Part II
Applications of Acid-Base Equilibria
Common Ion Effect
Calculate [H+] and the percent dissociation of HF in a
solution containing 1.0 M HF (Ka = 7.2x10-4) and 1.0 M NaF.
[ F  ]solution  [ F  ]HF  [ F  ]NaF
NaF

Na 
co
change
ceq
H
F
1.0 M
1.0 M
HF


 F

(1  x) x
K a
 7.2  104
(1  x)
x  7.2 104
pH   log[ H  ]   log( x)   log(7.2 104 )  3.14
x
7.2  10
 %   100 % 
 100 %  0.072 %
co
1
4
Previous Example:
Calculate [H+] and the percent dissociation of HF in a
solution containing 1.0 M HF (Ka = 7.2x10-4).
1 M HF
1 M HF + 1 M NaF
x
4.2x10-3
7.2x10-4
pH
1.57
3.14
%
0.42%
0.072%
Imagine 1 M HF solution at equilibrium, then NaF is added.
What happens?
Common Ion Effect
Buffered Solution
• is a solution that persists the change in its pH
when H+ or OH- is added.
• Solution of weak acid and the salt of its conjugate
base: HAc/NaAc
• Solution of weak base abd the salt of its
conjugate acid: NH3/NH4+
• Blood is a buffered solution.
A buffered solution contains 0.50 M acetic acid HC2H3O2,
Ka=1.8x10-5) and 0.50 M sodium acetate (NaC2H3O2).
Calculate the pH of this solution.
NaAc

Na

0.5 M
HAc

Ac

0.5 M
H

 Ac

co
change
ceq
[ H  ]  [ Ac  ] (0.5  x) x
K a

 1.8  10 5
[ HAc]
(0.5  x)
[ H  ]  x  1.8  10 5
pH  4.74
Calculate the change in pH that occurs when 0.010 mol solid
NaOH is added to 1.0 L of the above buffered solution.
NaOH

 OH 
Na 
0.01 mol
0.01 mol
HAc  OH 
Ac  H 2O

n = M x V
0.5x1.0
(before reaction) =0.5 mol

0.5x1.0
=0.5 mol
change
-0.01
+0.01
n = M x V
(before reaction)
0.49
mol
0.51
mol
HAc
H

 Ac

co
0.49
0
0.51
change
-x
x
x
ceq
0.49-x
x
0.49+x
K a
(0.51  x) x
 1.8  10 5
(0.49  x)
[ H  ]  x  1.7  10 5
pH  4.74
pH  0.02
Compare this pH change with that which occurs when
0.010 mol solid NaOH is added to 1.0 L water.

NaOH
Na
 OH

0.01 mol
0.01 mol
[OH ]solution  [OH ]NaOH



nOH
0.01 mol


 0.01 M
Vsolution
1.0 L

pOH   log[OH ]   log( 0.01)  2

pH  14  pOH  14  12  12
pH  pH final  pHinitial  12  7  5
HA  H
[ H  ]  [ A ]
K a
[ HA]
[ A ]
pK a pH  log
[ HA]

 A


[
A
]

log K a log[ H ]  log
[ HA]

[
A
]

 log K a  log[ H ]  log
[ HA]
[ A ]
pH  pK a log
[ HA]
[base]
pH  pK a log
[acid ]
Henderson-Hasselbalch Equation
HA  H

