\begin{align} \ce{NaOH &-> Na+ + OH-} & \Delta H^\circ &= \pu{-44.51 kJ/mol} \end{align}
The dissolution of sodium hydroxide in water is an exothermic process, and so, according to Le Chatelier’s principle, cooling the container should shift the reaction to the right. Shouldn’t this mean that cooling the container will increase the solubility?
According to a solubility chart for sodium hydroxide, heating the container will increase the solubility; why is this the case?