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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

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after adding of 0.12 M KOH only OH- concentration remains at 6.610^.pdf

  • 1. 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