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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."?

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Does that mean that I should disregard the statement "the more exothermic the solution, the more soluble the salt."?

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).

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    $\begingroup$ I saw your previous comments on the magnitudes, rather than the values, of ΔH(hydration) and ΔH(lattice E) being compared as well as the fact that both are usually negative values since energy is evolved. I can't believe I missed that. Thanks a lot for the clarification! And thanks for the explanation on Gibs free energy too. $\endgroup$ – Tetsu Dau Aug 22 at 11:39
  • $\begingroup$ @TetsuDau Yea, I rewrote the comment because I was using the wrong sign convention for the lattice enthalpy - it is the enthalpy of formation of the lattice from ions in the gas phase. $\endgroup$ – Buck Thorn Aug 22 at 11:45

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