King Saud University
College of Science
Disclaimer
Department of
Biochemistry
The texts, tables and images contained in this course presentation
are not my own, they can be found on:



References supplied
Atlases or
The web sites
BCH 231
Biochemical Calculations
Professor A. S. Alhomida
1
King Saud University
College of Science
Department of
Biochemistry
Instructors: Professor A. S. Ahomida
Office: Bldg 5, Room 2A 62, Tel: 467-5938, Email: alhomida@ksu.edu.sa
Schedule: Tuesday 8-9:50 am
Office Hour: Saturday and Monday 8-10 am, if you can’t come on this time you
should take an appointment
Textbooks:
1.
Biochemical calculations, by I. H. Segel
2.
The handout of the lectures
Objectives: To provide a basic understanding how to solve mathematical problems
in general biochemistry and how to prepare buffer solution and laboratory
agnets
Preparations: You are expected to prepare for each lecture by reading material to
be covered before it is to be presented in class
Examinations: Exams will consist of tow one-hour exam, tutorial exams and final
exam
Grading: One-hour exams will be 30%, tutorial exams will be 20% and final exam
will be 50%
Exam Schedule: One-hour exam will be on Tuesday, November XX, 2008 (Thu
Alqada XX, 1429 H )
2
3
Periodic Table
4
Solutions
Common solutions

Chemical solutions encountered in
everyday life:
Air
 Coffee
 Tap water
 Gasoline
 Shampoo
 Cough syrup

5
Solutions
A solution is a system of different
substances which is homogenous with
respect to its physical and chemical
properties
The components of a mixture are
uniformly intermingled (the mixture is
homogeneous)
6
Aqueous Solutions
Aqueous solutions are solutions in
which water is the solvent
Aqueous solutions are the most
common type of solution
7
Solutions
8
Solute
Dissolves in water (or other solvent)
Changes phase (if different from the
solvent)
Is present in lesser amount (if the same
phase as the solvent)
9
Solute
Retains its phase (if different from the
solute)
Is present in greater amount (if the
same phase as the solute)
10
Solute
When solid sodium
chloride dissolves,
the ions are
dispersed randomly
throughout the
solution
11
Some Properties of
Water
Water is able to dissolve so many
substances because:
Water is “bent” or V-shaped
 The O-H bonds are covalent
 Water is a polar molecule
 Hydration occurs when salts dissolve in
water

12
Some Properties of
Water
Water is a polar
molecule because it
is a bent molecule
The hydrogen end is
+ while the oxygen
end is Delta () is a partial
charge and less
than 1


04_40
H

2 O
105
H


13
Some Properties of
Water
Dipoles, bent geometry and H bond
14
Some Properties of
Water
Water dissolves many crystalline salts
by hydrating their component ions, e.g.,
NaCl
The ionic charges are partially
neutralized, and the electrostatic
attractions necessary for lattice
formation are weakened
15
Some Properties of
Water
16
Some Properties of
Water
17
Some Properties of
Water
Water is a remarkable solvent due to its
ability to dissolve a large variety of ionic
and polar substances as contributed by
its dipolar structure and capacity of
forming hydrogen bonds
18
Some Properties of
Water
In addition, its ionization and
amphipathic properties also helpful in
making water as an ideal solvent and
reactant
Colligative properties stem from the
reduction of the chemical potential of
the liquid solvent as a result of the
presence of solute
19
Some Properties of
Water
Water has an exceptionally high heat
capacity
Its boiling and melting points are
significantly higher than those
compounds with similar structure and
molecular weight
20
Some Properties of
Water
responsible for this anomalous behavior
and property
Water also has the highest surface
tension of all common liquid, except
mercury
21
Some Properties of
Water
22
Some Properties of
Water
Polar water
molecules interact
with the positive and
negative ions of a
salt, assisting in the
dissolving process
This process is
called hydration
04_41
+

+
H


 O H 
 + 
+  +
 + 
+  +
Cation


H
HO


+  +
 + 
+  +



Anion

23
Common Terms of
Solution Concentration
Saturated

When a solution contains as much solute
as will dissolve at that temperature
Unsaturated

When a solution has not reached its limit of
solubility
24
Common Terms of
Solution Concentration
Supersaturated

When a solution has more of the solute
dissolved than it normally would -very
unstable
Stock

Routinely used solutions prepared in
concentrated form
25
Common Terms of
Solution Concentration
Concentrated

Relatively large ratio of solute to solvent
Dilute

Relatively small ratio of solute to solvent
26
Standard Solution
A standard solution is a solution whose
concentration is accurately known
A volumetric flask is used to make a
standard solution
27
Standard Solution
04_44
Volume marker
(calibration mark)
Wash Bottle
Weighed
amount
of solute
(a)
(b)
(c)
(d)
28
Dilution of Stock
Solutions
When diluting stock solutions, the moles
of solute after dilution must equal the
moles of solute before dilution

Moles of solute before dilution = moles of
solute after dilution
Stock solutions are diluted using either:
A measuring or
 A delivery pipet and a volumetric flask

29
Dilution of Stock
Solutions
04_46
Rubber bulb
500 mL
(a)
(b)
(c)
30
Dilution Calculations
What volume of 12 M HCl must be
taken to prepare 15.0 mL of 0.25 M
HCl?
Solution



V1 x C1 = V2 x C2
V1 x 12 = (15.0 x 10-3) X 0.25
V1 =
(15.0 x10-3) X 0.25
12
= 0.31 x10-3 L= 0.31 mL
31
Dilution Calculations
What volume of 16 M sulfuric acid must
be used to prepare 1.5 L of 0.10 M
H2SO4 solution?
Solution



V1 x C1 = V2 x C2
V1 x 16 = 1.5 x 0.10
1.5 x 0.10
-3 L = 9.38 mL
V1 =
=
9.38
x
10
16
32
Concentration of
Solutions
33
Concentration of
Solutions
It is defined as the amount of dissolved
substance (solute) per unit amount of
solution or solvent
34
Ways of Expressing the
Solution Concentrations
The concentration is usually expressed
either:
(1) in physical units, i.e., weight and
volume units, or
 (2) in chemical units, i.e., in gram-formula
weights, gram-molecular weights, and
chemical equivalents (e.g., equivalent in
number of protons donated or accepted in
reacting)

35
Common Units of
Concentration of Solutions
Units of Quantity of
solute
solution or
solvent
g
mL
g
Concentration
units
Physical units
100 g solution
g/100 g solution
(weight/cent, w/w)
100 mL solvent
mL/100 mL solution
(volume/cent, v/v)
100 mL solution or
g/100 mL solution or
solvent
solvent (w-v/cent, w/v)
36
Common Units of
Concentration of Solutions
Units of Quantity of
solute
solution or
solvent
Concentration
units
Physical units
g
mg
mg
L solution
mL solution
L solution
μg
mL solution
g/L solution
g/L solution
mg/L solution (part per
million, ppm)
μg/mL solution (ppm)
37
Common Units of
Concentration of Solutions
Units of Quantity of
solute
solution or
solvent
mol
mmol
gfw
Concentration
units
Chemical units
L solution
mol/L solution (molarity,
M)
mL solution
mmol/mL solution
(molarity, M)
L solution
gfw/L or fw/L solution
(formality, F)
38
Common Units of
Concentration of Solutions
Units of Quantity of
solute
solution or
solvent
mfw
eq
Concentration
units
Chemical units
mL solution
mfw/mL solution
(formality, F)
L solution
eq/L solution
(normality, N)
39
Common Units of
Concentration of Solutions
Units of Quantity of
solute
solution or
solvent
meq
mol
Concentration
units
Chemical units
mL solution
meq/mL solution
(normality, N)
1000 g solvent
mol/1000 g solvent
(molality, m)
40
Concentration Based
on Volume
41
Concentration Based on
Volume
Concentration based on the amount of
dissolved solute per unit volume are the
most widely used in biochemistry
laboratories
The most common convention are
defined below:
42
Concentration Based on
Volume
Molarity (M)

It is the number of moles of solute per liter
(L) of solution
Molarity =
Number of moles of solute
Liter of solution
43
Concentration Based on
Volume
Molarity (M)
Molar concentrations are usually in [square
brackets], for example, [H+] = molarity of H+
ion
 To calculate M, we need to know the
weight (wt) of dissolved solute and its
molecular weight (MW):

44
Concentration Based on
Volume
Weight (g)
mole =
MW
Dilute solutions are often expressed in
terms of millimolarity (mM),
micromolarity (μM), and so on, where:
45
Concentration Based on
Volume
1 mmole = 10-3 moles
1 μmole = 10-6 moles
1 nmole = 1mμ mole = 10-9 moles
1 pmole = 1 μμmole = 10-12 moles
46
Concentration Based on
Volume
Therefore:



1 mM =
10-3
M=
1 uM = 10-6 M =
1 nM =
10-9
M=
1 mmol
L
1 μmol
L
1 nmol
L
=
1 μmol
mL
=
1 nmol
mL
=
1 pmol
mL
47
Concentration Based on
Volume
A 1 M solution contains one Avogadro’s
number of molecules per liter
Avogadro’s number
Number of molecules per g-mole
 Number of atoms per g-atom
 Number of ions per g-ion
 6.023 x 1023

48
Concentration Based on
Volume
In general practice, one Avogadro’s
number of particles (i.g., 1 g-mole or 1gatom or 1 g ion) is frequently called a
“mole” regardless of whether the
substance is ionic, monoatomic, or
molecluar in nature
For example, 35.5 g of Cl- ions may be
called a “mole” instead of a “g-ion”
49
Concentration Based on
Volume
Effective concentration (activity, a)

It is the effective or apparent molarity of a
solute
Activity and actual molarity are related
by:

a = g [M]


Where g = activity coefficient
The fraction of the actual concentration that is active
50
Concentration Based on
Volume
Because of interaction between solute
molecules that prevent their full
expression, g is usually less than unity
For example, HCl in a 0.1 M solution is
fully ionized, yet the solution behaves
as if it contains only 0.086 M H+
Thus g = 0.86
51
Concentration Based on
Volume
Normality (N)

It is the number of equivalents of solute per
liter of solution
Normality =
Number of equivalents of solute
Liter of solution
52
Concentration Based on
Volume
Normality (N)

To calculate N, we need to know the
weight (wt) of dissolved solute and its
equivalent weight (EW)
equivalent =
Weight (g)
EW
53
Concentration Based on
Volume
One equivalent (EW) of an acid or base
is the weight that contains one mole of
replaceable hydrogen or hydroxyl
The EW of a compound involved in a
redox reaction is the weight that
provides or accepts one Faraday (1
mole) of electrons
54
Concentration Based on
Volume
In general:
EW =
MW
n
Where n = the # of replaceable of H+ or
OH- per molecule (for acids and bases) or
 The # of electrons lost or gained per
molecule (for redox agents)

55
Concentration Based on
Volume
The molarity and normality are related
by:
N=nM
For example, What is the normality of a
0.01 M solution of H2SO4?

N = nM = 2 x 0.01 = 0.02
56
Concentration Based on
Volume
Weight percent, %w/w

It is the weight (w) in g of a solute in 100 g
of solution
%w/w =
mass solute (g)
mass solution (g)
X 100
57
Concentration Based on
Volume
Weight percent, %w/w

Weight per volume percent is often used
for routine laboratory solutions where exact
concentrations are not too important
58
Concentration Based on
Volume
Milligram percent (mg%)

It is the weight in mg of a solute per 100
mL of solution
mass solute (mg)

mg% =
volume solution (mL)
= x 100
59
Concentration Based on
Volume
Milligram percent (mg%)
It is often used in clinical laboratories
 For example, a clinical blood sugar value
of 225 means 225 mg of glucose per 100
mL of blood serum

60
Concentration Based on
Volume
Osmolarity
It is the molarity of particles in a solution
 A 1 M solution of a nondissociable solute is
also 1 Osmol that means the solution
contains 6.023 x 1023 particles/L
 A 1 M of a dissociable salt is n Osmol,
where n is the number of ions produced
per molecule

61
Concentration Based on
Volume
Osmolarity
Thus, a 0.03 M solution of KCl is 2 x 0.03 =
0.06 Osmol
 Osmolarity is often considered in
physiological studies where tissues or cells
must be bathed in a solution of the same
osmolarity as the cytoplasm in order to
prevent the uptake or release of water

62
Concentration Based on
Volume
Osmolarity
Blood plasma is 0.308 Osmol
 RBC suspended in a 0.308 Osmol NaCl
solution (0.154 M) would neither shrink nor
swell
 The 0.154 M NaCl solution is said to be
isotonic with respect to the RBC

63
Problems and Answers
(a) How many grams of solid NaOH are
required to prepare 500 mL of a 0.04 M
solution?
(b) Express the concentration of this
solution in terms of normality, g/L,
%w/v, mg%, and osmolarity?
64
Problems and Answers
Solution
(a) # of moles NaOH required = L x M
 # of moles NaOH required = 0.5 x 0.04 =
0.02 mole
Weight (g)
 # of mole =
MW


MW of NaOH = 40
65
Problems and Answers
Weight (g)

0.02 =
40
Then wt = 0.02 X 40 = 0.8 g
 Weight out 0.8 g NaOH, dissolve in water,
and dilute it 500 mL

66
Problems and Answers
(b) NaOH contains one OH/molecule
 Then N = M, and the solution is 0.04 N
 The solution contains 0.8 g/500 mL or 1.6
g/L
 1.6 g → 1000 mL
 x g → 100 mL

1.6 X 100
 %w/v =
= 0.16%
1000
67
Problems and Answers
mg% = 0.16 g/100 mL x 1000 mg = 160
mg%
 NaOH yields 2 particles (Na+ and OH-)
 Osmolarity of the solution = 2 x M = 0.08
Osmol

68
Problems and Answers
How many mL of 0.5 M H2SO4 are
required to make 1500 mL of a 0.002 M
H2SO4 solution?
Solution

The # of moles of H2SO4 in dilute solution
equals the # of moles of H2SO4 taken from
the concentrated solution
69
Problems and Answers
V1 x M1 = V2 x M2
 1.5 x 0.002 = V2 x 5

1.5 X 0.002
 V2 =
= 6.0 x 10-4 L = 0.6 mL
5

Take 0.6 mL of the concentrated solution
and complete the volume to 1.5 L with
water
70
Problems and Answers
What volume of 0.750 M NaOH solution
would be required to completely
neutralize 100 mL of 0.250 M H3PO4?
Solution
3 NaOH + H 3 PO 4  Na 3 PO 4 + 3 H 2 O
? L NaOH = 0.100 L H 3 PO 4 
0.250 mol H 3 PO 4

1 L H 3 PO 4
3 mol NaOH
1 L NaOH

 0.100 L NaOH
1 mol H 3 PO 4 0.750 mol NaOH
71
Problems and Answers
Calculate the molarity of a sulfuric acid
solution if 23.2 mL of it reacts with 0.212
g of Na2CO3?
Solution
Na 2 CO 3 + H 2SO 4  Na 2SO 4 + CO 2 + H 2 O
72
Problems and Answers
Solution
1 mol Na 2CO 3
? mol H 2SO 4 = 0.212 g Na 2CO 3 

106 g Na 2CO 3
1 mol H 2SO 4
 0.00200 mol H 2SO 4
1 mol Na 2CO 3
0.00200 mol H 2SO 4
? M H 2SO 4 
 0.0862 M H 2SO 4
0.0232 L H 2SO 4
73
Problems and Answers
30.0 mL of 0.0750 N nitric acid solution
required 22.5 mL of calcium hydroxide
solution for neutralization. Calculate the
normality and the molarity of the
calcium hydroxide solution?
Solution
2 HNO3  Ca(OH) 2  Ca(NO3 ) 2  2 H 2 O
74
Problems and Answers
Solution
# meq HNO3  # meq Ca(OH) 2
mL HNO3  N HNO3  mL Ca(OH)2  N Ca(OH)2
N Ca(OH)2 
mL HNO3  N HNO3
mL Ca(OH)2
30.0 mL  0.0750 N
22.5 mL
 0.100 N Ca(OH) 2

Because 1 mol Ca(OH) 2  2 eq the solution is 0.0500 M Ca(OH) 2
75
Problems and Answers
Typical blood serum is about 0.14 M
NaCl. What volume of blood contains
1.0 mg of NaCl?
Solution
(1.0 mg NaCl)(1 g/1000 mg)(1 mol/58.45 g)(1 L/0.14
mol) = 1.2 x 10-4 L blood serum
76
Concentration Based on
Volume
Ionic strength
It measures the concentration of charges in
solution
 As the ionic strength of a solution
increases, the activity of coefficient of an
ion decreases

77
Concentration Based on
Volume
Ionic strength

The relationship between the ionic strength
and the molarity of a solution of ionizable
salt depends on the number of ions
produced and their net charge
78
Concentration Based on
Volume
Ionic strength
 1
2
Ionic stringth ( )   M i Z i
2 2



Where Mi = the molarity of the ion
Zi = the net charge of the ion (regardless of
sign)
∑ = a symbol meaning of “the sum of”
79
Concentration Based on
Volume
Ionic strength
Salt
Ionic strength
Type
Example
1:1
KCl, NaBr
M
2:1
CaCl2, Na2HPO4
3XM
2:2
MgSO4
4XM
3:1
FeCl3, Na3PO4
6XM
2:3
Fe2(SO4)3
15 X M
80
Problems and Answers
Calculate the ionic strength of a 0.02 M
of Fe2(SO4)3?

Ionic strength = ½[MFe3+ Z2Fe3+ + MSO42Z2SO42-]
81
Problems and Answers

The 0.02 M (Fe2(SO4)3 yields 0.04 M Fe3+
and 0.06 M SO42Γ
2
(0.04)(3)2 + (0.06)(-2)2
=
2
=
(0.6) + (0.24)
2
= 0.30
Or from the precalculated relationship for
2:3 salts:
 Ionic strength = 15 x M = (15) x (0.02) =
0.30

82
Concentration Based on
Weight
Weight percent, %w/w
It is the weight (w) in g of a solute in 100 g
of solution
 The concentrations of many commercial
acids are given in terms of %w/w

83
Concentration Based on
Weight
Weight percent, %w/w

In order to calculate the volume of the
stock solution required for a given
preparation, we must know the density (d)
in g/mL, or specific gravity (SG) and the
percent (%) content by weight, are
interrelated by:
84
Problems and Answers
Describe the preparation of 2 L of 0.4 M
HCl starting with a concentrated HCl
solution, 28%w/w. SG = 1.15?
Solution


# of moles = volume (L) x Concentration (M)
# of moles HCl needed = 2 x 0.4 = 0.80 mole
85
Problems and Answers
mole =
Weight (g)
MW
MW HCl = 36.5
 Then, # of g HCl needed (wt) = moles x
MW = 0.80 x 36.5 = 29.2 g

86
Problems and Answers
The stock solution is not pure HCl but only
28% HCl by weight
 Therefore,
 28 g HCl → 100 g solution
 29 g HCl needed → x g solution
100 x 29.2
 x g HCl needed =
= 104.3 g

28
87
Problems and Answers
Instead of weighting out 104.3 g of stock
solution, we can calculate the volume
required
 1.15 g → 1 mL
 104.3 g → x mL
104.3
 # of x mL HCL needed =
= 90.7 mL
1.15
 Measure out 90.7 mL stock solution and
dilute to 2 liters with water

88
Problems and Answers

All the above relationships between weight
(wt), density (d) and %w/w can be
combined into a single
expression:
%
wv X d X
100
%
w v x d x
100
89
Problems and Answers
Cow’s milk typically contains 4.5 % by
mass of the sugar lactose, C12H22O11.
Calculate the weight (mass) of lactose
present in 175 g of milk?
Solution
mass of solute
w/w% =
(100%)
mass of solution
90
Problems and Answers
Solution
weight of solute = (weight%)(weight of
solution)/(100%)
 weight of solute = (4.5 % lactose)(175
g)/(100%)
 weight of solute = 7.9 g lactose

91
Concentration Based on
Weight
Molality (m)

It is the number of moles of a solute in
1000 g of solvent
Molality =
Number of moles of solute
Kg of solvent
92
Concentration Based on
Weight
Molality (m)
It is used in certain physical chemical
calculation (e.g., calculation of boiling-point
elevation and freezing-point depression)
 For dilute aqueous solutions, m and M will
be quite close
 In order to interconvert m and M, we need
to know %w/w

93
Concentration Based on
Weight
Mole fraction (MF)
It is the fraction of the total number of
moles present represented by the
compound in question
 For example, in a solution containing (n 1)
moles of compound 1, (n2) moles of
compound 2, and (n3) moles of compound
3, the mole fraction (FM) of compound 2, is
given by:

94
Concentration Based on
Weight
Mole fraction (MF)
MF2 =

n2
n1 + n2 + n3
MF of a compound is important in certain
physical chemical calculation, but is not
often used in biochemistry
95
Problems and Answers
Calculate (a) the molality of the
concentrated stock HCl solution
described in problem # 4 (b) calculate
the mole fraction of HCl in the solution?
Solution

(a) the solution containing 28% w/w HCl,
i.e., 28 g HCl/100 g total (HCl + water) or
28 g HCl/72 g water (100 – 28)
96
Problems and Answers
28 g HCl → 72 g water
 x g HCl → 1000 g water

28 x 1000
 x g HCl =
72

mole =
= 388.9 g HCl
Weight (g)
MW

# of moles HCl =
388.9
36.5
= 10.65 moles
97
Problems and Answers
The molality of the solution = 10.65
 (b) in 100 g of solution, for example, we
have:
28
 moles of HCl =
= 0.767 moles

