Chapter 17a
 Ionic Equilibria: Part II
 Buffers and Titration Curves
1
Chapter Goals
1.
2.
3.
4.
5.
6.
7.
8.
The Common Ion Effect and Buffer Solutions
Buffering Action
Preparation of Buffer Solutions
Acid-Base Indicators
Titration Curves
Strong Acid/Strong Base Titration Curves
Weak Acid/Strong Base Titration Curves
Weak Acid/Weak Base Titration Curves
Summary of Acid-Base Calculations
2
The Common Ion Effect and
Buffer Solutions
 If a solution is made in which the same ion is
produced by two different compounds the common
ion effect is exhibited.
 Buffer solutions are solutions that resist changes in
pH when acids or bases are added to them.

Buffering is due to the common ion effect.
3
The Common Ion Effect and
Buffer Solutions
 There are two common kinds of buffer solutions:
1 Solutions made from a weak acid plus a soluble ionic
salt of the weak acid.
2 Solutions made from a weak base plus a soluble
ionic salt of the weak base
4
The Common Ion Effect and
Buffer Solutions
1.

Solutions made of weak acids plus a soluble ionic
salt of the weak acid
One example of this type of buffer system is:


The weak acid - acetic acid CH3COOH
The soluble ionic salt - sodium acetate NaCH3COO
The weak acid reacts with bases.

 CH COO-  H 
CH 3COOH 
3
100%
Na  CH 3COO 
 Na   CH 3COO-

The salt anion (a base) reacts with acids.
5
The Common Ion Effect and
Buffer Solutions
 Example 19-1: Calculate the concentration of H+and
the pH of a solution that is 0.15 M in acetic acid and
0.15 M in sodium acetate.

This is another equilibrium problem with a starting
concentration for both the acid and anion.


CH 3COOH  H  CH 3COO
(0.15  x) M x M
100%
xM

NaCH3COO  Na  CH 3COO
0.15 M
0.15 M

0.15 M
6
The Common Ion Effect and
Buffer Solutions
 Substitute the quantities determined in the previous
relationship into the ionization expression.

H CH COO 

 1.8 10
+
Ka
-
3
CH3COOH
5

x 0.15  x 

0.15  x 
7
The Common Ion Effect and
Buffer Solutions
 Apply the simplifying assumption to both the
numerator and denominator.
0.15  x   0.15 and 0.15  x   0.15
Making these assumption s gives
0.15 x
5
 1.8 10
0.15
5

x  1.8 10 M  H
 
pH  4.74
8
The Common Ion Effect and
Buffer Solutions
 This is a comparison of the acidity of a pure acetic acid
solution and the buffer described in Example 19-1.
9
The Common Ion Effect and
Buffer Solutions
 Compare the acidity of a pure acetic acid solution and
the buffer described in Example 19-1.
Solution
0.15 M CH3COOH
[H+]
1.6 x 10-3
pH
2.80
0.15 M CH3COOH &
0.15 M NaCH3COO buffer
1.8 x 10-5
4.74
 [H+] is 89 times greater in pure acetic acid than in buffer solution.
10
The Common Ion Effect and
Buffer Solutions
 The general expression for the ionization of a weak
+


HA  H  A
monoprotic acid is:
 The generalized ionization constant expression for a
weak acid is:
H A 



Ka

 HA
11
The Common Ion Effect and
Buffer Solutions
 If we solve the expression for [H+], this relationship
results:
HA 


 H   K A 

a

acid
salt
 By making the assumption that the concentrations of
the weak acid and the salt are reasonable, the
expression reduces to:
H

acid 

  Ka  salt
12
The Common Ion Effect and
Buffer Solutions
 The relationship developed in the previous slide is
valid for buffers containing a weak monoprotic acid
and a soluble, ionic salt.
 If the salt’s cation is not univalent the relationship
changes to:
H

acid 

  Ka  n salt
where n = charge on cation
13
The Common Ion Effect and
Buffer Solutions
 Simple rearrangement of this equation and
application of algebra yields the
Henderson-Hasselbach equation.
 
