1. No category

Chapter 16: Equilibria in Solutions
of Weak Acids and Bases
Weak Acids
• All weak acids behave the same way in
aqueous solution: they partially ionize

HA  H 2O  H 3O  A





[ H 3O ][ A ] [ H ][ A ]
Ka 

[ HA]
[ HA]
Ka is called the acid ionization constant
pK a   log K a
• Table 18.1 and Appendix C list the Ka
and pKa for a number of acids
A “large” pKa,means a “small” value of Ka and
only a “small” fraction of the acid molecules
ionize
15.5
Weak bases in water



B  H 2O HB  OH

the base ionization constant is


[ HB ][OH ]
Kb 
[ B]
Consider the following:

B H 2O  BH  OH

[ HB ][OH ]
Kb 

[B ]

consider B  A

HF + H2O ↔ H3O+ + FF- + H2O ↔ HF + OHSolve Ka*Kb=

For the weak acid :
[ H 3O  ][ A ]
Ka 
[ HA]
HA  H 2O H 3O   A
for the conjugate base :
[ HA][OH  ]
Kb 

[A ]
A  H 2O HA  OH 
the product is
[ H 3O  ][ A ] [ HA][OH  ]
K a  Kb 


[ HA]
[A ]


 [ H 3O ][OH ]  K w
Problem Solving
• Calculate the pH of a 0.0200 M
solution of a weak monoprotic acid
which is 3% ionized at 25°C? What
is Ka for the acid?
Problem Solving
• Calculate the pH of a 0.0300 M
solution of a weak base that is 3%
ionized at 25°C? What is Kb for the
base
If Ka for a weak acid is 2.9E-5,
what is its pKa value?
a) 4.54
b) 4.82
c) 5.29
d) 6.82
e) 7.89
What is the value of Kb for the following weak
conjugate bases?
• NaF
• NaCN
16.1. Ionization constants can be defined for weak acids and bases
18
What is the value of Kb for the following weak
conjugate bases?
• NaF
– Kw=Ka×Kb
– 1.47×10-11
Ka for HF =6.8×10-4
• NaCN
– Kw=Ka×Kb
– 1.61×10-5
Ka for HCN =6.2×10-10
16.1. Ionization constants can be defined for weak acids and bases
19
What is the value of Ka for the following
weak conjugate acids?
• NH4Cl
• C6H5NH3NO3
16.1. Ionization constants can be defined for weak acids and bases
20
What is the value of Ka for the following
weak conjugate acids?
• NH4Cl
– Kw=Ka×Kb
– 5.56×10-10
Kb for NH3 =1.8×10-5
• C6H5NH3NO3
– Kw=Ka×Kb
– 2.44×10-5
Ka =4.1×10-10
16.1. Ionization constants can be defined for weak acids and bases
21
Solving weak acid ionization problems:
1. Identify the major species that can affect the pH.
•
In most cases, you can ignore the autoionization of
water.
•
Ignore [OH-] because it is determined by [H+].
2. Use ICE to express the equilibrium concentrations in terms
of single unknown x.
3. Write Ka in terms of equilibrium concentrations. Solve for x
w/ simplifying assumption. [A]>400*K If approximation is
not valid, solve for x exactly
4. Calculate concentrations of all species and/or pH of the
solution.
15.5
What is the pH of a 0.5 M HF solution (at 250C)?
+][F-]
[H
= 7.1 x 10-4
Ka =
HF (aq)
H+ (aq) + F- (aq)
[HF]
Solved next page
What is the pH of a 0.5 M HF solution (at 250C)?
+][F-]
[H
= 7.1 x 10-4
Ka =
HF (aq)
H+ (aq) + F- (aq)
[HF]
HF (aq)
Initial (M)
Change (M)
Equilibrium (M)
H+ (aq) + F- (aq)
0.50
0.00
0.00
-x
+x
+x
0.50 - x
x
x
Ka << 1
15.5
What is the pH of a 0.5 M HF solution (at 250C)?
+][F-]
[H
= 7.1 x 10-4
Ka =
HF (aq)
H+ (aq) + F- (aq)
[HF]
HF (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.50
0.00
0.00
-x
+x
+x
0.50 - x
x
x
x2
= 7.1 x 10-4
Ka =
0.50 - x
Ka 
H+ (aq) + F- (aq)
x2
= 7.1 x 10-4
0.50
[H+] = [F-] = 0.019 M
[HF] = 0.50 – x = 0.48 M
Ka << 1
0.50 – x  0.50
x2 = 3.55 x 10-4
x = 0.019 M
pH = -log [H+] = 1.72
15.5
Determine the pH of 0.1M solutions of:
• HC2H3O2
• Ka=1.8×10-5
I
0.1M
N/A
C
E
0
0
0
0
• HCN
• Ka=6.2×10-10
I
C
E
0.1M
N/A
16.2. Calculations can involve finding or using Ka and Kb
26
Determine the pH of 0.1M solutions of:
• HC2H3O2