 A

co
[HA]o
0
[A-]o
change
-x
x
x
ceq
[HA]o-x
x
[A-]o+x
[ A ]o  x
pH  pK a log
[ HA]0  x
[base]
pH  pK a log
[acid ]
[ A ]o
pH  pK a log
[ HA]0
Calculate the pH of a solution containing 0.75 M lactic
acid (Ka=1.4x10-4) and 0.25 M sodium lactate.
0.25
pH   log(1.4  10 )  log
 3.38
0.75
4
A buffered solution contains 0.25 M NH3 (Kb=1.8x10-5)
and 0.40 M NH4Cl. Calculate the pH of this solution.
[base]o
pH  pK a log
[acid ]o
NH3
[base]o
pH   log K a log
[acid ]o
K aK b K w
pK a pK b pK w
pKb   log(1.8 105 )  4.74
Kw
K a
Kb
NH4+
pK a 14  pK b
pK a 14  4.74  9.26
[base]o
0.25
pH  pK a log
 9.26  log
 9.05
[acid ]o
0.4
Calculate the pH of the solution that results when 0.10 mol
gaseous HCl is added to 1.0 L of the buffered solution
containing 0.25 M NH3 (Kb=1.8x10-5) and 0.40 M NH4Cl.
 pH 
before addition  pK a log
HCl
0.1 mol
NH3
[base]o
0.25
 9.26  log
 9.05
[acid ]o
0.4
 H
 Cl 
0.1 mol
 H  NH 4


n = M x V
(before reaction)
0.25x1.0
=0.25 mol
0.40x1.0
=0.40 mol
change
-0.1
+0.1
n = M x V
0.15
0.50
(before reaction)
mol
mol
[base]o
0.15
 pH after addition  pK a log
 9.26  log
 8.73
[acid ]o
0.50
pH  pH final  pHinitial  8.73  9.05  0.32
Acid addition
pH ↓
Base addition
pH ↑
Buffering Capacity
It represents the amount of H+ or OH the buffer can
absorb without a significant change in pH.
High buffering capacity: solution absorbs large amounts of
acid or base without significant change in pH.
Low buffering capacity: small amounts of acid or base can
produce a significant change in pH.
Factors affecting the buffering capacity:
1) The concentrations of buffer components
• The more concentrated the components of a buffer , the
greater the buffer capacity.
2) The ratio [A-]/[HA]
• The closer this ratio is to 1, the higher the buffering
capacity.
The relation between buffer capacity and pH change.
HAc
H

 Ac

[ A ]o
1
[ HA]0
0.01 mol
solid
NaOH
added to
1L
buffer.
Factors affecting the buffering capacity:
1) The concentrations of buffer components
2) The ratio [A-]/[HA]
• The closer this ratio is to 1, the higher the buffering
capacity.
[ A ]o
pH  pK a log
[ HA]0
[ A ]o
1
[ HA]0
pH  pK a
maximum buffer
capacity
Problem
A chemist needs a solution buffered at pH= 4.30 and can
choose from the following acids and their sodium salts:
1.
2.
3.
4.
pKa
2.87
4.89
4.19
7.46
chloroacetic acid, Ka= 1.35 x 10-3
propanoic acid, Ka= 1.3 x 10-5
benzoic acid, Ka= 6.4 x 10-5
hypochlorous acid, Ka= 3.5 x 10-8
Calculate the ratio [A-]/[HA] required for the best system
to yield a pH of 4.30.

[ A ]o
pH  pK a log
[ HA]0

[ A ]o
4.30  4.19  log
[ HA]0
[ A ]o
 1.29
[ HA]0
Strong acid – Strong base titration
HCl(aq) + NaOH(aq)  H2O + NaCl(aq)
net ionic equation: H+(aq) + OH-(aq)  H2O
Titration of
50 mL 0.200 M HNO3 with 0.100 M NaOH
A. No NaOH has been added.
pH   log[ H  ]   log[ HNO3 ]
pH   log( 0.200)  0.699
B. 10 mL NaOH has been added.
nNaOH  M NaOH  VNaOH  0.100 mol
L  10 mL  1 mmol
n 
HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)
HNO3 initially present
 M HNO  VHNO  0.200 mol
L  50 mL  10 mmol
n 
3
HNO3 remaining
3
 nHNO
3

initially present
 nHNO
3

reacted
n 

 n 
 n
n 
 10  1  9 mmol
HNO3 remaining
HNO3 initially present
NaOH added
HNO3 remaining
nHNO
nH
9 mmol
[H ] 