log H   log K a  log
acid 
salt 
multiply by - 1
 
 log H    log K a  log

salt 
pH  pK a  log
acid 
salt 
acid 
The Henderson-Hasselbach equation is one method to calculate the pH
of a buffer given the concentrations of the salt and acid.
14
Weak Bases plus Salts of
Weak Bases
2. Buffers that contain a weak base plus the salt of a
weak base
 One example of this buffer system is ammonia plus
ammonium nitrate.
NH3
+

 H 2 O  NH 4  OH
100%

3
NH 4 NO3  NH  NO
Kb
+
4

NH OH 

 1.8 10
+
4
NH3 

5
15
Weak Bases plus Salts of
Weak Bases
 Example 19-2: Calculate the concentration of OH-
and the pH of the solution that is 0.15 M in aqueous
ammonia, NH3, and 0.30 M in ammonium nitrate,
NH4NO3.
NH3
+


 H 2 O  NH 4  OH
(0.15  x) M
xM
100%
xM

3
NH 4 NO3  NH  NO
0.30 M
+
4
0.30 M 0.30 M
16
Weak Bases plus Salts of
Weak Bases
 Substitute the quantities determined in the previous
relationship into the ionization expression for
ammonia.
Kb

NH OH 

 1.8 10

4

5
NH3 

0.30  x  x 
Kb 
 1.8  10 5
0.15  x 
The simplifyin g assumption can be applied.

0.30  x 
Kb 
 1.8 10 5
0.15


x  9.0 10 6 M  OH 
pOH  5.05 and pH  8.95
17
Weak Bases plus Salts of
Weak Bases
 A comparison of the aqueous ammonia concentration
to that of the buffer described above shows the
buffering effect.

Solution
0.15 M NH3
[OH-]
1.6 x 10-3 M
pH
11.20
0.15 M NH3 &
0.15 M NH4NO3 buffer
9.0 x 10-6 M
8.95
The [OH-] in aqueous ammonia is 180 times greater than in the buffer.
18
Weak Bases plus Salts of
Weak Bases
 We can derive a general relationship for buffer
solutions that contain a weak base plus a salt of a
weak base similar to the acid buffer relationship.

The general ionization equation for weak bases is:
:B  H 2 O 
 BH   OH 
where B represents a weak base
19
Weak Bases plus Salts of
Weak Bases
 The general form of the ionization expression is:
BH OH 



Kb

 B
 Solve for the [OH-]
OH 

 B
base
 Kb 

BH salt


20
Weak Bases plus Salts of
Weak Bases
 For salts that have univalent ions:
OH

base

  K b  salt 
 For salts that have divalent or trivalent ions:
OH

base

  Kb  nsalt 
where n = charge on anion
21
Weak Bases plus Salts of
Weak Bases
 Simple rearrangement of this equation and
application of algebra yields the
Henderson-Hasselbach equation.

log OH


base 
  log K b  log salt 
multiply by - 1

salt 
 log OH    log K b  log
base

salt 
pOH  pK b  log
base

22
Buffering Action
 These movies show that buffer solutions resist
changes in pH.
23
Buffering Action
 Example 19-3: If 0.020 mole of gaseous HCl is
added to 1.00 liter of a buffer solution that is
0.100 M in aqueous ammonia and 0.200 M in
ammonium chloride, how much does the pH
change? Assume no volume change due to
addition of the HCl.
1 Calculate the pH of the original buffer solution.
24
Buffering Action
NH3 
OH   K  NH Cl
0.10M
OH   1.8 10  0.20M
OH   9.0 10 M
-
b
4
-
5
-
6
pOH  5.05
pH  8.95
25
Buffering Action
2 Next, calculate the concentration of all species
after the addition of the gaseous HCl.


