Calculate the pH of a 1.00 L solution that is 0.150 M in
Benzoic acid (HC7H5O2) and 0.050 M in potassium
benzoate (KC7H5O2). The pKa of benzoic acid is 4.20
This is a buffer
solution since it
contains a weak acid
and its conjugate
base (a salt of the
acid).
What is the equilibrium chemical equation?
KC7H5O2 dissociates in water to K+ and C7H5O2
-
The acid establishes an equilibrium.
HC7H5O2 + H2O ↔ H3O+ + C7H5O2-
Since this is a buffer solution, the Henderson-Hasselbalch equation
can be used: pH = pKa + Log{[conj. Base])/[Acid]}
0.050
pH = 4.20 + Log
= 3.72
0.150
1
Calculate the pH of a 1.00 L solution that is 0.150 M in Benzoic acid
(HC7H5O2) and 0.050 M in potassium benzoate (KC7H5O2). The pKa of
benzoic acid is 4.20)
Since this is a buffer solution, the Henderson-Hasselbalch equation
can be used: pH = pKa + Log{[conj. Base])/[Acid]}
0.050
pH = 4.20 + Log
= 3.72
0.150
This is the problem on slide 1
a. What will the pH be after the addition of 10.0 mL of 0.10 M HCl?
Approach: Determine the moles of each species and then decide which one
Increase in and which one decrease.
1.00 LBenzoic acid 0.150moles
 0.150molebenzoic acid
L
1.00 LKC H O 1moleC H O 1 0.050moles
7 5 2
7 5 2
 0.050moleC H O 1
7 5 2
1mol KC H O
L
7
5
2
0.0100 LHCl 0.10moles
 0.0010moleHCl
L
2
Calculate the pH of a 1.00 L solution that is 0.150 M in Benzoic acid
(HC7H5O2) and 0.050 M in potassium benzoate (KC7H5O2). The pKa of
benzoic acid is 4.20)
Since this is a buffer solution, the Henderson-Hasselbalch equation
can be used: pH = pKa + Log{[conj. Base])/[Acid]}
0.050
pH = 4.20 + Log
= 3.72
0.150
This is the problem on slide 1
a. What will the pH be after the addition of 100.0 mL of 0.10 M HCl?
Approach: Determine the moles of each species and then decide which one
Increase in and which one decrease.
 0.150molebenzoic acid
 0.050moleC H O 1
7
5
2
 0.001moleHCl
The HCl will increase the moles of acid and decrease the moles of base.
0.050-0.001
pH = 4.20 + Log
= 3.71
0.150+0.001
3
Calculate the pH of a 1.00 L solution that is 0.150 M in Benzoic acid
(HC7H5O2) and 0.050 M in potassium benzoate (KC7H5O2). The pKa of
benzoic acid is 4.20)
Since this is a buffer solution, the Henderson-Hasselbalch equation
can be used: pH = pKa + Log{[conj. Base])/[Acid]}
0.050
pH = 4.20 + Log
= 3.72
0.150
This is the problem on slide 1
b. What will the pH be after the addition of 100.0 mL of 0.10 M HCl?
Approach: Determine the moles of each species and then decide which one
Increase in and which one decrease.
 0.150molebenzoic acid
 0.050moleC H O 1
7
5
2
 0.010moleHCl
The HCl will increase the moles of acid and decrease the moles of base.
0.050-0.010
pH = 4.20 + Log
= 3.60
0.150+0.010
4
Calculate the pH of a 1.00 L solution that is 0.150 M in Benzoic acid
(HC7H5O2) and 0.050 M in potassium benzoate (KC7H5O2). The pKa of
benzoic acid is 4.20)
Since this is a buffer solution, the Henderson-Hasselbalch equation
can be used: pH = pKa + Log{[conj. Base])/[Acid]}
0.050
pH = 4.20 + Log
= 3.72
0.150
This is the problem on slide 1
c. What will the pH be after the addition of 100.0 mL of 0.10 M NaOH?
Approach: Determine the moles of each species and then decide which one
Increase in and which one decrease.
 0.150molebenzoic acid
 0.050moleC H O 1
7
5
2
 0.010mole NaOH
The NaOH will increase the moles of base and decrease the moles of acid.
0.050+0.010
pH = 4.20 + Log
= 3.83
0.150-0.010
5
Calculate the pH of a solution that is 0.150 M in hydrazine (N2H4) and 0.050
M in its conjugate acid (N2H5Cl ). The Kb of hydrazine is 1.7 x 10-6.
What the chemistry?
N2H5Cl → N2H5+ + Cl-
Dissociate the salt:
Write equilibrium expression for the weak base: N2H4 + H2O ↔ N2H5+ + OH-
[N2H5+ ][OH1- ]
Kb =
[N2H4 ]
I. 0.150
C. -x
E. 0.150 – x
[0.050 + x][x]
1.7 x 10 =
[0.150 - x]
0.050
+x
0
+x
0.050 + x
+x
-6
X = [OH1-] = 5.1 x 10-6
pOH = 5.29 and pH = 8.71
Or with the Henderson Hasselbach equation:
pH = pK a + log
Base
Conj.Acid

-14
pH = -log 10
1.7x10 -6

+ log
0.150
= 8.71
0.050
6
Calculate the pH of a 1.00 L solution that is 0.150 M in hydrazine (N2H4) and
0.050 M in its conjugate acid (N2H5Cl ). The Kb of hydrazine is 1.7 x 10-6.
What the chemistry?
Dissociate the salt:
N2H5Cl → N2H5+ + Cl-
Write equilibrium expression for the weak base: N2H4 + H2O ↔ N2H5+ + OHOr with the Henderson Hasselbach equation: (from slide number 6)
Base
pH = pK a + log
Conj.Acid

-14
10
pH = -log
1.7x10
-6

+ log
0.150
= 8.71
0.050
Determine the pH after the addition of 100.0 mL of 0.10 M HCl.
The HCl will increase the moles of conj. acid and decrease the moles of base.
moles of base = 0.150moles N2 H4
moles of conj. acid = 0.050molesN H H 
2 5
moles of HCl acid = 0.010 moles HCl
pH   log 5.88 x 10-9 + log
0.150-0.010
0.050+0.010
pH = 8.59
7
Calculate the pH of a 1.00 L solution that is 0.150 M in hydrazine (N2H4) and
0.050 M in its conjugate acid (N2H5Cl ). The Kb of hydrazine is 1.7 x 10-6.
What the chemistry?
Dissociate the salt:
N2H5Cl → N2H5+ + Cl-
Write equilibrium expression for the weak base: N2H4 + H2O ↔ N2H5+ + OHOr with the Henderson Hasselbach equation: (from slide number 6)
Base
pH = pK a + log
Conj.Acid

-14
10
pH = -log
1.7x10
-6

+ log
0.150
= 8.71
0.050
Determine the pH after the addition of 100.0 mL of 0.10 M NaOH.
The answer is on the next slide. Try working the problem first.
8
The NaOH will increase the moles of base and decrease the moles of conj. acid.
pH   log 5.88 x 10-9 + log
0.150-0.010
0.050+0.010
pH = 8.83
9