 0.15M
50  10 mL
V
V


3
pH   log[ H  ]   log(0.15)  0.82
C. 20 mL NaOH (total) has been added.
nNaOH  M NaOH  VNaOH  0.100 mol
L  20 mL  2 mmol
HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)
n 
 10 mmol
HNO3 initially present
n 

 n 
 n
n 
 10  2  8 mmol
HNO3 remaining
HNO3 initially present
HNO3 remaining
nHNO
nH
8 mmol
[H ] 


 0.11M
50  20 mL
V
V


3
pH   log[ H  ]   log(0.11)  0.94
D. 50 mL NaOH (total) has been added.
pH  1.31
NaOH added
So on until:
E. 100 mL NaOH (total) has been added.
nNaOH  M NaOH  VNaOH  0.100 mol
L  100 mL  10 mmol
HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)
n 
HNO3 initially present
 10 mmol
n 

 n 
 n
n 
 10  10  0 mmol
HNO3 remaining
HNO3 initially present
HNO3 remaining
nHNO
nH
0 mmol
[H ] 


 0.0M
50  50  mL
V
V


[ H  ]  10 7 M
3
( from water ionization )
pH   log[ H  ]   log(107 )  7
Stoichiometric point or point of equivalence
NaOH added
F. 150 mL NaOH (total) has been added.
nNaOH  M NaOH  VNaOH  0.100 mol
L  150 mL  15 mmol
HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)
n
n 
HNO3 initially present
 10 mmol
nOH  nNaOH in excess  5 mmol
n

NaOH in excess

NaOH in excess
n

nOH
nNaOH excess
5 mmol
[OH ] 


 0.025 M
50  150  mL
V
V

pOH   log[OH  ]   log( 0.025)  1.6
pH  14  pOH  12.4
 nNaOH added  nHNO
NaOH in excess


 nNaOH added  nNaOH reacted
3

initially present
 15  10  5 mmol
F. 200 mL NaOH (total) has been added.
nNaOH  M NaOH  VNaOH  0.100 mol
L  200 mL  20 mmol
HNO3(aq) + NaOH(aq)  H2O + NaCl(aq)
n
n 
HNO3 initially present
 10 mmol
nOH  nNaOH in excess  10mmol
n

NaOH in excess

NaOH in excess
n

nOH
nNaOH excess
10 mmol
[OH ] 


 0.04 M
50  200  mL
V
V

pOH   log[OH  ]   log( 0.04)  1.4
pH  14  pOH  12.6
 nNaOH added  nHNO
NaOH in excess


 nNaOH added  nNaOH reacted
3

initially present
 20  10  10 mmol
Weak acid – Strong base titration
Titration of
50 mL 0.100 M HAc with 0.100 M NaOH
A. No NaOH has been added.
pH of 0.10 M HAc?
[ H  ]  x  co  K a  0.1  1.8  10 5
pH  2.87
B. 10 mL NaOH has been added.
nNaOH  M NaOH  VNaOH  0.100 mol
L  10 mL  1 mmol
HAc(aq) + OH-(aq)  H2O + Ac-(aq)
n 
HAc remaining
 nHAc initiallypresent  nHAc reacted
n 
HAc
mol

M

V

0
.
100
L  50 mL  5 mmol
HAc
HAc
initially present
n 
HAc remaining
n 
 nHAc initially present  nNaOH added
HAc remaining
 5  1  4 mmol
nHAc
4 mmol
[ HAc] 

 0.067 M
50  10  mL
V
n 
Ac
formed
nAc
 nNaOH added  1 mmol
1 mmol
[ Ac ] 

 0.0167 M
50  10  mL
V


[ A ]o
0.0167
5
pH  pK a log
  log(1.8  10 )  log
 4.14
[ HA]0
0.067

n
A
nA
[ A ]o
V
pH  pK a log
 pK a log n  pK a log
[ HA]0
nHAc
V

HAc
C. 25 mL NaOH has been added.
nNaOH  M NaOH  VNaOH  0.100 mol
L  25 mL  2.5 mmol
HAc(aq) + OH-(aq)  H2O + Ac-(aq)
n 
HAc remaining
n 
 nHAc initially present  nNaOH added
HAc remaining
n 
Ac
 5  2.5  2.5 mmol
formed
 nNaOH added  2.5 mmol
nA
2.5
pH  pK a log
 4.74  log
 4.74
nHAc
2.5

