Question

Ethylamine, C2H5NH2, is a weak base. 250.0 mL of a 0.160 M C2H5NH2  solution is titrated with 0.500 M HCl (aq). Find the KbKb value in a lecture slide. Calculate the pH at the following points.

1. 30. mL of HCl solution added.
2. At the halfway point.
3. At the equivalence point. Verify any assumption you made.
4. 90. mL of HCl solution added.

Answer 1

ethylamine = Kb = 6.3×10^–4

pKb = 3.2

millimoles of ethylamine = 250 x 0.16 = 40

a) 30 mL titrant added :

millimoles of HCl = 30 x 0.5 = 15

C2H5NH2 + HCl --------------------> C2H5NHCl

40      15                                 0

25                   0                                        15

here salt and base remained . so it forms basic buffer

pOH = pKb + log [salt / base]

         = 3.2 + log [15 / 25]

          = 2.98

pH = 11.02

b) At half way point :

at half way point

pOH = pKb

         = 3.2

pH = 10.8

c) at equivalence volume :

millimoles of HCl = 40

volume of HCl = 40 / 0.5 = 80 mL

(CH3)3N    + HCl --------------------> (CH3)3NHCl

40    40                                  0

   0             0                                   40

here salt only remained .

salt = 40 / (total volume)

      = 40 / (250 + 80)

      = 0.121 M

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

       = 7 - 1/2 (3.2 + log 0.121)

       = 5.86

pH = 5.86

d) 90 ml added :

millimoles of HCl = 90 x 0.5 = 45

(CH3)3N    + HCl --------------------> (CH3)3NHCl

40      45                                     0

   0                5                                      40

here strong acid remained.

[HCl] = 5 / (250 + 90)

          = 0.0147

pH = -log [H+]

      = -log (0.0147)

      = 1.83

pH = 1.83