# Relationship between solubility and enthalpy of solution

I'm reading my lab manual and there is a part where it goes through Van't Hoff's isochore. That's fine but then it states

The variation of the solubility of a substance with temperature may thus be given by the relation

$$\dfrac{d \ln Sol}{dT}=\dfrac{\Delta H}{RT^2}$$

where do they derive this from. Is it from chemical potentials and mole fractions. I can't really find a good derivation and explanation online.

## 2 Answers

For equilibria such as $\ce{[C6H5COOH]_{(s)} <=> [C6H5COOH]_{(aq)} }$. The equilibrium constant $K_{eq}$ can be written in the very simple form:

$$K_{eq}\equiv \ce{[C6H5COOH]_{(aq)}} \equiv \text{solubility}\, (sol)$$

That's more than likely where the equation comes from. Then you get

$$\dfrac{d \ln Sol}{dT}=\dfrac{\Delta H}{RT^2}$$

when you separate the variables and integrate with respect to both sides you get:

$$\int_{\ln sol_1}^{\ln sol_2}d(\ln sol) = \int_{T_1}^{T_{2}}\dfrac{\Delta H}{RT^2}\, dT$$

$$\ln sol_2 - \ln sol_1 = \dfrac{\Delta H}{R}\Bigg{(}\dfrac{1}{T_1}- \dfrac{1}{T_2}\Bigg{)}$$

Which gives the useful result

$$\ln \dfrac{sol_2}{sol_1}= \dfrac{\Delta H}{R}\Bigg{(}\dfrac{1}{T_1}- \dfrac{1}{T_2}\Bigg{)}$$

Take this as a particular case of the relation between $\Delta G$ and equilibrium constant: $$K_{eq}=e^{-\frac{\Delta G}{RT}}$$ Really, for a process of dissolution, the solubility basically is the equilibrium constant. (That is, unless the compound dissociates in the solution.) Then $\ln K=-{\Delta G\over RT}$, and $\Delta G=\Delta H - T\Delta S$, and I'm sure you can plug one into another and take the derivative.

As to how the said relation came about, that's another question, and a great deal more difficult. It comes from the statistical mechanics. I'd rather not run ahead of your curriculum, but if the sudden appearance of natural logarithms out of nowhere will bother you to the point of insomnia, go read about the Gibbs paradox.