after adding of 0.12 M KOH only OH- concentration remains at 6.6*10^-3 so pH= 14-+log(6.6*10^-3) =11.82 indicator is phenopthalein or phenol red Solution after adding of 0.12 M KOH only OH- concentration remains at 6.6*10^-3 so pH= 14-+log(6.6*10^-3) =11.82 indicator is phenopthalein or phenol red.