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.