36.5


moles of water =
MFHCl =
nHCl
nHCl + nH2O
72
18
=
= 4.0 moles
0.767
= 0.161
0.767 + 4.0
98
Problems and Answers
A solution is prepared by mixing 15.0 g
of Na2CO3 and 235 g of H2O. The final
volume of solution is 242 mL. Calculate
the %w/w and %w/v concentration of
the solution?
99
Problems and Answers
Solution


g solution = 15.0 g Na2CO3 + 235 g H2O =
250.0 g
15.0 g solute
%w/w =
x 100 = 6.0% Na2CO3
250 g solution

15.0 g solute
%w/v =
x 100 = 6.20% Na2CO3
242 mL solution
100
Problems and Answers
What is the molarity of 500 mL NaOH
solution if it contains 6.0 g NaOH? How
many moles of NaOH?
Solution


# of mole =
M=
6.0 g
40
= 0.15 mole
# of mole of solute
Volume (L) solution
=
0.15
0.5
= 0.3 M
101
Problems and Answers
What is the m of a solution prepared by
dissolving 25.3 g Na2CO3 in 458 g
water?
Solution
25.3


# of moles of Na2CO3 =
Molality =
106
# of mole solute
Kg of water
= 0.24 mole
25.3
=
0.458
= 0.52
102
Problems and Answers
How many g of NaCl are contained in
250.0 mL of 0.22 M NaCl solution?
Solution
1 L → 0.22 mole NaCl
 0.25 L → x mole


# of mole of NaCl = 0.25 x 0.22 = 0.055 mole
103
Problems and Answers
How many g of NaCl are contained in
250.0 mL of 0.22 M NaCl solution?
Solution
1 mole NaCl → 58.4 g
 0.055 mole → x g
 # of g NaCl = 0.055 x 58.4 = 3.2 g

104
Equilibrium Constants
105
Equilibrium Constants
Many reactions that occur in nature are
reversible and don’t proceed to
completion
Instead, they come to an apparent halt
or equilibrium at some point between 0
and 100% completion
106
Equilibrium Constants
At equilibrium, the net velocity is zero
because the absolute velocity in the
forward direction exactly equals the
absolute velocity in the reverse direction
The position of equilibrium is
conveniently described by an
equilibrium constant, Keq
107
Equilibrium Constants
For example, consider the dissociation
of a weak acid:
HA
k1
H+ + A-
K -1

The forward velocity (vf) is proportional to
the concentration of HA:
108
Equilibrium Constants
vf α [HA] or vf = k1[HA]
 Whre k1 is a proportionality constant,
known as rate constant (first-order
constant)
 The reverse velocity (vr) is proportional to
the concentration of H+ and A- and to the
product of the concentration of A- and H+

109
Equilibrium Constants
Vr α [H+] and vr α [A-]
 Then, vr α [H+] [A-] or vr = k-1[H+][A-]
 Where k-1 is a second-order constant
 At the equilibrium: vf = vr

110
Equilibrium Constants
k1
[H+][A-]

k1[HA] = k-1[H+][A-] or

The ratio of two constants is defined as Keq
K-1
=
[HA]
[H+][A-]

Keq =

Keq is an acid dissociation constant and
would be indicated as Ka
[HA]
111
Equilibrium Constants
If the reaction in question is:
A+B
k1
2C
K -1
vf = k1[A][B] and vr = k-1[C][C] = k-1[C]2
Keq =
[C]2
[A][B]
112
Problems and Answers
Consider reaction below:
E+S
k1
ES
K -1
What are the units of (a) k1, (b) k-1, and (c)
Keq?
113
Problems and Answers
Solution
(a) vf = k1[E][S]
 Let vf = M X min-1, and [E] and [S] = M


k1 =
vf
[E][S]
M X min-1
=
[M][M]
= M-1 X min-1
114
Problems and Answers
Solution


(b) vr = k-1[ES]
k-1 =
vr
[ES]
M X min-1
=
[M]
= min-1
115
Problems and Answers
Solution
k1
M-1 X min-1
-1
(c) Keq =
=
=
M
k-1
min-1
 Or
[ES]
[M]
 Keq =
=
= M-1
[E][S]

[M][M]
116
Biochemistry of Acids
and Bases
117
Acids and Bases
118
Acids and Bases
119
Acids and Bases
120
Acids and Bases
Acids
Have a sour taste
 Vinegar is a solution of acetic acid
 Citrus fruits contain citric acid
 React with certain metals to produce
hydrogen gas

121
Acids and Bases
Bases
Have a bitter taste
 Feel slippery
 Many soaps contain bases

122
Some Properties of Acids
Produce H+ (as H3O+) ions in water (the
hydronium ion is a hydrogen ion
attached to a water molecule)
Taste sour
123
Some Properties of Acids
Corrode metals
Electrolytes
React with bases to form a salt and
water
pH is less than 7
Turns blue litmus paper to red
124
Some Properties of Bases
Produce OH- ions in water
Taste bitter, chalky
Are electrolytes
125
Some Properties of Bases
Feel soapy, slippery
React with acids to form salts and water
pH greater than 7
Turns red litmus paper to blue
126
Some Common Bases
NaOH

Sodium hydroxide: lye
KOH

Potassium hydroxide: liquid soap
Ba(OH)2

Barium hydroxide: stabilizer for plastics
127
Some Common Bases
Mg(OH)2

Magnesium hydroxide: “MOM” Milk of
magnesia
Al(OH)3

Aluminum hydroxide: Maalox (antacid)
128
Acid and Base Definitions
Arrhenius (traditional) definition
Look to the structure
 Acids



Produce H+ ions (or hydronium ions H3O+)
Bases
Produce OH- ions
 (problem: some bases don’t have hydroxide
ions!)

129
Acid and Base Definitions
Arrhenius:

Acid is a substance that produces H+
(H3O+) in water
130
Acid and Base Definitions
Arrhenius:

Base is a substance that produces OH- in
water
131
Acid and Base Definitions
Brønsted-Lowry definition
Look to the reactions
 Acids



Bases


Proton donor
Proton acceptor
A proton is really just a hydrogen atom that
has lost it’s electron!
132
Acid and Base Definitions
Brønsted-Lowry
Acid is a proton donor
 Base is a proton acceptor

Base
Acid
Conjugate
acid
Conjugate
base
133
Acid-base Theories
The Brønsted definition means NH3 is a
BASE in water and water is itself an
ACID
NH3
Base
+
H2O
Acid
NH4+ + OHAcid
Base
134
Conjugate Pairs
Conjugate base

The species formed from an acid when it
donates a proton to a base
Conjugate acid

The species formed from a base when it
accepts a proton from an acid
135
Conjugate Pairs
Acid-base reaction

A proton-transfer reaction
Conjugate acid-base pair

Any pair of molecules or ions that can be
interconverted by transfer of a proton
136
Conjugate Pairs
137
Conjugate Pairs
Label the acid, base, conjugate acid,
and conjugate base in each reaction?
HCl + OH-  Cl- + H2O
H2O + H2SO4  HSO4- + H3O+
138
Acids and Base Definitions
Lewis definition

Lewis acid

A substance that accepts an electron pair
139
Acids and Base Definitions
Lewis base definition

A substance that donates an electron pair
140
Lewis Acids and Bases
Formation of hydronium ion is also an
excellent example
H
+
ACID
•• ••
O—H
H
BASE
••
H O—H
H
Electron pair of the new O-H bond
originates on the Lewis base
141
Lewis Acids and Bases
142
Classification of Acids
and Bases
System
Acid
Base
Arrhenius
Bronsted
Lewis
H+ donor
H+ donor
Electron pair
acceptor
OH- donor
H+ acceptor
Electron pair
donor
143
Lewis Acid-Base Interactions
in Biochemistry
The heme group in
hemoglobin can
interact with O2 and
CO2
The Fe ion in
hemoglobin is a
Lewis acid whereas
O2 and CO2 can act
as Lewis bases
144
Lewis Acid-Base Interactions
in Biochemistry
145
Amphoterism
H 2O
Acts as either an acid or a base, it is an
amphoteric
H2O + H2O → H3O+ + OH OH- conjugate base of H2O
 H3O+ conjugate acid of H2O

146
Amphoterism
HCO3- is an amphoteric
HCO3- + H2O → H3O+ + CO32HCO3- + H2O → H2CO3 + OH-
147
Amphoterism
Water is amphiprotic

Because it acts both an acid and a base
Autoprotolysis or autoionization
When one molecule transfers a proton to
another molecule of the same kind
2 H2O →H3O+ + OH
An O-H bond is strong; the fraction of
protons transferred is very small
148
Ionization of Strong Acid
and Base
Strong acids
Transfers all protons to water
 Completely ionizes
 Strong electrolyte
 Conjugate base is weaker and has
negligible tendency to be protonated

149
Ionization of Strong Acid
and Base
Weak Acid
Fraction of protons transferred to water
 Partly ionized
 Weak electrolyte
 Conjugate base is stronger readily
accepting protons from water

150
Ionization of Strong Acid
and Base
As acid strength decreases, base
strength increases
The stronger the acid, the weaker its
conjugate base
The weaker the acid, the stronger its
conjugate base
151
Ionization of Strong Acid
and Base
Strong base
All molecules accept a proton
 Completely ionizes
 Strong electrolyte
 Conjugate acid is weaker with negligible
tendency to donate protons

152
Ionization of Strong Acid
and Base
Weak base
Fraction of molecules accept proton
 Partly ionized
 Weak electrolyte
 Conjugate acid is stronger that readily
donates protons

153
Ionization of Strong Acid
and Base
As base strength decreases, acid
strength increases
The stronger the base, the weaker its
conjugate acid
The weaker the base the stronger its
conjugate acid
154
Ionization of Strong Acid
and Base
HCl in solution is essentially 100%
ionized to H3O+ and Cl-:
HCl + H2O → H3O+ + Cl-
H3O+ (conjugate acid of water) is actual
form of H+ in solution
Strong base as NaOH ionizes
extensively in solution to give OH-:
NaOH → Na+ + OH155
Ionization of Water
156
Ionization of Water
The ionization of water can be
described by:
(1) a dissociation constant (Kd)
 (2) an ionization constant (Ki)
 (3) a specific constant for water (Kw)

157
Ionization of Water
Simple dissociation (ionization)
HOH

Kd =
H+ + OH-
[H+][OH-]
[HOH]
158
Ionization of Water
Conjugate acid-base pair


HOH + HOH
Ki =
H3O+ + OH-
[H3O+][OH-]
[HOH]2
159
Ionization of Water
Ionization constant of water (equilibrium
constant of water)
(1) Kw = Kd [H2O]
(2) Kw = KI [H2O]2
(3) Kw = [H+][OH-]
160
Ionization of Water
Water produces two conjugate acidbase pairs:
HOH


(1)
(2)
OHH3O+
HOH
161
Ionization of Water
The molarity of pure water can be
calculated as follows:
(1) Molarity =
# of mole
L
Weight (g)
(2) # of mole =
MW
162
Ionization of Water
# of mole of water =
Molarity of water =
1000
55.55
1L
18
= 55.55 moles
= 55.55 M
We substitute the above values into the Kd
and Ki expressions:
163
Ionization of Water
[H+][OH-]


Kd =
Ki =
[HOH]
(10-7) (10-7)
=
[H3O+][OH-]
[HOH]2
(55.55)
= 1.8 x 10-16
(10-7) (10-7)
=
(55.55)2
= 3.24 x 10-18
164
Ionization of Water
Kw = Kd [H2O] = (1.8 x 10-16) x (55.55) = 0-14
 Kw = Ki [H2O]2 = (3.2 x 10-18) x (55.6)2=10-14
 Kw = [H+][OH-] = 10-14

Kw = [H+][OH-] = 10-14
pKw = pH + pOH = 14
165
pH and pOH
166
pH and pOH
The potential of the hydrogen ion was
defined in 1909 as the negative of the
logarithm of [H+]
The pH scale is a way of expressing the
strength of acids and bases
pH = - log
[H+]
= log
1
[H+]
167
pOH
KW = [H3O+][OH-]= 10-14
-logKW = -log [H3O+]-log[OH-]= -log (10-14)
pKW = pH + pOH = -(-14)
pKW = pH + pOH = 14
168
pH and pOH
pOH = - log
[OH-]
= log
1
[OH-]
pH + pOH = 14
169
170
pH of Common Substances
171
pH and pOH Scales
172
[OH-]
[H+]
pOH
pH
173
Calculating the pH
pH = - log [H+]

Remember that the [ ] mean molarity
What are (a) H+ concentration, (b) pH,
(c) OH- concentration and (d) pOH of
0.001 M solution of HCl?
174
Calculating the pH
Solution
(a) HCl is a strong acid and it dissociated
into 10-3 M H+ and 10-3 Cl When we are dealing with strong acids, the
H+ contribution from the ionization of water
is neglected

175
Calculating the pH
Solution
(b) pH = - log [H+]
 pH = - log 10-3 = - (-3) = 3
 (c) Kw = [H+][OH-] = 10-14



[OH-] =
Kw
[H+]
=
10-14
10-3
= 10-11 M
(d) pOH = - log [OH-] = log 10-11 = 11
176
Calculating the pH
177
Calculating the pH
If the pH of Coke is 3.12, what is [H+]?
Solution
pH = - log [H+] then - pH = log [H+]
 Take antilog (10x) of both sides and get
 10-pH = [H+]
 [H+] = 10-3.12 = 7.6 x 10-4 M


To find antilog on your calculator, look for
“Shift” or “2nd function” and then the log button
178
Calculating the pH
A solution has a pH of 8.5. What is the
molarity of hydrogen ions in the
solution?
179
Calculating the pH
Solution
pH = - log [H+]
 8.5 = - log [H+]
 - 8.5 = log [H+]
 Antilog - 8.5 = antilog (log [H+])
 10-8.5 = [H+]
 3.16 x 10-9 M = [H+]

180
What is Acid Rain?
Dissolved carbon dioxide lowers the pH
CO2(g) + H2O  H2CO3  H+ + HCO3Atmospheric pollutants from combustion
NO, NO2 + H2O → HNO3
Both strong
acids
SO2, SO3 + H2O → H2SO4
Both strong acids
pH < 5.3
181
Calculating the pH
The pH of rainwater collected in a
certain region of the northeastern KSA
on a particular day was 4.82. What is
the H+ ion concentration of the
rainwater?
The OH- ion concentration of a blood
sample is 2.5 x 10-7 M. What is the pH
of the blood?
182
Calculating the pH and
pOH
A biochemist dilutes concentrated HCl
to make two solutions: (a) 3.0 M and (b)
0.0024 M. Calculate the [H3O+], pH,
[OH-], and pOH of the two solutions at
25°C?
What is the [H3O+], [OH-], and pOH of a
solution with pH = 3.67? Is this an acid,
base, or neutral?
183
Calculating the pH and
pOH
What is the pH of a10-8 M solution HCl?
Solution
The first tendency of many students is to
say pH = 8
 This is obviously incorrect
 NO matter how much one dilutes a strong
acid, the solution will never become
alkaline or neutral

184
Calculating the pH and
pOH
Solution
In this dilute solution, the contribution of H+
ions from H2O is actually greater than the
amount of contributed by HCl
 [H+] = 10-7 (from water) + 10-8 (from HCl)
 pH = - log [H+] = - log1.1 x 10-7 = 6.96
 But this is still not completely correct

185
Calculating the pH and
pOH
Solution
Actually, the slight increase in H+ ions from
HCl tends to depress the ionization of
water and shits to the equilibrium of the
HOH
H+ + OHthe reaction to the left, so both HOH and HCl
ionize to from H+ ions

186
Calculating the pH and
pOH
Solution
HOH
H+ + OH HCl → H+ + Cl Let x = [H+] M from H2O, then
 [OH-] = x
 x from HCl = 10-8
 [H+] = x + 10-8, [OH-] = x
187
Calculating the pH and
pOH
Solution
[H+][OH-] = 10-14 then by substituting these
values in the Kw of water, we have
 (x + 10-8) (x) = 10-14
 x2 + 10-8x = 10-14
 x2 + 10-8x -10-14 = 0, this a quadratic
equation

188
Calculating the pH and
pOH
x=

√
2
- b ± b – 4ac
2a
Where a =1, b = 10-8, c = -10-14
189
Calculating the pH and
pOH
Solution

-
x=
(10-8)
±
√
(10-8 )2 – 4(1) (-10-14)
2
x = + (9.5 x 10-8) and – (9.5 x 10-8)
 x = x + 10-8 = (9.5 x 10-8) + 10-8 = (10.5 x
10-8)

190
Calculating the pH and
pOH
Solution
x = x + 10-8 = (9.5 x 10-8) + 10-8 = (10.5 x
10-8)
 pH = - log (10.5 X 10-8) = 6.98

Calculate the pH of 10-7 M HCl
solution?
191
pH Testing
There are several ways to test pH
Blue litmus paper (red = acid)
 Red litmus paper (blue = basic)
 pH paper (multi-colored)

192
pH Testing
pH meter (7 is neutral, <7 acid, >7 base)
 Universal indicator (multi-colored)
 Indicators like phenolphthalein
 Natural indicators like red cabbage,
radishes

193
pH Paper Testing
Paper tests like litmus paper and pH
paper
Put a stirring rod into the solution and stir
 Take the stirring rod out, and place a drop
of the solution from the end of the stirring
rod onto a piece of the paper
 Read and record the color change
 Note what the color indicates

194
pH Paper Testing
You should only use a small portion of
the paper
You can use one piece of paper for
several tests
195
pH Paper Testing
196
pH Meter
Tests the voltage of
the electrolyte
Converts the voltage
to pH
Very cheap,
accurate
Must be calibrated
with a buffer solution
197
pH Indicators
Indicators are dyes that can be added
that will change color in the presence of
an acid or base
Some indicators only work in a specific
range of pH
198
pH Indicators
Once the drops are added, the sample
is ruined
Some dyes are natural, like radish skin
or red cabbage
199
Neutralization
200
Neutralization
Neutralization reaction is a reaction in
which an acid and a base react in an
aqueous solution to produce a salt and
water
acid + base → salt + water
201
Neutralization
The concentration of acid (or base) in
solution can be determined by
performing a neutralization reaction
An indicator is used to show when
neutralization has occurred
 Often use phenolphthalein that is colorless
in neutral and acid; turns pink in base

202
Steps of Neutralization
Reaction
A measured volume of acid of unknown
concentration is added to a flask
Several drops of indicator added
A base of known concentration is slowly
added, until the indicator changes colormeasure the volume
203
Steps of Neutralization
Reaction
The solution of known concentration is
called the standard solution

Added by using a buret
Continue adding until the indicator
changes color

Called the “end point” of the titration
204
Setup for Titrating an
Acid with a Base
buret
NaOH
205
Neutralization and
Indicators
206
Problems and Answers
(a) How many mL of 0.025 M H2SO4 are
required to neutralize exactly 525 mL of
0.06 M KOH? (b) What is the pH of the
neutralization solution?
207
Problems and Answers
Solution

# of moles (equivalents) of H+ required = #
of moles (equivalents) of OH- present
Nacid x Vacid = Nbase x Vbase
N=nM

Normality of H2SO4 = 0.025 x 2 = 0.05 N
208
Problems and Answers
Solution
Nacid x Vacid = Nbase x Vbase


0.05 x Vacid = 0.06 x 0.525
Vacid =
0.06 x 0.525
0.05
= 0.63 L = 630 mL
209
Problems and Answers
Solution

(b) The neutralization solution contains
only K2SO4, i.e., it is salts of strong acid
and strong base, so no effect on pH = 7
210
Problems and Answers
How many mL of 0.05 N HCl are
required to neutralized exactly 8.0 g of
NaOH (MW = 40)?
Solution
At the equivalent point, the # of moles H+
added = the # of moles OH- present
 Vacid x Nacid = # of moles (equivalents) of H+
added

211
Problems and Answers
Solution

wt
8.0
# of moles of NaOH (n) =
=
=
MW
40
0.02 mole
n


N=
L=
L
n
N
; then, n = N x L
=
0.02
0.05
= 0.4 = 400 mL
212
Titration
213
Titration
Titration is the process of adding a
known amount of solution of known
concentration to determine the
concentration of another solution
A method of determining the
concentration of one solution by
reacting it with a standard solution
214
Titration
Nacid x Vacid = Nbase x Vbase
Standard solution

A solution with known concentration
215
Titration
When titrating acids against bases, the
end point of the titration is at the
equivalence point
Equivalence point

The point where the H+ concentration is
equal to the OH- concentration
216
Strong Acid-strong
Base Titration
217
Strong Acid-strong Base
Titration
When a strong acid is titrated with a
strong base the pH at any point is
determined solely by the concentration
of untirted acid or excess base
The conjugate base that is formed (Cl-)
has no effect on the pH
218
Strong Acid-strong Base
Titration
NaOH + HCl → H2O + NaCl
OH- + H+ → H2O

100% ionization! No equilibrium
219
Strong Acid-strong Base
Titration
pH at equivalent
point
220
Features of Strong Acidstrong Base Titration Curve
The pH starts out low, reflecting the
high [H3O+] of the strong acid and
increases gradually as acid is
neutralized by the added base
Suddenly the pH rises steeply
221
Features of Strong Acidstrong Base Titration Curve
This occurs in the immediate vicinity of
the equivalence point
For this type of titration the pH is 7.0 at
the equivalence point
Beyond this steep portion, the pH
increases slowly as more base is added
222
Strong Base-strong Acid
Titration
223
Problem and Answer
Calculate the appropriate values and
draw the curve for the titration of 500
mL of 0.01 N HCl with 0.01 N KOH?
Solution
A titration curve is a plot pH vs mL or
(equivalents or moles) of standard titrant
(solution) added
 For the titration of a given amount of acid,

