Question

How does pKa relate to acidity?

Answer 1

Consider the general reaction of an acid in water:

`HA+H_2O rightleftharpoonsH_3O^+ + A^-`

Explanation:

Now it is well known that for a given temperature, the extent of and acid base reaction is a constant:

`K_a=([H_3O^+][A^-])/([HA])`

Clearly, for strong acids, `K_a` is large, and `K_a` is smaller for weaker acids.

Now this is a mathematical expression, which we are free to manipulate, provided that we do it to BOTH sides of the equation. Now we can take `log_10` of both sides of the equation:

`log_10K_a` `=` `log_10([H_3O^+]) +log_10{[[A^-]]/[[HA]]}`

On rearrangement:

`-log_10{[H_3O^+]}=-log_10K_a+log_10{[[A^-]]/[[HA]]}`

But by definition `-log_10{[H_3O^+]}=pH` and `-log_10K_a=pK_a`.

And so `pH=pK_a+log_10{[[A^-]]/[[HA]]}`

And thus, when we tritrate a weak acid with a base, and plot `pH` versus volume of titrant, at the point of half equivalence,

`pH=pK_a` because `log_10{[[A^-]]/[[HA]]}=log_10{1}=0,` i.e at half equivalence, by definition `[HA]=[A^-]`.