Question

50 mL of 0.100 M aniline (aminobenzene, Kb = 3.984 x 10-10) is titrated with 0.050 M HCl. Calculate the pH (to the nearest 0.1 pH unit) at 0, 25, 50, 75, 125, 150 mL of HCl added.

Answer 1

pKb = -logKb = 9.4

millimoles of aniline = 50 x 0.1 = 5

A ) 0 mL HCl added :

pH = 14 - 1/2 [pKb - logC]

pH = 14 - 1/2 [9.4 -log-0.1]

pH = 8.8

B) 25 mL HCl added

millimoles of HCl = 0.05 x 25 = 1.25

B + HCl -----------------------> BHCl

5 1.25 0

3.75 0 1.25

pOH = pKb + log [salt / base]

pOH = 9.4 + log (1.25 / 3.75)

pOH = 8.9

C) 50 mL HCl added

millimoles of HCl = 50 x 0.05 = 2.5

it is half equivalence point here pOH = pKb

pOH = 9.4

pH + pOH = 14

pH = 4.6

D ) 75 mL HCl added

millimoles of acid = 75 x 0.05 = 3.75

B + HCl ------------------> BHCl

5 3.75 0

1.25 0 3.75

pOH = 9.4 + log (3.75 / 1.25)

pOH = 9.9

pH = 4.1

E ) 125 mL HCl added

millimoles of acid added = 125 x 0.05 = 6.25

[H+] = 6.25 - 5 / (125 +50) = 7.12 x 10-3 M

pH = 2.2

F) 150 mL added

millimoles of HCl = 150 x 0.05 = 7.5

[H+] = 2.5 / (200) = 0.0125 M

pH = 1.9