Question

Consider the titration of 70.0 mL of 0.0300 M C₂H₅NH₂ (a weak base; K_b = 0.000640) with 0.100 M HClO₄. Calculate the pH after the following volumes of titrant have been added:

(a) 0.0 mL pH =	(b) 5.3 mL pH =	(c) 10.5 mL pH =
(d) 15.8 mL pH =	(e) 21.0 mL pH =	(f) 29.4 mL pH =

Answer 1

millimoles of weak base = 70 x 0.03 = 2.1

kb = 0.000640

pKb = -logKb = -log (0.000640) = 3.19

a) before the addition of any HClO4

C₂H₅NH₂   + H2O --------------> C₂H₅NH₃+    + OH-

0.03                                            0                   0

0.03 - x                                      x                       x

Kb = x^2 / 0.03 - x = 0.000640

x = 4.07 x 10^-3

[OH-] = 4.07 x 10^-3 M

pOH = -log ( 4.07 x 10^-3 ) = 2.39

pH = 11.61

b) after the addition of 5.3 mL HClO4

millimoles of HClO4 = 5.3 x 0.1 = 0.53

C₂H₅NH₂   + HClO4 --------------> C₂H₅NH₃+

2.1              0.53                               0

1.57              0                                0.53

pOH = pKb + log [salt / base]

        = 3.19 + log [0.53 / 1.57]

        = 2.72

pH = 11.28

c) after the addition of 10.5 mL HClO4

millimoles of HClO4 = 10.5 x 0.1 = 1.05

this is half equivalence point. so

pOH = pKb

pOH = 3.19

pH +pOH =14

pH = 10.81

d) after the addition of 15.8 mL HClO4

millimoles of acid = 15.8 x 0.1 = 1.58

C₂H₅NH₂   + HClO4 --------------> C₂H₅NH₃+

2.1             1.58                               0

0.52              0                                1.58

pOH = 3.19 + log [1.58 / 0.52]

        = 3.67

pH = 10.33