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Why dissolving sodium hydroxide in water is an exothermic process. What causes the evolution of heat when dissolving sodium hydroxide in water?

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  • $\begingroup$ The enthalpy of solvation of the ions is higher than the lattice energy in the solid, so it is exothermic. See e.g. chemguide.co.uk/physical/energetics/solution.html $\endgroup$ – Karsten Theis Jan 30 at 17:20
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    $\begingroup$ Actually it’s only exothermic at neutral pH because of the exothermic reaction of hydroxide with protons. Once the pH goes up (which takes only a very small amount of hydroxide), the dissolution is endothermic $\endgroup$ – Andrew Jan 30 at 23:50
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Let's start with your statement that "dissolving sodium hydroxide in water is an exothermic process." I assume it is based on the reported standard enthalpy of solvation, which is $-44.2$ kJ/mol.

If you go into the lab and try to make a 1 M solution of sodium hydroxide, however, you'll find that the solution gets quite cold. That process is endothermic.

Why the disagreement? The standard enthalpy of solvation is defined for infinite dilution in neutral pH water, so it applies only to the very first tiny amount of $\ce{NaOH}$ that is added to a finite volume of water.

It turns out that the reaction $$\ce{NaOH(s) -> Na+(aq) + HO-(aq)}$$ is actually endothermic. However, the reaction $$\ce{HO-(aq) + H3O+(aq) -> 2H2O}$$ is even more exothermic, so the combined reaction (which is what is occurring at infinite dilution) $$\ce{NaOH(s) + H3O+ -> Na+(aq) + 2 H2O}$$ is exothermic.

At infinite dilution, the added hydroxide ion immediately reacts with hydronium and we observe the net exothermic reaction. But as we add more hyroxide, the hydronium ion concentration is quickly reduced to the point that most of the added hydroxide does not react with hydronium, and we instead observe an endothermic reaction that is just the solvation of the hydroxide and sodium ions without the reaction of hydroxide and hydronium.

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This exothermic phenomena is due to the interaction of the ion $OH^-$ with water. The $OH^-$ ion has three free doublets on the Oxygen atoms, which may attract strongly the protons of three neighbouring $H_2O$ molecules. Don't forget that in $H_2O$, the cloud of the only electron of the $H$ atom is nearly entirely placed between the two nuclei $H$ and $O$. So the nucleus of the $H$ atom is practically "nude" in water, I mean not protected by its surrounding electron, as seen from outside. As a consequence, the $H$ nucleus is attracted by any negative charge to be found in the surroundings. And there are plenty available doublets around any $OH^-$ ion in water. Setting up a chemical bond always produces energy, or heat, even if this bond is not an ordinary covalence. When dissolved in water, each $OH^-$ ion may attract $ 3 H_2O$ molecules, which make $3$ new bonds and an ion with the Oxygen atom in the center of a tetrahedron. Three of the four apexes are occupied by a $H$ atom belonging to three different water molecule. The fourth is the $H$ atom coming from the original $OH^-$ ion. There has been long discussions in the literature about the real formula of the ion $OH^-$. It should be written $[OH(H_2O)_3]^-$. But some chemists think that there are more than $ 3 H_2O$ molecules around each $OH^-$ ion. The problem is not definitively solved. Meanwhile, the usual chemist prefers to ignore this difficulty, and to say that the formula of the ion is simply $OH^-$ .

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    $\begingroup$ -1 for terrible formatting in spite of my earlier comments. $\endgroup$ – Mithoron Jan 30 at 19:40
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    $\begingroup$ This answer is incorrect. Simple solvation of sodium cation and hydroxide anion is actually endothermic. The “exothermic” description applies only to infinite dilution in neutral water, where the acid-base reaction of hydroxide with protons occurs $\endgroup$ – Andrew Jan 30 at 23:54

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