D. 40 mL NaOH has been added.
nNaOH  M NaOH  VNaOH  0.100 mol
L  40 mL  4 mmol
HAc(aq) + OH-(aq)  H2O + Ac-(aq)
n 
HAc remaining
n 
 nHAc initially present  nNaOH added
HAc remaining
Ac
 5  4  1 mmol
pH  pK a log
nAc
n 

nHAc
formed
 nNaOH added  4 mmol
4
 4.74  log  5.34
1
E. 50 mL NaOH has been added.
nNaOH  M NaOH  VNaOH  0.100 mol
L  50 mL  5 mmol
HAc(aq) + OH-(aq)  H2O + Ac-(aq)
n 
HAc remaining
n 
 nHAc initially present  nNaOH added
HAc remaining
n 
Ac
 5  5  0 mmol
pH  pK a log
nAc

nHAc
HAc exists no more
[ HAc]  [OH  ]
Kb 
[ Ac  ]
co
ceq
 nNaOH added  5 mmol
not applicable because
Ac   H 2O  HAc  OH 
change
formed
[ Ac ]o 

nAc
V


5 mmol
 0.05 M
50  50  mL
x2
Kb 
0.05  x
x  co  Kb  co 
Kw
Ka
14
K
1
.
0

10
6
[OH  ]  x  co  w  0.05 

5
.
27

10
M
5
Ka
1.8  10
pOH   log[OH  ]   log(5.27  10 6 )  5.28
pH  14  pOH  8.72
F. 60 mL NaOH (total) has been added.
nNaOH  M NaOH  VNaOH  0.100 mol
L  60 mL  6 mmol
HAc(aq) + NaOH(aq)  H2O + NaAc(aq)
n
n 
 5 mmol
HAc initially present
nOH  nNaOH in excess  1 mmol

nOH
n

NaOH in excess

NaOH in excess
n
 nNaOH added  nNaOH reacted
 nNaOH added  nHAc initially present

NaOH in excess
 6  5  1 mmol
nNaOH excess
1 mmol
[OH ] 


 0.091M
50  60  mL
V
V


pOH   log[OH  ]   log( 0.091)  2.04
pH  14  pOH  11.96
G. 75 mL NaOH (total) has been added.
pH  12.30
Hydrogen cyanide gas (HCN), a powerful respiratory
inhibitor, is highly toxic. It is a very weak acid (Ka =6.2x10-10)
when dissolved in water. If a 50.0-mL sample of 0.100 M HCN
is titrated with 0.100 M NaOH, calculate the pH of the
solution
a. after 8.00 mL of 0.100 M NaOH has been added.
nNaOH  M NaOH  VNaOH  0.100 mol
L  8 mL  0.8 mmol
n 
HCN initially present
 M HAc  VHAc  0.100 mol
L  50 mL  5 mmol
HCN(aq) + OH-(aq)  H2O + CN-(aq)
n 
HCN remaining
n 
HCN remaining
n 
 nHCN initiallypresent  nHCN reacted
 nHCN initiallypresent  nNaOH added
HAc remaining
CN 
 5  0.8  4.2 mmol
pH  pK a log
nCN
n 

nHCN
formed
 9.21  log
 nNaOH added  0.8 mmol
0.8
 8.49
4.2
b. at the halfway point of the titration.
HCN(aq) + OH-(aq)  H2O + CN-(aq)
n
n
n



  5  2.5 mmol
NaOH added
NaOH added
n 
HCN remaining
n 
1
2
n 
 M NaOH  VNaOH
HCN initially present
VNaOH
1
2
 nHCN initiallypresent  nNaOH added
HCN remaining

