[UG]

I am slightly confused about this matter.
I was reading this table:


After the equivalence point, the last figure: Base added = 20 mL [OH^-] = 20/30
This is where I am confused. So at this point 20 mL of base has been added right? And 10mL has already reacted with H₃O⁺ to form H₂O right? Therefore, does that not mean that only 10mL of base is in excess? So shouldn't the hydroxide concentration be 10/30?

[UG]

Is it just a typo maybe? 

[Borek]

0.5/20.1 is wrong as well. Typos IMHO.

[UG]

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.

[Borek]

General idea seems OK.

http://www.titrations.info/titration-curve-calculation

http://www.titrations.info/acid-base-titration-curve-calculation

[UG]

That was basically what the first link said 

"After equivalence point titration curve shape is calculated from the excess of the titrant added - thus it will be just pH of excess titrant in the case of acid-base titration."