HC2H3O2 + H2O↔ H3O+(aq) + C2H3O2-(aq)
-5
• Ka=1.8×10
x2
I
 1.8  10 5
0.1
0.1M
N/A
0
0
C
X=1.34(10-3)M
E
-x
-x
+x
+x
pH=2.87
• HCN
• Ka=6.2×10-10
x2
0.1
 6.2  1010
X=7.87(10-6)M
(0.1-x)≈0.1
HCN
I 0.1M
C -x
E (0.1-x) ≈ 0.1
N/A
x
x
+ H2O↔ H3O+(aq) + CN-(aq)
N/A
-x
N/A
0
+x
x
0
+x
x
pH=5.10
16.2. Calculations can involve finding or using Ka and Kb
27
Determine the pH of 0.1M solutions of:
• N 2 H4
• Kb=1.7×10-6
I
C
E
• NH3
• Kb=1.8×10-5
I
C
E
0.1M
N/A
0
0
0.1M
N/A
0
0
16.2. Calculations can involve finding or using Ka and Kb
28
Determine the pH of 0.1M solutions of:
• N 2 H4
• Kb=1.7×10-6
x2
 1.7  106
0.1
X=4.12(10-4)M
pOH=3.38
pH=10.62
• NH3
• Kb=1.8×10-5
x2
 1.8  10 5
0.1
X=1.34(10-3)M
pOH=2.87
pH=11.13
N2H4
I
C
E
0.1M
-x
N/A
-x
(0.1-x) ≈ 0.1 N/A
NH3
I
C
E
+ H2O ↔ OH-(aq) + N2H5+(aq)
0.1M
-x
0
+x
x
0
+x
x
+ H2O↔ H3O+(aq) + CN-(aq)
N/A
-x
(0.1-x) ≈ 0.1 N/A
0
+x
x
16.2. Calculations can involve finding or using Ka and Kb
0
+x
x
29
What is the pH of a 0.30 M solution of phenol
(C6H5OH), an ingredient in some older
mouthwashes? Ka=1.3×10-10?
A. 9.2
B. 0.52
C. 9.4
D. none of these
16.2. Calculations can involve finding or using Ka and Kb 5.20
30
Determine the % ionization of 0.2M solution of
HC2H3O2
• Ka=1.8×10-5
I
C
E
0.2M
N/A
0
16.2. Calculations can involve finding or using Ka and Kb
0
31
Determine the % ionization of 0.2M solution of
HC2H3O2
• Ka=1.8×10-5
HC2H3O2 + H2O↔ H3O+(aq) + C2H3O2-(aq)
x2
 1.8  10 5
0.2
x=1.90×10-3M
I
C
E
0.2M
-x
N/A
-x
(0.2-x) ≈ 0.2 N/A
0
+x
x
0
+x
x
x
 100  %
moles ionized per liter
% ionization 
 100 0.2
available moles per liter
0.95 % ionized
16.2. Calculations can involve finding or using Ka and Kb
32
Determine the % ionization of 0.1M solution of
HC2H3O2
• Ka=1.8×10-5
HC2H3O2 + H2O↔ H3O+(aq) + C2H3O2-(aq)
x2
 1.8  10 5
0.1
I
C
E
0.1M
-x
N/A
-x
(0.1-x) ≈ 0.1 N/A
0
+x
x
0
+x
x
x=1.34 x 10-3M
x
moles ionized per liter
% ionization 
 100 0.1  100  %
available moles per liter
1.3 % ionized
16.2. Calculations can involve finding or using Ka and Kb
33
Determine the % ionization of 0.1M solution of
HC2H3O2
• Ka=1.8×10-5
I
C
E
0.1M
N/A
0
16.2. Calculations can involve finding or using Ka and Kb
0
34
What is the % ionization of 0.10M HOCl?
(Ka=3.5×10-8)
A. 0.6 %
B. 0.06%
C. 3.5×10-6 %
D. none of these
16.2. Calculations can involve finding or using Ka and Kb
35
What is the % ionization of 0.50M HOCl?
(Ka=3.5×10-8)
A. 0.13 %
B. 7.0×10-6 %
C. 0.06 %
D. none of these
16.2. Calculations can involve finding or using Ka and Kb 0.3%
36
The pH of a 0.50 M
solution of an acidic
drug, HD, is 3.5.
What is the value of
the Ka for the drug?
HD + H2O
↔
I 0.50 }
H3O+
+
0
D-
C -x
+x
+x
x
x
-x
E 0.50- }
x
0
16.2. Calculations can involve finding or using Ka and Kb
37
The pH of a 0.50 M
solution of an acidic
drug, HD, is 3.5.
What is the value of
the Ka for the drug?
HD + H2O
↔
I 0.50 }
C -x
-x
E 0.50- }
x
H3O+
+
0
Given the pH, we find
the value of x=10-pH
x=10-3.5 =3.16×10-4
D0
+x
+x
x
x
x2
 Ka
0.50  x
4 2
3.16  10   Ka
0.5  3.16  10  4 
Ka=2.0×10-7
16.2. Calculations can involve finding or using Ka and Kb
38
What is the pH of a 0.5 M HF solution (at 250C)?
Ka=7.1E-4 at this temperature.
When can I use the approximation?
Ka << 1
0.50 – x  0.50
When x is less than 5% of the value from which it is subtracted.
x = 0.019
0.019 M
x 100% = 3.8%
0.50 M
Less than 5%
Approximation ok.
What is the pH of a 0.05 M HF solution (at 250C)?
Ka 
x2
= 7.1 x 10-4
0.05
0.006 M
x 100% = 12%
0.05 M
x = 0.006 M
More than 5%
Approximation not ok.
Must solve for x exactly using quadratic equation or method of
