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In the case of water dissolving sodium chloride. I was under the impression that hydrogen bonds were much weaker than the ionic bonds in sodium chloride, so how do they overcome the ionic bond strength? Do the ions simply get swarmed with enough water molecules so that the combined dipole interaction forces break the ionic bond? Or is it possible that because the sodium chloride dissolution reaction is endothermic the hydrogen bonds on their own are not enough to break the ionic bonds so the reaction must draw energy from the surroundings to do so (the reaction must be entropically favorable to do so, right?)?

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  • $\begingroup$ Ion-dipole interactions, not hydrogen bonds. Hydrogen bonds are between water molecules. $\endgroup$
    – orthocresol
    Commented Apr 5, 2017 at 19:39

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Your presumption that it is hydrogen bonding that breaks the ionic bonds is wrong. Forget H-bonds. It is true that they exist in both pure water and in a sodium chloride solution. Water molecules surround both the Na+ and Cl- ions. The interaction there is dipole-ion. Na-OH2 and Cl-HOH. As is obvious, only one of those involve an H atom. (note that the "-" I used is not intended to suggest either a covalent or an ionic bond, rather it is intended to show the electronic interaction between the two oppositely charged species (the Cl & Na ions have full integer charge, the H & O atoms have only partial opposite charge, mostly induced.) This question can be answered just by hand-waving about the thermodynamics: the energy of the solvated Na and Cl ions is less than that of those two ions in the crystal. So, on average, the ions will leave the crystal and enter the liquid (until their concentration is so high that equilibrium is established between the two phases). So, while it is true that HOH...Cl bond can be called a H-bond, thinking about it this way has no obvious advantage as far as I can see. The dipole-ion interactions that are occurring are indiscriminate. Each ion is surrounded by lots of (oriented) water molecules. To learn more see https://en.wikipedia.org/wiki/Solvation_shell and especially https://en.wikipedia.org/wiki/Solvation

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  • $\begingroup$ You should be more concerned with formatting, IMO. $\endgroup$
    – Mithoron
    Commented Apr 5, 2017 at 21:56

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