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1
Chapter 17a
Ionic Equilibria: Part II Buffers and Titration Curves
2
Chapter Goals
1. The Common Ion Effect and Buffer Solutions
2. Buffering Action
3. Preparation of Buffer Solutions
4. Acid-Base Indicators
Titration Curves
5. Strong Acid/Strong Base Titration Curves
6. Weak Acid/Strong Base Titration Curves
7. Weak Acid/Weak Base Titration Curves
8. Summary of Acid-Base Calculations
3
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.
4
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
5
The Common Ion Effect and Buffer Solutions1. 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
acids. with reacts base) (aanion salt The
COOCH Na COOCHNa
H COOCH COOHCH
bases. with reacts acid weak The
-3
%1003
-33
6
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.
MMM
MxMxMx
15.0 15.0 15.0
COOCHNaCOONaCH
)15.0(
COOCHHCOOHCH
3100%
3
-33
7
The Common Ion Effect and Buffer Solutions Substitute the quantities determined in the previous
relationship into the ionization expression.
x
xx
15.0
15.0108.1
COOHCH
COOCH HK 5
3
-3
+
a
8
The Common Ion Effect and Buffer Solutions Apply the simplifying assumption to both the
numerator and denominator.
74.4pH
H108.1
108.10.15
0.15
gives sassumption theseMaking
15.015.0 and 15.015.0
5
5
Mx
x
xx
9
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.
10
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 [H+] pH
0.15 M CH3COOH 1.6 x 10-3 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.
11
The Common Ion Effect and Buffer Solutions The general expression for the ionization of a weak
monoprotic acid is:
The generalized ionization constant expression for a weak acid is:
HA H A+
KH A
HAa
12
The Common Ion Effect and Buffer Solutions If we solve the expression for [H+], this relationship
results:
By making the assumption that the concentrations of the weak acid and the salt are reasonable, the expression reduces to:
H KHA
A
acid
salta
H Kacidsalta
13
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 Kacid
n salt
where n = charge on cation
a
14
The Common Ion Effect and Buffer Solutions Simple rearrangement of this equation and
application of algebra yields the
Henderson-Hasselbach equation.
acid
saltlogpKpH
acid
saltlogKlogHlog
1-by multiply
salt
acidlogKlogHlog
a
a
a
The Henderson-Hasselbach equation is one method to calculate the pHof a buffer given the concentrations of the salt and acid.
15
Weak Bases plus Salts of Weak Bases2. Buffers that contain a weak base plus the salt of a
weak base One example of this buffer system is ammonia plus
ammonium nitrate.
5
3
+4
3+4
%10034
-+423
108.1NH
NH
NONHNONH
OHNHOH NH
OHKb
16
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.
MMM
MxMxMx
30.0 30.0 30.0
NONHNONH
)15.0(
OHNH OH NH
3+4
%10034
+423
17
Weak Bases plus Salts of Weak Bases Substitute the quantities determined in the previous
relationship into the ionization expression for ammonia.
8.95pH and 5.05pOH
OH100.9
108.115.0
30.0K
applied. becan assumption gsimplifyin The
108.115.0
30.0K
108.1NH
OH NHK
6
5b
5b
5
3
4b
Mx
x
x
xx
18
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 [OH-] pH
0.15 M NH3 1.6 x 10-3 M 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.
19
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 O BH OH
where B represents a weak base2
20
Weak Bases plus Salts of Weak Bases The general form of the ionization expression is:
Solve for the [OH-]
KBH OH
Bb
OH K
B
BHbase
salt
b
21
Weak Bases plus Salts of Weak Bases For salts that have univalent ions:
For salts that have divalent or trivalent ions:
OH Kbasesaltb
OH Kbase
n salt
where n = charge on anion
b
22
Weak Bases plus Salts of Weak Bases Simple rearrangement of this equation and
application of algebra yields the
Henderson-Hasselbach equation.
base
saltlogpKpOH
base
saltlogKlogOHlog
1-by multiply
salt
baselogKlogOHlog
b
b
b
23
Buffering Action
These movies show that buffer solutions resist changes in pH.
24
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.
25
Buffering Action
8.95pH 5.05pOH
100.9OH
20.0
10.0108.1OH
ClNH
NHKOH
6-
5-
4
3b
-
M
M
M
26
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.
mol 0.220 mol 0.080 mol 0 rxn.After
mol 0.020+ mol 0.020-mol 0.020- Change
mol 0.200 mol 0.100 mol 0.020 Initial
ClNH NH HCl 43
27
Buffering Action
HCl NH NH 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
mol1.0 L
mol1.0 L
3 4
NH
NH Cl
3
4
M M
M M
0 0800 080
0 2200 220
..
..
28
Buffering Action
3 Using the concentrations of the salt and base and the Henderson-Hassselbach equation, the pH can be calculated.
M
M
220.0
080.0108.1OH
ClNH
NHKOH
5
4
3b
29
Buffering Action
OH KNH
NH Cl
OH
OH
pOH 5.19 pH 8.81
b3
4
18 100 0800 220
6 5 10
5
6
...
