Chapter Nineteen - Abington Heights School District
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Transcript Chapter Nineteen - Abington Heights School District
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.
CH3COOH H CH3COO
(0.15 x) M x M
100%
xM
NaCH3COO Na CH3COO
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 NH4 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-]
B
OH K BH salt
b
base
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 nsalt
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
Initial
NH3 NH 4 Cl
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
0.220 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
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.
NH 4 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
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
0.120 mol
M NH3
0.120M
1.0 L
0.180 mol
M NH4Cl
0.180M
1.0 L
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.
NaOH + CH 3COOH Na CH 3COO + H 2O
Initial
Change
After rxn.
10.0 mmol 30.0 mmol
-10.0 mmol -10.0 mmol
0
20.0 mmol
+ 10.0 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
NH 4 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 Cloriginal
? 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
8.0 mmol 0.0 mmol
2.0 mmol
rxn.
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
5.0 mmol 0.0 mmol
5.0 mmol
rxn.
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
1.0 mmol 0.0 mmol
9.0 mmol
rxn.
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
10.0 mmol
rxn.
No acid or base neutral
pH 7.00
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:
KOH +
Initial:
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
+ 2.00 mmol
2.00 mmol
8.0 mmol
M CH 3COOH
0.067 M
120 mL
2.0 mmol
M CH COO-
0.017 M
3
120 mL
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
7.1 105 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:
Initial:
KOH +
10.0 mmol
CH 3COOH K +CH 3COO- H 2O
10.0 mmol
Chg. due to rxn:-10.0 mmol -10.0 mmol
After rxn:
0.0 mmol
0.0 mmol
+ 10.0 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 3COOHOH-
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