Why isn't CH3OH basic if it contains OH?

Question

I was asked to find the most basic solution from the following:

$\ce{K2SO4}$
$\ce{CH3OH}$
$\ce{Na2CO3}$
$\ce{(NH4)2CrO4}$
$\ce{BaCl2}$

I know that I have to identify the solution that contains the conjugate base of a weak acid.

Knowing this, I immediately eliminated $\ce{BaCl2}$ ($\ce{HCl}$), $\ce{K2SO4}$ ($\ce{H2SO4}$), and $\ce{(NH4)2CrO4}$ ($\ce{H2CrO4}$ — which I've heard is a strong acid).

This left $\ce{CH3OH}$ and $\ce{Na2CO3}$.

The correct answer is $\ce{Na2CO3}$ since $\ce{CO3^{2-}}$ is the conjugate base of $\ce{H2CO3}$.

However, why can the final answer not be $\ce{CH3OH}$, especially considering that $\ce{OH}$ is basic? Also, could $\ce{OH}$ be considered the conjugate base to a weak acid?

Answer 1

Unfortunately, chemistry is not as simple as going

Oh, I spotted $\ce{OH}$, it must be a basic compound!

I would argue that the vast majority of all compounds one could write with $\ce{OH}$ inside are either neutral or acidic and only a minority is basic.

Take, for example, sulphuric acid. As Ivan correctly mentioned in a comment, rather than writing the traditional $\ce{H2SO4}$, you could also write it as $\ce{O2S(OH)2}$ which lays more emphasis on the molecular structure (to hydroxy groups and two oxy groups all bound to sulphur). We all know that $\ce{H2SO4}$ is acidic, not basic, even though it has $\ce{OH}$ groups.

Or take alcohols such as methanol ($\ce{H3C-OH}$) or ethanol ($\ce{H3C-CH2-OH}$). These also feature $\ce{OH}$ groups, they are even commonly written that way, but are about as acidic or basic as water ($\ce{HOH}$) is.

So the question may be why did you get taught that $\ce{OH}$ is basic? Well strictly speaking, it is not the $\ce{OH}$ hydroxy group that is basic but the $\ce{OH-}$ hydroxide ion. Only those compounds that dissociate into some cation and a hydroxide ion can be considered basic. For that to happen, ‘some cation’ typically needs to be metallic, at least at your level. So recognise a metal bonded to $\ce{OH-}$ and you should most likely be right calling the compound basic.

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Some final notes:

1. You should always check the directly corresponding conjugate acid. So when analysing $\ce{K2SO4, Na2CO3}$ and $\ce{(NH4)2CrO4}$, trace them back to $\ce{HSO4- , HCO3-}$ and $\ce{HCrO4-}$, respectively. That still leaves us with two rather strong acids but the third ($\ce{HCO3-}$) is even weaker.

2. The conjugate acid to methanol $\ce{H3C-OH}$ would be the methyloxonium ion $\ce{H3C-OH2+}$, which is a strong acid. This goes for pretty much all organic hydroxy groups.

3. The $\mathrm{p}K_\mathrm{a}$ of water — hydroxide’s conjugate acid — is $15.7$, so yes, that is a very weak acid.

Answer 2

$\ce{CH3OH}$ is a compound known as methanol. Unlike the sodium found in $\ce{Na2CO3}$, the $\ce{CH3}$ group found in methanol will not separate from the $\ce{OH-}$. This means that the only way that methanol can act as a base is for the oxygen to pick up another hydrogen from solution, making the compound $\ce{CH3OH2+}$.

On the other hand, sodium will readily dissociate from $\ce{CO3^{2-}}$. This is because sodium is a metal, and it will complete its octet when it loses an electron. Once dissociated, $\ce{CO3^{2-}}$ will act as a base and pick up protons, becoming either $\ce{HCO3-}$ or $\ce{H2CO3}$. Since when $\ce{CO3^{2-}}$ gets protonated its net charge decreases, this will act as the stronger base.