224
Problem and Answer
Solution
(1) The pH at any position up to the
equivalent point is calculated from the
concentration of excess (untitrated) of H+
remaining (taking the increasing volume
into account)
225
Problem and Answer
Solution
(2) At the equivalent point, the solution
contains only KCl (salt) that has no effect
on the pH, i.e., pH = 7
(3) The pH at positions beyond the
equivalent point is calculated from the
concentration of excess of OH-
226
Problem and Answer
227
Weak Acid-strong
Base Titration
228
Weak Acid-strong Base
Titration
A weak acid (HA) ionizes (dissociates)
in an aqueous solution to yield a small
about of H+ ions:
HA
H+ + A When OH- ions are added, they are
neutralized by the H+ ions to form H2O
OH- + H+ → H2O
229
Weak Acid-strong Base
Titration
The removal of H+ ions disturbs the
equilibrium between the weak acid and
its ions
Consequently, more HA ionizes to
reestablish the equilibrium
230
Weak Acid-strong Base
Titration
The newly produced H+ can then be
neutralized by more OH- and so on until
all the hydrogen originally present is
neutralized
231
Weak Acid-strong Base
Titration
The overall result, the sum of reactions
above, is the titration of HA with OH-:
HA
H+ + AOH- + H+ → H2O
HA + OH-
H2O + A -
232
Weak Acid-strong Base
Titration
The number of equivalents of OHrequired equals the total number of
equivalents of hydrogen present (as H+
+ HA)
The pH at the exact end (equivalence)
point of the titration is not 7 but higher
because of the hydrolysis of A- ion
233
Weak Acid-strong Base
Titration
In the absence of any remaining of HA,
the A- ion reacts with H2O to produce
ions and the unionized HA
234
Weak Acid-strong Base
Titration
The H+ concentration and pH during the
titration can be calculated from the Ka
expression or from the HendersonHasselbalch equation provided the
concentration of conjugate acid and
conjugate base in known
235
Weak Acid-strong Base
Titration
We can calculate the concentration of
HA and A- during the titration by
assuming that:
moles HAremain = moles HAorig – moles HAtitrated
moles A- = moles HAtitrated
236
Weak Acid-strong Base
Titration
237
Weak Acid-strong Base
Titration
CH3COOH + NaOH  CH3COONa + H2O
CH3COOH + OH-  CH3COO- + H2O
At equivalence point (pH > 7):
CH3COO- + H2O  OH- + CH3COOH
238
Weak Acid-strong Base
Titrations
pH at equivalent
point
pKa
[HA] = [A-]
239
Differences Between Strong Acidstrong Base Titration Curve and Weak
Acid-strong Base Titration Curve
The initial pH is higher
A gradually rising portion of the curve,
called the buffer region, appears before
the steep rise to the equivalence point
The pH at the equivalence point is
greater than 7.0
The steep rise interval is less
pronounced
240
Strong Acid-weak Base
Titrations
HCl + NH3 → NH4Cl
H+ + NH3 → NH4Cl
At equivalence point (pH < 7):

NH4+ + H2O
NH3 + H+
241
Strong Acid-weak Base
Titrations
242
Weak Base-strong Acid
Titration
243
Differences Between Weak Acidstrong Base Titration Curve and Weak
Base-strong Acid Titration Curve
The initial pH is above 7.0
A gradually decreasing portion of the
curve, called the buffer region, appears
before a steep fall to the equivalence
point
244
Differences Between Weak Acidstrong Base Titration Curve and Weak
Base-strong Acid Titration Curve
The pH at the equivalence point is less
than 7.0
Thereafter, the pH decreases slowly as
excess strong acid is added
245
246
Types of Acid-base
Reactions: Summary
247
Acid-base Indicators
248
Acid-base Indicators
Which indicator(s) would you use for a
titration of HNO2 with KOH?
Weak acid titrated with strong base
At equivalence point, will have
conjugate base of weak acid
At equivalence point, pH > 7
Use cresol red or phenolphthalein
249
Acid-base Indicators
250
Finding the Equivalence Point
(Calculation Method)
Strong acid vs. strong base

100 % ionized! pH = 7 No equilibrium!
Weak acid vs. strong base

Acid is neutralized; Need Kb for conjugate
base equilibrium
251
Finding the Equivalence Point
(Calculation Method)
Strong acid vs. weak base

Base is neutralized; Need Ka for conjugate
acid equilibrium
Weak acid vs. weak base

Depends on the strength of both; could be
conjugate acid, conjugate base, or pH 7
252
Ionization of Weak
Acid-base
253
Ionization of Weak Acidbase
Weak acids
Much less than 100% ionized in water
 One of the best known is acetic acid =
CH3COOH

254
Ionization of Weak Acidbase
Weak base
Less than 100% ionized in water
 One of the best known weak bases is
ammonia
 NH3 + H2O
NH4+ + OH
255
Common Weak Bases
256
Equilibrium Constants
for Weak Acids
Weak acid has Ka < 1
Leads to small [H3O+] and a pH of 2 - 7
257
Equilibrium Constants
for Weak Bases
Weak base has Kb < 1
Leads to small [OH-] and a pH of 12 - 7
258
Ionization Constants of
Weak Acids and Bases
259
Relationship of Ka, Kb,
+
[H3O ] and pH
260
Relationship of Ka, Kb,
[H3O+] and pH
When weak acid (HA) is dissolved in
water, it ionizes to form H3O+ and A

HA + H2O
Ka =
H3O+ + A-
[H3O+][A-]
[HA]
261
Relationship of Ka, Kb,
[H3O+] and pH
If we start with A- and dissolve in water,
it ionizes to form OH- and HA


A- + H2O
Kb =
HA + OH-
[HA][OH-]
[A-]
The larger the Ka, the stronger the acid
and the larger Kb, the stronger the base
262
Relationship of Ka, Kb,
[H3O+] and pH
Solving the Ka and Kb expression for
[H3O+] and [OH-]:
[H3
O +]
=
[OH-] =
[HA][Ka]
[A-]
[A-][Kb]
[HA]
263
Relationship of Ka, Kb,
[H3O+] and pH
Substituting into [H3O+] [OH-] = Kw:

Kw =
[HA][Ka]
[A-]
x
[A-][Kb]
[HA]
= KaKb
Kw = KaKb
pKa + pKb = 14
264
Relationship of Ka, Kb,
[H3O+] and pH
265
pH of Solution of Weak
Acid-base
266
pH of Solution of Weak
Acid-base
When weak acid (HA) is dissolved in
water, it ionizes to form H3O+ and A
HA + H2O
H+ + A-
[H+][A-]

Ka =

The concentration of [H+] = [A-]
[HA]
267
pH of Solution of Weak
Acid-Base
[H+]2

Ka =

[H+] = √Ka [HA]
[HA]
pKa + p[HA]
pH =
2
268
pH of Solution of Weak
Acid-Base

Similar relationship can be derived for
weak bases
pOH =
pKb + p[A-]
2
269
Problems and Answers
HA is 0.1% ionized in 0.2 M solution. (a)
Calculate Ka of the acid? (b) the pH of
the solution? (c) How much the weaker
is the active acidity of the HA solution
compared to a 0.2 M solution of HCl?
(d) How many mL of 0.1 N KOH would
be required to neutralize completely 0.5
L of the 0.2 M HA solution?
270
Problems and Answers
Solution

(a) Degree of ionization =
[H+]ionized

0.1 =

[H+]ionized =
[H+]ionized
[HA]orig
x 100
x 100, then
0.20
0.1 x 0.2
100
= 2.0 x 10-4 M
271
Problems and Answers
HA
H+
A-
Start
Change
0.2 M
0
0
- (0.1% of 0.2) M =
- (2 x 10-4) M
+ (2 x 10-4) M
+ (2 x 10-4) M
EQM
(0.2 - 2 x 10-4) M
(2 x 10-4) M
(2 x 10-4) M
272
Problems and Answers
Solution



Ka =
[H+][A-]
[HA]
=
(2 x 10-4) (2 x 10-4)
0.2 – 2 x 10-4
Since HA has ionized in small amount
comparing to the original concentration, we
can omit (- 2 x 10-4)
Ka =
(2 x 10-4) (2 x 10-4)
0.2
= 2 x 10-7
273
Problems and Answers
Solution
(b) pH = - log [H+] = - log [2.0 x 10-4] = 3.7
 (c) A 0.2 M HCl would be ionized 100%
and yield 0.2 M H+
 pH = - log [H+] = - log [0.2] = 0.7
 HA solution is 3 pH units < HCl solution
 So HA is 1000 times < HCl, but not 3 times

274
Problems and Answers
Solution
(d) # of moles of OH- required = total # of
moles of H+ present
 Vbase x Nbase = Vacid x Nacid
 because HA is monoprotic acid, N = M
 Vbase x 0.1 = 0.5 x 0.2

0.5 x 0.2

Vbase =
0.1
= 1.0 L of KOH required
275
Problems and Answers
Ka for HA = 1.6 x 10-6. Calculate (a) pH,
(b) degree of ionization of HA in 10-3 M
solution, (c) pKa, (d) pKb?
Solution

Let x = M of HA ionizes; then H+ = A- = x M
276
Problems and Answers
HA
H+
A-
Start
Change
10-3 M
0
0
-xM
+xM
+xM
EQM
10-3 – x M
xM
xM
277
Problems and Answers
Solution



Ka =
[H+][A-]
[HA]
=
(x) (x)
10-3 - x
= 1.6 x 10-6
Assuming that HA is < 10% ionized comparing to
the original concentration, we can omit (- x) from
the denominator (if Ka ≤ 10-5)
1.6 x 10-6 =
(x)2
10-3
; x =√ 1.6 x 10-9 = 4.0 x 10-5
278
Problems and Answers
Solution
x = [H+] = 4 x 10-5 M
 pH = - log (4 x 10-5) = 4.4
[H+]ionized
 (b) Degree of ionization =

[HA]orig
=
x 100
4 x 10-5
[10-3]
x 100 = 4%
279
Problems and Answers
Solution
(c) pKa = - log Ka = - log 1.6 x 10-6 = 5.8
 (d) pKa + pkb = 14
 pKb = 14 – 5.8 = 8.2

280
Problems and Answers
Calculate (a) [H+] in 0.20 M solution of
moderate strong acid, HA, where Ka = 3
x 10-2, and (b) the degree of ionization
of the acid?
Solution

(a) Let x = M of HA ionizes, then H+ = A- =
xM
281
Problems and Answers
HA
H+
A-
Start
Change
0.02 M
0
0
-xM
+xM
+xM
EQM
0.02 – x M
xM
xM
282
Problems and Answers
Solution





Ka =
[H+][A-]
[HA]
=
(x) (x)
0.02 - x
= 3 x 10-2
Ka >> 10-5, so the denominator should not be
simplified
x2 = (3 x 10-2)(0.02 - x) = (6 x 10-4) – (3 x 10-2x)
x2 + (3 x 10-2x) – (6 x 10-4) = 0
Solve for x using the quadratic equations
283
Problems and Answers
x=
-b±
√
b2 – 4ac
2a
- (3 x 10-2) ±
x=
x=
√
; a = 1, b = 3 x 10-2, c = - 6 x 10-4
(3 x 10-2)2 – 4(-6 x 10-4)
2
+ (2.74 x 10-2)
2
; neglecting the –ve answer
[H+] = 1.4 x 10-2 M
284
Problems and Answers
Solution
(b) degree of ionization =
[H+]ionized
[HA]orig
x 100
1.4 x 10-2
=
0.02
X 100 = 70%
285
Problems and Answers
Calculate the pH of 3.5 x 10-2 M solution
of amine with a pKa of 9.6?
Solution

RNH2 + H2O
RNH3+ + OH-
[RNH3+][OH-]

Kb =

Kb = pKw - pKa = 14 – 9.6 = 4.4
[RNH2]
286
Problems and Answers
Solution

p[RNH2] = - log [RNH2] = 3.5 x 10-2 = 1.46
pKb + p[RNH2]

pOH =

pH = 14 – pOH = 14 – 2.93 = 11.07
2
=
4.4 + 1.46
2
= 2.93
287
Problems and Answers
The Kb for ammonia is 1.8 x 10-5. What
is the Ka?
Solution


RNH2 + H2O
Kb =
RNH3+ + OH-
[RNH3+][OH-]
[RNH2]
288
Problems and Answers

The ammonium ion, ionizing as an weak
acid:
NH4+
H+ + NH3
[H+][NH3]

Ka =

Kw = Ka Kb = 10-14
[NH4+]
289
Problems and Answers

Ka =
10-14
1.8 x 10-5
= 5.5 x 10-10
290
Problems and Answers
Calculate the ionic strength of a 0.1 M
of butyric acid, Ka = 1.5 x 10-5?
Solution

Butyric acid (HB) is only partially ionized,
the unionized molecules have no effect on
the ionic strength of the solution
291
Problems and Answers
HB
H+
B-
Start
Change
0.1 M
0
0
-xM
+xM
+xM
EQM
0.1 – x M
xM
xM
292
Problems and Answers
Solution

Ka =
[H+][B-]
[BH]
=
(x) (x)
(0.1 – x)
=
(x)2
0.1
= 1.5 x 10-5
x = [H+] = 1.3 x 10-3 M; [B-] = 1.3 x 10-3 M
 Ionic strength = ½∑MiZi2 = ½[MH+ZH+2 +MB+ ZB-2] = ½ [(1.3 x 10-3) (1)2 + (1.3 x 10-3) (1)2] = 1.3 x 10-3

293
Problems and Answers
The pKa of acetic acid and
trichloroacetic acid are 4.74 and 0.7,
respectively. Which is the stronger acid
and what are the corresponding
ionization constants?
294
Problems and Answers
Solution

For acetic acid


For trichloroacetic acid


pKa = - log Ka = - log 4.74 = 1.8 x 10-5
pKa – log Ka = - log 0.7 = 0.2
Thus trichloroacetic acid, Ka = 0.2, is more
completely ionized and thus a stronger
acid than acetic acid, Ka = 1.8 x 10-5
295
Problems and Answers
The Ka for formic acid (HCOOH) is 1.76
x 10-4. Calculate formate concentration
in a solution containing 100 mM H+ ions
and 28 mM formic acid?
296
Problems and Answers
Solution
HCOOH



H+ + HCOO-
Ka = 1.76 x 10-4 =
[H+][HCOO-]
[HCOOH]
Rearranging the equation gives:
[HCOO-] =
Ka [HCOOH]
[H+]
=
(1.76 x 10-4)(2.8 x 10-5)
(10-4)
= 4.9 x10-5 M
297
Effect of
Concentration on
Degree of Ionization
298
Effect of Concentration
on Degree of Ionization
The relationship between concentration,
Ka and degree of ionization can be
derived as follows:
299
Effect of Concentration
on Degree of Ionization
HA
H+
A-
Start
Change
C
0
0
- nC
+ nC
+ nC
EQM
C – nC
nC
nC
Where:
C = the original concentration of HA
 n = the fraction of ionized (as a decimal)

300
Effect of Concentration
on Degree of Ionization

Ka =
(nC) (nC)
(C- nC)
Rearranging the equation gives:
 KaC – Ka nC = n2C2
 Ka – Kan = n2C; thus Ka (1- n) = n2C

1-n

C=
n2
Ka
301
Effect of Concentration
on Degree of Ionization
If nC is very small compared to C, then
Ka = n2 C
 Solving for C or n:



C=
n=
Ka
n2
√
Ka
C
302
Effect of Concentration
on Degree of Ionization
At what concentration of weak acid
(HA), Ka = 1.6 x 10-6 will be (a) 10%
ionized, (b) 50% ionized, and (c) 90%
ionized?
303
Effect of Concentration
on Degree of Ionization
Solution
(1 – 0.1)
1-n

(a) C =

(b) C =

(c) C =

So, the degree of ionization increases as the initial
concentration of HA decreases
n2
1-n
n2
1-n
n2
Ka =
Ka =
Ka =
(0.1)2
(1 – 0.5)
x 1.6 x10-6 = 1.4 x 10-4
x 1.6 x 10-6 = 3.2 x 10-6
(0.5)2
(1 – 0.9)
(0.9)2
x1.6 x 10-6 = 2.0 x 10-7
304
Acid and Base
Properties of Salts
305
Acid and Base
Properties of Salts
I want you to be able to predict whether
the solution of a salt is acidic, basic or
neutral
How will you be able to do this?
306
Salt Hydrolysis
A salt:
from the anion of an acid (Cl-)
 Comes from the cation of a base
(Na+)
 Formed from a neutralization reaction
 Comes
307
Salt Hydrolysis
 Some
neutral; others acidic or basic
Salt hydrolysis
Salt reacts with water to produce acid or
base solution
 Hydrolyzing salts usually made from:

Strong acid + weak base, or
 Weak acid + strong base


What pH will result from the above
combinations?
308
Salt Hydrolysis
To see if the resulting salt is acidic or
basic, check the “parent” acid and base
that formed it:
NaCl
HCl + NaOH
 NH4OH
H2SO4 + NH4OH
 CH3COONa
CH3COOH + NaOH

309
Acid and Base
Properties of Salts
310
Acid and Base
Properties of Salts
(1) Salts which have cations from
strong bases and the anions of strong
acids have no effect upon the [H+] when
dissolved in water
311
Acid and Base
Properties of Salts
Example: NaCl
Na+ is the cation of NaOH, a very strong
base, therefore Na+ is a very weak
conjugate acid
 It does not react with water to take an OHfrom it

312
Acid and Base
Properties of Salts
Example: NaCl
Cl- is the anion of HCl, a very strong acid,
therefore Cl- is a very weak conjugate base
 It does not react with water to take an H+
from it

313
Acid and Base
Properties of Salts
Example: NaCl
We expect that the water solution of NaCl
to be neutral
 We have not changed the concentration of
H3O+ or the concentration of the OH- ions
by inserting NaCl into the water

314
Acid and Base
Properties of Salts
(2) Salts which have cations from strong
bases and the anions of weak acids
produce a basic solution when
dissolved in water
315
Acid and Base
Properties of Salts
Example: NaC2H3O2 sodium acetate
NaC2H3O2 in water dissociates 100% to
Na+ ions and CH3COO CH3COO- is the anion of the weak acid,
acetic acid, therefore its conjugate base is
stronger and will react with water to
produce more OH- ions


316
Acid and Base
Properties of Salts
Example: NaC2H3O2 sodium acetate
Na+ + H2O → N.R.
CH3COO- + H2O  CH3COOH + OH-
317
Acid and Base
Properties of Salts
(3) A salt which has the cation from a
weak base and the anion of a strong
acid produce an acidic solution
Example: NH4Cl
NH4Cl  NH4+ + ClCl- + H2O → N.R. (why?)
NH4+ + H2O  NH3 + H3O+
318
Acid and Base
Properties of Salts
(4) Salts derived from a weak acid and
a weak base have both ions reacting
with water
Whether the solution is acidic or basic
depends on the relative acid - base
strengths of the two ions
319
Acid and Base
Properties of Salts
To determine this, you will need to
compare the cation's Ka with the Kb of
the anion



If the Kb of the anion > the Ka of the cation, the
solution is basic
If the Kb of the anion < the Ka of the cation, the
solution is acidic
If the Kb of the anion = the Ka of the cation, the
solution is neutral
320
Hydrolysis of Salts of
Weak Acids and Bases
Salts of weak acids (the conjugate
base) react with water to produce the
weak acid and OHA- + HOH
HA + OHThe ionization of the conjugate base is
called hydrolysis and the hydrolysis
contstan (Kh) is identical to Kb
321
Hydrolysis of Salts of
Weak Acids and Bases
Keq =
[HA][OH-]
[A-][H2O]
; Keq[H2O] =Kh=
[HA][OH-]
[A-]
=Kb
Similarly, salts of weak bases react with
water to produce the weak base and H+
HA + HOH
H3O+ + A[H3O+][A-]
Keq = [HA][H O] ; Keq[H2O] =Kh=
2
[H3O+][A-]
[HA]
=Ka
322
Problems and Answers
(a) Calculate the pH of a 0.1 M solution
of NH4Cl, Kb for NH4OH = 1.8 x 10-5 and
(b) the degree of hydrolysis of the salt?
Solution
NH4Cl is salt of a weak base and a
strong acid:
HCl + NH3
NH4Cl
323
Problems and Answers
Solution

NH4Cl is salt of a weak base and a strong
acid:
HCl + NH3

NH4Cl
The solution of NH4Cl will be acidic
because of the hydrolysis of the NH4+:
NH4+ + H2O
H+ + NH4OH
324
Problems and Answers
Solution

Kh = Ka =
Start
Change
EQM
Kw
Kb
=
10-14
1.8 x 10-5
= 5.6 x 10-10
NH4+ + H2O
H+
NH4OH
0.1 M
0
0
-xM
+xM
+xM
0.1 – x M
xM
xM
325
Problems and Answers
Solution

Kh = Ka =
[NH4OH][H+]
[NH4+]
= 5.6 x 10-10
(x)2
 Kh = Ka =
=
= 5.6 x 10-10
(0.1 – x)
(0.1 – x)
(x) (x)

We can omit (- x) from the denominator
because Ka < 10-5), rearranging the
equation gives:

x = [H+] = 7.46 x 10-6; pH = - log [7.46 x 10-6] =
5.13
326
Problems and Answers
Solution
x = [H+] = 7.46 x 10-6
 pH = - log [7.46 x 10-6] = 5.13

