Can we dissolve salts with higher lattice enthalpy by providing heat?

It is well-known that



> 
> A salt with a higher lattice energy will not dissolve at all, because then $\Delta\_{\text{sol}}H^\ominus$ will be negative.
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So, the reaction should be endothermic, right? My question is:


**If then we provide heat to the reaction, then we should be able to dissolve such salts. Is this true?**




Heat itself it just indirect factor. It is temperature what determines the shift of solubility products. If the dissolution is endothermic process, the salt [solubility product](https://en.wikipedia.org/wiki/Solubility_equilibrium) would increase with higher temperature.


But if the solubility product is still small even at high temperature, like e.g. $\pu{e-16}$ instead of $\pu{e-20}$, then only a tiny extra salt amount would dissolve when heated.


The effective result depends on the particular case. The typical cases where significant solubility increase is observed are potassium salts $\ce{KNO3, KClO3, KClO4}$. But they have limited solubility even at low temperature - see the [solubility table](https://en.wikipedia.org/wiki/Solubility_table#P).


Generally, the dependence of an equlibrium constant of a reaction, including dissolving/precipitation process, is described by [van't Hoff equation](https://en.wikipedia.org/wiki/Van_%27t_Hoff_equation#Equation)



> 
> $$\frac{\mathrm{d}}{\mathrm{d}T} \ln K\_\mathrm{eq} = \frac{\Delta H^\ominus}{RT^2}$$
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For solubility products, being equlibrium constants, is valid the following reaction:


$$\Delta G^{\circ}\_\mathrm{dis} = \Delta H^{\circ}\_\mathrm{dis} - T \Delta S^{\circ}\_\mathrm{dis} = RT \ln K\_\mathrm{sp}$$


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