I'm wondering why some acid base reactions occur, let's say for example:

$$\ce{CH3CH2OH + H2O <<=> CH3CH2O- + H3O+}$$

Why does this reaction occur, because the alkoxide ion is a really strong base right?

Why should its conjugate acid give off a $\ce{H+}$ to water, if it was a strong base let's say $\ce{NaOH}$ it would make sense to me, because the $\ce{OH-}$ is probably a stronger base then the alkoxide ion.

Question: Why does the above reaction take place, considering the fact that water is a really weak base?

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    $\begingroup$ Certainly some of the amine will be protonated. The extent to which that happens depends on the equilibrium constant. $\endgroup$ – MaxW Apr 24 '16 at 17:17
  • $\begingroup$ so an acid base reaction will Always occur? @MaxW $\endgroup$ – KingBoomie Apr 24 '16 at 17:37
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    $\begingroup$ $$\ce{CH3CH2NHCH3 + H2O <=> NH2(CH3)(CH2CH3)^+ + OH^- }$$ $\endgroup$ – MaxW Apr 24 '16 at 17:54
  • $\begingroup$ If you're asking generally then it's much to broad. $\endgroup$ – Mithoron Apr 24 '16 at 20:08
  • $\begingroup$ I narrowed my question down, hopfully you can help me now @Mithoron $\endgroup$ – KingBoomie Apr 27 '16 at 7:05

The first thing you need to realise is that every reaction is in equilibrium, but some reactions have an equilibrium so far to one side that they are effectively complete (or they don't go at all). The position of equilibrium is determined by the relative stability of the products and the reactants. $$\ce{CH3CH2OH + H2O <<=> CH3CH2O- + H3O+}$$

The strength of acids can be compared by looking at $\mathrm{p}K_\mathrm{a}$ values. The $\mathrm{p}K_\mathrm{a}$ of $\ce{H3O+}$ in water is -1.7 and the $\mathrm{p}K_\mathrm{a}$ of ethanol in water is 16. So you are correct that hydronium is a much stronger acid than ethanol - or equivalently, ethoxide is a much stronger base than water. However, some of the water will deprotonate the ethanol and so the reaction will go to a small extent. This is why the equilibrium arrows are written with a big arrow on the reverse reaction and a small arrow on the forward reaction.

The equilibrium constant for the reaction is related to the standard Gibbs energy change for the reaction by: $$\Delta_\mathrm{r} G^\circ = -RT\ln K$$

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