1

2

3

4

0.2 M
CH
3
COOH ( aq )
K a
= 1.8 x 10 -5
H +
( aq ) + CH
3
COO
-
( aq )
Initial ( M ):
Change ( M ):
Equilibrium ( M ):
+
-3
AcOH( aq )
0.2
x
0.2 x
M
H + ( aq ) +
0.00
+ x x
Ac (aq )
0.00
+ x x
-3
3
Common Ion Effect !
5

The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.
Specific case of LCP
CH
3
COOH ( aq ) H +
( aq ) + CH
3
COO
Then we add some CH
3
COONa (a strong electrolyte).
-
( aq )
CH
3
COONa ( s ) Na +
( aq ) + CH
3
COO
-
( aq ) common ion
The presence of a common ion suppresses the ionization of a weak acid or a weak base. It will influence the pH.
What is the pH if I add 0.30 M CH
3
COONa?
6

Example 16.1
(a) Calculate the pH of a 0.20 M CH
3
COOH solution. (2.72)
(b) What is the pH of a solution containing both 0.20 M CH
3 and 0.30 M CH
3
COONa? The K a of CH
3
COOH is 1.8 x 10 -5 .
COOH
CH
3
0.20 M
COOH( aq )
?
H + ( aq ) + CH
3
COO ( aq )
CH
3
COONa(aq) → Na + (aq) + CH
3
COO (aq)
0.30 M 0.30 M
7

Example 16.1
(a) Calculate the pH of a 0.20 M CH
3
COOH solution. (2.72)
(b) What is the pH of a solution containing both 0.20 M CH
3 and 0.30 M CH
3
COONa? The K a of CH
3
COOH is 1.8 x 10 -5 .
COOH
Initial ( M
Change (
):
M ):
Equilibrium ( M ):
CH
3
COOH( aq )
0.20
x
0.20x
H + ( aq ) + CH
3
COO ( aq )
+
0 x x
0.30
+ x
0.30
+x
K a
=
+
3
-
[H ][CH COO ]
0.20x
[0.2]
If > 400
K a we can neglect x x
0.20x x
 x
0.20
x = [H + ] = 1.2 x 10 -5 M
-log (1.2 x 10 -5 ) = 4.92
8

Example 16.1
(a) Calculate the pH of a 0.20 M CH
3
COOH solution. (2.72)
(b) What is the pH of a solution containing both 0.20 M CH
3 and 0.30 M CH
3
COONa? The K a of CH
3
COOH is 1.8 x 10 -5 .
COOH
(4.92)
CH
3
COOH ( aq ) H +
( aq ) + CH
3
COO
Then we add some CH
3
COONa (a strong electrolyte).
-
( aq )
CH
3
COONa ( s ) Na +
( aq ) + CH
3
COO
-
( aq ) common ion
H + pH
There is an easier way to calculate the pH for a common ion solution!
9

Consider mixture of weak acid HA and salt NaA.
K a
=
[H + ][A
-
]
[HA]
HA (aq)
NaA (s)
H +
(aq) + A -
(aq)
Na +
(aq) + A
-
(aq)
[H + ] =
K a
[HA]
[A
-
]
Henderson-Hasselbalch equation
-log [H + ] = -log K a
- log
[HA]
[A
-
] pH = pK a
+ log
[conjugate base]
[acid]
-log [H + ] = -log K a
+ log
[A
-
]
[HA] pH = pK a
[A ]
+ log
[HA] pK a
= -log K a
10
Assumption: that the [A ] from NaA >> [A ] from HA

Example 16.1
(a) Calculate the pH of a 0.20 M CH
3
COOH solution. (2.72)
(b) What is the pH of a solution containing both 0.20 M CH
3 and 0.30 M CH
3
COONa? The K a of CH
3
COOH is 1.8 x 10 -5 .
COOH
CH
3
COOH ( aq ) H +
( aq ) + CH
3
COO
-
( aq )
Then we add some CH
3
COONa (a strong electrolyte).
CH
3
COONa ( s ) Na +
( aq ) + CH
3
COO
-
( aq ) pH = pK a
[A
-
]
+ log
[HA] pH = -log (1.8 x 10 -5 ) + log
[0.30]
[0.20]
= 4.92
11

12

Sacrificial portion of the car that protects the passengers.
“They don’t make em’ like they use to!”
2000 Chevrolet Malibu vs. 1959 Chevrolet Bel Air (30 mph)
Buffer- the chemical equivalent to a crumple zone.
13

