Thanks Borek,

Just to clear something up, if I am titrating a weak acid with a strong base, lets say [HA]=0.15 mol L^{-1} and [NaOH]=0.15 mol L^{-1}. At the equivalence point, 10mL of NaOH has been added to 10mL of HA. Here is my question I want to clear up: after the equivalence point, when 20mL of NaOH has been added altogether, how would you calculate the pH?

This is what I did; I assumed that all of the OH^{-} ions had been neutralised so therefore the amount of NaOH in excess is 10mL, since the total volume of the solution has been tripled, the concentration has been diluted 3 fold, so [NaOH] is now 0.05 mol L^{-1}, is this correct? The pH would be 12.7.