1. Salts
neutralization reactions
acids
bases
strong acid+
strong base
non-hydrolyzing salt
weak acid+
strong base
hydrolyzing salt
strong acid+
weak base
hydrolyzing salt
weak acid+
weak base
hydrolyzing salt
CH3COOH
NaOH
1.00 M
1.00 M
500 mL

2. CH3COOH  CH3COO- + H+ 1.00 M Ka = 1.8 x 10-5 = x2 500 mL 1.00 - x I C
0.00
0.00 C -x +x +x E x x x
x =
4.24 x 10-3
= [H+] .
pH = 2.37

3. Equivalence Point
CH3COOH
NaOH
1.00 M
mol acid
= mol base
1.00 M
500 mL
1.00 mol L
x L
= mol acid
0.500 mol base 
1.00 mol L
= L base
CH3COOH 
CH3COO-
+ H+
+ OH- 
H2O
+ CH3COO-
[CH3COO-] =
0.500
mol
= 0.50 M
0.500 L
CH3COO-
+ H2O 
CH3COOH
+ OH-
strong conjugate base

4. Equivalence Point
CH3COOH
NaOH
1.00 M
1.00 M
500 mL
CH3COO-
+ H2O 
CH3COOH
+ OH-
Kb = Kw
= 1 x 10-14
= 5.56 x 10-10
1.8 x 10-5 . Ka .
x =
[OH-]
= 1.67 x 10-5
pH =
9.22
5.56 x =
[CH3COOH]
[OH-]
= x2
[CH3COO-] x

5. CH3COOH NaOH 1.00 M 0.100 mol H+ = 0.100 mol OH- 1.00 M 500 mL
add 100 mL
0.500 mol
mol
= .400 mol CH3COOH
= M .
0.600 L
0.100 mol
= M CH3COO-
0.600 L .
CH3COOH 
CH3COO-
+ H+
[CH3COOH]
[CH3COO-]
[H+] I
0.667
0.167
0.00
Ka = 1.8 x 10-5 =
( x)
(x) C -x +x +x
(0.667 – x) E x x x
x =
7.20 x 10-5
pH =
4.14

6. presence of conjugate base inhibits dissociation
Common ion effect
presence of conjugate base
inhibits dissociation
weak acid or base +
conjugate base or acid . . .
CH3COOH 
CH3COO-
+ H+ .
[CH3COOH]
[CH3COO-]
[H+] I
0.667
0.00
0.00
Ka = 1.8 x 10-5 =
(x)
(x) C -x +x +x
(0.667 – x) E x x x
x =
3.46 x 10-3
pH =
2.46

7. Henderson-Hasselbalch Equation H+
+ A-
Ka =
[H+]
[A-]
[H+] = Ka
[HA]
[A-]
[HA]
- log [H+]
= - log Ka
+ log
[A-]
[HA] pH
= pKa
+ log
[A-]
[HA]

8. Henderson-Hasselbalch Equation
half-way point
CH3COOH
NaOH
1.00 M
0.250 mol H+
= mol OH-
1.00 M
500 mL
add 250 mL
0.500 mol
mol
= .250 mol CH3COOH
= M .
0.750 L
0.250 mol
= M CH3COO-
0.750 L .
Henderson-Hasselbalch Equation pH
= pKa
+ log
[A-]
[HA]
pH = pKa
pH =
4.74
+ log
.333
= 4.74
.333

9. Henderson-Hasselbalch Equation
CH3COOH
NaOH
1.00 M
0.375 mol H+
= mol OH-
1.00 M
500 mL
add 375 mL
0.500 mol
mol
= .125 mol CH3COOH
= M .
0.875 L
0.375 mol
= M CH3COO-
0.875 L .
Henderson-Hasselbalch Equation pH
= pKa
+ log
[A-]
[HA]
pH =
4.74
+ log
.429
= 5.22
.143

10. pH
= pKa
+ log
[A-]
[HA]
Kb = 1.8 x 10-3
What is the pH of 100 mL of a 0.10 M solution of NH3
a) b) c) d) e) 12.63
What volume of 1.00 M HCl is needed to reach the
equivalence point?
a) 1000 mL b) 100 mL c)10 mL d) 1 mL e) 0.1 mL
What is the pH at the equivalence point?
a) b) c) 3.25 d) e)
What is the pH at the half-way point?
a) b) c) 7.00 d) e)