(b) Degree of ionization =
[H+]ionized
[HA]orig
x 100
7.46 X 10-6
=
10-1
x 100 = 7.4 x 10-3%
327
Problems and Answers
Calculating concentrations of species in
a solution of oxalic acid, (COOH)2 (a
diprotic acid), Ka1 = 5.9 x 10-2, Ka2 = 6.4
x 10-5?
Solution
H2C2O4 +H2O
H3O+ + HC2O4- Ka1 = 5.90 x 10-2
HC2O4- + H2O
H3O+ + C2O42- Ka2 = 6.40 x 10-5
328
Problems and Answers
Solution
If the Ka's are far apart, we can make the
problem easier
 We first compare the Ka's to see which one
is more important in delivering H3O+'s to
the solution
 In this problem Ka1 >> Ka2

329
Problems and Answers
Solution

We will assume that all the H3O+'s came
from the first equation, when calculating
the [H3O+], [HC2O4-], and the [C2O42-] in the
0.10 M H2C2O4 solution
330
Problems and Answers
H2C2O4+ H2O
H3O+
[HC2O4-]
Start
Change
0.1 M
10-7
0
-xM
+xM
+xM
EQM
0.1 – x M
xM
xM

Ka1 >> 10-5 we cant omit (- x), we have to
use the quadratic equation
331
Problems and Answers
Solution





Ka=
[H3O+][HC2O4-]
[H2C2O4]
=
(x) (x)
0.1 - x
= 5.9 x 10-2
x2 = (5.9 x 10-2)(0.1-x) = (6 X 10-4) – (3 x 10-2x)
x2 + 5.9 X 10-2x - 5.9 X 10-3 = 0
√
2
- b ± b – 4ac
x=
; a = 1, b = 5.9 x 10-2, c = - 5.9
2a
x 10-3
x = [HC2O4-] = [H3O+] = 0.053 M
332
Problems and Answers
Solution
HC2O4- + H2O
H3O+
[C2O42-]
Start
0.053 M
0.053 M
0
Change
EQM
-xM
+xM
+xM
0.053 – x M
0.053 + x M x M

Ka2=
[H3O+][C2O42-]
[HC2O4-]
(0.053 + x) (x)
=
= 6.4 x 10-5
(0.053 – x)
333
Problems and Answers
Solution

6.4 x 10-5 =
(0.053 + x) (x)
(0.053 - x)
Ka2 < 10-5, we can omit (± x), rearrange the
equation to get the value for x:
 x = [C2O42-] = 6.4 x 10-5 M

334
Problems and Answers
Solution

We obtain the following values for the
concentrations of the following species in
solution:
[H3O+] = [HC2O4-] = 0.053 M
 [C2O42-] = 6.4 x 10-4 M
 [H2C2O4] = 0.10 - 0.053 = 0.05 M

10-14
-13 M
 [OH-] =
=
1.9
x
10
0.053
335
Common Ion
336
Common Ion
337
Common Ion
Common Ion:
Two dissolved solutes that contain the
same ion (cation or anion)
 The presence of a common ion suppresses
the ionization of a weak acid or a weak
base

338
Common Ion Effect
Common-ion effect:

It is the shift in equilibrium caused by the
addition of a compound having an ion in
common with the dissolved substance
339
Preparing Laboratory
Solutions and
Reagents
340
Where Do Solution
"Recipes" Come From?
Original scientific literature
Lab manuals

Instructional
Lab manuals

Professional
Handbooks
Manufacturers and suppliers
341
How Solutions are
Prepared
Preparing dilute solutions from
concentrated ones
Biological buffers
Assuring the quality of a solution
342
Preparing Dilute Solutions
From Concentrated Ones
Concentrated solution = stock solution

Use this equation to decide how much
stock solution you will need:
C1V1= C2V2




C1 = Concentration of stock solution
C2 = Concentration you want your dilute solution to be
V1 = How much stock solution you will need
V2 = how much of the dilute solution you want to make
343
“X” Solutions
The concentration of a stock solution is
sometimes written with an “X”
The “X” is how many more times the
stock is than normal
You generally want to dilute such a
stock to 1X, unless told otherwise
344
Problems and Answers
(1) A can of frozen orange juice is
labeled 4X. How would you dilute it to
make 1 L of drinkable drinkable juice?
Solution

Using the following equation:
C1V1= C2V2
345
Problems and Answers
C1 V1 = C2 V2
 4X (V1) = 1 x (1 L)
 V1 = 0.25 L
 Use 0.25 L of orange juice, BTV 1 L

346
Problems and Answers
(2) How many g of NaCl are contained
in 250.0 mL of 0.22 M NaCl solution?
Solution
250.0 mL ÷ 1000 = 0.25 L
 Now, weight as a conversion factor

n
 M=
; then
L
347
Problems and Answers

# of moles (n) = M x L = 0.22 x 0.25 =
0.055 moles
weight
 n=
; then
MW


weight (g) = n x MW = 0.055 x 58.44 = 3.21 g
Dissolve 3.2 g NaCl in water and dilute to a total
volume of 250 mL
348
Biological Buffers
Laboratory buffers

Solutions to help maintain a biological
system at proper pH
pKa of a buffer

The pH at which the buffer experiences
little change in pH with addition of acids or
bases = the pH at which the buffer is most
useful
349
Factors Affecting the
Buffers
Temperature
Some buffers change pH as their
temperature and/or concentration changes
 Tris buffer, widely used in molecular
biology, is very sensitive to temperature

Dilution
Some buffers are sensitive to dilution
 Phosphate buffer is sensitive to dilution

350
Adjusting the pH of a
Buffer
This is done to set the buffer to a pH
value which is:
Somewhat close to its pKa
 Useful for the biological system the buffer
is to be used with

Often adjust pH using NaOH or HCl

Not method used for phosphate buffer
351
Bringing a
Solution to the Proper pH
Adjust the pH when the solution is at the
temperature at which you plan to use it
Mix the solute(s) with most, but not all,
the solvent
Do not bring the solution to volume
Stir solution
352
Be careful Not to
crash the
electrode with the
stir bar
Stir
bar
353
Assuring The
Quality of A Solution
Documentation, labeling, recording what
was done
Traceability
SOPs
354
Assuring The
Quality of A Solution
Maintenance and calibration of
instruments
Stability and expiration date recorded
Proper storage
355
HendersonHasselbalch Equation
356
Henderson-Hasselbalch
Equation
Lawrence Joseph Henderson wrote an
equation, in 1908, describing the use of
carbonic acid as a buffer solution
Karl Albert Hasselbalch later reexpressed that formula in logarithmic
terms, resulting in the Henderson–
Hasselbalch equation
357
Henderson-Hasselbalch
Equation
Hasselbalch was using the formula to
study metabolic acidosis, which results
from carbonic acid in the blood
358
Henderson-Hasselbalck
Equation
The Henderson–Hasselbalch (often
misspelled as Henderson–Hasselbach)
equation describes the derivation of pH
as a measure of acidity (using pKa, the
acid dissociation constant) in biological
and biochemical systems
359
Henderson-Hasselbalck
Equation
The equation is also useful for
estimating the pH of a buffer solution
and finding the equilibrium pH in acidbase reactions (it is widely used to
calculate isoelectric point of the
proteins)
360
HendersonHasselbalck Equation
Derivation
361
Henderson-Hasselbalck
Equation Derivation
A useful expression relating the (a) Ka
of a weak acid (HA) and the pH of a
solution of the HA or (b) Kb of weak
base and the pOH of the a solution of
the weak base can be derived as
follows:

(a) For weak acid:
HA
H+ + A362
Henderson-Hasselbalck
Equation Derivation
[H+] [A-]

(a) Ka =

Rearrangement the above equation:
 [H+]

= Ka
[HA]
[HA]
[A-]
Taking the logarithms of both sides:
363
Henderson-Hasselbalck
Equation Derivation




log
[H+]
= log Ka + log
[HA]
[A-]
Multiplying both sides by – 1
- log
[H+]
= - log Ka – log
pH = pKa – log
[HA]
[A-]
[HA]
[A-]
364
Henderson-Hasselbalck
Equation Derivation
pH = pKa + log
[A ]
[HA]
365
Henderson-Hasselbalck
Equation Derivation



(b) For weak base:
MOH
M+ + OHKb =
[M+] [OH-]
[MOH]
[OH-] = Kb
[MOH]
[M+]
[MOH]
 - log [OH-] = - log Ka - log [M+]
366
Henderson-Hasselbalck
Equation Derivation
pOH = pKb + log
+
[M ]
[MOH]
367
Henderson-Hasselbalck
Equation Derivation
These are so called HendersonHasselbalch equation (or buffer
equation)
When the concentrations of conjugate
acid and conjugate base are equal, pH
= pKa and pOH = pKb
368
Henderson-Hasselbalck
Equation Derivation
The same relationship can be seen from
the original Ka or Kb expression; when
[A-] = [HA], [H+] = Ka and when [OH-] =
[MOH], [OH-] = Kb
369
Henderson-Hasselbalch
Equation
It combines the equilibrium constant
expression and the pH resulted
In other words, the dissociation of an
ion is related to the pH and the
concentrations of cations and anions in
the solution
370
Henderson-Hasselbalck
Equation Derivation
According to Henderson-Hasselbalck
equation, the pH of a solution containing
HA and A- is independent of
concentration; pH is established solely
by the ratio of the conjugate base to
conjugate acid

This is not true, as we will see later
Henderson-Hasselbalck equation limitation
371
Henderson-Hasselbalck
Equation Derivation
The assumption is valid as long as [A-]
and [HA] are high compared to pKa, but
not so high as to warrant correction for
the activity coefficients
Under the usual laboratory conditions,
the concentrations might be 0.1 M or
less with pKa values of 10-3 or less, so
this condition is met
372
Henderson-Hasselbalck
Equation Derivation
The equation is used mostly to calculate
pH of solution created mixing known
amount of acid and conjugate base (or
neutralizing part of acid with strong
base)
The equation is valid when it contains
equilibrium concentrations of acid and
conjugate base
373
Henderson-Hasselbalck
Equation Derivation
It will be noted that many buffer
problems involve a determination of the
ratio of the [A-] to [HA] and total of the
[A-] and [HA] together
[A-]
 a=
[HA]

b = [A-] + [HA]
374
Henderson-Hasselbalck
Equation Derivation
It is essential in any buffer problem to
identify from the given description the
values of pKa, pH, a, and b
The most comprehensive buffer
problem asks for the preparation of a
volume of buffer of a certain
concentration and pH from a weak acid
for given pH and a strong base
375
Henderson-Hasselbalck
Equation Derivation
The general approach may be by the
following steps:
(1) Since pH and pKa are given, determine
the [A-]/[HA] ratio
(2) The concentration of the buffer is usually
expressed as the sum of A- and HA forms
(3) If the [A-] (or [HA]) is given, then the
corresponding HA (or A-) can be calculated
from the ratio in step # 1
376
Henderson-Hasselbalck
Equation Derivation
(4) If the total concentration, [A-] + [HA], is
given, then the solution of the two
simultaneous equation is required, e.g.,
[A-]
 If
= 2………..………………………(i)
[HA]
 [A-] + [HA] = 0.3 M………………………...(ii)
 Then, since [A-] = 2 [HA] and by substitute
this value in (i)
 2 [HA] + [HA] = 0.3 M
377
Henderson-Hasselbalck
Equation Derivation
0.3
 [HA] =
=
0.1
M
3
Now, if we substitute this value in (i)
 [A-] = 2 x 0.1 = 0.2 M
(5) From the [A-] and [HA] the actual
amounts of A- and HA can calculated for
the volume required from their molecular
weights

378
HendersonHasselbalck Equation
Limitations
379
Henderson-Hasselbalck
Equation Limitations
The most significant is the assumption
that the concentration of the acid and its
conjugate base at equilibrium will
remain the same as the formal
concentration
380
Henderson-Hasselbalck
Equation Limitations
This neglects the dissociation of the
acid and the hydrolysis of the base
The dissociation of water itself is
neglected as well
381
Henderson-Hasselbalck
Equation Limitations
These approximations will fail when
dealing with relatively strong acids or
bases, dilute or very concentrated
solutions (< 1 mM or >1 M), or heavily
skewed acid/base ratios (> 100 to 1)
382
Henderson-Hasselbalck
Equation Limitations
In case of solutions containing not-soweak acids (or not-so-weak bases)
equilibrium concentrations can be far
from concentrations of substances put
into solution
Let's replace acetic acid from the
example with something stronger, e.g.,
dichloroacetic acid, with pKa = 1.5
383
Henderson-Hasselbalck
Equation Limitations
The same reasoning leads to result pH
= 1.5 - which is wrong
The reason is simple because
dichloroacetic acid is strong enough to
dissociate on its own and equilibrium
concentration of conjugate base is not
0.05 M but 0.03 M
384
Henderson-Hasselbalck
Equation Limitations
As a rule of thumb you may remember
that acids with pKa < 2.5 dissociate too
easily and use of HendersonHasselbalch equation for pH prediction
can give wrong results, especially in
case of diluted solutions
385
Henderson-Hasselbalck
Equation Limitations
For solutions above 10 mM and acids
weaker than pKa ≥ 2.5, HendersonHasselbalch equation gives results with
acceptable error
The same holds for bases with pKb ≥ 2.5
386
Henderson-Hasselbalck
Equation Limitations
However, the same equation will work
perfectly regardless of the pKa value if
you are asked to calculate ratio of acid
to conjugated base in the solution with
known pH
Similar problem is present in calculation
of pH of diluted buffers
387
Henderson-Hasselbalck
Equation Limitations
Let's see what happens when you dilute
acetic buffer 50/50
The more diluted the solution is, the
more solution pH is dominated not by
the presence of acetic acid and its
conjugated base, but by the water
autodissociation (autoionization)
388
Henderson-Hasselbalck
Equation Limitations
pH of 1 mM solution is close enough to
the expected (from pKa) value, more
diluted solutions deviate more and more
It is worth of noting here that 1 mM
buffer solution has so low capacity, that
it has very limited practical use
389
Henderson-Hasselbalck
Equation Limitations
Henderson-Hasselbalch equation can
be also used for pH calculation of
polyprotic acids, as long as the
consecutive pKa values differ by at least
2 (better 3)
390
Henderson-Hasselbalck
Equation Limitations
Thus it can be safely used in case of
phosphoric buffers (pKa1 = 2.15, pKa2 =
7.20, pKa3 = 12.4), but not in case of
citric acid (pKa1 = 3.13, pKa2 = 4.76, pKa3
= 6.40)
In the latter case to calculate pH you
should use full equation 11.16 - or pH
calculator
391
Henderson-Hasselbalck
Equation Limitations
If you are looking for a way to calculate
buffer composition, you can reverse the
equation
Using known pH and known pKa you
can calculate ratio of concentrations of
acid and conjugated base, necessary to
prepare the buffer
392
Henderson-Hasselbalck
Equation Limitations
Further calculations depend on the way
you want to prepare the buffer
393
Problem and Answer
(3) Calculate the appropriate values and
draw the curve for the titration of 0.5 L
of 0.1 M acetic acid (Ka = 1.8 x10-5) with
0.1 M KOH?
Solution

(a) At the start, pH depends of mainly on
the concentration of the acid and the value
of Ka
394
Problem and Answer
Solution

pH =
pKa + p[CH3COOH]
2
=
4.76 + 1
2
= 2.88
(b) At any point during the titration the pH
can be calculated from
[CH3COO-]
 pH = pKa + log

[CH3COOH]
395
Problem and Answer
Solution




By adding 100 mL of 0.1 M KOH
# of moles KOH added = 0.1 x 0.1 = 0.01 mole
0.01 mole CH3COOH have been converted into
0.01 mole of CH3COO# moles CH3COOH remaining = moles CH3COOH
originally present – moles of CH3COOH titrated to
CH3COO-
396
Problem and Answer
Solution



# moles CH3COOH remaining = moles CH3COOH
originally present – moles of CH3COOH titrated to
CH3COO# moles CH3COOH remaining = (0.5 x 0.1) – 0.01
= 0.04 mole
The volume has changed, but the ratio of mole
CH3COO-/mole CH3COOH is the same as the
ratio of [CH3COO-]/[CH3COO-]
397
Problem and Answer
Solution




pH = 4.76 + log
0.01
0.04
= 4.76 + log 0.25 = 4.16
(c) When 250 mL of 0.1 M KOH has been added,
the original CH3COOH is half titrated
pH = pKa + log
[CH3COO-]
[CH3COOH]
[CH3COO-]
[CH3COOH]
= 1 then [CH3COO-] = [CH3COOH]
398
Problem and Answer
Solution

pH = pKa = 4.76

(d) Beyond the halfway point, the pH is still given
by:
[CH3COO-]
pH = pKa + log
[CH3COOH]

399
Problem and Answer
Solution




By adding 375 mL of 0.1 M KOH
# of moles KOH added = 0.375 x 0.1 = 0.0375
mole
0.0375 mole CH3COOH have been converted into
0.0375 mole of CH3COO# moles CH3COOH remaining = moles CH3COOH
originally present – moles of CH3COOH titrated to
CH3COO-
400
Problem and Answer
Solution


# moles CH3COOH remaining = 0.05 –
0.0375 = 0.0125 mole
0.0375
pH = 4.76 + log
= 4.76 + 0.48 =
0.0125
5.24
401
Problem and Answer
(e) What is the pH of the solution after
adding exactly 500 mL of 0.1 M KOH?
Solution

Theoretically, all the CH3COOH has been
titrated to CH3COO-, but pH is not 7
because CH3COO- ionizes into:
CH3COO- + H2O
HA + OH402
Problem and Answer
The pH at the endpoint can be calculated
from Kb
 pKw = pKa + pKb

14

pKb =

At the endpoint of titration, the addition of one
mole of KOH to one mole of CH3COOH didn’t
produce exactly one mole of CH3COO-
4.76
= 9.24
403
Problem and Answer
At the endpoint of titration, the addition of
one mole of KOH to one mole of
CH3COOH didn’t produce exactly one mole
of CH3COO So the # of moles of KOH = the # of moles
of CH3COO [CH3COO-] = 0.5 x 0.1 = 0.05 M

404
Problem and Answer

pOH =
pKa + p[CH3COO-]
2
=
9.24 + 1.3
2
= 5.27
pH = pKw – pOH = 14 – 5.27 = 8.73
 We started with CH3COO- (weak base) and
titrated with HCl (strong acid), the titration
curve would be identical with the above
curve but y-axis would be read from right to
left

405
Weak Acid-strong Base
Titrations
(e)
(c)
(d)
(b)
(a)
406
Buffers
407
Buffers
Buffer solutions are solutions that resist
change in proton, H+ (hydronium) ion
and the hydroxide (OH-) ion
concentration (and consequently pH)
upon addition of small amounts of acid
or base, or upon dilution
408
Buffers
Buffer solutions consist of a weak acid and its
conjugate base (more common) or a weak
base and its conjugate acid (less common)
The resistive action is the result of the
equilibrium between the weak acid (HA) and
its conjugate base (A−):
HA + H2O
H3O+ + A−
409
Buffers
410
Buffers
Since it is a pair of chemicals:
One chemical neutralizes any acid added,
while the other chemical would neutralize
any additional base
 They make each other in the process!

