Question

Consider the titration of 40.0 mL of 0.0600 M (CH₃)₃N (a weak base; K_b = 6.40e-05) with 0.100 M HNO₃. 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) 26.4 mL pH =

Answer 1

millimoles of weak base = 40 x 0.06 = 2.4

kb = 6.4 x 10^-5

pKb = -logKb = -log ( 6.4 x 10^-5) = 4.19

a) before the addition of any HNO3

(CH3)3N + H2O -------------->(CH3)3N H+    + OH-

0.06                                            0                   0

0.06 - x                                      x                       x

Kb = x^2 / 0.06 - x = 6.4 x 10^-5

x = 1.93 x 10^-3

[OH-] = 1.93 x 10^-3 M

pOH = -log ( 1.93 x 10^-3 ) = 2.71

pH = 11.29

b) after the addition of 6 mL HClO4

millimoles of HNO3 = 6 x 0.1 = 0.6

(CH3)3N + HNO3 --------------> (CH3)3NH+

2.4              0.6                               0

1.8             0                                0.6

pOH = pKb + log [salt / base]

        = 4.19 + log [0.6 / 1.8]

        = 3.71

pH = 10.29

c) after the addition of 12 mL HNO3

millimoles of HNO3 = 12 x 0.1 = 1.2

this is half equivalence point. so

pOH = pKb

pOH = 4.19

pH +pOH =14

pH = 9.81

d) after the addition of 18 mL HNO3

millimoles of acid = 18 x 0.1 = 1.8

(CH3)3N + HNO3 --------------> (CH3)3NH+

2.4              1.8                              0

0.6              0                                1.8

pOH = 4.19 + log [1.8 / 0.6]

        = 4.67

pH = 9.33