Chapter 10
Monoprotic Acid-Base Equilibria
1
Measuring pH in a Single Cell
2
1
pH Calculation for a Various Solution Strengths
What’
s the pH of a 0.10 M HBr solution?
HBr H 2O  Hg 3O Br -
K a 105.8

pH -log H  log(0.10) 1.00
What’
s the pH of 0.10 M KOH solution?
KOH(s)  K OH -
H K /OH 110


14
w
pH 13.00
/ 0.10 1.0 10 13
What’
s the pH of 1x10-8 M KOH solution?
H K /OH 110


14
w
pH 6.00
/ 110 8 110 6
–How can a solution of OH- be acidic (pH<7)?
3
Using the Systematic Approach
Clearly OH- solution cannot be acidic. We must account
for the OH- from water.
Reactions
KOH(s)  K OH H 2 O  H OH -
Charge Balance
Mass Balance
Equilibrium
K 
H 

OH 
K 1.0 10 M
OH 
K 
H 




8


w
  
if x H  , then OH  1.0 108 x
x 2 (1.0 108 ) x (1.0 1014 ) 0
x 9.6 10 8 M or - 1.110 7 M

pH log H  7.02
4
2
General Rules
(A) At “
high”concentration (>10-6
M) pH has a value based on added
H+ or OH(B) At “
low”concentration (<10-8
M), pH is 7.00 because we have not
added enough acid or base to affect
the pH of the water
(C) In between (10-6-10-8 M) the Kw
and added species are comparable.
The systematic treatment is
necessary.
5
A Generic Weak Acid Problem
In a monoprotic weak acid problem, you will always
have 4 equations/4 unknowns.
Charge Balance 
H 

A 
OH 
Mass Balance
F 
A 

HA
Equilibria
  
K w H  OH 
H A 
Ka 



HA 
–But when you do the algebra, you will find a cubic
equation! Since chemists don’
t want to do that math,
we usually make an approximation.
–Any respectable weak acid will produce far more H+
than the water dissociation
6
3
Approximation
When you make that assumption:
 H A 
x H 


H A  x( x)
Ka 



HA 
x2

F x F x
x2 x 
K a F 
K a 0
–Then of course you have to verify the approximation!
–If you see a problem like this, you should immediately
see that [H+]=[A-]=x and setup and solve the equation:
x2
Ka 
F x
7
Fraction of Dissociation
Fraction of dissociation, α, is the fraction of acid in the
A- form:

A   x x
A HA x ( F x) F


If 0.050 M of a weak acid yielded 6.8x10-3 M conjugate
base in solution:
x 6.8 10 3
 
0.136
F
0.050
The acid is 13.6% dissociated in solution at a F
concentration of 0.050 M
8
4
Weak Acid Example
Find the pH of 0.100 M trimethylammonium chloride.
pKa of trimethylammonium=9.800
H
+
N
H3C
Cl-
CH3
CH3
Ka
(CH 3 )3 NH  (CH 3 )3 N H 
(F-x)
x
2
x
K a 1.58 1010 
0.100 x
x
x 3.97 106 M
pH 5.40
–This equation can always be solved with the quadratic
formula, or you can approximate x = 0 in the
denominator. If you solve for x with the
approximation, it is considered good if x <0.01F 9
Generic Weak Base Setup
A weak base is treated almost identically as a weak acid:
BH OH 
Kb
Kb 
B H 2O  BH OH 



B
–If we assume all the
comes from the basic reaction and
not dissociation of H2O, then [OH-]=[BH+]=x and
F=[B]+[BH+]

B F x
OH-
BH OH 
Kb 


B
-
x2
F x
–This is similar to a weak acid problem except x=[OH-]
10
5
Weak Base Example
Find the pH of 0.10 M ammonia. pKa of NH4+= 9.244
Kb

