How many milliliters of 0.202 M KOH should be added to 500.0 mL of 0.0233 M
tartaric acid (C4H6O6; FW 150.087) to adjust the pH to 2.75? Ka1 = 1.0 x 10-13 and Ka2 = 4.6 x10-5.
I ended up with KOH mL being about 1.5mL. The equiv. point would be around 57mL. I see the answer is to be around 20mL. I have no idea how this comes about.