.
MM
M
30
Buffering Action
4 Finally, calculate the change in pH.
-0.14=8.95-8.81=pH
pHpH pH originalnew
31
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!You do it!
32
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.
mol 0.120 mol 0 mol 0.180 rxn.After
mol 0.020+ mol 0.020- mol 0.020- Change
mol 0.100 mol 0.020 mol 0.200 Initial
NaCl OH NH NaOH ClNH 234
33
Buffering Action
MM
MM
180.0L 1.0
mol 180.0
120.0L 1.0
mol 120.0
mol 0.120 mol 0 mol 0.180 rxn.After
mol 0.020+ mol 0.020- mol 0.020- Change
mol 0.100 mol 0.020 mol 0.200 Initial
NaCl OH NH NaOH ClNH
ClNH
NH
234
4
3
34
Buffering Action
2 Calculate the pH using the concentrations of the salt and base and the Henderson-Hasselbach equation.
9.08pH 4.92pOH
102.1OH
180.0
120.0108.1OH
ClNH
NHKOH
5
5
4
3b
M
M
M
35
Buffering Action
3 Calculate the change in pH.
0.13=8.95-9.08=pH
pHpH =pH originalnew
36
Buffering Action
This table is a summary of examples 19-3 and 19-4.
Notice that the pH changes only slightly in each case.
Original SolutionOriginal
pH
Acid or base
added
New pH
pH
1.00 L of solution containing
0.100 M NH3 and 0.200 M NH4Cl
8.95
0.020 mol NaOH
9.08 +0.13
0.020 mol HCl
8.81 -0.14
37
Preparation of Buffer Solutions
This move shows how to prepare a buffer.
38
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.
NaOH mmol 0.10mL
mmol 0.100mL 100=NaOH mmol ?
COOHCH mmol 0.30mL
mmol 0.15mL 200=COOHCH mmol ? 33
39
Preparation of Buffer Solutions
Sodium hydroxide and acetic acid react in a 1:1 mole ratio.
NaOH + CH COOH Na CH COO + H O
Initial 10.0 mmol 30.0 mmol
Change -10.0 mmol -10.0 mmol +10.0 mmol
After rxn. 0 20.0 mmol 10.0 mmol
3 3 2
40
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:
COONaCH 0333.0mL300
mmol 0.10
COOHCH 0667.0mL300
mmol 0.20
3COONaCH
3COOHCH
3
3
MM
MM
41
Preparation of Buffer Solutions
Substitution of these values into the ionization constant expression (or the Henderson-Hasselbach equation) permits calculation of the pH.
KH CH COO
CH COOH
H
pH
a3
3
18 10
18 10 0 0667
0 03333 6 10
4 44
5
55
.
. .
..
.
M
42
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 14 104 85 5. . M
43
Preparation of Buffer Solutions
The appropriate equilibria representations are:
MxMxMx
MMM
Cl NH ClNH
104.1 104.1 104.110.0
OH NH OH + NH
44
555
-423
44
Preparation of Buffer Solutions
Substitute into the ionization constant expression (or Henderson-Hasselbach equation) for aqueous ammonia
assumption gsimplifyin apply the
104.110.0
104.1 104.1K
108.1NH
OH NHK
5
55
b
5
3
4b
x
45
Preparation of Buffer Solutions
g/L 9.6mol
g 53
L
mol 13.0
L
ClNH g ?
ClNH=ClNH 13.0
108.110.0
104.1 K
applied. becan assumption gsimplifyin The
104.110.0
104.1 104.1K
108.1NH
OH NHK
4
original44
55
b
5
55
b
5
3
4b
Mx
x
x
46
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
47
Acid-Base Indicators
The equilibrium constant expression for an indicator would be expressed as:
HIn
In HKa
48
Acid-Base Indicators
If the preceding expression is rearranged the range over which the indicator changes color can be discerned.
In
HInK
H
-a
49
Acid-Base Indicators
Color change ranges of some acid-base indicators
IndicatorColor in
acidic range pH rangeColor in
basic range
Methyl violet Yellow 0 - 2 Purple
Methyl orange Pink 3.1 – 4.4 Yellow
Litmus Red 4.7 – 8.2 Blue
Phenolphthalein Colorless 8.3 – 10.0 Red
50
Titration CurvesStrong 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.
OHKClOKOHHClO 244
51
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%.
00.1log(0.100)pH
100.0H
10000.100 100.0
ClOHHClO 4%100
4
M
M.M M
52
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 ___?
1.17pH 067.0H
067.0mL 120
HClO mmol 8.0
rxn.
mmol 2.0 mmol 0.0 mmol 8.0 After
mmol 2.0 mmol 2.0- mmol 2.0- :Change
mmol 2.0 mmol 10.0 :Start
OHKClO KOH HClO
4HClO
244
4
M
MM
53
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 ___?
