# Why does the solubility of LiCl increase with temperature if the dissolution is exothermic?

In an answer to another question, the following data on the solubility of $$\ce{LiCl}$$ at different temperatures is presented.

This suggests that dissolving LiCl is endothermic (according to the van't Hoff relationship). However, the enthalpy of dissolution has been measured and extrapolated to infinite dilution, and it is negative (the value given in one of the comments is $$\Delta H = \pu{-37 kJ/mol}$$, reference).

What is the reason for this discrepancy? Is it because $$\ce{LiCl}$$ is so soluble that there is some non-ideality effect? Maybe dissolving the first couple of molecules of $$\ce{LiCl}$$ is endothermic, but then when you dissolve more it becomes exothermic? Or is the activity of $$\ce{LiCl}$$ at increasing temperatures actually decreasing even though the concentration is increasing? Or is the van't Hoff relationship not a general one?

There is another question ($$\ce{NaOH}$$ solubility with temperature) which claims that Le Chatelier's principle sometimes does not apply, but it does not mention van't Hoff (and the answer might be incomplete).

So in short it's the opposite of #2: dissolving the first couple of molecules of $$\ce{LiCl}$$ is exothermic, but then if you dissolve more it become endothermic. The heat of dissolution to form an infinitely dilute solution is not the same as the heat associated with formation of a saturated solution.
Alternately you might say that the heat of dilution of $$\ce{LiCl}$$ is also exothermic (see e.g. Reference 1).
• For NaOH, that is definitely the case. As the concentration increases, the pH increases and the exothermic $\ce{H+ + HO- -> H2O}$ reaction ceases, leaving only the endothermic $\ce{NaOH(s) -> Na+(aq) + HO-(aq)}$ reaction. Anyone who has made a concentrated NaOH solution can tell you that it gets quite cold if not continually heated. I don't know the details of the LiCl dissolution. – Andrew Jun 3 '19 at 21:24