successive approximation.
15.5
The method of successive approximations
What is the pH of a 0.122 M monoprotic acid whose
Ka is 5.7 x 10-4?
HA (aq)
H+ (aq) + A- (aq)
Solved next page
What is the pH of a 0.122 M monoprotic acid whose
Ka is 5.7 x 10-4?
HA (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.122
0.00
0.00
-x
+x
+x
0.122 - x
x
x
x2
= 5.7 x 10-4
Ka =
0.122 - x
Ka 
H+ (aq) + A- (aq)
x2
= 5.7 x 10-4
0.122
0.0083 M
x 100% = 6.8%
0.122 M
Ka << 1
0.122 – x  0.122
x2 = 6.95 x 10-5
x = 0.0083 M
More than 5%
Approximation not ok.
15.5
x2
= 5.7 x 10-4
Ka =
0.122 - x
ax2 + bx + c =0
x = 0.0081
HA (aq)
Initial (M)
Change (M)
Equilibrium (M)
x2 + 0.00057x – 6.95 x 10-5 = 0
-b ± b2 – 4ac
x=
2a
x = - 0.0081
H+ (aq) + A- (aq)
0.122
0.00
0.00
-x
+x
+x
0.122 - x
x
x
[H+] = x = 0.0081 M
pH = -log[H+] = 2.09
15.5
Few substances are more effective in relieving
intense pain than morphine. Morphine is an
alkaloid – an alkali-like compound obtained
from plants – and alkaloids are all weak bases.
In 0.010 M morphine, the pH is 10.10.
Calculate the K b and pK b
Nicotinic acid, HC2H4NO2, is a B vitamin. It is
also a weak acid with Ka = 1.4e-5. What is
the [H+], pH and percent ionization of a
0.050 M solution of nicotinic acid?
Pyridine, C5H5N, is a bad smelling liquid for
which Kb = 1.5e-9. What is the pH and
percent ionization of a 0.010 M aqueous
solution of pyridine?
Phenol is an organic compound that in
water has Ka = 1.3e-10. What is the
pH and percent ionization of a 0.15M
solution of phenol in water?
Codeine, a cough suppressant
extracted from crude opium, is a
weak base with a pKb of 5.79. What
will be the pH of a 0.020 M solution of
codeine?
What is [H3O+] in a 1.1 E-1 M
solution of HCN?
Ka for HCN is 4.0E-10.
a) 4.0E-10 M
b) 3.6E-9 M
c) 6.0E-5 M
d) 6.6E-6 M
e) 1.1E-1 M
What is the [OH-] concentration of a
0.19 M solution of NH3 in water?
Kb for NH3 is 1.8E-5.
a) 1.8E-5 M
b) 1.8E-3 M
c) 0.19 M
d) 1.54 M
e) 12.45 M
What is the pH of a 0.125 M pyruvic
acid? (Ka = 3.2E-3)
Phenol, also known as carbolic acid, is
sometimes used as a disinfectant. What
are the concentrations of all of the
substances in a 0.050 M solution of
phenol, HC6H5O?
What percentage of phenol is ionized,
for this acid Ka = 1.3E-10?
Nicotinic acid is a monoprotic acid and another
name for niacin which is a vitamin. Minute
quantities are found in all living cells.
When 0.10 mole of nicotinic acid HC6H4NO2 is
dissolved in enough water to make 1.50 L of
solution the pH of the solution is found to be
2.92. What is the Ka of nicotinic acid?
Propionic acid which occurs in dairy products in
small amounts is a weak monoprotic acid
CH3CH2COOH. If 0.10 mole of the acid is
mixed with sufficient water to make 250 mL of
solution, calculate the pH of the solution.
Ka = 1.13E-5
Aspirin is acetylsalicylic acid HC9H7O4 (HAsp). It
is a moderately weak monoprotic acid. If you
have a 0.0075 M solution of the acid, what is
the pH of the solution?
What is the concentration of HAsp at equilibrium?
Ka for HAsp is 3.3e-4.
Hydrazine, N2H4 is used as a rocket fuel and it is
a weak base with a Kb of 8.5e-5. If 0.15
moles of hydrazine are dissolved in enough
water to make 750.0 mL of solution, what is
the pH of the solution?
Example: Calculate the pH of a 0.0010
M solution of dimethylamine, a base,
for which Kb=9.6x10-4.
Example: Calculate the pH of a 0.0010 M
solution of dimethylamine for which
Kb=9.6x10-4.
B (aq )  H 2O BH  (aq )  OH  (aq ) K b  9.6  10 4
I 0.0010
0
0
x
x
x
E 0.0010  x
x
x
C
[ BH  ][OH - ]

