2 Consider the titration of 500 mL of 0100 M ethylene diammi

2. Consider the titration of 50.0 mL of 0.100 M ethylene diammine (en), NH2CH2CHINHz (Kal = 1.42.107, Ka2 = 1.18 0.10) with 0.200 M HCl. The titration reaction occurring is: Determine the pH after the following volumes of HCI have been added: a. At 0.0 m (no added acid) b. 13.00 ml

Solution

pKb1 = 4.07

pKb2 = 7.15

base molarity = 0.100 M

(a) before addition of any HCl

B   +   H2O <----------------------> BH+   + OH-

0.10 0             0   ------------------> initial

0.10 -x                                      x              x ------------------------> equilibrium

Kb1 = [BH+][OH-]/[B]

8.475 x 10^-5 = x^2 / 0.10 -x

x^2 + 8.475 x 10^-5 x - 8.475 x 10^-6 = 0

x = 2.87 x 10^-3

[OH-] = 2.87 x 10^-3 M

pOH = -log (2.87 x 10^-3 )

pOH = 2.54

pH + pOH = 14

pH = 11.46

(b) after additon of 13 mL HCl

mmoles of base = 50 x 0.1 = 5

mmoles of HCl = 13 x 0.2 = 2.6

Base +    HCl    --------------> BH+  

5            2.6                            0

2.4           0                            2.6

pOH = pKb + log [2.6 / 2.4]

       = 4.07 + log [2.6 / 2.4]

pOH = 4.10

pH + pOH = 14

pH = 9.90