1. CH 17 #42 – weak base titration
CH 17 #42 a– weak base ionization in pure water.
H20 + NH3  NH4+ + OH-
THIS IS BEFORE THE TITRATION, NO H+ ADDED YET.
Kb = [NH4+][OH-]
[NH3]
1.8 * 10-5 = [X][X]
[0.030-X]
1.8 * 10-5 = [X][X]
[0.030]
X = * 10-4 = [OH-] ; pOH- =3.14 ; pH = 10.86

2. NEUTRAIZATION STOICHIOMETRY
CH 17 #42 b– weak base TITRATION AFTER L OF H+ IS ADDED TO THE BASE SOLUTION IN PART b.
H+ ADDED = M * L = 2.5 * 10-4 mol
NH3 INITIAL = M ( 9.0 * 10-4 MOLE)
Kb = [NH4+][OH-]
[NH3]
NEUTRAIZATION STOICHIOMETRY
(MOLES)
NH3 H+
 NH4+
9.0*10-4
2.5*10-4 -X X
6.5*10 -4
2.5*10-4 MOL /0.040L=6.25*10-3 MOLAR
1.8 * 10-5 = [6.25*10-3][X]
[ ]
X = 4.6*10-5 MOLAR : pH = 9.66
6.5*10-4/0.040L =

3. NEUTRAIZATION STOICHIOMETRY
CH 17 #42 c– weak base TITRATION AFTER L OF H+ IS ADDED TO THE BASE SOLUTION IN PART c.
H+ ADDED = M * L = 5.0 * 10-4 mol
NH3 INITIAL = M ( 9.0 * 10-4 MOLE)
Kb = [NH4+][OH-]
[NH3]
NEUTRAIZATION STOICHIOMETRY
(MOLES)
NH3 H+
 NH4+
9.0*10-4
5.0*10-4 -X X
4.0*10 -4
5.0*10-4 MOL /0.050L=0.010 MOLAR
1.8 * 10-5 = [0.010][X]
[0.008]
OH- = X = 1.44*10-5 MOLAR : pH = 9.19
4.0*10-4/0.050L = 0.008M

4. NEUTRAIZATION STOICHIOMETRY
CH 17 #42 d– weak base TITRATION AFTER L OF H+ IS ADDED TO THE BASE SOLUTION IN PART d.
H+ ADDED = M * L = 8.75 * mol
NH3 INITIAL = M ( 9.0 * 10-4 MOLE)
Kb = [NH4+][OH-]
[NH3]
NEUTRAIZATION STOICHIOMETRY
(MOLES)
NH3 H+
 NH4+
9.0*10-4
8.75*10-4 -X X
0.25*10 -4
8.75*10-4 MOL /0.065L= MOLAR
1.8 * 10-5 = [ ][X]
[4.0*10-4]
OH- = X = 5.349*10-7 MOLAR : pH = 7.73
0.25*10-4/0.065L = 4.0*10-4M

5. NEUTRAIZATION STOICHIOMETRY WILL NOW UNDERGO HYDROLYSIS
CH 17 #42 e– weak base TITRATION AFTER L OF H+ IS ADDED TO THE BASE SOLUTION IN PART e.
H+ ADDED = M * L = 9.0 * 10-4 MOLE
NH3 INITIAL = M ( 9.0 * 10-4 MOLE)
♪ NOTE: THE MOLES OF H+ ADDED = THE MOLES OF BASE INITIAL, YOU ARE AT EQUIVALENCE!
NEUTRAIZATION STOICHIOMETRY
(MOLES)
NH3 H+
 NH4+
9.0*10-4 -X X
WILL NOW UNDERGO HYDROLYSIS

6. CH 17 #42 e(continued)– weak base TITRATION AT EQUIVALANCE
CH 17 #42 e(continued)– weak base TITRATION AT EQUIVALANCE. H2O + NH4+  NH3 + H+
NH4+ 
 NH3 H+
9.0*10-4 -X X
9.0*10-4 -X
Ka = [NH3][H+]
[NH4+]
5.56*10-10 = [X][X]
[9.0*10-4mol/0.660 L]
5.56*10-10 = [X][X]
[ M]
X = 2.75 * M H+
pH = 5.56