411
Buffers
Any alkali added to the solution is
consumed by the H3O+ ions
These ions are mostly regenerated as
the equilibrium moves to the right and
some of the acid dissociates into H3O+
ions and the conjugate base (A-)
412
Buffers
If a strong acid is added, the conjugate
base (A-) is protonated, HA, and the pH
is almost entirely restored
This is an example of Le Chatelier's
principle and the common ion effect
413
Buffers
This contrasts with solutions of strong
acids or strong bases, where any
additional strong acid or base can
greatly change the pH
414
Buffers
By comparing two graphs when an
strong acid is titrated with a strong base
the curve will have a large gradient
throughout showing that a small
addition of base/acid will have a large
effect compared to a weak acid/strong
base titration curve which will have a
smaller gradient near the pKa
415
Strong Acid-strong Base
Titration
pH at equivalent
point
416
Weak Acid-strong Base
Titrations
pH at equivalent
point
pKa
[HA] = [A-]
417
Buffers
When writing about buffer systems they
can be represented as salt of conjugate
base/acid, or base/salt of conjugate acid
It should be noted that here buffer
solutions are presented in terms of the
Brønsted-Lowry notion of acids and
bases, as opposed to the Lewis acidbase theory
418
Preparation of Buffer
Solutions
Buffers are prepared in three ways by:
Mixing a weak acid and its conjugate base
or a weak base and its conjugate acid
 Mixing an excess of a weak acid with a
limited amount of strong base
 Mixing an excess of a weak base with a
limited amount of a strong acid

419
Buffer Preparation
Use the Henderson-Hasselbalch
equation in reverse
Choose weak acid with pKa close to
required pH
 Substitute into Henderson-Hasselbalch
equation
 Solve for the ratio of [conjugate
base]/[acid]

420
Physical Criteria of
Biochemical Buffers
421
Physical Criteria of
Biochemical Buffers
pKa should reside between 6.0 and 8.0

Because most biological reactions take
place at near-neutral pH between 6 and 8,
ideal buffers would have pKa values in this
regime to provide maximum buffering
capacity there
422
Physical Criteria of
Biochemical Buffers
They should exhibit high water solubility
and minimal solubility in organic
solvents

For ease in handling and because
biological systems are in aqueous
systems, good solubility in water was
required
423
Physical Criteria of
Biochemical Buffers

Low solubility in nonpolar solvents (fats,
oils, and organic solvents) was also
considered beneficial, as this would tend to
prevent the buffer compound from
accumulating in nonpolar compartments in
biological systems such as cell membranes
and other cell compartments
424
Physical Criteria of
Biochemical Buffers
They should not permeate cell
membranes

This will reduce the accumulation of buffer
compound within cells
They should not exhibit any toxicity
towards cells
425
Physical Criteria of
Biochemical Buffers
They should not interfere with any
biological process
The salt effect should be minimal,
however, salts can be added as
required

Highly ionic buffers may cause problems or
complications in some biological systems
426
Physical Criteria of
Biochemical Buffers
Well-behaved cation interactions
If the buffers form complexes with cationic
ligands, the complexes formed should
remain soluble
 Ideally, at least some of the buffering
compounds will not form complexes

427
Physical Criteria of
Biochemical Buffers
Ionic composition of the medium and
temperature should have minimal effect
of buffer capacity
Buffers should be stable and resistant to
enzymatic degradation

The buffers should resist enzymatic and
non-enzymatic degradation
428
Physical Criteria of
Biochemical Buffers
Buffer should not absorb either visible
or UV light

Buffers should not absorb visible or
ultraviolet light at wavelengths longer than
230 nm so as not to interfere with
commonly-used spectrophotometric
assays
429
Buffer Capacity and
Buffer Range
430
Buffer Capacity and
Buffer Range
There is a limit to the capacity of a
buffer solution to neutralize added acid
or base
In general, the more concentrated the
buffer components in a solution, the
more added acid or base the solution
can neutralize
431
Buffer Capacity and
Buffer Range
As a rule, a buffer is most effective if the
concentrations of the buffer acid and its
conjugate base are equal
432
Buffer Capacity and
Buffer Range
Maximum buffering capacity is found
when pH = pKa, and buffer range is
considered to be at a pH = pKa ± 1
The buffer capacity (buffer index or
buffer intensity) is the amount of acid or
base that can be added before a
significant change in pH
433
Buffer Capacity and
Buffer Range
Buffer solution is able to retain almost
constant pH when small amount of acid
or base is added
Quantitative measure of this resistance
to pH changes is called buffer capacity
Buffer capacity can be defined in many
ways
434
Buffer Capacity and
Buffer Range
You may find it defined as maximum
amount of either strong acid or strong
base that can be added before a
significant change in the pH will occur
This definition, instead of explaining
anything, raises a question what is a
significant change?
435
Buffer Capacity and
Buffer Range
Sometimes even change of 1 unit
doesn't matter too much, sometimes,
especially in biochemical systems, 0.1
unit change is a lot
Buffer capacity can be also defined as
quantity of strong acid or base that must
be added to change the pH of one liter
of solution by one pH unit
436
Buffer Capacity and
Buffer Range
Such definition, although have its
practical applications, gives different
values of buffer capacity for acid
addition and for base addition (unless
buffer is equimolar and its pH = pKa)
437
Buffer Capacity and
Buffer Range
This contradicts intuition, for a given
buffer solution its resistance should be
identical regardless of whether acid or
base is added
438
Buffer Capacity and
Buffer Range
Buffer capacity definition that takes this
intuition into account is given by
dn
β = dpH

Where n is number of equivalents of added
strong base
439
Buffer Capacity and
Buffer Range
Note that addition of dn moles of acid
will change pH by exactly the same
value but in opposite direction
We will derive formula connecting buffer
capacity with pH, pKa and buffer
concentration
440
Buffer Capacity and
Buffer Range
To make further calculations easier let's
assume that the strong base added is
monoprotic, we also assume volume of
1 which will allow us to treat
concentration and number of moles
interchangeably
441
Buffer Capacity and
Buffer Range
Charge balance of the solution is given
by the equation:
[A-] + [OH-] = [B+] + [H+]
[B+] is concentration of the strong base
present
442
Buffer Capacity and
Buffer Range
Total concentration of the buffer, Cbuf, is
given by:
Cbuf = [HA] + [A-]
From dissociation constant definition we
have:
[H+] [A-]
[HA] =
, so
Ka
443
Buffer Capacity and
Buffer Range
Cbuf =
[A-] =
[H+] [A-]
Ka
+ [A-], or
[H+] Ka
Ka + [H+]
The above equations and Kw when combined
give us the amount of the strong base:
444
Buffer Capacity and
Buffer Range
n=

Kw
[H+]
- [H+] +
Cbuf Ka
Ka + [H+]
Now we are ready to calculate derivative:
β=
β = (-
dn
dpH
Kw
[H+]2
=
dn
d[H+]
-1-
x
d[H+]
dpH
Cbuf Ka
(Ka + [H+])2
)(- 2.3[H+])
445
Buffer Capacity and
Buffer Range
So finally buffer capacity is given by
β = 2.3(
Kw
[H+]
+ [H+] +
Cbuf Ka[H+]
(Ka +
[H+])2
)
For buffer having Cbuf > 10-3 M, then
[A-] [HA]
β = 2.3 [A-] + [HA]
446
Buffer Capacity and
Buffer Range
Note that first two terms in the buffer
capacity formula are not dependent on
the buffer presence in the solution
They reflect the fact that solutions of
high (or low) pH are resistant to pH
changes
447
Buffer Capacity and
Buffer Range
As it was already signaled in the pH of
buffer section, such solutions have high
buffer capacity regardless of the
presence (or lack of the presence) of a
classic buffer
448
Buffer Capacity and
Buffer Range
β
pH
449
Buffer Capacity and
Buffer Range
Above plot shows how the buffer
capacity changes for the 0.1 M solution
of acetic buffer
As expected buffer exhibits the highest
resistance to acid and base addition for
the equimolar solution (when pH = pKa)
450
Buffer Capacity and
Buffer Range
From the plot it is also obvious that
buffer capacity has reasonably high
values only for pH close to pKa value
The further from the optimal value, the
lower buffer capacity of the solution
451
Buffer Capacity and
Buffer Range
Solution containing only conjugated
base (pH 8-10) has buffer capacity of
zero, for the higher pH presence of the
strong base starts to play an important
role
452
Buffer Capacity and
Buffer Range
In the case of pure acetic acid solution
(pH below 3) pH is already low enough
to be resistant to changes due to the
high concentration of H+ cations
453
Buffer Solution
Applications
454
Buffer Solution
Applications
Their resistance to changes in pH
makes buffer solutions very useful for
biochemical manufacturing and
essential for many biochemical
processes
455
Buffer Solution
Applications
The ideal buffer for a particular pH has
a pKa equal to the pH desired, since a
solution of this buffer would contain
equal amounts of acid and base and be
in the middle of the range of buffering
capacity
456
Buffer Solution
Applications
Buffer solutions are necessary to keep
the correct pH for enzymes in many
organisms to work
457
Buffer Solution
Applications
Many enzymes work only under very
precise conditions; if the pH strays too
far out of the margin, the enzymes slow
or stop working and can denature, thus
permanently disabling its catalytic
activity
458
Buffer Solution
Applications
A buffer of carbonic acid (H2CO3) and
bicarbonate (HCO3−) is present in blood
plasma, to maintain a pH between 7.35
and 7.45
Industrially, buffer solutions are used in
fermentation processes and in setting
the correct conditions for dyes used in
coloring fabrics
459
Buffering Agents
460
Buffering Agents
A buffering agent adjusts the pH of a
solution
The function of a buffering agent is to
drive an acidic or basic solution to a
certain pH state and prevent a change
in this pH
461
Buffering Agents
Buffering agents have variable
properties:
Some are more soluble than others
 Some are acidic while others are basic

As pH managers, they are important in
many biochemical applications,
including agriculture, food processing,
medicine and photography
462
Buffering Agents
Buffering agents can be either the weak
acid or weak base that would comprise
a buffer solution
Buffering agents are usually added to
water to form buffer solutions
They are the substances that are
responsible for the buffering seen in
these solutions
463
Buffering Agents
These agents are added to substances
that are to be placed into acidic or basic
conditions in order to stabilize the
substance
Buffered aspirin has a buffering agent,
such as MgO, that will maintain the pH
of the aspirin as it passes through the
stomach of the patient
464
Buffering agents
Another use of a buffering agent is in
antacid tablets, whose primary purpose
is to lower the acidity of the stomach.
465
Mechanism of Buffering
Agents
The way buffering agents work is seen
in how buffer solutions work
Using Le Chatelier's principle we get an
equilibrium expression between the acid
and conjugate base
466
Mechanism of Buffering
Agents
As a result we see that there is little
change in the concentrations of the acid
and base so therefore the solution is
buffered
A buffering agent sets up this
concentration ratio by providing the
corresponding conjugate acid or base to
stabilize the pH of that it is added to
467
Mechanism of Buffering
Agents
The resulting pH of this combination can
be found by using the HendersonHasselbalch equation
468
Buffering Agents Vs.
Buffer Solutions
Buffering agents are similar to buffer
solutions in that buffering agents are the
main components of buffer solutions
They both regulate the pH of a solution
and resist changes in pH
469
Buffering Agents Vs.
Buffer Solutions
A buffer solution maintains the pH for
the whole system which is placed into it,
whereas a buffering agent is added to
an already acidic or basic solution,
which it then modifies and maintains a
new pH
470
Buffering Agents Vs.
Buffer Solutions
Buffering agents and buffer solutions
are similar except for a few differences:
Solutions maintain pH of a system,
preventing large changes in it, whereas
agents modify the pH of what they are
placed into
 Agents are the active components of a
buffer solutions

471
Buffering Agents
Examples
Monopotassium phosphate is an example
of a buffering agent
 It has a mildly acidic reaction; when
applied as a fertilizer with urea or
diammonium phosphate, it minimizes pH
fluctuations which can cause nitrogen loss

472
Buffering Agents
In humans

Buffering agents in humans, functioning in
acid base homeostasis, are extracellular
agents (e.g., bicarbonate, ammonia) as
well as intracellular agents (including
proteins and phosphate)
473
Buffer Preparations
Problems and Answers
474
Problem and Answers
(4) Calculate the pH and the buffer
capacity of a solution obtained by
mixing 112 mL of 0.24 M HOAc and 136
mL of 0.11 M NaOH? pKa = 4.74
Solution

Upon mixing, the following neutralization
reaction takes place:
CH3COOH + NaOH
CH3COONa
475
Problem and Answers
Initially, we have
 NaOH = 0.11 x 0.136 = 0.015 M
 HOAc = 0.24 x 0.112 = 0.027 M
 Total volume = 0.112 + 0.136 = 0.248 L

476
Problems and Answers
Solution

After the addition, whereupon:
Start
Change
EQM

HOAc + NaOH
OAcNa
0.027 M
0
0.015 M
(0.027 – 0.015) - 0.015 M
0.015 M
0.012 M
0.015 M
0
[OAc-]
pH = pKa + log
[HOAc]
477
Problems and Answers
Solution




[OAc-] =
0.015
= 0.06 M
0.248
0.012
[HOAc] =
= 0.05 M
0.248
0.06
pH = 4.74 + log 0.05 = 4.74 + 0.079 = 4.82
[A-][HA]
(0.06)(0.05)
β = 2.3
= 2.3
= 0.063
(0.06
+
0.06)
([A-] + [HA])
478
Problems and Answers
(5) If you added 27 mL of a 0.1 M lactic
acid (Ka = 1.85 x 10-5) solution to 75 mL
of 0.033 M NaOH solution. What would
be the pH of the resultant solution?
479
Problems and Answers
Solution
Initially, we have
 NaOH = 0.033 x 0.075 = 2.48 x 10-3 mole
 Lactic acid = 0.10 x 0.027 = 2.7 x 10-3 mole

480
Problems and Answers
Solution
During the neutralization, 2.48 x 10-3 mole
of lactate are produced. Hence, 0.22 x 10-3
mole of lactic acid remain (2.7 – 2.48 =
0.22)
 Total volume = 0.075 + 0.027 = 0.102 L

481
Problems and Answers
Solution

After the addition, whereupon:
LA +
NH4OH
Lactate
Start
Change
2.7 x 10-3 m
2.48 x 10-3 m
0
(2.7 x 10-3 –
2.48 x 10-3)
- 2.48 x 10-3 m
2.48 x 10-3 m
EQM
0.22 x 10-3 m
0
2.48 x 10-3 m

pH = pKa + log
[Lactate]
[Lactic acid]
482
Problems and Answers
Solution

2.48 x 10-3
pH = 4.73 + log
= 4.73 + 1.05 = 5.78
0.22 x 10-3
483
Problems and Answers
(6) 25 mL of 0.12 M HA solution are
titrated with 0.1 M NaOH solution. After
the addition of 17.5 mL NaOH, the pH =
5.8. Calculate the pKa of the HA?
484
Problems and Answers
Solution



After the addition of 17.5 NaOH, then
The # of mole of A- = 0.0175 x 0.1 = 1.75 x 10-3
The remaining of HA = (0.025 x 0.12) – 1.75 x 10-3
= 1.25 x 10-3 M

[A-]
pH = pKa + log
[HA]

-]
[A
pKa = pH – log
[HA]
= 5.8 – log
1.75 x 10-3
= 5.65
1.25 x 10-3
485
Problems and Answers
(7) How many mL of 0.5 M NaOH
solution should be added to 40 mL of
0.10 M H3PO4, (pka1 = 2.12 and pKa2 =
7.21)1to prepare a buffer of pH 7.0?
Solution

Let V mL = the volume of the buffer
486
Problems and Answers
Solution

Since pH 7.0, pka1 2.12 and pKa2 7.21, it
follows that the buffer consists of NaH2PO4
and Na2HPO4
H3PO4 + NaOH
NaH2PO4 + NaOH
NaH2PO4 + H2O….1
Na2HPO4 +H2O..2
487
Problems and Answers
Solution


The # of mole of NaOH added to reaction # 1 =
0.04 x 0.1= 4 x 10-3 and additionally x mole for
reaction # 2 (4 x 10-3 + x) = total mole of NaOH
added
The [Na2HPO4] = x/V are formed in reaction # 2,
so the [NaH2PO4] = (4 x 10-3 - x)/V
488
Problems and Answers
Solution

[HPO42-]
pH = pKa + log
, then
[H2PO4 ]

[HPO42-]
log
= 7.0 – 7.21 = - 0.21 then anti log
[H2PO4-]

[HPO42-]
[H2PO4-] = 0.62
489
Problems and Answers
Solution
[x/V]
 0.62 = [(4 x 10-3 - x)/V ] , then x = 1.53 x 10-3
mole/V, so the total moles of NaOH added
= 4 x 10-3 + 1.53 x 10-3 = 5.53 x 10-3 moles
n
n 5.53 x 10-3
 M=
,thus L = = 0.5 =11.06 x 10-3 L
L
M

The volume of NaOH added = 11.06 mL
490
Problems and Answers
(8) Describe how to prepare 1 L of 0.2
M NaH2PO4 (MW 122)-Na2PO4 (MW
142) buffer, pH 7.0, pKa 6.86?
Solution

The ionization of the buffer components in
an aqueous solution will be:
491
Problems and Answers
Solution
NaH2PO4 → Na+ + H2PO4


H+ + HPO42-
Na2HPO4 → 2 Na+ + HPO42pH = pKa + log
7.0 = 6.86 + log
[HA]
[A-]
[HPO42-]
[H2PO4-]
492
Problems and Answers
Solution

[HPO42-]
[H2PO4-]


= 1.38 and [HPO42-] + [H2PO4-] = 0.2 M
1.38
The # of mole of HPO4 =
x 0.2 = 0.116 mole
2.38
1
The # of mole of H2PO4 =
x 0.2 = 0.084 mole
2.38
2-
493
Problems and Answers
Solution
Since the MW of Na2HPO4 = 142; thus
 0.116 x 142 = 116.47 g
 The MW of NaH2PO4 = 122; thus
 0.084 x 122 = 10.25 g
 So to make this phosphate buffer, 10.24 g
of NaH2PO4 and 116.47 g of Na2HPO4 are
dissolved to water and complete to 1 L

494
Problems and Answers
(9) How many mL HCl solution (d = 1.18
g/mL, containing 35.6% HCl by weight)
should be added to 100 mL of 1.0 M
CH3COONa solution, pH = pKa = 4.75?
Solution
g
Normality of HCl = (
) x ( w/w ) x ( 1 )
L
100
EW
= (1180) x (0.356) x (0.027) = 11.51
495
Problems and Answers
Solution
A- +
HCl
HA
Start
0.1 M
xM
xM
Change
EQM
(0.1 – x) M
-xM
+xM
(0.1 – x) M
0M
+xM
[A-]
pH = pKa + log
[HA]
[x]
[x]
0 = log
,by taking anti log, then 1 =
[0.1 - x]
[0.1 - x]
496
Problems and Answers
Solution
x = 0.1 – x
2 x = 0.1
x = 0.05 mole
0.05 mole of HCl should be added
0.05
Because N = M = 11.51, then 11.51 = L
0.05
Volume (L) =
= 4.34 x 10-3 L = 4.34 mL
11.51
497
Problems and Answers
(10) Calculate the change in pH of a
0.17 M NH3-0.19 M NH4Cl buffer, when
(a) 20 mL of 0.1 M HCl, (b) 20 mL of 0.1
M NaOH are added to 200 mL of
buffer? (c) what is the buffer capacity of
the buffer? (Kb 1.8 x 10-5)
498
Problems and Answers
Solution
NH3 + H2O
NH4+ + OH-
[NH4+]
 (a) Initially, pOH = pKb + log
[NH3]
( 0.19)
 pOH= 4.74 + log
= 4.79
(0.17)

pH = 14 – 4.79 = 9.21
499
Problems and Answers
Solution



When HCl is added to the buffer then
NH3 + HCl
NH4+ + OHLet x = [OH-] after the addition of HCl, whereupon:
NH3
+
HCl
NH4+
Start
0.17 M
9 x 10-3 M
0.19 M
Change
[(0.15 – 9 x 10-3) – x] M
- 9 x 10-3 M
[(0.172 + 9 x 10-3) + x]
EQM
(0.141 – x) M
0M
(0.181 + x) M
500
Problems and Answers
Solution
[NH4+]
 pOH = pKb + log
[NH3]




(0.181 + x)
pOH= 4.74 + log
, we might cancel (±
(0.141 – x)
x)
pOH= 4.74 + log 1.28 = 4.74 + 0.11 = 4.85
pH = 14 – 4.85 = 9.15
Hence, the pH is decreased by 9.21 – 9.15 = 0.06
unit
501
Problems and Answers
Solution



(b) When NaOH is added to the buffer then
NH4+ + OHNH3 + H2O
Let x = [OH-] after the addition of NaOH,
whereupon:
NH4+
+
NaOH
NH3
Start
0.19 M
9 x 10-3 M
0.17 M
Change
[(0.172 – 9 x 10-3) – x] M
- 9 x 10-3 M
[(0.15 + 9 x 10-3) + x]
EQM
(0.163 – x) M
0M
(0.159 + x) M
502
Problems and Answers
Solution
[NH4+]
 pOH = pKb + log
[NH3]




(0.163 – x)
pOH= 4.74 + log
, we can cancel (± x)
(0.159 + x)
pOH= 4.74 + log 1.025 = 4.74 + 0.011 = 4.751
pH = 14 – 4.751 = 9.25
Hence, the pH is increased by 9.25 – 9.21 = 0.04
unit
503
Problems and Answers
(11) (a) How many g of NaHCO3 should
be added to 3.2 g of Na2CO3 to prepare
0.5 L of buffer of pH 10.7? (b) What will
be the pH of the buffer if 1 mL of 1 M
HCl solution is added? (c) How many
mL of 1 M NaOH should be added to
2.5 g of NaHCO3 to prepare 0.5 L buffer
of pH 10.2? pKa2 10.3.
504
Problems and Answers
Solution
3.2
= 0.03 mole
106

(a) # of moles of Na2CO3 =

pH = pKa + log [Na2CO3]
[NaHCO3]

[Na2CO3]
[Na2CO3]
10.7 -10.3 = log
,thus
= 2.51
[NaHCO3]
[NaHCO3]
505
Problems and Answers
Solution



2.51 =
0.03/0.5
thus,
[NaHCO3/0.5]
0.03
[NaHCO3] =
= 0.012 M
2.51
# of g of NaHCO3 = 0.012 x 84 = 1.01 g
506
Problems and Answers
Solution



(b) When HCl is added to the buffer then
Na2CO3 + HCl
NaHCO3 + NaCl
After the addition of HCl, whereupon:
Na2CO3
HCl
NaHCO3
Start
0.03 M
1M
0.012 M
Change
(0.03 –10-3) M
- 10-3 M
(0.012 + 10-3)
EQM
(0.029) M
0M
(0.013) M
507
Problems and Answers
Solution
[Na2CO3]
[NaHCO3]

pH = pKa + log

0.029
pH = 10.3 + log
= 10.3 + 0.35 = 10.65
0.013
508
Problems and Answers
Solution



(c) When mL NaOH is added to the buffer then
NaHCO3 + NaOH
Na2CO3 + H2O
After the addition of HCl, whereupon:
NaHCO3
NaOH
Na2CO3
Start
0.029 M
1M
0M
Change
(0.029 – x) M
- xM
+xM
EQM
(0.029 – x) M
0M
xM
509
Problems and Answers
Solution
[Na2CO3]
[NaHCO3]

pH = pKa + log

(x)
10.2 = 10.3 + log
(0.029 – x)

- 0.1 = log


(x)
, by taking antilog, then
(0.29 – x)
(x)
0.79 =
, then
(0.029 – x)
x = 0.013 M
510
Problems and Answers
Solution
[Na2CO3] produced = [NaOH] added =
0.013 M
n
0.013
 Volume of NaOH added = M = 1
=
0.013 L
 # of mL of NaOH added = 13 mL

