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The breaking of bonds usually absorbs energy and forming bonds usually releases energy.

Then, why is the dilution of acids and bases exothermic?

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  • $\begingroup$ Except when it is endothermic; I have water soluble fertilizer that is strongly endothermic. $\endgroup$ – blacksmith37 Dec 10 '20 at 16:49
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Dissolution can be both exothermic like for $\ce{NaOH}$ or $\ce{H2SO4}$, or endothermic like for $\ce{CaCl2 . 6 H2O}$, $\ce{KNO3}$ or $\ce{KClO3}$.

The net thermal effect sign is determined by combination of endothermic breaking of a ionic lattice or chemical bonds and of exothermic ion hydration.

$$\Delta H_\mathrm{dissol} = \Delta H_\mathrm{ioniz} + \Delta H_\mathrm{hydr}$$

Compounds with the greater absolute value of lattice or bond energy ( the former term ) than hydration energy ( the latter term ) have endothermic dissolution. Compounds with the opposite relation have exothermic dissolution.

To address dilution after original misreading:

Dilution of concentrated solutions is exothermic, as the exothermic ion hydration is gradual, depending on H2O : ion molar ratio.

For liquid concentrated acids like $\ce{H2SO4}$ or $\ce{HNO3}$, the process of dissolution gradually switch to dilution, with acid molecules being gradually ionized and hydrated.

I suppose a complicated process with a complicated PE graph, as it is not a single reaction like e.g. SN2 nucleophilic substitution. E.g. hydration of protons comes with multiple steps and several layers.

$$\ce{H+ -> H3O+ -> H3O+ . H2O -> H3O+.(H2O)2 -> H3O+.(H2O)3 -> \\ .. -> H3O+.(H2O)6 -> .. -> H3O+.(H2O)20 }$$

See Wikipedia - hydronium - solvation. Additionally, dissociation and hydration are not subsequent processes, but are ongoing simultaneously.

Similar would happen with anions.

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    $\begingroup$ The question was about dilution, pretty sure that's no typo. $\endgroup$ – Karl Dec 10 '20 at 7:46
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    $\begingroup$ @Karl Ah, I see. I know difference, of course, I have just read something else than what was written. :-) It was before my morning coffee. $\endgroup$ – Poutnik Dec 10 '20 at 7:50
  • $\begingroup$ Probably all the steps described by Poutnik 2 days ago are exothermic. I repeat them H+⟶H3O+⟶H3O+⋅H2O⟶H3O+⋅(H2O)2⟶H3O+⋅(H2O)3⟶⋅⋅⟶H3O+⋅(H2O)6⟶⋅⋅⟶H3O+⋅(H2O)20 $\endgroup$ – Maurice Dec 12 '20 at 12:10
  • $\begingroup$ @Poutnik In the case of bases, I think, there is no hydration going on during dilution. So, how is that exothermic? $\endgroup$ – Shub Dec 28 '20 at 5:22
  • $\begingroup$ @Shub Because you think wrong here. $\endgroup$ – Poutnik Dec 28 '20 at 6:26

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