Question
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
(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)?
(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)?
Answers
I would add mols NaOH to neutralize all of the acetic acid and add enough to neutralize 1/2 benzoic acid.
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