511
Problems and Answers
(12) How many g of Tris (MW 121) and
how many mL 1 M HCl should be added
to prepare 500 mL of0.05 M Tris buffer
of pH 7.6 and pKa 8.4?
512
Problems and Answers
Solution
When HCl is added to the buffer then
 Tris0 + HCl
Tris+ + NaCl
 The total moles of buffer came from Tris0 = 0.05 x
0.5 = 0.025 moles
 Then Tris0 + Tris+ = 0.025 moles
0]
[Tris
 pH = pKa + log
[Tris+]

[Tris0]
 7.6 = 8.4 + log [Tris+]
513
Problems and Answers
Solution
[Tris0]

= 0.16
+
[Tris ]
 [Tris0]
0.16
=
x 0.05 = 6.9 x 10-3 M
1.16
1
= 1.16 x 0.05 = 0.043 M
Then the # of moles Tris+ = the # of moles HCl
added = 0.043 x 0.5 = 0.022 moles
 [Tris+]

514
Problems and Answers
Solution
# of mL HCl = 0.022 L = 22 mL should be
added to prepare 0.5 L of 0.05 M Tris
buffer
 The total moles of buffer came from Tris0 =
0.025 moles
 # of g Tris0 = # of moles x MW = 0.025 x
121 = 3.03 g

515
Problems and Answers
(13) Compare the buffer capacities of
the following three acetate buffers, (a)
0.01 M HOAc-0.1 M OAc-, (b) 0.01 M
HOAc-0.004 M OAc- and (c) 0.01 M
HOAc-0.001 M OAc-?
Solution

β = 2.3
[A ][HA]
([A-] + [HA])
516
Problems and Answers

(a) β = 2.3

(b) β = 2.3
0.01 x 0.10
0.01 + 0.10
0.01 x 0.04
0.01 + 0.04
0.01 x 0.001
= 0.02
= 0.007

(c) β = 2.3

It can be seen that the buffer capacity increases
with increasing the concentrations of the buffer
components
0.01 + 0.001
= 0.002
517
Problems and Answers
(14) What are the concentrations of
acetic acid (HOAc) and acetate (OAc-)
in a 0.2 M acetate buffer, pH 5, Ka for
HOAc = 1.7 x 10-5, pKa = 4.77?
Solution

0.2 M acetate buffer contains a total of 0.2
mole of acetate/L, some of the total acetate
is in HOAc, and some of in OAc518
Problems and Answers
HOAc
[OAc-]
[H+]
Start
Change
0.2 M
0M
0M
-xM
+xM
+xM
EQM
0.2 – x M
xM
xM


pH = pKa + log
[OAc-]
[HOAc]
(x)
5.0 = 4.77 + log
(0.2 – x)
519
Problems and Answers
Solution


5.0 – 4.77 = log
0.23 = log
(x)


(0.2 – x)
(x)
(0.2 – x)
(x)
; taking antilog for both sides
(0.2 – x)
= 1.7, then x = 0.126 M = [OAc-]
[HOAc]= 0.2 – 0.126 = 0.074 M
520
Problems and Answers
(15) Describe the preparation of 3 L of
0.2 M acetate buffer, pH 5, starting with
from solid OAcNa trihydrate (MW 136)
and 1 M solution of HOAc?
Solution
521
Problems and Answers
HOAc
[OAc-]
[H+]
Start
Change
0.2 M
0M
0M
-xM
+xM
+xM
EQM
0.2 – x M
xM
xM


pH = pKa + log
[OAc-]
[HOAc]
5.0 = 4.77 + log
(x)
(0.2 – x)
522
Problems and Answers
Solution


5.0 – 4.77 = log
0.23 = log
(x)
(0.2 – x)
(x)
; taking antilog for both sides
(0.2 – x)
(x)
-]
=
1.7,
then
x
=
0.126
M
=
[OAc
(0.2 – x)
 # of moles of OAc- in 3 L = 0.126 x 3 =
0.378 moles

523
Problems and Answers
Solution
[HOAc] = 0.2 – 0.126 = 0.074 M
 # of moles of HOAc in 3 L = 3 x 0.074 =
0.222 moles
 We need 3 L of the 0.2 M acetate buffer
 Total # of moles (HOAc + OAc-) = 3 x 0.2 =
0.6 moles

524
Problems and Answers
Solution
The 0.378 moles of OAc- comes from solid
NaOAc
wt
 # of moles =

MW
wt
 0.378 =
; then wt = 51.41 g
136
525
Problems and Answers
Solution
The 0.222 mole of HOAc comes from 1 M
stock solution
 # of moles = volume x molarity
 0.22 mole = volume (L) x 1 M
0.22
 Volume =
= 0.22 L = 222 mL

1
526
Problems and Answers
Solution

To prepare the buffer solution, dissolve
51.4 g of NaOAc in some water, add 222
mL of the 1 M stock HOAc solution, and
then complete the volume to 5 L
527
Problems and Answers
528
Problems and Answers
(16) If 10 mL of 0.09 M acid, when
mixed with 20 mL of 0.15 M potassium
salt of that acid, gave a solution with a
pH 5.85. Calculate the pKa for the acid?
529
Problems and Answers
Solution
The total volume of the final mixture is 30
mL
 C1 x V1 = C2 x V2
 0.09 x 10 = C2 x 30
10
 The final concentration of acid =
x 0.09
30
= 0.03 M

530
Problems and Answers
Solution
10
 The final concentration of salt =
x 0.15
30
= 0.10 M
[salt]
 pH = pKa + log
[acid]
[salt]
0.10
 pKa = pH – log
= 5.85 – log
[acid]
0.03

pKa = 5.85 – 0.52 = 5.33
531
Problems and Answers
(17) Describe the preparation of 5 L of
0.3 M acetate buffer, pH 4.47, starting
from 2 M solution of HOAc, pKa = 4.77
and 2.5 M solution of KOH?
532
Problems and Answers
Solution
[OAc-]
 pH = pKa + log
[HOAc]
[OAc-]
 4.47 – 4.77 = log
[HOAc]
[OAc-]
 - 0.30 = log
; taking antilog
[HOAc]
[OAc-]
1.5

= 0.5 =
= ratio
[HOAc]
1
533
Problems and Answers
Solution

0.5
1.5
OAc-
of the total acetate is present as
1
and
1.5
of total the acetate is
present as HOAc
534
Problems and Answers
Solution
The final solution contains:
0.5
x 0.3 M = 0.1 M OAc- (0.5 mole in 5 L)

1.5
1
x 0.3 M = 0.2 M HOAc (1.0 mole in 5 L)
1.5
In this buffer, all of the acetate must be
provided by the HOAc
535
Problems and Answers





This buffer is prepared by converting the proper
amount of the HOAc to OAc- by adding KOH
We need 5 L x 0.3 m = 1.5 total moles
How much mL we need from the stock 2.0 M
acetate:
n
n = V x M; V =
M
1.5
The volume =
= 0.75 L = 750 mL of acetate
2.0
solution
536
Problems and Answers
The # of moles of OAc- = the # of moles of
KOH = 0.5 moles
n
n
0.5
 M=
;V=
=
= 0.2 L of KOH

V

M
2.5
Now mix 750 mL of acetate with 200 mL of
KOH and complete the final volume into 5
Liters
537
Problems and Answers
(18) Calculate the pH of the 0.30 M
NH3-0.36 M NH4Cl buffer system. What
is the pH after the addition of 20.0 mL of
0.050 M NaOH to 80.0 mL of the buffer
solution, Ka = 5.6 x 10-10?
Solution
NH4+
NH3 + H+
538
Problems and Answers
NH4+
NH3
H+
Start
Change
0.36 M
0.30 M
0M
-xM
+xM
+xM
EQM
0.36 - x M
0.30 + x M
xM
Common ion effect
(0.36 – x)  0.36
(0.30 + x)  0.30
539
Problems and Answers
[NH3]

pH = pKa + log

Final volume = 80.0 mL + 20.0 mL = 100 mL

[NH4+] = 0.36 M x 0.080 L = 0.029 mol /0.1 L = 0.29 M
[OH-] = 0.050 x 0.020 L = 0.001 mol /0.1 L = 0.01 M
[NH3] = 0.30 M x 0.080 = 0.024 mol /0.1 L = 0.24 M


[NH4
+]
= 9.25 + log
0.25
0.28
= 9.17
540
Problems and Answers
NH4+
+ OH-
0.29 M
0.01
Start
Change 0.29 – 0.01 M - 0.01
EQM
0.28 M
pH = pKa + log
0.0 M
[NH3]
[NH4+]
NH3
H2O
0.24 M
+ 0.01 M
0.25 M
= 9.25 + log
0.25
0.28
= 9.20
541
pH Changes in Buffers
In general, a buffer is used to maintain
the pH relatively constant during the
course of the reaction that produces or
uses H+ ions
The ability of a buffer to maintain a
near-constant pH increases as the
concentration of buffer increases
542
pH Changes in Buffers
It is not always possible to use
concentrated buffer, because enzyme,
tissue, or cell under investigation may
be sensitive to high ionic strength, or
the assay may require that pH be
adjusted easily to some higher or lower
value at the end of the reaction
543
Problems and Answers
(19) Show mathematically why 0.01 M
HOAc-0.01 M OAc- buffer, Ka 1.7 x 10-5
can’t maintain an absolutely constant
pH upon the addition of H+?
Solution

Because [HOAc] = [OAc-]
[OAc-]
 Then the ratio of
=1
[HOAc]
544
Problems and Answers
Solution

pH = pKa + log 1, log 1 = 0

pH = pKa, then [H+] = Ka = 1.7 x 10-5
[H+] [OAc-] (1.7 x 10-5) (10-2)
-5
 Ka =
=
=
1.7
x
10
(10-2)
[HOAc]
545
Problems and Answers
Now suppose 10-3 M HCl is added to the
buffer
 If all the H+ reacts with OAc- to give HOAc
(thus maintaining [H+] at 10-5 M), the new
concentration of HOAc would be 0.011 M
and the new concentration of OAc- would
be 0.009 M, then by substituting these
values in Ka expression:

546
Problems and Answers
Solution
[H+] [OAc-] (1.7 x 10-5) (0.009)
 Ka =
=
= 1.7 x 10-5
[HOAc]
(0.011)
547
Problems and Answers
(20) Solutions corresponding to 0.6
mole HCl, 0.05 mole H3PO4, 0.08 mole
NaH2PO4, 0.35 mole Na3PO4, and 0.11
mole NaOH are mixed in a flask.
Calculate the pH of the resulting
solution?
548
Problems and Answers
Solution
Let x L = the volume of the resulting
solution
 First, we assume complete ionization of all
the electrolytes and we calculate the total
of amount of each on in the solution
 Then, we assume that the H+ ions react
with the base of the solution, to decrease
the strength

549
Problems and Answers
Solution
We have:
 # of mole of ionizable H+ = 0.6 + (0.05 x 3)
+ (0.08 x 2) = 0.9
 # of mole of ionizable OH- = 0.11
 # of mole of ionizable PO4-3 = 0.05 + 0.08 +
0.35 = 0.48

550
Problems and Answers
Start
Change
EQM
OH- H+
H2PO4- HPO42-
PO43-
0.11
0.91
0.08
-
0.35
- 0.11
0.9 – 0.11 = 0.8
- 0.32
-
0.48
0.8 – 0.48 = 0.32
+ 0.32
+ 0.48
0
0.48 – 0.32 = 0.16
0
0.16
0
0
0
0.32
551
Problems and Answers
Solution



The solution contains 0.32 mole H2PO4-,
0.16 mole HPO42- and 0.11 mole NaCl;
therefore, it is buffer with a pH equal to:
pH = pKa + log
pH = 7.21 + log
[HPO42-]
[H2PO4-]
[0.16]
[0.32]
= 6.91
552
Polyprotic Acids and
Bases
553
Polyprotic Acids and
Bases
A polyprotic acid can donate more than
one H+

Carbonic acid


Sulfuric acid


H2CO3 ; dissolved CO2 in water
H2SO4
Phosphoric acid

H3PO4
554
Polyprotic Acids and
Bases
A polyprotic base: can accept more than
one proton

Carbonate ion


Sulfate ion


CO32SO42-
Phosphate ion

PO43555
Polyprotic Acids and
Bases
Treat each step of protonation or
deprotonation sequentially
H2CO3 + H2O → H3O+ + HCO3HCO3- + H2O → H3O+ + CO32-
Ka1 = 4.3 x 10-7
Ka2 = 4.8 x 10-11
Typically
Ka1 >> Ka2 >> Ka3 >>…
 Harder to loose a positively charged proton
from a negatively charged ion, because of
attraction between opposite charges

556
Polyprotic Acids and
Bases
The Ka values are numbered in order of
decreasing acid strength (Ka1, Ka2, etc)
The Kb values are numbered in order of
decreasing base strength
Remember that the conjugate base of
the strongest acid group is the weakest
basic group, and vice versa
557
Polyprotic Acids and
Bases
The Ka and Kb values are numbered
accordingly as shown below:
Ka1
H2A
Kb2
H+ + HAKa2 Kb1
H+
+
A2558
Polyprotic Acids and
Bases


Ka1
[H+][HA-]
=
[H2A]
Ka2
[H+][A2-]
=
[HA-]
We must be sure to use the Ka and Kb
(pKa and pKb) of the same ionization
Kw = Ka1Kb2 or pKw = pKa1 + pKb2
 Kw = Ka2Kb1 or pKw = pKa2 + pKb1

559
Polyprotic Acids and
Bases
560
Ionization Constants of
Some Polyprotic Acids
561
Amino Acids are
Polyprotic
562
Amino Acids are
Polyprotic
563
Amino Acids are
Polyprotic
564
565
566
Finding the pH
Titration of H2PO4 (H3A)
pH  pKa 2
pH
[H2 A  ]  [HA 2 ]
50%
pKa 1  pKa 2
pH 
2
pH  pKa1
50%
[H2 A  ]
100%
[H3 A]  [H2 A  ]
50%
H3 A 100%
50%
V of NAOH
567
Finding the pH
Solutions containing amphoteric anions
(H2A-) as the only acid and base major
species
H 2 A  (aq)  H 2 A  (aq)  H 3 A (aq)  HA 2 (aq)

H A HA  K
K

K
H A 
H A  HA 
2
3
 2
2
a2
a1
2-
3
568
Finding the pH
1
K
H 3 A 2
H

2A

2

Ka2
K a1
H 3 A (aq)  H   H 2 A 
from
H H


K a1
H H


2
K a1
2
A-
H 3 A 

2
A-
H 3 A 

H   H A 

H A 
K

3

a1
2
569
Finding the pH

K a2
K a1
 
H

 Ka
 1
H  

 pH 

2


K a1 K a2
pK a1  pK a2
2
570
Finding the pH
571
Titration of Weak
Polyprotic Acid-strong
Base
572
Titration of Weak Polyprotic
Acid-strong Base
Determine the molarity of phosphoric
acid solution
Identify Ka1, and Ka2 of phosphoric acid
573
Experimental Technique
Titrate the weak polyprotic H3PO4 with
the strong base NaOH
Monitor the pH of the titration solution
as a function of volume of NaOH added
pH meter
 Drop counter

574
Titration of Weak
Polyprotic Acid
575
Titration of Weak
Polyprotic Acid
Example of a
diprotic acid (Lue)



Two equivalence
points
Sharp increases in pH
Volume to first
equivalence point = ½
volume to second
equivalence point
576
Titration of Phosphoric
Acid
H3PO4
Weak acid



Ka1 H3PO4
Ka2 H2PO4Ka3 HPO42-
Triprotic
Three equivalence
points
577
Titration of H3PO4 with
Strong Base
Stock NaOH in buret
25 mL H3PO4 acid titrated into 400 mL
beaker with stir bar

Note:
Obtain 75-100 mL of NaOH from stock
Obtain 50 mL of H3PO4 from stock
578
Titration of Phosphoric
Acid
H3PO4 + OH-  H2PO4- + H2O
H2PO4- + OH-  HPO42- + H2O
HPO42- + OH-  PO43- + H2O
579
Titration of H3PO4 with
Strong Base
Find the pH for three
different triprotic
acids at points A-C
580
Titration of Phosphoric
Acid
Point A: 1st end point:

moles NaOH = moles H3PO4
Point B: 2nd end point
Point C: ½ distance from point A to
point B (obtain pH at point C from the
titration curve
581
Titration of Phosphoric
Acid
Buffer
582
Titration of Phosphoric
Acid
Phosphoric acid is a good example of a
titration where the first two equivalence
points, corresponding to base reaction
with the first and second protons,
respectively, are clearly visible
By clearly visible, we mean that there is
a large change in pH at the equivalence
point
583
Titration of Phosphoric
Acid
The acid dissociation constants for
phosphoric acid are quite different from
each other with pKa's of 2.15, 7.21, and
12.15
Because the pKa are so different, the
protons are reacted at different pH‘s
584
Titration of Phosphoric
Acid
The relationship to pH is most easily
found by recognizing that all principle
species
There are three such points for
phosphoric acid
pH = pKa1
 pH = pKa2
 pH = pKa3

585
Titration of Phosphoric
Acid
These points are important in the
prediction of the titration curves
They correspond to points where half of
an equivalent of proton has been
consumed by addition of strong base
Thus, the point where pH = pKa1 is
halfway to the first equivalence point
586
Titration of Phosphoric
Acid
Where pH = pKa2 is halfway between
the first and second equivalence points,
etc
The solution has maximum buffer
capacity at these points
In other words, there is maximum
resistance to changes in pH
587
Features of the Titration of
Polyprotic Acid with Strong Base
The loss of each mole of H+ shows up
as separate equivalence point (but only
if the two pKas are separated by more
than 3 pK units)
The pH at the midpoint of the buffer
region is equal to the pKa of that acid
species
588
Features of the Titration of
Polyprotic Acid with Strong Base
The same volume of added base is
required to remove each mole of H+
589
Isoelectric Point (pI)
590
Isoelectric Point (pI)
The isoelectric point, pI, is the pH of an
aqueous solution of an amino acid at
which the molecules have no net charge
In other words, the positively charged
groups are exactly balanced by the
negatively charged groups
591
Isoelectric point (pI)
The isoelectric point (pI) is the pH at
which a particular molecule or surface
carries no net electrical charge
Amphoteric molecules called zwitterions
contain both positive and negative
charges depending on the functional
groups present in the molecule
592
Isoelectric point (pI)
They are affected by pH of their
surrounding environment and can
become more positively or negatively
charged due to the loss or gain of
protons (H+)
593
Isoelectric point (pI)
The pI value can also affect the
solubility of a molecule at a given pH
Such molecules have minimum
solubility in water or salt solutions at the
pH which corresponds to their pI and
often precipitate out of solution
594
Isoelectric point (pI)
Biochemical amphoteric molecules such
as proteins contain both acidic and
basic functional groups
Amino acids which make up proteins
may be positive, negative, neutral or
polar in nature, and together give a
protein its overall charge
595
Isoelectric point (pI)
At a pH below their pI, proteins carry a
net positive charge; above their pI they
carry a net negative charge
596
Isoelectric point (pI)
pH = pI → proteins are precipitated out
of solution
pH < pI → proteins carry a net positive
charge
pH > pI → proteins carry a net negative
charge
597
Isoelectric point (pI)
Proteins can thus be separated
according to their isoelectric point
(overall charge) on a polyacrylamide gel
using a technique called isoelectric
focusing, which utilizes a pH gradient
to separate proteins
598
Isoelectric point (pI)
Isoelectric focusing is also the first step
in 2-D gel polyacrylamide gel
electrophoresis
599
Isoelectric Focusing
pI of a protein: net
charge = 0
A pH gradient is
established by allowing
a mixture of organic
acids and bases
(ampholytes)
Protein migrates until it
reaches the pH that
matches its pI
600
Isoelectric point (pI)
For an amino acid with only one amine
and one carboxyl group, the pI can be
calculated from the pKas of this
molecule
pK1  pK 2
pI 
2
601
Isoelectric point (pI)
For amino acids with more than two
ionizable groups, such as lysine, the
same formula is used, but this time the
two pKa's used are those of the two
groups that lose and gain a charge from
the neutral form of the amino acid
602
Isoelectric point (pI)
Lysine has a single carboxylic pKa and
two amine pKa values (one of which is
on the R-group), so fully protonated
lysine has a +2 net charge
603
Isoelectric point (pI)
To get a neutral charge, we must
deprotonate the lysine twice , and
therefore use the R-group and amine
pKa values

They are found at list of standard amino
acids
604
Isoelectric point (pI)
pI for lysine:
9.06  10.54
pI 
 9.80
2
605
Isoelectric point (pI)
The pH of an electrophoretic gel is
determined by the buffer used for that
gel
If the pH of the buffer is above the pI of
the protein being run, the protein will
migrate to the positive pole (negative
charge is attracted to a positive pole)
606
Isoelectric point (pI)
If the pH of the buffer is below the pI of
the protein being run, the protein will
migrate to the negative pole of the gel
(positive charge is attracted to the
negative pole)
If the protein is run with a buffer pH that
is equal to the pI, it will not migrate at all
607
pI and Plasma Proteins
This is also true for individual amino
acids
Protein has many negative charges

Requires H+ to neutralize, therefore low pI
Consider a protein with pI = 4
If pH increases above pI protein becomes?
 If pH decreases below pI protein
becomes?

608
pI and Plasma Proteins
Higher the pI the more + ve charge,- ve
charge is protein at physiological pH?
Proteins are net negative charge at pH
7.4, all have pI less than 7.4
Therefore pH is above pI, protein is
negative
 If pH is less than pI, protein is positive.