The HCl will react with some of the ammonia and
change the concentrations of the species.
This is another limiting reactant problem.
HCl 
NH3  NH 4 Cl
Initial
0.020 mol 0.100 mol 0.200 mol
Change - 0.020 mol - 0.020 mol + 0.020 mol
After rxn. 0 mol
0.080 mol
0.220 mol
26
Buffering Action
HCl 
NH 3  NH 4 Cl
Initial
0.020 mol 0.100 mol 0.200 mol
Change - 0.020 mol - 0.020 mol + 0.020 mol
After rxn. 0 mol
0.080 mol
0.080 mol
M NH 3 
 0.080 M
1.0 L
0.220 mol
M NH 4Cl 
 0.220 M
1.0 L
0.220 mol
27
Buffering Action
3 Using the concentrations of the salt and base and
the Henderson-Hassselbach equation, the pH can
be calculated.
NH3 
OH   K  NH Cl
0.080M
OH   1.8 10  0.220M

b
4

5
28
Buffering Action
OH   K

OH 
OH 


b

 NH 3 
 NH 4Cl
0.080 M
 18
.  10 
0.220 M
6
 6.5  10 M
pOH  5.19
5
pH  8.81
29
Buffering Action
4 Finally, calculate the change in pH.
pH  pH new  pH original
pH = 8.81 - 8.95 = -0.14
30
Buffering Action
 Example 19-4: If 0.020 mole of NaOH is added to
1.00 liter of solution that is 0.100 M in aqueous
ammonia and 0.200 M in ammonium chloride, how
much does the pH change? Assume no volume
change due to addition of the solid NaOH.
You do it!
31
Buffering Action
 pH of the original buffer solution is 8.95, from above.
1. First, calculate the concentration of all species after
the addition of NaoH.


NaOH will react with some of the ammonium chloride.
The limiting reactant is the NaOH.
NH4 Cl 
Initial
0.200 mol
Change - 0.020 mol
After rxn. 0.180 mol
NaOH  NH3  H 2 O  NaCl
0.020 mol 0.100 mol
- 0.020 mol + 0.020 mol
0 mol
0.120 mol
32
Buffering Action
NH 4 Cl 
Initial
0.200 mol
Change - 0.020 mol
NaOH  NH3  H 2 O  NaCl
0.020 mol 0.100 mol
- 0.020 mol + 0.020 mol
After rxn. 0.180 mol
0 mol
0.120 mol
M NH3 
 0.120 M
1.0 L
0.180 mol
M NH4Cl 
 0.180 M
1.0 L
0.120 mol
33
Buffering Action
2 Calculate the pH using the concentrations of the
salt and base and the Henderson-Hasselbach
equation.
NH3 
OH   K  NH Cl
0.120 M
OH   1.8 10  0.180M
OH   1.2 10 M

b
4

5

5
pOH  4.92
pH  9.08
34
Buffering Action
3 Calculate the change in pH.
pH = pH new  pH original
pH = 9.08 - 8.95 = 0.13
35
Buffering Action
 This table is a summary of examples 19-3 and 19-4.
Original Solution
1.00 L of solution
containing
0.100 M NH3 and
0.200 M NH4Cl
Original Acid or New
base
pH
pH
pH
added
0.020
mol
9.08 +0.13
NaOH
8.95
0.020
mol
8.81 -0.14
HCl
 Notice that the pH changes only slightly in each case.
36
Preparation of Buffer Solutions
 This move shows how to prepare a buffer.
37
Preparation of Buffer Solutions
 Example 19-5: Calculate the concentration of H+ and
the pH of the solution prepared by mixing 200 mL of
0.150 M acetic acid and 100 mL of 0.100 M sodium
hydroxide solutions.
 Determine the amounts of acetic acid and sodium
hydroxide prior to the acid-base reaction.
0.15 mmol
? mmol CH3COOH = 200 mL 
 30.0 mmol CH3COOH
mL
0.100 mmol
? mmol NaOH = 100 mL 
 10.0 mmol NaOH
mL
38
Preparation of Buffer Solutions
 Sodium hydroxide and acetic acid react in a 1:1 mole
ratio.
Initial
Change
After rxn.
NaOH + CH 3COOH  Na CH 3COO + H 2O
10.0 mmol 30.0 mmol
-10.0 mmol -10.0 mmol + 10.0 mmol
0
20.0 mmol
10.0 mmol
39
Preparation of Buffer Solutions
 After the two solutions are mixed, the total volume of
the solution is 300 mL (100 mL of NaOH + 200 mL of
acetic acid).