[ B]
x2

(0.0010  x)
Put in standard form
2
Kb 
x
(0.0010  x)
(0.0010  x) K b  x 2
0  x 2  K b x  0.0010 K b  x 2  9.6 10 4 x  9.6 10 7
Solve for x and the equilibrium concentrations
Only positive solutions are allowed, so
 9.6 10  4  (9.6 10  4 ) 2  4(1)( 9.6 10 7 )
x
2(1)
 6.110  4 M and
[ BH  ]  [OH  ]  x  6.110  4 M
[ B]  (0.0010  x) M  3.9 10  4 M
A 0.17 M solution of HAsO3 was
found to have a pH of 1.09 at 25oC.
What is the Ka value for this acid?
a) 1.2E-2
b) 3.2E-2
c) 7.5E-2
d) 1.7E-1
e) 4.8E-1
A 0.53 M solution of a weak base,
R3N was found to have a pH of 10.01
What is the Kb value for this base?
a) 1.8E-20
b) 7.5E-19
c) 2.0E-8
d) 3.2E-8
e) 2.0E20
Codeine, a cough suppressant extracted
from crude opium, is a weak base with a
pKb of 5.79. What will be the pH of a
0.020 M solution of codeine?
What is the pH of a 0.122 M monoprotic acid
whose Ka is 5.7 x 10-4?
HA (aq)
Solved next page
H+ (aq) + A- (aq)
Learning Check
•
•
•
•
•
•
•
0.1M solutions of the following are acid/ base/ neutral
or amphoteric? …
•acid
HCl
•neutral
NaCl
•base
NaCN
•acid
HCN
•base
•base
Na2S
•acid
Na3PO4
NH4Cl
16.3. Salt solutions are not neutral if the ions are weak acids or bases
65
The salt, KC2H3O2, dissociates in
water as follows:
KC2H3O2  K+ + C2H3O2What pH value should it produce?
a) less than 7
b) greater than 7
c) about 7
The salt, NaHSO4, dissociates in
water as follows:
NaHSO4  Na+ + HSO4What pH value should it produce?
a) less than 7
b) greater than 7
c) about 7
The salt, Na2CO3, dissociates in
water as follows:
Na2O3  2Na+ + CO32What pH value should it
produce?
a) less than 7
b) greater than 7
c) about 7
The salt, NaClO4, dissociates in
water as follows:
+
2NaClO4  Na + ClO4
What pH value should it
produce?
a) less than 7
b) greater than 7
c) about 7
A basic salt
If 0.10 mole of sodium carbonate, Na2CO3, is
added to enough water to make 1.0 L of
solution, what is the pH of the solution?
Kb1 of CO3- is 2.1e-4.
• Solved on next pages . . .
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of Na2CO3.
Na+ + H2O  neutral
CO32- +
H2O
HCO3+
OHbase
acid
acid
base
Kb = 2.1 x 10-4
Step 1.
Set up concentration table
[CO32-]
[HCO3-]
[OH-]
initial
change
equilib
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of Na2CO3.
Na+ + H2O ---> neutral
CO32- +
H2O
HCO3+
OHbase
acid
acid
base
Kb = 2.1 x 10-4
Step 2.
Solve the equilibrium expression
2
- ][OH- ]
[HCO
x
3
Kb = 2.1 x 10-4 =

