# I was told that the more exothermic the solution, the more soluble the salt. Is this statement wrong?

I was also told that $$\Delta H(\text{solution}) = \Delta H(\text{hydration}) - \Delta H(\text{lattice E})$$

However, for $$\Delta H(\text{solution})$$ to be exothermic, it has to have a negative value <0. Which means lattice energy is going to have to be greater than $$\Delta H(\text{solution})$$ to produce a negative value, based on the equation above.

Yet I am also told that hydration energy released needs to be enough to compensate for the lattice energy required to break down the solid. So it can't be lesser than $$\Delta H(\text{lattice E})$$.

Isn't this contradictory? I am confused.

I have read a similar question at Enthalpy of solution but the answer given is that dissolving is dependent on Gibb's free energy instead. Does that mean that I should disregard the statement "the more exothermic the solution, the more soluble the salt."?

• – Buck Thorn Aug 22 '19 at 10:48

No, based on the Gibbs free energy equation $$\Delta G = \Delta H -T \Delta S$$ and the condition that $$\Delta G<0$$ for spontaneous dissolution, if you assume $$\Delta S$$ is constant then the statement is true.
The point of that other post is that as a general rule you should not simply ignore the $$\Delta S$$ term since sometimes it is the decisive factor determining solubility (such as in the case of endothermic dissolution reaction).