The concentrations of the acid and base are:
M CH 3COOH
M NaCH 3COO
20.0 mmol

 0.0667 M CH3COOH
300mL
10.0 mmol

 0.0333M NaCH3COO
300mL
40
Preparation of Buffer Solutions
 Substitution of these values into the ionization
constant expression (or the Henderson-Hasselbach
equation) permits calculation of the pH.
H CH COO 



Ka  18
.  10
H 

5

3
CH 3COOH 
18
.  10 0.0667


 3.6  10
pH  4.44
5
0.0333
5
M
41
Preparation of Buffer Solutions
 For biochemical situations, it is sometimes important
to prepare a buffer solution of a given pH.
 Example 19-6:Calculate the number of moles of solid
ammonium chloride, NH4Cl, that must be used to
prepare 1.00 L of a buffer solution that is 0.10 M in
aqueous ammonia, and that has a pH of 9.15.

Because pH = 9.15, the pOH can be determined.
pOH = 14.00 - 9.15 = 4.85
OH   10
-
4.85
5
 14
.  10 M
42
Preparation of Buffer Solutions
 The appropriate equilibria representations are:
NH3 + H 2 O
0.10  1.4 10 M
 NH 

4
5

OH-
1.4 10 5 M 1.4 10 5 M
NH4 Cl  NH4  Cl
xM
xM
xM
43
Preparation of Buffer Solutions
 Substitute into the ionization constant expression (or
Henderson-Hasselbach equation) for aqueous
ammonia
Kb

NH OH 

 1.8 10

4

NH3 

1.4 10  x 1.4 10 

0.10  1.4 10 
5
Kb
5
5
5
apply the simplifyin g assumption
44
Preparation of Buffer Solutions
Kb

NH OH 

 1.8 10

4

NH3 

1.4  10  x 1.4  10 

0.10  1.4 10 
5
Kb
5
5
5
The simplifyin g assumption can be applied.

x  1.4  10 5 
Kb 
 1.8  10 5
0.10
x  0.13 M NH 4 Cl = NH 4 Cloriginal
? g NH 4 Cl 0.13 mol 53 g


 6.9 g/L
L
L
mol
45
Acid-Base Indicators
 The point in a titration at which chemically equivalent amounts
of acid and base have reacted is called the equivalence point.
 The point in a titration at which a chemical indicator changes
color is called the end point.
 A symbolic representation of the indicator’s color change at the
end point is:



HIn  H  In
Color 1
Color 2
46
Acid-Base Indicators
 The equilibrium constant expression for an indicator
would be expressed as:

H In 


Ka

HIn 
47
Acid-Base Indicators
 If the preceding expression is rearranged the range
over which the indicator changes color can be
discerned.
In  
-
 HIn
Ka

H
 
48
Acid-Base Indicators
Color change ranges of some acid-base indicators
Indicator
Methyl violet
Color in
acidic range pH range
Yellow
0-2
Color in
basic range
Purple
Methyl orange
Pink
3.1 – 4.4
Yellow
Litmus
Red
4.7 – 8.2
Blue
Phenolphthalein
Colorless
8.3 – 10.0
Red
49
Titration Curves
Strong Acid/Strong Base Titration Curves
 These graphs are a plot of pH vs. volume of acid or
base added in a titration.
 As an example, consider the titration of 100.0 mL of
0.100 M perchloric acid with 0.100 M potassium
hydroxide.
 In this case, we plot pH of the mixture vs. mL of
KOH added.
 Note that the reaction is a 1:1 mole ratio.
HClO 4  KOH  KClO 4  H2O
50
Strong Acid/Strong Base
Titration Curves
 Before any KOH is added the pH of the HClO4
solution is 1.00.