2
0.10 - x
[CO3 ]
Assume 0.10 - x = 0.10, because 100•Kb < Co
x = [HCO3-] = [OH-] = 0.0046 M
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of Na2CO3.
Na+ + H2O ---> neutral
CO32- +
H2O
HCO3+
OHbase
acid
acid
base
Kb = 2.1 x 10-4
Step 3.
Calculate the pH
[OH-] = 0.0046 M
pOH = - log [OH-] = 2.34
pH + pOH = 14,
so pH = 11.66, and the solution is ________.
Problem Solving
Calculate the pH of 0.40 M KNO2
Problem Solving
Calculate the pH of 0.20 M NaCN
Problem Solving
Calculate the number of grams of
NH4Br that have to be dissolved in
1.00 L of water at 25C to have a
solution with a pH of 5.16.
unbuffered
What is the pH of 1.00 L of pure water at
25ºC?
What is the pH of 1.00 L of pure water after
adding 1 mL of 1.00 M HCl?
Buffers introduced
QUESTION: What is the pH of 0.25 M NH3(aq)?
NH3(aq) + H2O
NH4+(aq) + OH-(aq)
How does the pH change if we add NH4Cl to the
system at equilibrium? We are adding an ion
COMMON to the equilibrium.
Le Chatelier** predicts that the equilibrium will
shift to the __________.
The pH will go _____________.
After all, NH4+ is an acid!
The Common Ion Effect
What is the pH of the 0.25 M NH3(aq) when
mixed so that it also contains 0.10 M NH4Cl?
NH3(aq) + H2O
NH4+(aq) + OH-(aq)
The Common Ion Effect
Problem: What is the pH of a solution with 0.10
M NH4Cl and 0.25 M NH3(aq)?
NH3(aq) + H2O
NH4+(aq) + OH-(aq)
+ ][OH- ]
[NH
x(0.10 + x)
-5
4
Kb  1.8 x 10 =
=
[NH3 ]
0.25 - x
[OH-] = x = (0.25 / 0.10)Kb = 4.5 x 10-5 M
This gives pOH = 4.35 and pH = 9.65
pH drops from 11.33 to 9.65 on
adding a common ion.
Buffers “work” because the weak acid can
react with added base and the weak base
can react with added acid
Consider the general buffer made so that both
HA and A- are present in solution
• When base (OH-) is added:


HA(aq)  OH (aq)  A (aq)  H 2O
• When acid (H+) is added:


A (aq)  H (aq)  HA(aq)
– Net result: small changes in pH
The Common Ion Effect
What is the pH after adding 1.00 mL of 1.00 M HCl to a
solution which is 0.10 M NH4Cl and 0.25 M NH3(aq)
[initial pH=9.65]
NH3(aq) + H2O
NH4+(aq) + OH-(aq)
What is the [H3O+] in a solution
that is 0.16 M in HC2H3O2,
WITHOUT any added NaC2H3O2?
a) 1.3E-3 M
b) 1.1E-2 M
c) 3.3E-2 M
d) 2.2E-1 M
e) 0.16 M
What is the [H3O+] in a solution
that is 0.16 M in HC2H3O2,
but, WITH NaC2H3O2
added to make [C2H3O2-] = 0.88M
a) 3.3E-6 M
b) 3.1E-5 M
c) 1.1E-2 M
d) 0.16 M
e) 0.88 M
For the general weak acid HA:


HA(aq ) 
H
(
aq
)

A
(aq )

[ H  ][ A ]
Ka 
[ HA]
rearrangin g gives
[ HA]
mol HA

[ H ]  K a  or [ H ]  K a

[A ]
mol A

– Thus both the value of Ka and the ratio of the
molarities (or the ratio of moles) affect the pH
– These last two relations are often expressed in
logarithmic form