609
pI and Plasma Proteins
Need to go to a higher pH to neutralize
or compensate for + ve charges
Minimum solubility occurs at pI since
there is no intermolecular repulsion
At pH 7.4 (blood pH), all blood proteins
are negative charge and therefore have
pI’s less than 7.4
610
pI and Plasma Proteins
611
Isoelectric Point (pI)
For simple amino acids such as Ala, the
pI is an average of the pKa's of the –
COOH (2.34) and –NH3+ (9.69) groups
Thus, for Ala
2.34 + 9.69
 pI =
= 6.02
2

Same as the experimentally determined value
612
Titration of Alanine
613
Titration of Amino Acids
with an Ionizable -R
An amino acid with an ionizable -R
group has three pKa’s
To calculate the pI, identify the two
values that saddle the point on the
titration curve when the net charge of
the amino acid is zero
614
Titration of Amino Acids
with an Ionizable -R
The differences in pI values for different
amino acids are an indication of the
different acidic and basic properties
they have
The pKa’s for the carboxyl group of most
amino acids falls in the range of 1.8 –
2.4 while the pKa’s for the amino groups
tend to fall in the range of 8.8 – 11.0
615
Titration of Amino Acids
with an Ionizable -R
The pKa’s of the -R groups of Cys, His,
Asp, and Glu fall between that of their COOH and –NH3+ groups
The pKa’s of the -R groups of Tyr, Lys,
and Arg are above that of their amino
pKa’s
616
Titration of Amino Acids
with an Ionizable -R
If additional acidic or basic groups are
present as side-chain functions, the pI is
the average of the pKa's of the two most
similar acids
To assist in determining similarity we
define two classes of acids
617
Titration of Amino Acids
with an Ionizable -R
The first consists of acids that are
neutral in their protonated form (e.g. COOH and -SH)
The second includes acids that are
positively charged in their protonated
state (e.g., -NH3+)

-NH3+ is a Lewis acid
618
Titration of Amino Acids
with an Ionizable -R
The second includes acids that are
positively charged in their protonated
state (e.g. -NH3+). [-NH3+ is a Lewis
acid]
619
620
Titration of Arginine
621
Titration of Arginine
The similar acids are the guanidinium
species on the side-chain (pKa = 13.2)
and the -NH3+ group (pKa = 9.0)

So the calculated
13.2 + 9.0
 pI =
= 11.1
2
622
Titration of Histidine
623
Titration of Aspartic
Acid
Aspartic acid is another triprotic amino
acid
In this case the pKas are; 2.12, 3.86,
and 9.82
The first two are carboxylic acid
protons; the last is the ammonium
proton
624
Titration of Aspartic
Acid
In this case we might expect that the
first two equivalence point would be
obscured by the fact that the two acidic
pKas are relatively close
The titration curve plot is shown below
625
626
Titration of Aspartic
Acid
627
Titration of Aspartic
Acid
In the case of aspartic acid, the similar
acids are the -COOH (pKa = 2.01) and
the –COOH on the –R group (pKa=
3.83)
2.01 + 3.83
 pI =
= 2.92
2
628
Titration of Glutamic
Acid
629
630
UV-visible
Spectroscopy
631
Introduction of
Spectrometric Analyses
The study how the chemical compound
interacts with different wavelengths in a
given region of electromagnetic
radiation is called spectroscopy or
spectrochemical analysis
632
Introduction of
Spectrometric Analyses
The collection of measurements signals
(absorbance) of the compound as a
function of electromagnetic radiation is
called a spectrum
633
Energy Absorption
The mechanism of absorption energy is
different in the Ultraviolet, Infrared, and
Nuclear magnetic resonance regions
However, the fundamental process is
the absorption of certain amount of
energy
634
Energy Absorption
The energy required for the transition
from a state of lower energy to a state
of higher energy is directly related to the
frequency of electromagnetic radiation
that causes the transition
635
What is Spectroscopy?
The study of molecular structure and
dynamics through the absorption,
emission and scattering of light
636
What is Light?
According to Maxwell,
light is an
electromagnetic field
characterized by a
frequency (f), velocity
(v), and wavelength (λ)
Light obeys the
relationship

f=
v
λ
637
The Electromagnetic
Spectrum
638
The Electromagnetic
Spectrum
E = hn
n=c/l
639
Electromagnetic
Radiation
c = lv
Where l = wavelength (cm), v =
frequency (sec-1), c = the speed of the
light (3 x 1010 cm/sec)
The units of (v) are sec-1, which means
vibrations per second
640
Problems and Answerws
An optical filter passes only far red light
with an average wavelength of 6500 Ǻ.
Calculate (a) the wavelength in
nanometers and cm, (b) the wave
number in cm-1, and (c) the frequency?
Solution

(a) 6500 Ǻ = 6500 x 10-10 m = 6.5 x 10-7 m
641
Problems and Answerws
Solution
(a) 6500 Ǻ = 6500 x 10-10 m = 6.5 x 10-7 m
 l = 650 nm or l = 6.5 x 10-5 cm

1
1
=
-5
6.5 x 10 cm
(b) wave number =
=
l
-1
15384 cm
c
3 x 1010
 (c) lv = c, then v =
= 6.5 x 10-5 = 4.61 x
l
14
-1
10 sec

642
Electromagnetic
Radiation
The energy of photon:
h (Planck's constant) = 6.62 x 10-27 (Erg x
sec)
C
 E=h u=h

l

u=
C
l
then C = u x l
643
Spectroscopy
Spectral distribution of radiant energy
Wave number (cycles/cm)
X-Ray
UV
200 nm
Visible
400 nm
IR
Microwave
800 nm
Wavelength (nm)
644
Transmission and Color
The human eye sees the
complementary color to that which is
absorbed
645
Transmission and Color
646
Absorbance and
Complementary Colors
647
Two-Component Mixture
Example of a two-component mixture
with little spectral overlap
648
Two-Component Mixture
Example of a two-component mixture
with significant spectral overlap
649
Influence of 10%
Random Error
Influence on the calculated
concentrations

Little spectral overlap:


10% error
Significant spectral overlap:

Depends on similarity, can be much higher
(e.g. 100%)
650
Influence of 10%
Random Error
651
Absorption Spectra of
Hemoglobin Derivatives
652
Light Sources
UV spectrophotometer
Hydrogen gas lamp
 Mercury lamp

Visible spectrophotometer

Tungsten lamp
Infrared (IR) spectrophotometer

Carborundum (SIC)
653
Dispersion Devices
Non-linear
dispersion

Temperature
sensitive
Linear dispersion

Different orders
654
Dispersion of Polychromatic
Light with a Prism
Prism

Spray out the spectrum and choose the
certain wavelength (l) that you want by
moving the slit
655
Dispersion of Polychromatic
Light with a Prism
Infrared
monochromatic
Ray
Polychromatic
Ray
PRISM
Red
Orange
Yellow
Green
SLIT
Blue
Violet
Ultraviolet
Polychromatic Ray
Monochromatic Ray
656
Photomultiplier Tube
Detector
High sensitivity at
low light levels
Cathode material
determines spectral
sensitivity
Good signal/noise
Shock sensitive
Cathode
Anode
657
The Photodiode
Detector
Wide dynamic range
Very good
signal/noise at high
light levels
Solid-state device
658
Schematic Diagram of a
Photodiode Array
Same
characteristics as
photodiodes
Solid-state device
Fast read-out cycles
659
Conventional
Spectrophotometer
Schematic of a conventional single-beam spectrophotometer
660
Conventional
Spectrophotometer
Optical system of a double-beam
spectrophotometer
661
Conventional
Spectrophotometer
Optical system of a split-beam
spectrophotometer
662
Definition of Resolution
Spectral resolution is a measure of the
ability of an instrument to differentiate
between two adjacent wavelengths
663
Instrumental Spectral
Bandwidth
The spectral bandwith (SBW) is defined
as the width, at half the maximum
intensity, of the band of light leaving the
monochromator
664
Natural Spectral
Bandwidth
The natural spectral bandwidth (NBW)
is the width of the sample absorption
band at half the absorption maximum
665
Transmission Characteristics
of Cell Materials
All materials exhibit at least
approximately 10% loss in
transmittance at all wavelengths
666
Transmission Characteristics
of Cell Materials
667
Spectrophotometric
Cells (Cuvettes)
UV spectrophotometer

Quartz (crystalline silica)
Visible spectrophotometer

Glass
IR Spectrophotometer

NaCl
668
Cell Types I
Open-topped rectangular standard cell
(a) and apertured cell (b) for limited
sample volume
669
Cell Types II
Micro cell (a) for very small volumes
and flow-through cell (b) for automated
applications
670
Transmittance and Concentration
The Bouguer-Lambert Law
T  I / I 0  e  ConstPathlength
671
Transmittance and Path
Length: Beer’s Law
Concentration
T  I / I 0  e  ConstConcentration
672
The Beer-BouguerLambert Law
A   log T   log I / I 0   log I 0 / I     b  c
673
Beer Lambert Law
As the cell thickness increases, the
intensity of I (transmitted intensity of
light ) decreases
Light
I0
I
Glass cell filled with
concentration of solution (C)
674
Beer Lambert Law
R=
I
I0
where,
R:Transmittance
 I0: Original light intensity
 I: Transmitted light intensity


% Transmittance = 100 x
I
I0
675
Beer Lambert Law
Absorbance (A) or optical density (OD)
= log 1
T
= log
I0
I
= 2 - log%T
676
Beer Lambert Law
log =

Io
I
= scl
It is proportional to (c) (concentration of
solution) and is also proportional to (l)
(length of light path through the solution)
677
Beer Lambert Law
A  cl = cl

By definition and it is called the Beer
Lambert law
A = scl

s = Specific extinction coefficient (1 g of
solute per liter of solution)
678
Beer Lambert Law
A = mcl


m = Absorbancy index or molar extinction
coefficient
Extinction coefficient of a solution
containing 1g-molecule of solute per 1 liter
of solution
679
Problems and Answers
Calculate the molar absorption
coefficient (m) at 351 nm for
aquocobalamin in 0.1 M phosphate
buffer, pH 7 from the following data,
which were obtained in 1-cm cell?
Solution c x 105 M Io
I
A
B
24.4
32.8
2.23
1.90
93.1
94.2
680
Problems and Answers
Solution

for solution A, log
1
Io
Io
I
= log
m =

2.38 x 104 liter cm-1mol-1
cl
I
=
27.4
1

x log
9.31
1 x 2.3 x
10-3
= 0.53
x 0.53 =
681
Problems and Answers
Solution


for solution B, log
m =
1
cl
x log
Io
I
=
Io
I
= log
94.2
32.8
1
1 x 1.9 x
10-3
= 0.46
x 0.46 =
2.4 x 104 liter cm-1mol-1
 The average e from these two
measurements is 2.39 x 104 liter cm-1mol-1

682
Problems and Answers
A solution containing 2 g/L of a lightabsorbance in 1-cm cuvette transmits
75% of the incident light of a certain
wavelength. (a) Calculate the
transmission of the solution containing 4
g/L, (b) if the MW = 250, calculate the
m?
683
Problems and Answers
Solution
Io
1
 s =
x log
cl
I
1
1
 s =
x log
= 0.062
1x2
0.75
684
Problems and Answers
Solution
(a) the value transmission (I) for the
solution at any concentration may be
calculated as follows:
I
 log o = scl

I
100
 log
= (0.062)(4)(1) = 0.25
I

log I = 1.75 then I = 56.23% = 0.56%
685
Problems and Answers
Solution

(b) m = s x MW = 0.062 x 250 = 15.50
686
Problems and Answers
The specific absorption (s) of a
glycogen-iodine complex at 450 nm is
0.2. Calculate the concentration of
glycogen in a solution of the iodine
complex, which has absorbance of 0.38
in a 3 cm cuvette?
687
Problems and Answers
Solution

A = scl then 0.38 = (0.2) (c) (3)
0.38
 c=
= 0.63%
0.6
688
Steps in Developing A
Spectrophotometric Analytical
Method
Run the sample for spectrum
Obtain a monochromatic wavelength for
the maximum absorption wavelength
Calculate the concentration of your
sample using Beer Lambert equation:

A = cl
689
Steps in Developing A
Spectrophotometric Analytical
Method
Absorbance
2.0
0.0
200
250
300
350
400
450
Wavelength (nm)
690
Steps in Developing A
Spectrophotometric Analytical
Method
There is some A vs. C where graph is
linear
NEVER extrapolate beyond point known
where becomes non-linear
691
Steps in Developing A
Spectrophotometric Analytical
Method
Absorbance at 280 nm
1.0
Slope of Standard Curve =
0.5
1
4
2
3
Concentration (mg/ml)
A
C
5
692
Spectrometric Analysis
Using Standard Curve
Avoid very high or low absorbencies
when drawing a standard curve
The best results are obtained with 0.1 <
A<1
Plot the absorbance vs. concentration to
get a straight line
693
Spectrometric Analysis
Using Standard Curve
Absorbance at 540 nm
1.2
0.8
0.4
1
2
Concentration (g/l) glucose
3
4
694
Spectrometric Analysis
Using Standard Curve
Every instrument has a useful range for
a particular analyte
Often, you must determine that range
experimentally
695
Spectrometric Analysis
Using Standard Curve
This is done by making a dilution series
of the known solution
These dilutions are used to make a
working curve
696
Spectrometric Analysis
Using Standard Curve
697
Spectrometric Analysis
Using Standard Curve
Make a dilution series of a known
quantity of analyte and measure the
absorbance
Plot concentrations v. absorbance
698
Spectrometric Analysis
Using Standard Curve
699
Spectrometric Analysis
Using Standard Curve
What concentration do you think the
unknown sample is?
700
Spectrometric Analysis
Using Standard Curve
In this graph, values above A = 1.0 are
not linear
If we use readings above A = 1.0, graph
isn’t accurate
701
Spectrometric Analysis
Using Standard Curve
702
Spectrometric Analysis
Using Standard Curve
The best range of this
spectrophotometer is A = 0.1 to A = 1.0,
because of lower errors A = 0.4 is best
703
Spectrometric Analysis
Using Standard Curve
704
Relating Absorbance
and Transmittance
Absorbance rises linearly with
concentration
Absorbance is measured in units
Transmittance decreases in a non-linear
fashion
Transmittance is measured as a %
Absorbance = log10 (100/%
transmittance)
705
Relating Absorbance
and Transmittance
706
Precision and Accuracy
(a) Precision
(b) Precision
(c) Precision
(d) Precision
–, Accuracy
+, Accuracy
–, Accuracy
+, Accuracy
–
–
+
+
707
Protein Determinations
Most proteins exhibit a distinct
ultraviolet light absorption maximum at
280 nm wavelength, due primarily to the
presence of Tyrosine and Tryptophan
708
Protein Determinations
Since the Tyrosine and Tryptophan
content of various enzymes varies
within narrow limits, the absorption peak
at 280 nm has been used as a rapid
and fairly sensitive measure of protein
concentration
709
Protein Determinations
Unfortunately, nucleic acids, which are
apt to be present in enzyme
preparations, absorb light at wavelength
280 nm
Nucleic acids, however, absorb much
more strongly at 260 nm wavelength UV
light
710
Protein Determinations
For protein it is the reverse situation
This assay uses this reverse
relationship to calculate the interference
of nucleic acids in the estimation of
protein concentration
711
Protein Determinations
Absorbance assays are fast and
convenient, since no additional reagents
or incubations are required
No protein standard need be prepared.
The assay does not consume the
protein
712
Protein Determinations
The relationship of absorbance to
protein concentration is linear
Because different proteins and nucleic
acids have widely varying absorption
characteristics there may be
considerable error, especially for
unknowns or protein mixtures
713
Protein Determinations
Any non-protein component of the
solution that absorbs ultraviolet light will
interfere with the assay
[Protein]mg/mL = 1.55 A280 – 0.76A260
[Protein]mg/mL = 144 (A215 – A225)
714
Problems and Answers
Estimate the protein concentration of
the undiluted and diluted solution shown
below?
Solution
A280
A260 A225 A215
Undiluted
Diluted (1:10)
0.35
-
0.20
-
0.20
0.47
715
Problems and Answers
Solution





(1) [Protein]mg/mL = 1.55A280 – 0.76A260
[Protein]mg/mL = (1.55)(0.35) – (0.76)(0.20) = 0.38
mg/mL
(2) [Protein]mg/mL = 144 (A215 – A225)
[Protein]mg/mL = 144 (0.47 - 0.20) = 38.88 mg/mL
The diluted solution contained = 38.88 x 10 =
388.8 mg/mL = 38.88 mg/mL. Thus, the two
methods yield essentially the same result
716
Problems and Answers
A solution containing NAD+ and NADH
had OD of 0.31 at 340 nm and at 1.2 at
260 nm. Both NAD+ and NADH absorb
at 260 nm but only NADH absorbs at
340 nm. Calculate the concentration of
NAD+ and NADH? The m are given in
the Table below:
717
Problems and Answers
Compound
m (M-1 x cm-1)
Absorbance
260 nm
340 nm
260 nm
340 nm
NAD+
18000
0
x
0
NADH
15000
6220
y
0.31
Solution


The concentration of each form may be calculated
as follows:
First calculate the concentration of NADH from its
absorbance at 340 nm where the NAD+ doesn’t
absorb
718
Problems and Answers
Solution

A340(NADH) = mc(NADH)l, then
0.31
A340(NADH)
=
=
= 4.9 x 10-5 M
m
6220
C(NADH)
 The absorbance of NADH at 260 nm
 A260(NADH) = mcNADHl = 15 x 103 x 4.9 x 10-5
= 0.74
 A260(NAD+) = 1.2 – 0.74 = 0.46

719
Problems and Answers
Solution


A260(NAD+) = mc(NAD+)l
c(NAD+)
0.46
A260(NAD+)
=
= 18000 = 2.6 x 10-5 M
m
720
Radiation in
Biochemistry
721
Radiation
Radiation is the emission of energy from
any source
For example, light from sun, heat from
body, emissions from radioactive
material, etc
Radiation can be divided into two kinds:
Ionizing
 Non-ionizing

722
Ionizing Radiation
Ionizing radiation consists of high
energy waves that are able to penetrate
cells and cause ionization in different
parts of cells
Ionization is caused by the removal of
an electron out of the orbit and creation
of a positive charge on the atom or
molecule
723
Ionizing Radiation
Ionized molecules are unstable and
quickly undergo chemical changes,
including the formation of free radicals
Ionizing radiation can lead to a mutation
in DNA, which could contribute to
cancer
724
Ionizing Radiation
Cellular changes from radiation takes
only a fraction of second, other changes
such as carcinogenesis may take years
to develop
725
Ionizing Radiation
The amount of cellular damage is
related to the dose of radiation receive
by the cell
Examples:
X-rays
 Cosmic rays
 Particles from radioactive materials

726
Exposure to Ionizing
Radiation
We are constantly exposed to ionizing
radiation from natural sources including:
Cosmic rays from the solar system
 Radioactive elements present in the soil
 Radioactive gases such as radon

727
Exposure to Ionizing
Radiation
Nonmedical radiation, this includes
nuclear radiation coming from:
Nuclear weapon explosions in war
(Hiroshima and Nagasaki)
 Testing of nuclear weapons (India,
Pakistan, Korea, etc)
 Leakage of fallout of nuclear material by
accidents (e.g, Chernobyl)

728
Exposure to Ionizing
Radiation
Medical radiation, the use of radiation in
medicine includes:
Diagnostic radiology such as X-rays.
 Nuclear medicine which involves
administration of radioactive substance in
the body for diagnosis or treatment of
disease
 Radiotherapy which uses high energy rats
to kill cancer cells

729
X-Rays
X-rays are part of the electromagnetic
spectrum
They have a short wavelength but it is
longer than the wavelength of gamma
radiation
They have less energy and are less
penetrating as compared to gamma
rays
730
X-Rays
They have no charge and travel at the
speed of light
They pass through matter and body
tissues (hence we use them to get
pictures of bones, etc)
They are partially absorbed and cause
some ionization
731
X-Rays
They are stopped by lead or thick
pieces of other metals
Like gamma rays, X-rays are often
given out when unstable nuclei decay
They are also given out by atoms of
heavy elements (such as tungsten)
when they are bombarded with high
speed electrons
732
X-Rays
Hospital X-ray machine use this method
X-rays are hazardous when they are
very intense
This is certainly the case when you
have an X-ray in hospital. This is why
they are only given when necessary
733
Radioactivity
734
Radioactivity Alpha ()
radiation
-Radiation is a stream of particles
They are made from two protons and
two neutrons, same as a nucleus of
helium 2He4
They have a charge of +2 and mass of
4
Because they have a large charge, 
particles ionize other atoms strongly
736
Radioactivity Alpha ()
radiation
-Particles are relatively slow and
heavy
They have a very short range in the air
(a few centimeters)
They have a low penetrating power and
can be stopped by a sheet of paper
-Particles are given out by the nuclei
of unstable atoms
738
Radioactivity Alpha ()
radiation
Main source in nature is radon gas
which is produced from naturally
occurring uranium
There are also  sources in radioactive
and medical wastes
Airborne radiation is not a great hazard
because it is so short range and will be
stopped by clothing
739
Radioactivity Alpha ()
radiation
However, contamination of food, water
or air supplies is hazardous as the
source might then be breathed in or
swallowed, giving out the highly ionizing
radiation inside a person's body
740
Beta () Radiation
-Radiation is a stream of fast moving
electrons, which have a negative charge
-Particles have a charge of -1, are
same as electrons
They are fast and have very little mass
(about 7000 times lighter than an particle)
They are weakly ionizing
741
Beta () Radiation
-Particles have a medium penetrating
power
They can be stopped by a sheet of
aluminum
-Radiation is given out by unstable
atoms
742
Beta () Radiation
As with -radiation, airborne radiation is
not a great hazard:
 is weakly ionizing and not very long
range
 However, contamination of food, water or
air supplies is hazardous as the source
might then be breathed in or swallowed

