# Why do weak acids partially dissociate ?

Jan 25, 2016

Consider the typical equilibrium:

$H A + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {A}^{-}$

If the $H - A$ bond is STRONG, then the acid should be weak, and the equilibrium lies to LEFT.

#### Explanation:

All acid-base behaviour is qualified by the identity of the solvent. In water (the most common and convenient solvent) we rationalize acid/base behaviour by invoking ${H}_{3} {O}^{+}$ and $H {O}^{-}$ ions as the characteristic cation and anion of the SOLVENT.

For a given acid series $H - A$, we might consider the strength of the $H - A$ bond as the determinant. Let's look at the strength of the $H - X$ bond:

$H - F$; $570$ $m o {l}^{-} 1$

$H - C l$; $432$ $m o {l}^{-} 1$

$H - B r$; $366$ $m o {l}^{-} 1$

$H - I$; $298$ $m o {l}^{-} 1$

The weakest bond, $H - I$, is also the STRONGEST acid. The weakest acid, $H - F$, has the strongest bond. Of course, entropy approximations are important too, as the small fluoride ion is definitely entropically DISFAVOURED (as it is smaller and more polarizing). So the answer to your question is that, for weak acids, the hydrogen-hetero-atom bond should be relatively strong. (This assumes "all things being equal" , which they never are!)