[ A ]initial
pH  pK a  log
[ HA]initial
or
[ salt ]
pH  pK a  log
[acid ]
• The first is called the Henderson-Hasselbalch
equation, and is frequently encountered in
biology courses
• When preparing a buffer, the concentration
ratio is usually near 1, so the pH is mostly
determined by the pKa of the acid
Henderson-Hasselbalch
Equation
[Conj. base]
pH  pK a + log
[Acid]
This shows that the pH is determined
largely by the pKa of the acid and then
adjusted by the ratio of acid and
conjugate base.
• Typically, the weak acid is selected so the
the desired pH is within one unit of the pKa
For a useful buffer : pH  pK a  1
• A buffer’s capacity is determined by the
magnitudes of the molarities of its
components
• Generally, the pH change in an experiment
must be limited to about  0.1 pH unit
Adding an Acid to a Buffer
Problem: What is the pH when 1.00 mL of 1.00
M HCl is added to
a) 1.00 L of pure water (before HCl, pH = 7.00)
b) 1.00 L of buffer that has [HOAc] = 0.700 M
and [OAc-] = 0.600 M Ka(HOAc)=1.8E-5
Adding an Acid to a Buffer
Problem: What is the pH when 1.00 mL of 1.00 M HCl
is added to
a) 1.00 L of pure water (before HCl, pH = 7.00)
b) 1.00 L of buffer that has [HOAc] = 0.700 M and
[OAc-] = 0.600 M Ka(HOAc)=1.8E-5
Solution to Part (a)
Calc. [HCl] after adding 1.00 mL of HCl to 1.00 L of
water
M1•V1 = M2 • V2
M2 = 1.00 x 10-3 M
pH = 3.00
(pH = 4.68)
A buffer made from 0.10 mol HA (pKa=7.20)
and 0.15 mol NaA in 2.0 L has 0.02 mol of
HCl added to it with no volume change.
What is the pH change?
ANALYSIS: This buffer problem is “best” solved in terms of moles.
HCl is a strong acid. The H+ it contributes to solution increases the
amount of HA present at the expense of A-.
SOLUTION: The pH before addition of HCl was:
A buffer made from 0.10 mol HA (pKa=7.20)
and 0.15 mol NaA in 2.0 L has 0.02 mol of
HCl added to it with no volume change.
What is the pH change?
ANALYSIS: This buffer problem is “best” solved in terms of moles.
HCl is a strong acid. The H+ it contributes to solution increases the
amount of HA present at the expense of A-.
SOLUTION: The pH before addition of HCl was:
[ A ]
0.15 mol
pH  pK a  log
 7.20  log
 7.38
[ HA]
0.10 mol
After the HCl ionizes and reacts:
[ A ] final  (0.15  0.02) mol  0.13 mol
[ HA] final  (0.10  0.02) mol  0.12 mol
the pH of the new solution is :
0.13 mol
pH  7.20  log
 7.23, and the pH change is
0.12 mol
pH  pH finial  pH initial  7.23  7.38  0.15
• The pH change was greater than –0.1 unit. The buffer
effectively resisted the pH change, however, because if
the HCl had been added to pure water, the pH change
would have been much larger:
0.02 

pH    log
  7.00  5.00
2.0 

What is the pH of a dihydrogen
phosphate buffer
in which [H2PO4-] = 0.31 M,
and [HPO42-] = 0.35 M?
a) 5.3E-2
b) 7.15
c) 7.20
d) 7.26
e) 7.31
What is the pH of a monohydrogen
phosphate buffer
in which [HPO42-] = 0.55 M,
and [PO43-] = 0.51 M?
a) 12.25
b) 12.30
c) 12.35
d) 12.45
e) 12.50
• Acids that can donate more than one H+ to
solution are called polyprotic acids
• Table 18.3 (and Appendix C) list the
ionization constants of a number of
polyprotic acids
• The ionization constants for these acids are
numbered to keep tract of the degree of ionization
• Note that, for a given polyprotic acid, the
magnitudes of the ionization constants are always:
Ka1 > Ka2 (> Ka3, if applicable)
Suppose some H3PO4 is added to water, using
the constants in Table 18.3:




4
H 3 PO4 (aq ) H (aq )  H 2 PO
[ H  ][ H 2 PO4 ]
K a1 
[ H 3 PO4 ]
 7.1 10 3

4



H 2 PO (aq ) H (aq )  HPO
2
4
Ka2
[ H  ][ HPO42 ]

[ H 2 PO4 ]
 6.3 10 8

3
HPO42 (aq ) 
H
(
aq
)

PO

4
K a3
[ H  ][ PO43 ]

[ HPO42 ]
 4.5 10 13
– The total [H+] is then
[ H  ]total  [ H  ]first step  [ H  ]second step  [ H  ]third step
Since
K a1  K a 2  K a 3 , it follows that
[ H  ]first step  [ H  ]second step  [ H  ]third step so that
[ H  ]total  [ H  ]first step
– This is generally true when any polyprotic acid
is added to water
• This greatly simplifies the determination of
the pH in solutions of polyprotic acids
Example: What is the pH and [CO32-] in 0.10 M
carbonic acid (H2CO3)?
ANALYSIS: Carbonic acid is a diprotic acid. The
pH will depend on the [H+] generated from the
first ionization. Ionization constants can be
obtained from Table 18.3
SOLUTION: Solve the first ionization first, then
substitute the results into the second reaction.