Remember perchloric acid is a strong acid that ionizes
essentially 100%.
100%

HClO 4  H  ClO
0.100M

4
0.100M 0.100 M
H   0.100M

pH  log(0.100)  1.00
51
Strong Acid/Strong Base Titration
Curves
 After a total of 20.0 mL 0.100 M KOH has been
added the pH of the reaction mixture is ___?
HClO 4  KOH  KClO 4  H 2 O
Start : 10.0 mmol 2.0 mmol
Change : - 2.0 mmol - 2.0 mmol  2.0 mmol
After
rxn.
8.0 mmol 0.0 mmol
2.0 mmol
8.0 mmol HClO 4
M HClO4 
 0.067 M
120 mL
H   0.067 M pH  1.17
 
52
Strong Acid/Strong Base
Titration Curves
 After a total of 50.0 mL of 0.100 M KOH has been
added the pH of the reaction mixture is ___?
HClO 4  KOH  KClO 4  H 2 O
Start : 10.0 mmol 5.0 mmol
Change : - 5.0 mmol - 5.0 mmol  5.0 mmol
After
rxn.
5.0 mmol 0.0 mmol
5.0 mmol
5.0 mmol HClO 4
M HClO4 
 0.033M
150 mL
H   0.033M pH  1.48
 
53
Strong Acid/Strong Base
Titration Curves
 After a total of 90.0 mL of 0.100 M KOH has been
added the pH of the reaction mixture is ____?
HClO 4  KOH  KClO 4  H 2 O
Start : 10.0 mmol 9.0 mmol
Change : - 9.0 mmol - 9.0 mmol  9.0 mmol
After
rxn.
1.0 mmol 0.0 mmol
9.0 mmol
1.0 mmol HClO 4
M HClO4 
 0.0053M
190 mL
H   0.0053M pH  2.28
 
54
Strong Acid/Strong Base
Titration Curves
 After a total of 100.0 mL of 0.100 M KOH has been
added the pH of the reaction mixture is ___?
HClO 4  KOH  KClO 4  H 2 O
Start :
10.0 mmol 10.0 mmol
Change : - 10.0 mmol - 10.0 mmol  10.0 mmol
After
0.0 mmol 0.0 mmol
rxn.
No acid or base  neutral
pH  7.00
10.0 mmol
55
Strong Acid/Strong Base
Titration Curves
 We have calculated only a few points on the titration
curve. Similar calculations for remainder of titration
show clearly the shape of the titration curve.
56
Weak Acid/Strong Base
Titration Curves
 As an example, consider the titration of 100.0 mL of
0.100 M acetic acid, CH3 COOH, (a weak acid) with
0.100 M KOH (a strong base).

The acid and base react in a 1:1 mole ratio.
1 mol
1mol
1mol
CH 3COOH + KOH  K +CH 3COO- + H 2O
1mmol
1mmol
1mmol
57
Weak Acid/Strong Base
Titration Curves
 Before the equivalence point is reached, both
CH3COOH and KCH3COO are present in solution
forming a buffer.
 The KOH reacts with CH3COOH to form KCH3COO.
 A weak acid plus the salt of a weak acid form a buffer.
 Hypothesize how the buffer production will effect the
titration curve.
58
Weak Acid/Strong Base
Titration Curves
1. Determine the pH of the acetic acid solution before
the titration is begun.

Same technique as used in Chapter 18.


CH 3COOH  CH 3COO  H
0.10  x M
xM

H CH COO 

 1.8 10
+
Ka
xM
-
3
CH3COOH

x  x 
Ka 
 1.8 10 5
0.10  x 
5
59
Weak Acid/Strong Base
Titration Curves
 CH COO-  H 
CH 3COOH 
3
0.10  x  M
xM

H CH COO 

 1.8 10
+
Ka
xM
-
3
CH3COOH

x x 
Ka 
 1.8 10 5
0.10  x 
5
The simplifyin g assumption can be applied.
x 2  1.8 10 6  x = 1.3 10-3
H   1.3 10

-3
 pH  2.89
60
Weak Acid/Strong Base
Titration Curves
 After a total of 20.0 mL of KOH solution has been
added, the pH is:
Initial:
KOH +
2.00 mmol
CH 3COOH  K +CH 3COO-  H 2O
10.0 mmol
Chg. due to rxn:-2.00 mmol - 2.00 mmol
After rxn:
0.00 mmol 8.00 mmol
8.0 mmol
M CH 3COOH 
 0.067 M
120 mL
2.0 mmol
M CH COO- 
 0.017 M
3
120 mL
+ 2.00 mmol
2.00 mmol
61
Weak Acid/Strong Base
Titration Curves
Ka  18
.  10
 
