# HCl + NaOH: Why H prefers to move to O instead of staying with Cl?

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

• This "replacement" teaching should be banned, it seems. This is acid-base reaction! Mar 29 '17 at 13:55
• I know that this is an acid-base reaction. The question is why (fundamentally, at low level) it happens. Mar 29 '17 at 19:11

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 $\ce{Cl-}$.
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.
• 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. Mar 29 '17 at 11:44