[
H
][
HCO



3 ]
H 2CO3 (aq )  H (aq )  HCO3 (aq ) K a1 
[ H 2CO3 ]
 4.3 10 7
– Applying the usual procedures:
x  400  K a1 so that
x2
K a1 
and x  2.110  4 M  [ H  ]  [ HCO3 ]
0.10
– The pH is 3.68. Substituting these results into
the second ionization equation:

2
HCO3 (aq) 
H
(
aq
)

CO
Ka2

3 ( aq )
but [ H  ]  [ HCO3 ] so that
K a 2  [CO32 ]  4.7 10 11
[ H  ][CO32 ]

[ HCO3 ]
• Titration of a strong acid by a strong base
Titration curve for the titration of 25.00 mL of 0.2000 M
HCl (a strong acid) with the 0.2000 M NaOH (a strong
base). The equivalence point occurs at 25.00 mL added
base with a pH of 7.0 (data from Table 18.4).
•
Titration of a weak acid by a strong base
–
This can be divided into four regions
1) Before the titration begins: this is simple a
solution of weak acid
2) During the titration, but before the equivalence
point: the solution is a buffer
3) At the equivalence point: the solution contains a
salt of the weak acid, and hydrolysis can occur
4) Past the equivalence point: the excess added OHis used to determine the pH of the solution
–
Data for the titration of acetic acid with
sodium hydroxide is tabulated in Table 18.5
The titration curve for the titration of 25.00 mL of 0.200 M
acetic acid with 0.200 M sodium hydroxide. Due to hydrolysis,
the pH at the equivalence point higher than 7.00 (data from
Table 18.5).
QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic
acid with 0.100 M NaOH to the equivalence point. What is the pH
of the final solution?
Equivalence point
pH of solution of benzoic
acid, a weak acid
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
EQUILIBRIUM PORTION
Bz- + H2O
HBz + OH[Bz-]
[HBz]
equilib 0.020 - x x
Kb = 1.6 x 10-10
[OH-]
x
2
x
Kb  1.6 x 10-10 =
0.020 - x
Solving in the usual way, we find
x = [OH-] = 1.8 x 10-6, pOH = 5.75, and pH = 8.25
QUESTION: You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the equivalence
point. What is the pH of the final solution?
Half-way
point
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH
What is the pH at the half-way point?
HBz + H2O
H3O+ + Bz-
Ka = 6.3 x 10-5
[H3O+] = { [HBz] / [Bz-] } Ka
At the half-way point, [HBz] = [Bz-], so
[H3O+] = Ka = 6.3 x 10-5
pH = 4.20 NOTE: pH = pKa at 1/2 point
•
Titration of a weak base by a strong acid
–
–
This is similar to the titration of a weak acid
by strong base
Again dividing into four regions
1) Before the titration begins: this is a solution of a
weak base in water
2) During the titration, but before the equivalence
point: the solution is a buffer
3) At the equivalence point: the solution contains the
salt of the weak base, and hydrolysis can occur
4) Past the equivalence point: excess added H+
determines the pH of the solution
Titration curve for the titration of 25.00 mL of 0.200 M NH3
with 0.200 M HCl. The pH at the equivalence point is below
7.00 because of the hydrolysis of NH4+.
Problem Solving
• For the titration of 25.00 mL of 0.1000 M
ammonia with 0.1000 M HCl, calculate the
pH (a) before the addition of any HCl
solution, (b) after 10.00 mL of the acid has
been added, (c) after half of the NH3 has
been neutralized, and (d) at the equivalence
point.
Problem Solving
• Determine the pH for the titration of 25.00
mL of 0.200 M acetic acid with 0.200 M
sodium hydroxide after
the addition of 0.00 mL of NaOH
the addition of 10.00 mL of NaOH
the addition of 24.99 mL of NaOH
the addition of 25.00 mL of NaOH
the addition of 25.01 mL of NaOH
the addition of 26.00 mL of NaOH
This applies to the next 4 questions
Exactly 100 mL of 0.24 M HCl is
pipetted into a 300 mL flask. What
is the pH before any base is added?
a) 0.24
b) 0.48
c) 0.62
d) 0.97
e) 1.62
Exactly 100 mL of 0.24 M HCl is
pipetted into a 300 mL flask.
What is the pH after 17 mL of
0.30M NaOH have been added to
the flask?
a) 0.62
b) 0.80
c) 0.97
d) 1.06
e) 7.00
Exactly 100 mL of 0.24 M HCl is pipetted
into a 300 mL flask.
How many total mL of 0.30 M NaOH must be
added to the flask to reach the equivalence
point?
a) 60
b) 80
c) 75
d) 87
e) 93
Exactly 100 mL of 0.24 M HCl is
pipetted into a 300 mL flask.
What is the pH at the equivalence point
in the titration of HCl by NaOH?
a) 4
b) 5
c) 6
d) 7
e) 8
• Titration curves for diprotic acids
– The features are similar to those for monoprotic
acids, but two equivalence points are reached
The titration of
the diprotic acid
H2A by a strong
base. As each
equivalence
point is reached,
the pH rises
sharply.
• A few general comments about indicators
can be made
– Most dyes that are acid-base indicators are
weak acids, which can be represented as HIn
– The color change can be represented as:


HIn (aq) 
H
(
aq
)

In
(aq)

acid form
(one color)
K HIn
base form
(another color)
[ H  ][ In ]

[ HIn ]
– The color change will “appear” to the human
eye near the equivalence point of the indicator
– At the equivalence point, the concentration of
the acid and base form are equal, so that
pH at the equivalence point  pK HIn
– The best indicators have intense color(s) so
only a small amount will produce an intense
color change that is “easy” to see and won’t
consume too much of the titrant
Exactly 100 mL of 0.20 M HC2H3O2
are pipetted into a 300 mL flask.
What is the pH before any base is
added?
a) 0.20
b) 0.70
c) 0.89
d) 1.73
e) 2.72
This applies to next 3 questions also
Exactly 100 mL of 0.20 M HC2H3O2
are pipetted into a 300 mL flask.
What is the pH after 23 mL of
0.30M NaOH have been added to
the flask?
a) 3.48
b) 3,72
c) 4.45
d) 5.97
e) 7.00
Exactly 100 mL of 0.20 M HC2H3O2 are
pipetted into a 300 mL flask.
What is the total mL of 0.30 M NaOH
that must be added to the flask to reach
the equivalence point in this titration?
a) 47
b) 58
c) 61
d) 67
e) 97
Exactly 100 mL of 0.20 M HC2H3O2 are
pipetted into a 300 mL flask.
What is the pH at the equivalence point,
when 67 mL of 0.30 M NaOH have
been added?
a) 6.35
b) 7.00
c) 7.90
d) 8.91
e) 9.12
What is the pH of a 2.0E-1 M
solution of NH4+?
Ka for NH4+ is 5.6E-10.
a) 4.74
b) 4.97
c) 6.98
d) 8.55
e) 9.25
What is the pH of a 0.06 M solution
of NH3 in water?
(Kb for NH3 is 1.8E-5.)
a) 1.15
b) 4.74
c) 9.25
d) 11.02
e) 12.85
What is the % ionization in a 1.27E3 M solution of H2CO3?
Ka for H2CO3 is 4.2E-7.
a) 0.033 %
b) 1.8%
c) 3.0 %
d) 3.3 %
e) 30%
The pH of a HC2H3O2 solution is
3.37. What is the
[C2H3O2-]/[HC2H3O2]
ratio at this pH?
Ka for HC2H3O2 is 1.8E-5.
a) 1.0 E-2
b) 3.0 E-2
c) 4.2E-2
d) 1.2 E-1
e) 1.23
When 0.0300 moles of a
monoprotic acid is dissolved in 500.
ml water, the pH is 1.90.
What is the Ka for this acid?
a) 3.3E-3
b) 3.6E-3
c) 6.6E-2
d) 1.1 E-2
e) 2.1E-1
The Ka for H2PO4- is 6.2E-8.
What is the Kb value for the conjugate
base, HPO42-, at 298 K?
a) 6.2E-8
b) 1.6E-7
c) 4.38
d) 7.21
e) 6.2E6
What is the [OH-] of a 0.034 M
solution of NaCN in water
a) 1.0E-7 M
b) 3.3E-6 M
c) 4.3E-6 M
d) 8.3E-5 M
e) 9.2E-4 M
What is the pH of a 0.031 M
solution of NaC2H3O2 in water
a) 6.82
b) 7.89
c) 8.31
d) 8.62
e) 9.10