asked, so I wouldn't be surprised if I just failed to find a question with my search that does answer this question.)

I dissolved some $\ce{K3PO4}$ in water a while back, and the solution became very hot. I commented on this to my labmate, and he said "So that means it's more soluble at low temperature, right?" I evaded response because I disagreed but didn't have a thought-out response yet.

I think he thought of the reaction as: $$\ce{K3PO4 -> 3K^+ + PO4^{3-} + heat}$$

But I think that's missing an important step. I think it's more like this: $$\ce{K3PO4 -> 3K^+ + PO4^{3-}}$$ $$\ce{PO4^{3-} + H2O <=> HPO4 + OH- + heat}$$

Phosphoric acid's third pKa is pretty high, so the second reaction is pretty significant. This, I would say, is why this particular reaction is so exothermic. So, in this case, I don't really think the heat plays into the solubility of potassium phosphate.

Sometimes, though, solvation comes with a positive change in temperature. For instance, the enthalpy of solvation of sodium iodide is $\mathrm{-7.53\ kJ/mol}$. Sodium iodide, though, is still more soluble at high temperatures. I could be missing something here, though.

In the cases where it is specifically the solvation reaction that causes a temperature change, is it the true that lower temperatures always increase solubility of salts with exothermic solvation reactions (and vice-versa)?

I'd say that the answer is no because the solvation should be dependent on $\Delta H - T\Delta S$ and, since the enthalpy change half of that formula is temperature independent, as $T$ goes up, solvation should be favored by any solvation that has a positive change in entropy. However, I can't really think of any solvation that has a negative change in entropy.

Answer 1

Think of this in terms of equilibrium constant.

$$K=e^{-{\Delta G\over RT}}=e^{-{\Delta H-T\Delta S\over RT}}=e^{-{\Delta H\over RT}}\cdot e^{{\Delta S\over R}}$$

Now it is pretty clear that exothermic reaction means negative $\Delta H$, which in turn means positive coefficient in $-{\Delta H\over RT}$, which consequently is a decreasing function, as is the whole exponential expression. The entropy change (or indeed the sign of it) does not matter.

The example of $\ce{CaCl2}$ is irrelevant, for anhydrous $\ce{CaCl2}$ is never the phase in equilibrium with the solution. Roughly speaking, one simply cannot just up and dissolve $\ce{CaCl2}$. In effect, there would be two reactions: first the hydration, which produces $\ce{CaCl2.6H2O}$ (that's the exothermic part), and then the dissolution of that.

The implicit approximation we relied upon is that our $\Delta H$ and $\Delta S$ won't change much throughout the dissolution process. This fails when it comes to insanely well soluble things (think of $\ce{NaOH},\;\ce{KOH}$) where by the time we reach equilibrium, we might even have more solute than water. Naturally, the properties of such solution would be quite different from those of pure water.

Answer 2

No. Here is the data for the solubility of $\ce{CaCl2}$ from Wkipedia which dissolves exothermically.

     Temp    g/100 g Water
       0          59.5
      10          64.7
      20          74.5
      30         100
      40         128
      60         137
      80         147
      90         154
     100         159

So the solubility of $\ce{CaCl2}$ increases as the temperature increases even though the dissolution is exothermic.