19
Ionic Equilibria離子平衡: Part II
Buffers 緩衝液and Titration
Curves 滴定曲線
花青素
1
Chapter Goals
1. The Common Ion Effect and Buffer Solutions
(共同離子效應及緩衝溶液)
2. Buffering Action (緩衝作用)
3. Preparation of Buffer Solutions (緩衝溶液的製備)
4. Acid-Base Indicators (酸鹼指示劑)
Titration Curves (滴定曲線)
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
2
The Common Ion Effect and
Buffer Solutions
• Common ion effect 共同離子效應
– When a solution of a weak electrolyte is
altered by adding one of its ions from
another source, the ionization of the weak
electrolyte is suspressed.(當弱電解質溶液中加
入具有相同離子的強電解質時，弱電解質的電離平衡
會移動，使弱電解質的電離度下降，這種現象叫共同
離子效應)
• 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
•Buffer solution 緩衝溶液
– Resists changes in pH when strong acids or
strong bases are added.(當加入強酸或強鹼時,pH
的變化不大)
– contains a conjugate acid-base pair in
reasonable concentrations. It can react
with added base or acid (具共軛酸鹼對)
– Buffer solution contain (緩衝溶液含有:)
• A weak acid and a soluble ionic salt of the
weak acid (弱酸和具此弱酸根的可溶性鹽類)
– CH3COOH plus NaCH3COO
• A weak base and a soluble ionic salt of the
weak base (弱鹼和具此弱鹼根的可溶性鹽類)
– NH3 plus NH4Cl
4
The Common Ion Effect and
Buffer Solutions
1. Solutions made of weak acids plus a soluble
ionic salt of the weak acid
– Solution that contain a weak acid plus a salt of
weak acid are always less acidic than solutions that
contain the same concentration of weak acid
alone
– One example of this type of buffer system is:
• The weak acid - acetic acid CH3COOH
• The soluble ionic salt - sodium acetate NaCH3COO
CH3COOH  H+ + CH3COO- 部分解離
Na+CH3COO- 100%
Na+ + CH3COO- 完全解離
[CH3COO-] 增加  反應向左  減少[H+]
 pH增加
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.
Ka=
CH3COOH  H+ + CH3COO(0.15-x) M x M
xM
100%
Na+CH3COONa+ + CH3COO0.15 M
0.15 M 0.15 M
[H+] [CH3COO-]
(x)(0.15+x)
-5
=1.8x10 =
[CH3COOH]
(0.15-x)
(0.15+x)  0.15 and (0.15-x)  0.15
(x)(0.15)
= 1.8x10-5
(0.15)
x =1.8x10-5= [H+]
pH=4.74
6
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
0.15 M CH3COOH &
0.15 M NaCH3COO buffer
1.6 x 10-3
2.80
1.8 x 10-5
4.74
 [H+] is 89 times greater in pure acetic acid than in buffer
solution.
7
The Common Ion Effect and
Buffer Solutions
• The general expression for the ionization of a
weak monoprotic acid is:
HA  H+ + A• The generalized ionization constant
expression for a weak acid is:
Ka=
[H+] [A-]
[HA]
8
The Common Ion Effect and
Buffer Solutions
• If we solve the expression for [H+], this relationship
results:
[HA] acid
+
[H ] = Ka x
[A-] salt
• By making the assumption that the concentrations of
the weak acid and the salt are reasonable, the
expression reduces to:
[acid]
[H+] = Ka x
[salt]
• 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 x
n[salt]
Where n=charge on cation
9
The Common Ion Effect and
Buffer Solutions
• Simple rearrangement of this equation
and application of algebra yields the
Henderson-Hasselbach equation
log
[H+]
[acid]
=log Ka + log
[salt]
Multiply by -1
[salt]
-log
=-log Ka + log
[acid]
[salt]
pH= pKa + log
[acid]
[H+]
The Henderson-Hasselbach equation is one method to calculate
the pH of a buffer given the concentrations of the salt and acid.