A buffer solution consists of:
1. A weak acid (or weak base) and
HA (or B -
2. The salt of the weak conjugate base (or weak acid)
Both must be present!
A (or BH)
)
• If there is a high concentration of A and HA , the buffer solution has the ability to resist changes in pH upon the addition of small amounts of either strong acids or bases.
• The presence of a conjugate acid-base pair of a weak acid “ties up” small amounts of H + or OH – ions added to the solution.
14

15

Without Buffer
H +
H +
[HCl] = [H + ]
Without Buffer
HA
A -
A -
A -
HA
HA
OH -
OH -
HA
AH
AH
A -
HA
HA
A -
A -
A -
HA
A -
A -
[NaOH] = [OH ]
A – (aq) + H + (aq)

HA(aq)
[HCl] > [H + ]
Some tied up as HA
HA(aq) + OH – (aq)

A – (aq) + H
2
O(l)
[NaOH] > [OH ]
Some tied up as H
2
O

Water vs.
Buffer w/
1.0 M CH
3
COOH
1.0 M CH
3
COONa
In Water
HCl ( aq ) H +
( aq ) + Cl (aq)
In Buffer
H +
( aq ) + CH
3
COO -
( aq )
If you add enough HCl, all of the
CH
3
COO
HCl
will be consumed.

H + + Cl Wins!
CH
3
COOH ( aq )
17

Example 16.2
Which of the following solutions can be classified as buffer systems?
(a) KH
2
PO
4
/H
3
PO
4
1) Conjugate acid-base?
2) Weak acid/weak base?
(b) NaClO
4
/HClO
4
(c) C
5
H
5
N/C
5
H
5
NHCl
18

Example 16.2
Which of the following solutions can be classified as buffer systems?
(a) KH
2
PO
4
/H
3
PO
4
H
3
PO
4 is a weak acid, and its conjugate base,
H
2
PO
4
,is a weak base. It is a buffer.
(b) NaClO
4
/HClO
4
HClO
4 is a strong acid, its conjugate base, ClO is an extremely weak base. It is not a buffer.
4
,
(c) C
5
H
5
N/C
5
H
5
NHCl
C
5
H
5
N is a weak base and its conjugate acid, C
5
H
5
N + H
, is a weak acid. Therefore, this is a buffer system.
19

H + or OH more
H + or OH -
H +
20
OH -


Buffer capacity is its ability to resist pH changes.

The more concentrated the components of a buffer, the greater the buffer capacity.

A buffer also has the highest capacity when the component concentrations are equal: pH
 p K a
 log
[ A

]
[ HA ]
 p K a
 log 1
 p K a

A buffer whose pH is equal to or near the pK a component has the highest buffer capacity.
of its acid

The addition of strong acid or base weakens the buffer capacity.
21

Example 16.3
(a) Calculate the pH of a buffer system containing 1.0 M CH
3 and 1.0 M CH
3
COONa. The K a of CH
3
COOH
COOH is 1.8 x 10 -5 (Table
15.3).
(b) What is the pH of the buffer system after the addition of 0.10 mole of gaseous HCl to 1.0 L of the solution? Assume that the volume of the solution does not change when the HCl is added.
22

Example 16.3
(a) Calculate the pH of a buffer system containing 1.0 M CH
3 and 1.0 M CH
3
COONa. The K a of CH
3
COOH
COOH is 1.8 x 10 -5 (Table
15.3).
[A
-
] pH = pK a
+ log
[HA] pK a
= -log K a pH = -log (1.8 x 10 -5 ) + log
[1.0]
[1.0]
= 4.74
23

Example 16.3
(a) Calculate the pH of a buffer system containing 1.0 M CH
3 and 1.0 M CH of CH
3
COOH
COOH is 1.8 x 10 -5 (Table
15.3). (4.74)
3
COONa. The K a
(b) What is the pH of the buffer system after the addition of 0.10 mole of gaseous HCl to 1.0 L of the solution?
0.10 M
0.10 M
Initial (mol):
Change (mol):
Final (mol):
CH
3
COO ( aq ) + H + ( aq ) → CH
3
COOH( aq )
1.0
0.10
1.0
-0.10
-0.10
+0.10
0.90
0 1.1
pH = -log (1.8 x 10 -5 ) + log
[0.9]
[1.1]
= 4.66
24


Choose a target pH. Depends on what you want to study.

