Question

Consider the titration of 60.0 mL of 0.0400 M CH₃NH₂ (a weak base; K_b = 0.000440) with 0.100 M HBrO₄. Calculate the pH after the following volumes of titrant have been added:

(a) 0.0 mL pH =	(b) 6.0 mL pH =	(c) 12.0 mL pH =
(d) 18.0 mL pH =	(e) 24.0 mL pH =	(f) 28.8 mL pH =

Answer 1

a)

mmoles of CH3NH2 = 60.0 x 0.040 = 2.4

pKb = - log Kb = 3.356

CH3NH2   +   H2O   --------------> CH3NH3+ +   OH-

0.0400                                                0                  0

0.0400-x                                             x                  x

Kb = x^2 / 0.0400 - x

0.00044 = x^2 / 0.0400 - x

x = 3.98 x 10^-3

[OH-] = 3.98 x 10^3 M

pOH = -log ( 3.98 x 10^-3 ) = 2.40

pH = 11.60

b)

mmoles of acid = 6 x 0.100 = 0.6

CH3NH2   + H+    -------------> CH3NH3+

2.4              0.6                           0

1.8               0                            0.6

pOH = pKb + log [0.6 / 1.8]

      = 3.356 + log (0.6 / 1.8)

     = 2.88

pH = 11.12

c)

mmoles of acid = 12 x 0.1 = 1.2

this is half - equivalence point .

here pOH = pKb

pOH = 3.356

pH = 10.64

d)

mmoles of acid = 1.8

pOH = 3.356 + log [1.8 / 0.6]

         = 3.83

pH = 10.17

e)

salt concentraiton = 2.4 / 24 + 60 = 0.0286 M

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

    = 7 - 1/2 (3.356 = log 0.0286)

pH = 6.09

f)

pH = 2.27