10
The Common Ion Effect and
Buffer Solutions
Example 19-1-1: Use the Henderson-Hasselbalch equation to
calculate the pH of the buffer solution in Example 19-1
[salt]
pH= pKa + log
[acid]
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.
[acid]
pKa =-logKa
+
[H ] = Ka x
-5
=-log(1.8x10 )
[salt]
(0.15)
= 4.74
[H+] = 1.8x10-5 x
(0.15)
pH= 4.74 + log 0.15M
0.15M
[H+] = 1.8x10-5
pH=4.74
pH= 4.74 + log1
= 4.74 + 0
= 4.74
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 + H2O  NH4+ + OHNH4NO3
100%
NH4+ + NO3-
[NH4+] [OH-]
Kb=
=1.8x10-5
[NH3]
12
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 + H2O  NH4+ + OH-
(0.15-x) M
NH4NO3
100%
0.30 M
xM
xM
0.30 M
0.30 M
NH4+ + NO3-
(x)(0.30+x)
[NH4+] [OH-]
=
=1.8x10-5
Kb=
(0.15-x)
[NH3]
(0.30+x)  0.30 and (0.15-x)  0.15
(x)(0.30)
= 1.8x10-5
(0.15)
x =9.0x10-6= [OH-]
pOH=5.05, pH=8.95
13
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
0.15 M NH3 &
0.15 M NH4NO3 buffer
1.6 x 10-3 M
11.20
9.0 x 10-6 M
8.95
The [OH-] in aqueous ammonia is 180 times greater than in
the buffer.
14
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 + H2O  BH + OH
Where B represents a weak base
• The general form of the ionization expression is:
[BH+] [OH-]
Kb=
[B]
• Solve for the [OH-]
[OH-] = Kb x
[B] base
[BH+] salt
15
Weak Bases plus Salts of
Weak Bases
• For salts that have univalent ions:
[OH-]
[base]
= Kb x
[salt]
• For salts that have divalent or
trivalent ions:
[OH-]
[base]
= Kb x
n[salt]
Where n= charge on anion
16
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 Kb + log
[salt]
Multiply by -1
-log
[OH-]
=-log Kb + log
pOH= pKb + log
[salt]
[base]
[salt]
[base]
17
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.
[NH3]
[NH4Cl]
[OH-] = 9.0x10-6
[OH-] = Kb x
[OH-] = 1.8x10-5 x 0.10M
0.20M
pOH=5.05 pH=8.95
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 NH3=0.1M x 1L=0.1mol ?mol NH4Cl=0.2M x 1L=0.2mol
HCl + NH3 → NH4Cl
0.08mol
=0.08M
M
=
Initial
0.02 mol 0.1 mol 0.2 mol
NH3
1.0L
change -0.02 mol-0.02 mol+0.02 mol
0.22mol
=0.22M
After rxn 0 mol 0.08 mol 0.22 mol MNH4Cl= 1.0L
[NH3]
0.08M
= 1.8x10-5 x
[OH ] = Kb x
= 6.5x10-6
0.22M
[NH4Cl]
pH=pHnew-pHoriginal
pOH=5.19 pH=8.81
18
=8.81-8.95=-0.14
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.
•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.
NH4Cl + NaOH → NH3 + H2O + NaCl
Initial 0.2 mol 0.02 mol 0.1 mol
MNH3= 0.12mol =0.12M
1.0L
change -0.02 mol -0.02 mol+0.02 mol
0.18mol
After rxn 0.18 mol 0 mol
0.12 mol MNH4Cl= 1.0L =0.18M
[NH3]
-5 x 0.12M
-5
[OH ] = Kb x
=
1.8x10
=
1.2x10
[NH4Cl]
0.18M
pH=pHnew-pHoriginal
pOH=4.92 pH=9.08
19
=9.08-8.95=0.13
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
pH
8.95
Acid or
base
added
New
pH
pH
0.020 mol
NaOH
9.08
+0.13
0.020 mol
HCl
8.81
-0.14
• Notice that the pH changes only slightly in each case.