H
5



H  CH 3COO 
CH 3COOH 
CH 3COOH 

5
 18
.  10 
 

CH 3COO 


0.067
H  18
.  10 
 7.1  105 M
0.017
pH  4.15

5
 Similarly for all other cases before the equivalence
point is reached.
62
Weak Acid/Strong Base
Titration Curves
 At the equivalence point, the solution is 0.500
M in KCH3COO, the salt of a strong base
and a weak acid which hydrolyzes to give a
basic solution.


This is a solvolysis process as discussed in
Chapter 18.
Both processes make the solution basic.
 The solution cannot have a pH=7.00 at
equivalence point.
 Let us calculate the pH at the equivalence
point.
63
Weak Acid/Strong Base
Titration Curves
1. Set up the equilibrium reaction:
KOH + CH 3COOH  K +CH 3COO-  H 2O
Initial:
10.0 mmol 10.0 mmol
Chg. due to rxn:-10.0 mmol -10.0 mmol + 10.0 mmol
After rxn:
0.0 mmol
0.0 mmol
10.0 mmol
64
Weak Acid/Strong Base
Titration Curves
2. Determine the concentration of the salt in solution.
M KCH3COO
M KCH3COO
10.0 mmol
=
 0.0500M
200 mL
 0.0500M  0.0500M CH3COO
65
Weak Acid/Strong Base
Titration Curves
3. Perform a hydrolysis calculation for the potassium
acetate in solution.

CH 3COO  H 2 O  CH 3COOH  OH-
0.0500  x M
xM

CH 3COOHOH- 
K =
 5.6 10 11
b
xM
CH COO 

3
x x 
x2
Kb =

 5.6 10 10
0.0500  x  0.0500

x 2  2.8 10 11  x  5.27 10 6  OH 
pOH  5.28  pH  8.72

66
Weak Acid/Strong Base
Titration Curves
4. After the equivalence point is reached, the pH is
determined by the excess KOH just as in the strong
acid/strong base example.
KOH + CH 3COOH  K + CH 3COO-  H 2 O
Initial :
11.0 mmol 10.0 mmol
Chg. due to rxn : -10.0 mmol - 10.0 mmol
After rxn :
1.00 mmol 0.00 mmol
1.0 mmol
M KOH 
 4.8 10 3 M KOH
210 mL
OH   4.8 10 3 M
 pOH  2.32 and pH = 11.68

+ 10.00 mmol
10.00 mmol

67
Weak Acid/Strong Base Titration
Curves
 We have calculated only a few points on the titration
curve. Similar calculations for remainder of titration
show clearly the shape of the titration curve.
68
Strong Acid/Weak Base
Titration Curves
 Titration curves for Strong Acid/Weak Base
Titration Curves look similar to Strong
Base/Weak Acid Titration Curves but they are
inverted.
69
Weak Acid/Weak Base
Titration Curves
 Weak Acid/Weak Base Titration curves have
very short vertical sections.
 The solution is buffered both before and after
the equivalence point.
 Visual indicators cannot be used.
70
Synthesis Question
 Bufferin is a commercially prepared medicine
that is literally a buffered aspirin. How could
you buffer aspirin? Hint - what is aspirin?
71
Synthesis Question
 Aspirin is acetyl salicylic acid. So to buffer it
all that would have to be added is the salt of
acetyl salicylic acid.
72
Group Question
 Blood is slightly basic, having a pH of 7.35 to
7.45. What chemical species causes our
blood to be basic? How does our body
regulate the pH of blood?
73
Group Question
74
End of Chapter 19
 We have examined :
1
2
3
Gas phase equilibria in Chapter 17
Hydrolysis equilibria in Chapter 18
Acid/base equilibria in Chapter 19
 Chapter 20 is the last equilibrium chapter.

It involves solid/solution equilibria.
75