Choose the appropriate choice of a weak acid and conjugate base.

Most effective buffer when the pK a is closest to the target pH.
Either:

Mix the Acid/salt in the appropriate ratio.
or

The acid and its salt are mixed in a 1 to 1 ratio.

The pH is then adjusted by adding a strong acid or base until it equals your target pH.

Example 16.4
Describe how you would prepare a “phosphate buffer” with a pH of 7.40.
[A
-
] pH = pK a
+ log
[HA]
Known (7.40) Find the ratio
Choose the right acid/salt
Most Effective when pH = pK a
7.4 = -log (6.2 x 10 -8 ) + log
[A ]
[HA] 2-
[HPO ]
2
-
[H PO ]
0.19
= 10 = 1.5
26

Example 16.4
Describe how you would prepare a “phosphate buffer” with a pH of 7.40.
[A
-
] pH = pK a
+ log
[HA]
Known (7.40)
4
2
-
[H PO ]
0.19
= 10 = 1.5
Find the ratio
Choose the right acid/salt
Make the buffer:
Dissolve disodium hydrogen phosphate (Na
2
HPO
4
) and sodium dihydrogen phosphate (NaH
2
PO
4
) in a mole ratio of 1.5:1.0 in water. or
Dissolve disodium hydrogen phosphate (Na
2
HPO
4
) and sodium dihydrogen phosphate (NaH
2
PO
4
) in a mole ratio of 1.0:1.0 in water. Then add 0.2 equivalence of a strong base.
7.40 = 7.21 + log
[1+x]
[1-x] x = 0.2
27

28

1) K w
= [H + ][OH ] = 1.0 x 10 -14
2) pH =
log [H + ]
3) pOH =
log [OH]
B ( aq ) + H
2
O(l) BH( aq ) + OH ( aq )
[BH][OH
-
]
K b
=
[B ]
4) pH + pOH = 14
5) K a
K b
= K w
HA
( aq )
K a
=
H +
( aq )
+ A -
( aq )
[H + ][A
-
]
[HA]
Henderson-Hasselbalch equation
HA (aq) H +
(aq) + A -
(aq)
NaA (s) Na +
(aq) + A
-
(aq) pH = pK a
+ log
[A ]
[HA]
Assumption: that the [A ] from NaA >> [A ] from HA
29

30

In a titration , a solution of accurately known concentration is gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
• Redox titration
• Complexometric indicator
• Acid-base titration

Acid-Base Titration is a neutralization process aimed at adding a necessary amount of acid/H + (or base/OH ) to a solution of base/OH (or acid/H + ) in order to reach the equivalence point.

At the equivalence point , the number of moles of added H
3
O + (or
OH – ) ions equals the number of moles of OH – (or H
3
O + ) ions originally present.
31

Strong base to a strong acid.
• Start with [unknown] solution.
• Add known amounts of an acid or base.
• Monitor pH with respect to amount added.
• Graph pH vs concentration.
• Find equivalence point.
• Find pK a of weak acid or base.
Strong base to a weak acid.
Strong acid to a weak base.
32

NaOH HCl
NaOH ( aq ) + HCl ( aq ) H
2
O ( l ) + NaCl ( aq )
Na + + OH + H + + Cl H
2
O ( l ) + Na + + Cl -
OH ( aq ) + H + ( aq ) H
2
O ( l )
Strong means they completely dissociate.
25.0 mL of
0.1 M HCl adding 0.1 M NaOH
33

NaOH ( aq ) + HCl ( aq ) H
2
O ( l ) + NaCl ( aq )
Start with:
25.0 mL of
0.1 M HCl
Since they completely react we can easily calculate the pH when:
1) After the addition of 0 mL of 0.1 M NaOH .
0.1 M HCl = 0.1 M H + pH = -log [H + ] pH = 1
Adding:
0.1 M NaOH
34

NaOH ( aq ) + HCl ( aq ) H
2
O ( l ) + NaCl ( aq )
Start with:
25.0 mL of
0.1 M HCl
Since they completely react we can easily calculate the pH when:
1) After the addition of 10 mL of 0.1 M NaOH .
moles of NaOH/OH conc of HCl/H + after the reaction moles of HCl/H + total volume (25 + 10 mL)
Adding:
0.1 M NaOH
35