20
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.
?mmol CH3COOH =200ml x 0.15M
= 30.0 m mol
?mmol NaOH =100ml x 0.1M
= 10.0 m mol
NaOH + CH3COOH → NaCH3COO + H2O
Initial
10.0 mmol 30.0 mmol
change -10.0 mmol -10.0 mmol +10.0 mmol
After rxn 0.0 mmol 20.0 mmol
10.0 mmol
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:
20 mmol
10 mmol
MCH3COOH=
=0.0667M
300mL
[H+] [CH3COO-]
Ka=
[CH3COOH]
[H+]=3.6x10-5M
MNaCH3COO=
300mL
[H+] [0.0333]
=1.8x10-5= [0.0667]
pH =4.44
=0.0333M
21
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 Then pOH = 14.00 - 9.15
-] = 10-4.85
= 4.85
[OH
NH3 + H2O  NH4+ + OH-5M
=
1.4x10
(0.1-1.4x10-5) M (1.4x10-5) M (1.4x10-5) M
NH4Cl → NH4+ + Cl-
xM
xM
[NH4+] [OH-]
Kb=
[NH3]
xM
= 1.8x10-5
(1.4x10-5+x)(1.4x10-5)
(x)(1.4x10-5)
Kb=
=
= 1.8x10-5
-5
(0.1-1.4x10 )
(0.1)
x= 0.13 M = [NH4Cl]original
0.13Mx1L =0.13 mol
22
Acid-Base Indicators
• The point in a titration at which chemically
equivalent amounts of acid and base have reacted
is called the equivalence point當量點.
• The point in a titration at which a chemical indicator
changes color is called the end point 終點.
23
a. Methyl red 甲基紅
•Red at pH4 and below
•Yellow at pH 7 and above
•Red → orange → yellow
Three common indicators in solutions
that cover the pH range 3 to 11.
b. Bromthymol blue 溴瑞香草酚藍
•yellow at pH6 and below
•blue at pH 8 and above
•yellow → green → blue
c. Phenolphthalein酚酞
(most common use)
•Colorless below pH8
•Bright pink above pH10
Acid-Base Indicators
• Many acid-base indicators are weak organic acid,
HIn, where “In” represents various complex organic
gups
• A symbolic representation of the indicator’s color
change at the end point is:
HIn  H+ + In-
color1
color2
HIn represents nonionized acid molecules
In- represents the anion (conjugate base) of
HIn
• The equilibrium constant expression for an indicator
would be expressed as:
[H+] [In-]
Ka=
[HIn]
[In-]
[HIn]
Ka
=
[H+]
25
Acid-Base Indicators
Color change ranges of some acid-base
indicators
Indicator
Methyl violet甲基紫
Color in
acidic range pH range
Color in
basic range
Yellow
0-2
Purple
Methyl orange
Pink
3.1 – 4.4
Yellow
Litmus石蕊
Red
4.7 – 8.2
Blue
Phenolphthalein
Colorless
8.3 – 10.0
Red
26
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.
HClO4 + KOH → KClO4 + H2O
28
Strong Acid/Strong Base
Titration Curves
• Before any KOH is added the pH of the HClO4
solution is
.
– Remember perchloric acid is a strong acid that
ionizes essentially 100%.
HClO4
0.10M
100%
H+ + ClO4-
0.10M
0.10M
[H+] =0.10M
pH =-log(0.10)=1.0
29
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 ___?
? mmol HClO4 = 100ml x (0.1M) =10.0 mmol
? mmol KOH = 20ml x (0.1M) =2.0 mmol
HClO4 + KOH → KClO4 + H2O
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
MHClO4=
8.0 mmol
120mL
=0.067M
[H+]=0.067M
pH = 1.17
30
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 ___?
