\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?
Answer
The question to the Newton - Ask a Scientist program by Danna C Griffiths (to be found via the Internet Archive) has an answer for it.
What I understood from the explanation on that site:
Solubility is defined as the concentration of the solute in a saturated solution. So when considering the increase in solubility with temperature, you have to check the enthalpy of solution of $\ce{NaOH}$ in a saturated solution of $\ce{NaOH}$.
When you dissolve $\ce{NaOH}$ in pure water, the process is highly exothermic, but as the concentration of $\ce{NaOH}$ increases, the process becomes less exothermic and eventually near the saturation point it becomes endothermic (it becomes less favourable for more $\ce{NaOH}$ to dissolve).
So according to Le Chatelier’s Principle, although the dissolution process is overall exothermic, since solubility is determined only at the saturation point, the solubility increases with increase in temperature.
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