# Solubility increases as Temperature increases.. Ks increases with temperature meaning dissolving is endothermic… But what about NaOH?

Question: I was self-studying and came across some confusion across a question. My confusion lies in this: Solubility increases as temperature (T) increases, Ks increases with T, meaning dissolving is endothermic. But what about $\ce{NaOH}$?

The problem I am putting forward is more to do with the equilibrium Ks, especially using the Le Châtelier's principle. We all know that the solubility of ionic substances increases as increases in temperature. That is a fact, we don't have to argue with that. We also know that dissolving $\ce{NaOH}$ is exothermic, so we don't need to discuss that. However the two statements (observations) contradict the Le Chatelier principle, because if the dissolution of $\ce{NaOH}$ is exothermic, the solubility should decreases with increase in temperature! Why doesn't it?

• You should also consider self-dissociation whe considering hydroniums and hydroxides. – Caprica Aug 11 '16 at 9:11

While this is frequently the case, there are plenty exceptions, and only one exception is required to prove the statement false. Lets consider the solubility of $\ce{Ca(OH)2}$.
$\begin{array}{rr} \pu{0^oC}: & \pu{1.89g/L}\\ \pu{20^oC}: & \pu{1.73g/L}\\ \pu{100^oC}: & \pu{0.66g/L} \end{array}$
$\ce{Li2SO4}$ is another example that follows this trend of decreasing solubility with increasing temperature, and there are plenty others.
Although the heat of a chemical reaction often follows Le Chatelier's principle, there are exceptions. Based on the facts as laid out above and in the OP, the dissolution of $\ce{NaOH}$ is a case in which Le Chatelier's principle does not apply to the heat of the reaction; the dissolution of $\ce{NaOH}$ is exothermic, yet it's solubility increases with temperature.