? mmol KOH = 50ml x (0.1M) =5.0 mmol
HClO4 + KOH → KClO4 + H2O
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
MHClO4=
5.0 mmol
150mL
=0.033M
[H+]=0.033M
pH = 1.48
31
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 ____?
? mmol KOH = 90ml x (0.1M) =9.0 mmol
HClO4 + KOH → KClO4 + H2O
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
MHClO4=
1.0 mmol
190mL
=0.0053M
[H+]=0.0053M
pH = 2.28
32
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 ___?
? mmol KOH = 100ml x (0.1M) =10.0 mmol
HClO4 + KOH → KClO4 + H2O
Start
10.0 mmol 10.0 mmol
change -10.0 mmol -10.0 mmol +10.0 mmol
After rxn 0.0 mmol 0.0 mmol 10.0 mmol
No acid or base
Neutral
pH=7.0
33
Strong acid-strong base
34
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.
35
Weak Acid/Strong Base
Titration Curves
• As an example, consider the titration of 100.0 mL of
0.100 M acetic acid, CH3COOH, (a weak acid) with
0.100 M KOH (a strong base).
– The acid and base react in a 1:1 mole ratio.
1mol
1mol
1mol +
CH3COOH + KOH → K CH3COO + H2O
1mmol
1mmol
1mmol
• 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.
36
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.
37
1. Determine the pH of the acetic acid solution before
the titration is begun.
CH3COOH  CH3COO- +H+
(0.1-x) M
[CH3COO-]
xM
xM
(x)(x)
[CH3COOH]
(0.1-x)
x= 1.3x10-3=[H+]
x2= 1.8x10-6
pH= 2.89
Ka=
[H+]
= 1.8x10-5 =
• After a total of 20.0 mL of KOH solution has been
added, the pH is:
? mmol CH3COOH = 100ml x (0.1M) =10.0 mmol
? mmol KOH = 20ml x (0.1M) =2.0 mmol
KOH + CH3COOH → K+CH3COO- + H2O
Start
2.0 mmol
change -2.0 mmol
After rxn 0.0 mmol
8.0 mmol
10.0 mmol
-2.0 mmol
8.0 mmol
+2.0 mmol
2.0 mmol
2.0 mmol
MCH3COO-=
=0.017M
=0.067M
120mL
120mL
0.067
[CH3COOH]
+
-5
[H ] = Ka x
= (1.8x10 ) x
= 7.1x10-5
0.017
[CH3COO ]
pH= 4.15
38
MCH3COOH=
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.
39
Weak Acid/Strong Base
Titration Curves
• Set up the equilibrium reaction:
KOH + CH3COOH → K+CH3COO- + H2O
Start
10.0 mmol 10.0 mmol
change -10.0 mmol -10.0 mmol +10.0 mmol
After rxn 0.0 mmol 0.0 mmol
10.0 mmol
MKH3COOH=
10.0 mmol
200mL
=0.05M
0.05M CH3COO-
CH3COO- +H2O  CH3COOH +OH(0.05-x) M
xM
xM
(x)(x)
[OH-] [CH3COOH]
-10
=
Kb=
=5.6x10
(0.05-x)
[CH3COO-]
x2=2.8x10-11
x=5.27x10-6= [OH-]
pOH=5.28
pH=8.72 the equivalence point
40
Weak Acid/Strong Base
Titration Curves
After the equivalence point is reached, the pH is
determined by the excess KOH just as in the strong
acid/strong base example.
KOH + CH3COOH → K+CH3COO- + H2O
Start
11.0 mmol 10.0 mmol
change -10.0 mmol -10.0 mmol +10.0 mmol
After rxn 1.0 mmol 0.0 mmol
10.0 mmol
MKOH=
1.0 mmol
210mL
=4.8x10-3 M
[OH-]=4.8x10-3 M
pOH=2.32
pH=11.68
41
Weak acid-strong base
42
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.
44
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.
45
Table 19-7a, p. 763
Table 19-7b, p. 763