NaOH added
pH  pK a log
n 
CN 
 5  2.5  2.5 mmol
nCN
nNaOH
2.5 mmol


 25 mL
mol
M NaOH
0.1 L

nHCN
formed
 nNaOH added  2.5 mmol
2.5
 9.21  log
 9.21
2.5
halfway point of the titration: pH  pK a
c. at the equivalence point of the titration.
HCN(aq) + OH-(aq)  H2O + CN-(aq)
n

NaOH added
n
 nHCN initially present

NaOH added
n 
HCN remaining
n 
n
 5 mmol
VNaOH
 nHCN initiallypresent  nNaOH added
HCN remaining
 5  5  0 mmol

NaOH added
nNaOH 5 mmol


 50 mL
mol
M NaOH
0.1 L
n 
CN 
formed
 nNaOH added  5 mmol
[ HCN ]  [OH  ]
Kb 
[CN  ]
CN   H 2O  HCN  OH 
co
change
 M NaOH  VNaOH
nAc
5 mmol
[CN ]o 

 0.05 M
50  50  mL
V

ceq
[OH  ]  x  co  Kb  co 
Kw
Ka

pH=10.96
Titration Curves of Polyprotic Acids
pH 
pK a1  pK a2
2
pH = pKa2
pOH from Kb
equil’m of A2–
pH = pKa1
V/2
39
V
3V/2
2V
A chemist dissolves 2.00 mmol of a solid acid in 100.0 mL
water and titrates the resulting solution with 0.0500 M
NaOH. After 20.0 mL NaOH has been added, the pH is
6.00. What is the Ka value for the acid?
n 
HA initially present
 2 mmol
n

NaOH added
 M NaOH  VNaOH  0.050  20  1 mmol
n 
HAc remaining
n 
HA remaining
n 
HA remaining
 5  0.8  4.2 mmol
 nHA initially present  nHA reacted
 nHA initially present  nNaOH added  2  1  1 mmol
nA
pH  pK a log
nHAc

1
pH  pK a log
1
pK a 6
K a 1.0  10 6
Weak base – Strong acid titration
The pH curve for the titration of 100.0 mL of 0.050 M NH3 with
0.10 M HCl. Note the pH at the equivalence point is less than 7,
since the solution contains the weak acid NH4+.
.
Acid –Base Indicators
• Organic Dyes
• weak acid-weak base conjugate pair
HIn (aq)  H+ (aq) + In– (aq)
Acid form
(one color)
base form
(another color)
– If you put a small amount of HIn in solution
• Will be one color if acidic (phenolphthalein = colorless)
• Another color if basic (phenolphthalein = bright magenta)
• Color will change when pH rises or falls
42
[H  ][In  ]
K In 
[HIn ]
Color Change
• Bromothymol Blue
HIn (aq)  H+ (aq) + In– (aq)
Acid form
base form
(one color)
(another color)
– If
[ In  ]
 10
[ HIn]
, the color of In- will be observed.
– If
[ In  ] 1

[ HIn] 10
, the color of HIn will be observed.
[ In  ]
pH  pK In log
[ HIn]
pK In pK a

[ In ]
pH  pK In log
[ HIn]
– If
[ In  ]
 10
[ HIn]
, the color of In- will be observed.
[ In  ]
pH  pK In log
[ HIn]
pH  pK In1



1
– If
[ In  ] 1

[ HIn] 10
, the color of HIn will be observed.
[ In  ]
pH  pK In log
[ HIn]



pH  pK In1
 1
Color Change occurs in the range pH=pKIn1
Bromothymol Blue
KIn=1.0x10-7
Color Change occurs in the range pH=pKa1
pH=71=6-8
pH<6
pH=6-8
pH>8
Colors and approximate pH range of some common acid-base indicators.
pH
Is this indicator suitable for your titration?
100.0 mL of 0.10 M HCl
with 0.10 M NaOH.
To find if a given indicator is suitable,
calculate pH at the point of equiv.
titration of 50 mL of 0.1 M
HC2H3O2 with 0.1 M NaOH.
Best indicator when pH at the point
of equiv. is nearest to pKIn