Question

Consider the titration of 30.0 mL of 0.0700 M NH3 (a weak base; Kb = 1.80e-05) with 0.100 M HBrO4. 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) 35.7 mL pH =

Answer 1

millimoles of NH3 = 30 x 0.07 = 2.1

pKb = 4.74

a)

NH3 + H2O --------------------> NH4+ + OH-

0.07 0 0

0.07 -x x x

Kb = [NH4+][OH-]/[NH3]

Kb = x^2 / 0.07 -x

1.8 x 10^-5 =  x^2 / 0.07 -x

x^2 + 1.8 x 10^-5 x - 1.26 x10^-6 = 0

x = 1.11 x 10^-3

x = [OH-] = 1.11 x 10^-3 M

pOH = -log[OH-] = 2.96

pH + pOH = 14

pH = 11.04

b) 5.3 mL acid added:

millimoles of acid = 5.3 x 0.1 = 0.53

H+ + NH3 -------------> NH4+

0.53 2.1 0

0 1.57 0.53

pOH = pKb + log [ NH4+ / NH3]

pOH = 4.74 + log (0.53 / 1.57)

pOH = 4.27

pH + pOH = 14

pH = 9.73

(c) 10.5 mL

millimoles of acid = 10.5 x 0.1 = 1.05

H+ + NH3 -------------> NH4+

1.05 2.1 0

0 1.05 1.05

pOH = pKb + log [ NH4+ / NH3]

pOH = 4.74 + log (1.05 / 1.05)

pOH = 4.74

pH + pOH = 14

pH = 9.26

(d) 15.8 mL

millimoles of acid = 0.1 x 15.8 = 1.58

H+ + NH3 -------------> NH4+

1.58 2.1 0

0 0.52 1.58

pOH = pKb + log [ NH4+ / NH3]

pOH = 4.74 + log (1.58 / 0.52)

pOH = 5.22

pH + pOH = 14

pH = 8.78