Consider a weak acid in water:
HA +H_2Orightleftharpoons H_3O^+ + A^-
Now the equilbrium constant, K_a, describes this equilibrium:
K_a = ([H_3O^+][A^-])/[[HA]]
Now, as long as we do it to both sides, we can manipulate this equation. We take logarithms to the base 10 to make it a bit easier:
log_10K_a = log_10{[H_3O^+]} + log_10{[[A^-]]/[[HA]]}
(Note, when we write log_ab = c, we ask to what power do we raise the base a to get b; so a^c=b; likewise log_(10)100 = 2, and log_(10)1000 = 3. Before the days of ready access to electronic calculators, everyone would use log tables to perform multiplication and division.)
Rearranging this equation:
-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.
So finally, we get:
pH = pK_a + log_10{[[A^-]]/[[HA]]}.
Note that this tells us that at the point of half equivalence of a titration, when [A^-]=[HA], log_10(1)=0 (clearly), and thus pH = pK_a at half equivalence.
