Benzoic acid = HB
millimols HB = 50 x 0.015 = 0.750
mmols KOH = 30 x 0.015 = 0.45
............HB + KOH ==>KB + H2O
initial...0.75...0.45....0....0
change...-0.45..-0.45.+0.45..+0.45
equil.....0.30....0.....0.45..0.45
So what you end up with is benzoic acid (0.30 mmols--a weak acid) and potassium benzoata (0.45 mmols--the salt of a weak base) which is a buffered solution.
Use the Henderson-Hasselbalch equation to solve for pH.
what is the ph of a solution that results from adding 30 mL of .015M KOH to 50 mL of .015 M benzoic acid?
1 answer