NH 3 H 2 O  NH 
4 OH
(F-x)
x
K w 10 14.00
K b   9.244 1.75 10 5
K a 10
x
x2
1.75 105 
0.10 x
 
x OH  1.3 103 M
K
H OH
7.6 10

w


12
M
pH 11.12
11
Buffers
A buffered solution resists changes in pH when acids and
bases are added or when diluted.
The buffer is a mixture of an acid and its conjugate base
–Concentrations of HA and A- must be comparable (factor
of 10) to exert buffering
Buffers are extremely important, especially in biochemistry
12
6
pH Dependence of an Enzymatic Reaction
The rate of reaction of this
particular enzyme reaction
(amide bond cleavage by
Chymotrypsin) is twice as great
at pH=8 than at pH=7 or pH=9.
Chymotrypsin digests proteins
in the intestine
13
Making a Buffer
If you mix weak acid with its conjugate base, the
concentrations will remain close to what you mixed. If you
have 0.1 M of HA with a pKa=4:
Ka
x2
Ka 
 x 3.1103
0.10 x
HA  H A 
( 0.1-x)
x
x
0.031
For a 0.1 M solution of the conjugate base:
Kb
A  H 2 O  HA OH (0.1 - x)
x
x
x2
Kb 
 x 3.2 106 3.2 10 5
0.10 x
HA and A- do not react very much, and they react even less
when their conjugates are present because of LeChatlier’
s
principle
14
7
Henderson-Hasselbalch
The central equation for buffers is the HendersonHasselbalch, a form of the Ka eqn
H A 


Ka
-
HA
log K a log
pH pK a log
H A logH log A 
HA

HA 

-
-

A 
-

HA 
From this equation, we can know the pH of a solution if we
have the ratio of A- to HA and the pKa for the acid. For a
buffer made from weak base:
pH pK a log

B
BH 

15
Buffers-General Information
Whenever pH=pKa, [A-] must =[HA]
If there are multiple acids/bases in solution, every form of
the Henderson-Hassalbach must give the same pH
For every power of 10 change in [A-]/[HA], pH changes
by 1
16
8
Ratio Effect on pH
17
Henderson-Hasselbalch Example
Sodium hypochlorite (NaOCl, used in bleach) was dissolved
in a solution buffered at pH 6.20. Find the ratio of [OCl]/[HOCl]. pKa=7.53
HOCl  H  OCl -
pH pK a log
6.20 7.53 log
1.33 log
OCl 
-

HOCl
OCl 
-

HOCl
OCl 
-
HOCl
OCl 0.047
-

HOCl
18
9
Buffer Example
12.43 g of tris (MW=121.135) and 4.67 tris hydrochloride
(MW=157.596) are dissolved in 1.00 L water. What is the
pH? pKa=8.075
CH2OH
H 3N
+
C
CH2OH
CH2OH
Ka

H2N
C
CH2OH
H 
CH2OH
CH2OH
Assuming everything stayed in the same form:
4.67 g/L
.43 g/L
B 
0.0296 M
BH 12112.135
0.1026 M 
157.596 g/mol
g/mol

pH pK a log

B
0.1026
8.075 log
8.61
BH 

0.0296
19
Adding Acid to Buffer Solution
If you add 12.0 mL 1 M HCl to the solution What will be
the new pH?
B

H 
BH 
Conc i
0.1026
0.0120
0.0296
Conc f
0.0906
0
0.0416
pH pK a log

B
0.0906
8.075 log
8.41
BH 

0.0416
If you added 12.0 mL 1 M HCl to unbuffered solution, the
pH would be ~ 1.9
20
10
Preparing a Buffer Solution
How much (mL) of 0.500 M NaOH should be added to 10 g
of tris hydrochloride to give a pH of 7.60 in a volume of
250 mL?
10.0 g
mol TrisHCl 
0.0635
157.596 g/mol
Conc i
BH 
0.0635
Conc f
0.0635 - x
pH pK a log

OH  
x
B
0
0
x
mol B
x
7.60 log

mol BH
0.0635 x
x 0.0159 mol
0.0159 mol
mL NaOH 
0.0318 L 31.8 mL
0.500 mol/L
21
Buffer Capacity
Buffer capacity, ß, is the ability of a solution to resist
changes in pH when acid or base is added:
dC
dC
 b  a
dpH
dpH
where Ca and Cb are the moles of strong acid or base per
liter needed to change pH by 1
Use a buffer with a pKa as close as possible to the desired
pH. Useful range is ± 1 pH
22
11
Buffer Capacity Plot
A plot of Cb against pH and its
first derivative (ß).
This plot is for a 0.100 F solution
of an acid with pKa of 5.00
Most buffers also exhibit a
significant temperature dependence.
e.g., the pKa of tris changes about
0.031 units per degree temperature
change. A solution made up at room
temperature for pH=8.08 will have
pH~8.7 at 4°C and pH~7.7 at 37°C.
23
Commonly Used Buffers
24
12