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The question is - fundamentally why does H "decides" to replace Na? Why is it more preferable than staying with Cl?

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    $\begingroup$ This "replacement" teaching should be banned, it seems. This is acid-base reaction! $\endgroup$ – Mithoron Mar 29 '17 at 13:55
  • $\begingroup$ I know that this is an acid-base reaction. The question is why (fundamentally, at low level) it happens. $\endgroup$ – Stanislav Bashkyrtsev Mar 29 '17 at 19:11
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Because hydroxide is a stronger base than chloride. Chloride is a much larger ion, and thus its negative charge is spread over more volume, stabilizing it. The negative charge on the oxygen in hydroxide is spread out over less volume, leaving it more reactive. Thus, when a $\ce{H^+}$ ion is available, it will preferentially be retained by the hydroxide rather than the chloride.

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    $\begingroup$ We can also turn to the final arbiter of spontaneity: thermodynamics. This reaction at the standard state (1 M concentration, 1 bar pressure) is spontaneous to the tune of $\Delta_r G^\circ = -80.4\ \mathrm{kJ/mol}$ at room temperature. $\endgroup$ – Ben Norris Mar 29 '17 at 11:44
  • $\begingroup$ The quantitative information is nice, thanks for sharing. I think the spontaneity can be accounted for (qualitatively) by simply saying which base is stronger, because basicity is defined in terms of thermodynamics in the first place. $\endgroup$ – electronpusher Mar 29 '17 at 13:02
  • $\begingroup$ Thanks! One piece that I still don't get though: why does NaOH give away Na in exchange for H? $\endgroup$ – Stanislav Bashkyrtsev Mar 29 '17 at 19:25
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    $\begingroup$ Honestly, the most practical way to think of it is the Na+ wasn't there in the first place. Na+ acts here as what we call a spectator ion, playing essentially no role other than to balance the charge of its negative partner. When in solid form as NaOH(s), this role is important (see Ionic Bonding and Crystal Lattice). However, when NaOH hits water, it instantly splits up into Na+ and OH- (we say the sodium hydroxide compound dissociates into these ions). We rationalize this by supposing the ions would rather interact with water molecules than each other (see Solvation and Lattice Enthalphy). $\endgroup$ – electronpusher Mar 30 '17 at 6:52
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    $\begingroup$ However, OH- would rather interact with an H+ ion than with water molecules, because it can form a chemical bond and thus both ions (hydroxide OH- and the proton H+) become a more stable product: water. The forming of a covalent bond is not observed with Na+, not by hydroxide or any other compound that I've heard of. So you could think of when you put hydroxide in a solution of water and H+ (and Cl-, also a nonreactive spectator ion here), the hydroxide ditches the sodium ion it was settling for (dissociates), and also happens to then find a partner "for life" (reacts chemically with H+). $\endgroup$ – electronpusher Mar 30 '17 at 7:00
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Adding to electronpusher's answer, one must keep in mind that when HCl is dissolved in water, it doesn't exist as HCl molecules anymore. Rather, HCl ionises to form $\ce{H+}$ and $\ce{Cl-}$ ions. So in the aqueous state, the H isn't actually bonded to Cl, rather you can think of it existing as $\ce{H+}$ ions floating freely in water (its not really like that). The same goes for NaOH, in the aqueous state it really exists as $\ce{Na+}$ and $\ce{OH-}$ ions. However, when $\ce{H+}$ and $\ce{OH-}$ ions meet each other, they react to form $\ce{H2O}$, which actually exists as a molecule rather than as separate ions. And like mentioned in electronpusher's answer, the reason why they react in the first place is because $\ce{OH-}$ is a stronger base than $\ce{Cl-}$.

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