NaOH ( aq ) + HCl ( aq ) H
2
O ( l ) + NaCl ( aq )
Start with:
25.0 mL of
0.1 M HCl
Since they completely react we can easily calculate the pH when:
2) After the addition of 25 mL of 0.1 M NaOH .
NaOH (aq) + HCl (aq) H
2
O (l) + NaCl (aq)
25.0 mL of
0.1 M NaOH
25.0 mL of
0.1 M HCl
50.0 mL of
H
2
O
At the equivalence point!
K a
= K b
= [H + ] = [OH ] = 1.0 x 10 -7 pH = 7
Adding:
0.1 M NaOH
36

NaOH ( aq ) + HCl ( aq ) H
2
O ( l ) + NaCl ( aq )
Start with:
25.0 mL of
0.1 M HCl
Adding:
0.1 M NaOH
Since they completely react we can easily calculate the pH when:
3) After the addition of 35 mL of 0.1 M NaOH .
At 25 mL, NaOH has consumed all HCl. Anything beyond 25 mL is just adding OH to water.
10 mL x 0.1 mol/L NaOH x 1L/1000 mL = 1.0 x 10 -3 mol NaOH
= 0.0167 M NaOH
0.060 L
[NaOH]= [OH ] = 0.0167 M NaOH
37

25.0 mL of
0.1 M
CH
3
COOH
NaOH
CH
3
COOH (aq) + NaOH (aq)
CH
3
COOH
CH
3
COONa (aq) + H
2
O (l)
CH
3
COO
-
(aq) + H
2
O (l) CH
3
COOH (aq) + OH
-
(aq) adding 0.1 M NaOH
38

CH
3
COOH (aq) + NaOH (aq) CH
3
COONa (aq) + H
2
O (l)
Start with:
25.0 mL of
0.1 M
CH
3
COOH
Adding:
0.1 M NaOH
Since there is an equilibrium calculating the pH is a little more difficult:
1) After the addition of 0 mL of 0.1 M NaOH .
K a
=
CH
3
COOH (aq) CH
3
COO (aq) + H
+
(aq)
[H + ][A
-
]
[HA]
ICE table + Math
[H + ] = 1.3 x 10 -3
K a
= 1.8 x 10 -5 pH = -log (1.3 x 10 -3 ) = 2.87
39

Start with: 25.0 mL of 0.1 M CH
3
COOH
1) After the addition of 10 mL of 0.1 M NaOH .
moles of NaOH/OH -
10.0 mL ×
0.100 mol NaOH
×
1 L NaOH soln
1 L
1000 mL moles of CH
3
COOH
25.0 mL ×
0.100 mol CH COOH
3
1 L
1 L CH COOH soln 1000 mL
3
×
[A
-
] pH = pK a
+ log
[HA] pH = -log (1.8 x 10 -5 ) + log
0.001
0.0015
pH = 4.57
40

Start with: 25.0 mL of 0.1 M CH
3
COOH
1) After the addition of 25 mL of 0.1 M NaOH .
moles of NaOH/OH -
25.0 mL ×
0.100 mol NaOH
1 L NaOH soln
1 L
×
1000 mL
Initial (mol):
Change (mol):
Final (mol): moles of CH
3
COOH
25.0 mL ×
0.100 mol CH COOH
3
1 L
1 L CH COOH soln 1000 mL
3
×
CH
3
COOH ( aq ) + NaOH ( aq )
2.50 x 10 -3 2.50 x 10 -3
-2.50 x 10 -3 -2.50 x 10 -3
0 0
CH
3
COONa( aq ) + H
2
O( l )
0
+2.50 x 10 -3
2.50 x 10 -3
2.50 × 10 mol 1000 mL
×
50.0 mL 1 L
= 0.0500 mol/L = 0.0500 M
CH
3
COO -
( aq ) + H
2
O ( l ) OH -
( aq ) + CH
3
COOH ( aq )
41

Start with: 25.0 mL of 0.1 M CH
3
COOH
1) After the addition of 35 mL of 0.1 M NaOH .
moles of NaOH/OH -
35.0 mL ×
0.100 mol NaOH 1 L
×
1 L NaOH soln 1000 mL
-3
= 3.50 × 10 mol
Initial (mol):
Change (mol):
Final (mol):
CH
3
COOH ( aq )
2.50 x 10 -3
-2.50 x 10 -3
0 moles of CH
3
COOH
25.0 mL ×
0.100 mol CH COOH
3
1 L
1 L CH COOH soln 1000 mL
3
×
+ NaOH ( aq )
3.50 x 10 -3
-2.50 x 10 -3
1.00 x 10 -3
→ CH
3
COONa( aq )
0
+2.50 x 10 -3
2.50 x 10 -3
+ H
2
O( l )
42

