Question

Determine the pH during the titration of 38.3 mL of 0.163 M HI by 0.163 M KOH at the following points:

(a) Before the addition of any KOH

(b) After the addition of 19.2 mL of KOH

(c) At the equivalence point

(d) After adding 47.9 mL of KOH

Answer 1

millimoles of HI = 38.3 x 0.163 = 6.3244

a) initially only HI present

as HI is strong acid [H⁺] = [HI] = 0.163 M

pH = - log [H⁺] = - log[0.163]

pH = 0.79

b) millimoles of KOH added = 19.2 x 0.163 = 3.1296

6.3244 - 3.1296 = 3.1948 millimoes HI left

[HI] = 3.1296 / 38.3+19.2 = 3.1296 / 57.5 = 0.0544 M

pH = - log [H⁺]

pH = - log [0.0544]

pH = 1.26

c) strong acid strongbase titration so at equivalence point

pH = 7.0

d) millimoles of KOH = 47.9 x 0.163 = 7.8077

7.8077 - 6.3244 = 1.4833 millimoles KOH left

[KOH] = 1.4833 / 38.3 + 47.9 = 1.4833 / 86.2 = 0.0172 M

as KOH is strong base

[KOH] = [OH^-] = 0.0172 M

pOH = - log [OH^-]

pOH = - log [0.0172]

pOH = 1.76

pH = 14 - 1.76

pH = 12.24