Question

I did an experiment using 20ml of 0.1M acetic acid and added 8ml of 0.1M NaOH. pH obtained after adding NaOH was 5.2. How do i calculate the mols of NaOH that reacted? mols of acetate formed, acetic acid initially present and acetic acid unreacted. Also how do i determine the [H+], [CH3CO2-] and [CH3CO2H] and Ka

Answer 1

To save some typing, I will call acetic acid, HAc and acetate is Ac^-
millimoles HAc = mL x M = 20 x 0.1 = 2.0
mmoles NaOH = 8 x 0.1 = 0.8

...............HAc + NaOH ==> NaAc + H2O
initially......2.0....0........0.......0
added.................0.8...............
change.........-0.8...-0.8.....+0.8..+0.8
equil.......... 1.2.....0......0.8...0.8

(HAc) = mmoles/mL = 1.2/28 = ??
NaAc = same process = mmoles/mL.

pH = pKa + log [(NaAc)/(HAc)] = ??