Consider a solution made by mixing 100 mL of 2.0 M acetic acid (Ka= 1.8e-5) with 100 mL of 0.00020 M benzoic acid

Question

(Ka= 6.4e-5). How many moles of solid NaOH must be added to the solution so that
half of the benzoic acid is protonated (i.e. such that [HA] = [A-] for benzoic acid)?

Answer 1

I would add mols NaOH to neutralize all of the acetic acid and add enough to neutralize 1/2 benzoic acid.