Strong base to a strong acid.
Strong base to a weak acid.
Strong base to a strong acid.
Strong base to a weak acid.
At high [NaOH], the pH curve is dominated by OH .
43

25.0 mL of
0.1 M NH
3
HCl NH
3
HCl (aq) + NH
3
(aq) NH
4
Cl (aq)
NH
4
+
(aq) NH
3
(aq) + H +
(aq) adding 0.1 M HCl
44

Example 16.6
Calculate the pH at the equivalence point when 25.0 mL of 0.100 M
NH
3 is titrated by a 0.100 M HCl solution.
45

Example 16.6
Calculate the pH at the equivalence point when 25.0 mL of 0.100 M
NH
3 is titrated by a 0.100 M HCl solution.
At equivalent point:
HCl (aq) + NH
3
1
+
1
(aq) NH
4
Cl (aq)
1
25.0 mL ×
0.100 mol NH
3
1 L NH
3
1 L
×
1000 mL
NH4 + (aq)
Initial (M):
Change (M):
0.0500
-x
Equilibrium (M): (0.0500-x)
2.50 x 10 -3 mol
0.05 L
= 0.05 M
NH
3
(aq) +
0.000
+x x
H + (aq)
0.000
+x x
K a
=
3
4 x
2
5.6 × 10
-10
=
0.0500 - x
[0.05]
If > 400
K a we can neglect x pH = -log (5.3 x 10 -6 ) = 5.28
46


There are two buffer regions and two equivalence points corresponding to each deprotonation step.

Each of these steps involves a conjugate acid
– conjugate base equilibrium.
Titration of H
2
SO
3 with NaOH p K a2
= 7.19

The pH at the intermediate equivalence point depends on the pK a of the corresponding species.
p K a1
= 1.85
47

48

We want to know when we have reached the equivalence point.
pH Meter pH Indicator
>$1,500
Strips = $0.05
Solution = $0.01
49

Indicator – substance that changes color with respect to [H + ]
HIn ( aq
Color 1
) H +
( aq ) + In
-
( aq )
Color 2
[HIn]
[In
-
]

10
[HIn]
[In
-
]

10
Color of acid (HIn) predominates
Color of conjugate base (In
-
) predominates
50

Indicator – substance that changes color with respect to [H + ]
HIn ( aq
Color 1
) H +
( aq ) + In
-
( aq )
Color 2
Add acid (H + ) the equilibrium shifts left.
Remove H + the equilibrium shifts right.
-H +
+H +
It changes from red (at pH 3.1) to orange-yellow (at pH 4.4): 51

Slowly add base to unknown acid
UNTIL the indicator changes to pink
52

53

54

Indicator – substance that changes color with respect to [H + ]
HIn ( aq
Color 1
) H +
( aq ) + In
-
( aq )
Color 2
[HIn]
[In
-
]

10
[HIn]
[In
-
]

10
Color of acid (HIn) predominates
Color of conjugate base (In
-
) predominates
55

Some indicators transition through several colors.
Usually composed of many color changing molecules.
pH
56

57

Example 16.7
Which indicator or indicators would you use for the acid-base titrations here:
Near the equivalence point, the pH of the solution changes abruptly from 4 to 10. Therefore, all the indicators except thymol blue, bromophenol blue, and methyl orange are suitable for use in the titration.
58

Example 16.7
Which indicator or indicators would you use for the acid-base titrations here:
All the indicators except thymol blue, bromophenol blue, and methyl orange are suitable for use in the titration.
59

Example 16.7
Which indicator or indicators would you use for the acid-base titrations here:
Here the steep portion covers the pH range between 7 and 10; therefore, the suitable indicators are cresol red and phenolphthalein.
60

Example 16.7
Which indicator or indicators would you use for the acid-base titrations here:
Phenolphthalein can be used for this titration, but methyl red will not work.
61

Example 16.7
Which indicator or indicators would you use for the acid-base titrations here:
Here the steep portion covers the pH range between 3 and 7; therefore, the suitable indicators are methyl orange, methyl red and chlorophenol blue.
62

Example 16.7
Which indicator or indicators would you use for the acid-base titrations here:
Here the steep portion covers the pH range between 3 and 7; methyl red is more appropriate than phenolphthalein.
63

64

65