Question

You are given a 30.0mL sample of 0.20M C6H5COOH(aq) (benzoic acid). Ka for benzoic acid is 6.5x10^-5.

b) what is the pH after the addition of 15.0mL of 0.30M KOH(aq)?

c)if you start with a new 30.0mL sample of 0.20M C6H5COOH(aq), what volume of 0.20M NaOH(aq) is required to reach the halfway to the equivalence point?

Answer 1

millimoles of acid = 30 x 0.2 = 6.0

pKa = - log Ka = - log [6.5 x 10^-5] = 4.19

b) millimoles of KOH added = 15.0 x 0.3 = 4.5

6.0 - 4.5 = 1.5 millimoles acid left

4.5 millimoles of salt formed

[acid] = 1.5 / 45 = 0.033 M

[salt] = 4.5 / 45 = 0.1 M

pH = pKa + log [salt] / [acid]

pH = 4.19 + log [0.1] / [0.033]

pH = 4.67

c) half way of equivalence point means 50 % acid must be consume

6.0 / 2 = 3.0

3.0 millimoles NaOH must be added

3.0 = V x 0.20

V = 15 mL

15 mL NaOH must be added to reach half way of equivalence point.