# Why do weak acids like ethanol not act as a strong base in water?

In water, ethanol has a $$\mathrm{p}K_\mathrm{a} = 15.9$$ which means that its $$\mathrm{p}K_\mathrm{b}$$ is $$-1.9$$. Which implies that all weak acids in water are in fact strong bases. But this is not true, weak acids in fact do act like weak acids. That would mean that this equation isn't true always.

$$14 = \mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b}$$

Considering this is just the ionic product of water it should be valid always?

• It's the $\mathrm pK_\mathrm b$ of the conjugate base, not ethanol itself. $$\mathrm pK_\mathrm a(\ce{EtOH}) + \mathrm pK_\mathrm b(\ce{EtO-}) = 14$$ May 9, 2019 at 16:14
• And because pKa > 14, $\ce{EtO-}$ isn't going to exist in aqueous solution to any practical measure, making it a very strong base. $$\ce{EtO- + H2O -> EtOH + OH-}$$
– MaxW
May 9, 2019 at 17:33
• and ethanol is a very weak acid, hence it's conjugate base is strong. A weak acid like EtOOH also has a weak conjugate base.
– Karl
May 9, 2019 at 19:46
• @Karl did you mean strong conjugate base in the last line of your comment? May 9, 2019 at 20:30
• @WilliamR.Ebenezer ;-) Confuses me again every few years.
– Karl
May 9, 2019 at 21:21

There seems to be a misunderstanding: The conjugate of a strong base like EtONa can't be a weak acid, because weak acids, like acetic acid for example, are still quite acidic.

• Strong acid (HCl) - very weak conjugate base (chloride anion)

• strong base (EtONa) - very weak conj. acid (Ethanol)

A strong acid or base reacts strongly (with water), the resulting conjugate is rather inert, with the equillibrium fully on the side of the latter.

• weak acid (acetic acid) - weak conj. base (acetate anion)

• weak base (ammonia) - weak conj. base (ammonium)

A weak acid or base just reacts partly towards its conjugate, so does the pure conjugate: They are both rather weak. The equillibrium in an equimolar mixture is rather stable: A buffer solution.

The ideal weak base or acid has a $$pKa$$ of 7 (in water), that is $$1/2$$ of the -log ion product of the given solvent. If it's even weaker, the conjugate becomes stronger again.