Question

You have 15.00 mL of a 0.100 M aqueous solution of the weak base C6H5NH2 (Kb = 4.00 x 10-10). This solution will be titrated with 0.100 M HCl.

(a) How many mL of acid must be added to reach the equivalence point?

(b) What is the pH of the solution before any acid is added?

(c) What is the pH of the solution after 5.00 mL of acid has been added?

(d) What is the pH of the solution at the equivalence point of the titration? (e) What is the pH of the solution when 20.00 mL of acid has been added?

Answer 1

6)

a)

mmoles of weak base = 15 x 0.100 = 1.5

At equivalence point : mmoles of acid = mmoles of base

volume of acid added to reach equivalence point = 15.00 mL

b)

Kb = 4.00 x 10^-10

pKb = 9.40

pOH = 1/2 (pKb - log C)

        = 1/2 (9.40 - log 0.100)

       = 5.20

pH = 8.80

c)

mmoles of acid = 5x 0.10 = 0.5

NH2OH   +   H+   ------------> NH2OH2+

1.5               0.50                        0

1.0                0                         0.50

pOH = pKb + log [salt / base]

         = 9.40 + log [0.50 / 1.00]

         = 9.10

pH = 4.90

d)

at equivalence point , salt only remains.

salt concentration = 1.50 / 30 = 0.05 M

pH = 7 - 1/2 (pKb + log C)

     = 7 - 1/2 (9.40 + log 0.05)

pH = 2.95

e)

mmoles of acid = 20 x 0.1 = 2

pH = 1.84