1.48pH 033.0H
033.0mL 150
HClO mmol 5.0
rxn.
mmol 5.0 mmol 0.0 mmol 5.0 After
mmol 5.0 mmol 5.0- mmol 5.0- :Change
mmol 5.0 mmol 10.0 :Start
OHKClO KOH HClO
4HClO
244
4
M
MM
54
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 ____?
2.28pH 0053.0H
0053.0mL 190
HClO mmol 1.0
rxn.
mmol 9.0 mmol 0.0 mmol 1.0 After
mmol 9.0 mmol 9.0- mmol 9.0- :Change
mmol 9.0 mmol 10.0 :Start
OHKClO KOH HClO
4HClO
244
4
M
MM
55
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 ___?
7.00pH
neutral baseor acid No
rxn.
mmol 10.0 mmol 0.0 mmol 0.0 After
mmol 10.0 mmol 10.0- mmol 10.0- :Change
mmol 10.0 mmol 10.0 :Start
OHKClO KOH HClO 244
56
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.
57
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 COOH + KOH K CH COO + H O
1mmol 1mmol 1mmol3
+3
-2
58
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.
59
Weak Acid/Strong Base Titration Curves1. Determine the pH of the acetic acid solution before
the titration is begun. Same technique as used in Chapter 18.
5
a
5
3
-3
+
a
-33
108.110.0
K
108.1COOHCH
COOCH HK
10.0
HCOOCHCOOHCH
x
xx
xMxMMx
60
Weak Acid/Strong Base Titration Curves
89.2pH 101.3H
101.3= 108.1
applied. becan assumption gsimplifyin The
108.110.0
K
108.1COOHCH
COOCH HK
10.0
HCOOCHCOOHCH
3-
3-62
5a
5
3
-3
+
a
-33
xx
x
xx
MxMxMx
61
Weak Acid/Strong Base Titration Curves After a total of 20.0 mL of KOH solution has been
added, the pH is:
KOH + CH COOH K CH COO H O
Initial: 2.00 mmol 10.0 mmol
Chg. due to rxn:-2.00 mmol - 2.00 mmol + 2.00 mmol
After rxn: 0.00 mmol 8.00 mmol 2.00 mmol
8.0 mmol120 mL
2.0 mmol120 mL
3+
3-
2
CH COOH
CH COO
3
3-
M M
M M
0 067
0 017
.
.
62
Weak Acid/Strong Base Titration Curves
KH CH COO
CH COOH
HCH COOH
CH COO
H
pH
a3
3
3
3
18 10
18 10
18 100 0670 017
7 1 10
4 15
5
5
5 5
.
.
...
.
.
M
Similarly for all other cases before the equivalence point is reached.
63
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.
64
Weak Acid/Strong Base Titration Curves1. Set up the equilibrium reaction:
KOH + CH COOH K CH COO H O
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
3+
3-
2
65
Weak Acid/Strong Base Titration Curves2. Determine the concentration of the salt in solution.
COOCH 0500.00500.0
0500.0mL 200
mmol 10.0=
3COOKCH
COOKCH
3
3
MMM
MM
66
Weak Acid/Strong Base Titration Curves3. Perform a hydrolysis calculation for the potassium
acetate in solution.
8.72pH5.28pOH
OH1027.5108.2
106.50500.00500.0
=K
106.5COOCH
OHCOOHCH=K
0500.0
OH COOHCHOHCOOCH
6112
102
b
11
3
-3
b
-323
xx
x
x
xx
xMxMMx
67
Weak Acid/Strong Base Titration Curves4. After the equivalence point is reached, the pH is
determined by the excess KOH just as in the strong acid/strong base example.
11.68=pH and 2.32pOH
108.4OH
108.4mL 210
mmol 0.1
mmol 10.00 mmol 0.00 mmol 1.00 :rxnAfter
mmol 10.00+ mmol 10.0- mmol -10.0:rxn todue Chg.
mmol 10.0 mmol 11.0 :Initial
OHCOOCHKCOOHCH + KOH
3
KOH3
KOH
2-
3+
3
M
MM
68
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.
69
Strong Acid/Weak BaseTitration Curves Titration curves for Strong Acid/Weak Base
Titration Curves look similar to Strong Base/Weak Acid Titration Curves but they are inverted.
70
Weak Acid/Weak BaseTitration 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.
71
Synthesis Question
Bufferin is a commercially prepared medicine that is literally a buffered aspirin. How could you buffer aspirin? Hint - what is aspirin?
72
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.
73
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?
74
Group Question
75
End of Chapter 19
We have examined :1 Gas phase equilibria in Chapter 172 Hydrolysis equilibria in Chapter 183 Acid/base equilibria in Chapter 19
Chapter 20 is the last equilibrium chapter. It involves solid/solution equilibria.