# How to justify this contradiction in acid-base equilibrium of methanol and water? [duplicate]

Let's talk about the following compounds - water and methanol.

1. We know that methanol is slightly more acidic than water, because water's ability to donate a proton as an acid is reduced due to extensive hydrogen bonding. The $\mathrm{p}K_\mathrm{a}$ values are approximately $15.5$ and $15.7$, for methanol and water respectively.

2. Now consider their conjugate bases, i.e. methoxide ion and hydroxide ion. It is very clear that methoxide is a stronger base compared to hydroxide, due to the +I (inductive) effect of the methyl group (electron releasing).

How is this possible? Isn't it contradictory? It is a well known fact that a strong acid has a weak conjugate base, and a weak acid has a strong conjugate base.

This is a consequence of $\mathrm{p}K_\mathrm{a}$ + $\mathrm{p}K_\mathrm{b}$ = $\mathrm{p}K_\mathrm{w}$, where Kw is the ionic product of water, a constant at a given temperature.