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The solubility constant of $\ce{AgCl}$ at $25\mathrm{\,^{\circ}C}$ degrees is $1.6\cdot 10^{-10}$.

At $100\mathrm{\,^{\circ}C}$, the $K_{\mathrm{sp}}$ increases to $2.15\cdot 10^{-8}$.

Why is the dissolution of $\ce{AgCl}$ at $100\mathrm{\,^{\circ}C}$ considered endothermic? I would have thought it was exothermic.

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The typical way of explaining this is using Le Chatelier's Principle. That is, we could consider the reaction,$$\ce{AgCl(s) + heat<=>Ag+(aq) + Cl^-(aq)}$$This then demonstrates that energy in the form of heat is required to have the reaction proceed. Thus, at a higher temerature, the reaction will shift to the right as predicted by Le Chatelier's principle.

There are obviously more thorough explanations possible, but this is a good place to start.

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  • $\begingroup$ brief yet understandable. Much appreciated. $\endgroup$ – Darragh H Dec 8 '16 at 18:31
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Look at the answers to the question I asked about enthalpy change, temperature, and equilibrium.

Basically, the equilibrium constant depends on temperature in a way that is dependent on the sign of the enthalpy change of the reaction. If $\Delta H > 0$, then as temperature increases, the equilibrium constant increases, and vice versa.

You observe the relationship where increasing temperature correlates with increasing equilibrium constant, suggesting the reaction is endothermic.

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