How does pKa relate to acidity?

1 Answer
Jun 30, 2016

Answer:

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^-]#.