743
Gamma (g) Radiation
Gamma (g) rays are waves, not
particles
This means that they have no mass and
no charge
gRadiation is at the high frequency end
of the electromagnetic spectrum
744
Gamma (g) Radiation
It has a very short wavelength and will
pass through atoms with very little
chance of being deflected or absorbed
As part of the electromagnetic
spectrum, it travels at the speed of light
It will tend to pass through matter
without causing much ionization
745
Gamma (g) Radiation
It has an extremely long range in air but
gets weaker with distance
g Rays have a high penetrating power
and it takes a thick sheet of metal such
as lead, or concrete to reduce them
significantly
746
Gamma (g) Radiation
g Radiation is given out when unstable
radioactive nuclei decay and release
energy
It is usually given out in conjunction with
alpha and beta radiation
747
Comparison of Various
Radioactive Radiations
Mass


g
Radiation
Radiation
Radiation
4
1/2000
0
Charge
+2
-1
0
Speed
Slow
Fast
Vary fast
(AMU)
748
Comparison of Various
Radioactive Radiations
Ionizing


g
Radiation
Radiation
Radiation
High
Medium
0
Penetrating Low
Medium
High
ability
power
Stopped by Paper
Aluminum Lead
749
Non-ionizing Radiation
Non-ionizing radiation is low-frequency
radiation that does not have enough
energy to cause ionization
Examples:





UV rays
Visible light,
IR
Microwaves
Radio waves
750
Non-ionizing Radiation
Only UV rays have been associated
with cancer (skin cancer)
751
Ultraviolet Rays
UV-rays are also part of the
electromagnetic spectrum
They have a shorter wavelength than
visible light but longer than X-rays
They have no charge and travel at the
speed of light
The main source of UV rays is sun
752
Ultraviolet Rays
UV radiation is less energetic, and
therefore non-ionizing
But its wavelengths are preferentially
absorbed by bases of DNA and by
aromatic amino acids of proteins, so it
has important biological and genetic
effects
753
Ultraviolet Rays
The harmful biochemical effects of UV
include:
Sunburn
 Skin cancer

754
Ultraviolet Rays
UV is normally classified in terms of its
wavelength:

UV-C (180-290 nm)
Germicidal, most energetic and lethal
 It is not found in sunlight because it is absorbed
by the ozone layer


UV-B (290-320 nm)
Major lethal/mutagenic fraction
 Produces pyrimidine dimers

755
Ultraviolet Rays

UV-A (320 nm-visible)
Near UV
 Also has deleterious effects primarily because it
creates oxygen free radicals
 It produces very few pyrimidine dimers

The major lethal lesions are pyrimidine
dimmers in DNA (produced by UV-B
and UV-C)
756
Ultraviolet Rays
These are the result of a covalent
attachment between adjacent
pyrimidines in one strand
This can be a thymine-thymine dimer or
a thymine-cytosine dimer
These dimers block transcription and
DNA replication and are lethal if not
repaired
757
Ultraviolet Rays
758
Ultraviolet Rays
759
Ultraviolet Rays
Health concerns for UV exposure are
mostly for the range 290-320 nm in
wavelength, the range called UVB
The most effective biological
wavelength for producing skin burns is
297 nm
Whereas 330 nm being only 0.1% as
effective as 297 nm for biological effects
761
Nucleotide Excision
Repair
First, a UV-specific endonuclease
recognizes the dimer, and cleaves the
damaged strand at the 5-side of the
dimer
763
Nucleotide Excision
Repair
Next, the excision exonuclease
recognizes the incision made by
endonuclease and the gap is filled by
nucleotides addition using DNA
polymerase
Finally, DNA ligase will join the 3-OH of
newly synthesized DNA with 5-P of
remaining stretch of original DNA
764
Xeroderma
Pigmentosum
Pyrimidine dimers can be formed in the
skin cells of humans exposed to
sunlight
Xeroderma pigmentosum is a genetic
disease in which the cells cannot repair
the damaged DNA, resulting in the
development of skin cancer
765
Xeroderma
Pigmentosum
The most common form of this disease
is caused by the absence of the UVspecific endonuclease enzyme
766
Ozone Layer
Ozone is a form of oxygen
Each ozone molecule is made of three
oxygen atoms (O3)
Unlike oxygen, ozone is a poisonous
gas
767
Ozone Layer
But the ozone layer in atmosphere is
important for human health because it
absorbs most of the harmful UV
radiation from the sun before it reaches
the surface
768
Ozone Layer
Ozone layer sits at an altitude of about
10 to 50 km, with a maximum
concentration in the stratosphere at an
altitude of approximately 25 km
769
Ozone Layer
In recent years, scientists have
measured a seasonal thinning of the
ozone layer primarily at the Poles
This phenomenon is called as ozone
hole
770
Ozone Layer
Antarctica is the worst affected area.
Small amounts of ozone are constantly
being made by the action of sunlight on
oxygen
At the same time, ozone is being broken
down by natural processes
771
Ozone Layer
The total amount of ozone usually stays
constant because its formation and
destruction occur at about the same
rate
772
Ozone Layer
Ozone is created naturally in the
stratosphere by the combining of atomic
oxygen (O) with molecular oxygen (O2)
This process is activated by sunlight
Ozone is destroyed naturally by the
absorption of ultraviolet radiation, and
by the collision of ozone with other
atmospheric atoms and molecules
773
Ozone Layer
O3 + UV  O2 + O
O3 + O  2 O2
O3 + O3  3 O2
It appears that human activities are
altering the amount of stratospheric O3
The main agent responsible for this
destruction are human-made
chlorofluorocarbons (CFCs)
774
Ozone Layer
First produced by General Motors
Corporation in 1928, CFCs were
created as a replacement to the toxic
refrigerant ammonia
775
Ozone Layer
CFCs have also been used as a
propellant in spray cans, cleaner for
electronics, sterilant for hospital
equipment, and to produce the bubbles
in styrofoam
776
Ozone Layer
CFCs are cheap and are very stable
compounds, lasting up to 200 years in
the atmosphere
CFCs created at the Earth's surface drift
slowly upward to the stratosphere
where UV radiation from the sun causes
their decomposition and the release of
chlorine (Cl)
777
Ozone Layer
Chlorine in turn attacks the molecules of
ozone chemically converting them into
oxygen molecules
Cl + O3  ClO + O2
ClO + O  Cl + O2
778
Ozone Layer
A single Cl atom removes about
100,000 O3 molecules before it is taken
out of operation
Chlorine is removed from the
stratosphere by the following chemical
reactions:
ClO + NO2  ClONO2
CH4 + Cl  HCl + CH3
779
Ozone Layer
The ozone-destroying reactions take
place most rapidly only under certain
conditions in the stratosphere
These conditions (extreme cold,
darkness, isolation), followed by
exposure to light occur over the polar
regions after the long polar winter has
finished and the first spring sun appears
780
Radiation Dose
Curie (Ci) is the number of nuclear
disintegrations per second in 1 g of
radium
1 Ci = 37 billion disintegrations per sec
1 mCi = 37 million disintegrations per
sec
1 mCi = 37000 disintegrations per sec
781
Radiation Dose
Roentgen (R) is the intensity of radiation
that produces two billion ion-pairs in 1
mL of air
Roentgen does not accurately indicate
the amount of radiation on tissue
782
Radiation Dose
Radiation absorbed dose (Rad) is the
amount of radiation energy absorbed by
tissue
1 Rad = 100 ergs of energy per gram
tissue
783
Radiation Dose
Gray (Gy) is the most commonly used
unit of radiation exposure
1 Gy = 100 Rad
1 Gy = 1 Joule/kg
784
Isotopes in
Biochemistry
785
Milestones of History of
Isotopes
1896
A.A. Becquerel (1852 – 1908, Nobel price
of 1903)
 Discovery of radioactivity, start of nuclear
physics - base for the isotope physics

786
Milestones of History of
Isotopes
1910
F. Soddy (1877 - 1956, Nobel price of
1921)
 Foreign Correspondent Member of RAS
from 1924
 Term “ISOTOPE”
 Investigation of the isotopes properties and
origin

787
Milestones of History of
Isotopes
1911
J.J. Thomson (1856 – 1940, Nobel price of
1906
 Foreign Correspondent Member of RAS
from 1913
 Foreign Advisory Member of RAS from
1925
 The first direct observation of the isotopes
in experiments with the “cathode rays”

788
Milestones of History of
Isotopes
1919
F.W. Aston (1877 - 1945, Nobel price of
1922)
 Foreign Correspondent Member of RAS
from 1924
 Research for isotope phenomenon
 The first mass-spectrometer
 Curve of the nuclear “packing factors”

789
Isotopes in Biochemistry
Isotopes are any of the different species
of atom (Nuclides) of a chemical
element each having different atomic
mass (mass number)
Isotopes of an element have nuclei with
the same number of protons (the same
atomic number) but different numbers of
neutrons
790
Isotopes in Biochemistry
Therefore, isotopes have different mass
numbers, which give the total number of
nucleons (the number of protons plus
neutrons)
791
Isotopes in Biochemistry
A nuclide is any particular atomic
nucleus with a specific atomic number Z
and mass number A; it is equivalently
an atomic nucleus with a specific
number of protons and neutrons
Collectively, all the isotopes of all the
elements form the set of nuclides
792
Isotopes in Biochemistry
The distinction between the terms
isotope and nuclide has somewhat
blurred, and they are often used
interchangeably
793
Isotopes in Biochemistry
Isotope is better used when referring to
several different nuclides of the same
element; nuclide is more generic and is
used when referencing only one
nucleus or several nuclei of different
elements
794
Isotopes in Biochemistry
For example, it is more correct to say
that an element such as fluorine
consists of one stable nuclide rather
than that it has one stable isotope
795
Isotopes in Biochemistry
In IUPAC nomenclature, isotopes and
nuclides are specified by the name of
the particular element, implicitly giving
the atomic number, followed by a
hyphen and the mass number (e.g.
helium-3, carbon-12, carbon-13, iodine131 and uranium-238)
796
Isotopes in Biochemistry
In symbolic form, the number of
nucleons is denoted as a superscripted
prefix to the chemical symbol (e.g. 3He,
12C, 13C, 131I and 238U)
About 339 nuclides occur naturally on
Earth, of which 269 (about 79%) are
stable
797
Isotopes in Biochemistry
Counting the radioactive nuclides not
found in nature that have been created
artificially, more than 3100 nuclides are
currently known
798
Applications of
Isotopes
799
Applications of Isotopes
Several applications exist that capitalize
on properties of the various isotopes of
a given element
Use of chemical and biological
properties
800
Applications of Isotopes
Isotope analysis is the determination of
isotopic signature, the relative
abundances of isotopes of a given
element in a particular sample
801
Applications of Isotopes
For biochemical substances in
particular, significant variations of
isotopes of C, N and O can occur
Analysis of such variations has a wide
range of applications, such as the
detection of adulteration of food
products
802
Applications of Isotopes
The identification of certain meteorites
as having originated on Mars is based
in part upon the isotopic signature of
trace gases contained in them
Another common application is isotopic
labeling, the use of unusual isotopes as
tracers or markers in chemical reactions
803
Applications of Isotopes
Normally, atoms of a given element are
indistinguishable from each other.
However, by using isotopes of different
masses, they can be distinguished by
mass spectrometry or infrared
spectroscopy
804
Applications of Isotopes
For example, in 'stable isotope labeling
with amino acids in cell culture (SILAC)'
stable isotopes are used to quantify
proteins
If radioactive isotopes are used, they
can be detected by the radiation they
emit (this is called radioisotopic
labeling)
805
Applications of Isotopes
A technique similar to radioisotopic
labelling is radiometric dating: using the
known half-life of an unstable element,
one can calculate the amount of time
that has elapsed since a known level of
isotope existed
806
Applications of Isotopes
The most widely known example is
radiocarbon dating used to determine
the age of carbonaceous materials
Isotopic substitution can be used to
determine the mechanism of a reaction
via the kinetic isotope effect
Use of nuclear properties
807
Applications of Isotopes
Several forms of spectroscopy rely on
the unique nuclear properties of specific
isotopes
For example, nuclear magnetic
resonance (NMR) spectroscopy can be
used only for isotopes with a nonzero
nuclear spin
808
Applications of Isotopes
The most common isotopes used with
NMR spectroscopy are 1H, 2D,15N, 13C,
and 31P
Mössbauer spectroscopy also relies on
the nuclear transitions of specific
isotopes, such as 57Fe
Radionuclides also have important uses
809
Applications of Isotopes
Nuclear power and nuclear weapons
development require relatively large
quantities of specific isotopes
810
Applications of Isotopes
The process of isotope separation
represents a significant technological
challenge, but more so with heavy
elements such as uranium or plutonium,
than with lighter elements such as
hydrogen, lithium, carbon, nitrogen, and
oxygen
811
Applications of Isotopes
The lighter elements are commonly
separated by gas diffusion of their
compounds such as CO and NO
Uranium isotopes have been separated
in bulk by gas diffusion, gas
centrifugation, laser ionization
separation, and by a type of production
mass spectroscopy
812
Radioactive Isotope
Decay
813
Radioactive Decay
dN
lN = dt
dN
 Where = the number of atoms
dt
decaying per small increment of time (i.e.,
the count rate)
 N = the total number of radtioactive atoms
present at any given time
 l = a decay constant, different for each
isotope
814
Radioactive Decay

The negative sign indicates that the
number of radioactive atoms decreases
with the time
dN/N
l= dt
2.3
N
l=
log
t
No
815
Radioactive Decay

No and N can be expressed in any
consistent manner
(1) For example, No = 100%, N = % remaining
after time interval (t)
(2) No = 1, N = fraction remaining (as a decimal)
after time interval (t)
(3) No = original CPM in sample, N = CPM
remaining after interval (t)
(4) No= SA of sample at a certain time, N = SA of
the sample after an elapsed time (t)
816
Half-life, Mean-life and Effective
Half-life of Radioactive Isotopes
The half-life (t½) of a radioactive
isotopes is the time required for half of
the original number of atoms to decay
The relationship between t½ and l is
shown below:
0.693
t½=
l
817
log No
100
log N (linear scale)
Recent of original radioactivity
Half-life of Radioactive
Isotopes
50
Slope = - l/2.3
0
Number of half-life elapsed
Time
818
Mean-life of Radioactive
Isotopes
The Mean-life (m)is the average of a
radioactive isotopes of atom
It is the reciprocal of the disintegration
constant (l):
1
m=
= 1.44 t½
l
819
Effective Half-life of
Radioactive Isotopes
The radioactive isotopes in a
biochemical compound is subject to
elimination or turnover from the
organism or cell according to the rate
reflected in the biochemical half-life of
that compound
820
Effective Half-life of
Radioactive Isotopes
Assuming this rate to follow first-order
kinetics, then the biochemical rate
constant (lb) is related to the tb½ by:
0.693
tb½=
lb
821
Effective Half-life of
Radioactive Isotopes
The overall or effective half-life of the
radioactive isotope, teff½ is:
1
1
1
=
+
teff½ ta½ tb½

Where ta½ is the physical half-life of decay
of radioactive isotope
822
The Curie
The curie (Ci) is a standard unit of
radioactive decay
It was originally defined as the rate at
which 1 g of 226radium decays
Because of the relatively long half-life of
226Ra, the isotope served as a
convenient standard
823
The Curie
The curie is now defined as the quantity
of any radioactive substance in which
the decay rate is 3.7 x 1010
disintegrations per second, DPS (2.22 x
1012 DPM)
824
The Curie
Because the efficiency of most radiation
detection devices is less than 100%, a
given number of curies almost always
yields a lower than theoretical count
rate
825
The Curie
There is distinction between DPM and
CPM, for example, a sample containing
1 mCi of radioactive material has a
decay rate of 2.22 x 106 DPM
If only 30% the disintegrations are
detected, the observed count rate is
6.66 x 105 CPM
826
Specific Activity
Specific activity (SA) refers to the
amount of radioactivity per unit amount
of substance
Specific activity may be given in terms
of curies per gram (Ci/g), mCi/mg,
mCi/mmol, DPM/mmol, counts per mint
per micromole, CPM/mmol or in any
other convenient way
827
Specific Activity
Once the specific activity of a
compound is known, any given count
rate can be equated to the amount of
the compound in a sample
828
Specific Activity
In most studies with radioactive
compounds, it is assumed that there is
no isotope effects, i.e., it is assumed
that the radioactive molecules are
randomly distributed among the total
molecules of the compound and behave
identically to the nonradioactive
molecules
829
Isotopic Dilution
By the use of radioactive tracer
techniques, it is possible to analyze very
small amounts of compounds
This method is termed isotope dilution
and can involve either the determination
of an inactive compound by dilution with
the radioactive compound or vice versa
830
Isotopic Dilution
If a weight (Wa) of a radioactive
compound with a known specific activity
(SA) is mixed with a weight (Wi) of inert
compound, then SAr of the reisolated
compound is:
Wa
SAr = SA
Wa + Wi
831
Problems and Answers
45Ca
has a t½ of 163 days. Calculate (a)
the decay constant (l) in terms of day-1
and sec-1, (b) the % of the initial
radioactivity remaining in a sample after
90 days, (c) the activity in mCi, and (d)
the mean-life in sec?
832
Problems and Answers
Solution
0.693
0.693
 a l =
=
= 4.25 x 10-3 day-1
163
t½
1h
1 min
4.25 x 10-3 1 day

x
x
x
=
day
24 h 60 min 60 sec

4.92 x 10-8 sec-1
833
Problems and Answers

(b) l x t = 2.3 log

10-3
4.25 x
N0
N
, let N0 = 100% then
100
(90) = 2.3 log
N
100
 0.38 = 2.3 log
N
100
log N = 0.167, by taking anti log
 N = 68.07%

834
Problems and Answers
3.7 x 1010
4 DPS
 (c) 1 mCi =
DPS
=
3.7
x
10
106

x → 4.92 DPS, then
4.92 x 10-8
-4 mCi
 x=
=
1.33
x
10
3.7 x 104
835
Problems and Answers
(d) t½ = 163 x 24 h x 60 min x 60 sec =
14.1 x 106 sec
 m = 1.44 x t½ = 1.44 x 14.1 x 106 = 20.3 x
106 sec

836
Problems and Answers
What is the t½ of a isotope in a sample
with 5000 CPM and then 3500 CPM 5
hour later?
Solution
N0
2.3
 l=
log
t
N
2.3
3500
=
log
= 0.071 h
5
5000
0.693 0.693
 t½ =
=
= 9.7 h
0.071
l
837
Problems and Answers
14C
has t½ of 5700 years. Calculate the
fraction of 14C atoms that decay (a) per
year, (b) per minute, (c) the abundance
of 14C in the carbon that is participate in
carbon cycle and (d) the age of a
sample of biochemical material that
contains 3 DPM/g carbon?
838
Problems and Answers
Solution

0.693 0.693
(a) l = t
=
= 1.2 x 10-4 y-1
½
5700

1.2 x 10-4 atoms per atom decays per year or 1
1
atom out of
= 8.225 x 103
1.2 x 10-4

Thus 1 out of every 8225 radioactive atoms
decays per year
839
Problems and Answers
Solution
1.2 x 10-4
 (b) l =
= 2.31 x 10-10 min
(365)(24)(60)


1
= 4.32 x 109
-10
2.31 x 10
1 out of 4.32 x 109 radioactive atoms
decays per min
840
Problems and Answers
Solution
DPM
 (c)
g = lN, where N = the # of
14C
atoms per g of carbon
 13 = 2.31 x 10-10 N, then
 N = 5.6 x 1010 atoms 14C/g carbon
6.023 x 1023
 Total atoms of carbon =
= 5.02
12
x 1022
841
Problems and Answers
Solution
N
 Abundance =
Total atoms carbon x 100%

5.63 x 1010 14C
=
x 100 = 1.12
22
5.02 x 10
10-10%
13
 (d) lt = 2.3 log
3
13
-4
 (1.22 x 10 )t = 2.3 log
,then
3
 t = age = 12,029 years
842
Problems and Answers
40K
(t½ = 1.3 x 109) constitutes 0.012%
in nature. The human body contains
about 0.35% K by weight. Calculate the
total radioactive resulting from 40K
decay in a 75 kg human?
843
Problems and Answers
Solution
Total g of 40K = 0.012 x 0.35 75 x 103 =
3.15 10-2 g
3.15 x 10-2
20
 # of 40K atoms =
=
4.74
x
10
40
atoms
0.693
 l = 1.3 x 109 x 365 x 24 x 60 = 1.014 x 10-15
min-1

844
Problems and Answers
Solution
dN
-15)(4.74 x
 DPM = =
lN
=
(1.014
x
10
dt
1020) = 4.18 x 105
 1 mCi → 2.22 x 106 DMP

x → 4.81 x 105 DMP

x=
4.81 x 105
=
0.22
mCi
6
2.22 x 10
845
Problems and Answers
If the biochemical t½ of Fe in RBC is 60
day, what is the teff½ of 59Fe in RBC, ta½
= 46 day?
Solution

1
=
teff½
1
+
ta½
1
1
1
=
+
= 26 days
tb½
46
60
846
Problems and Answers
A bottle contains 1 mCi Phe-14C
(uniformly labeled) in 2 mL of solution.
The specific activity of the label Phe is
given as 150 mCi/mmol. Calculate (a)
the concentration of Phe in solution, and
(b) the activity of the solution in
CPM/mL at a counting efficiently of
80%?
847
Problems and Answers
Solution
(a) 150 mCi → 1 mmol → 2 mL

1 mCi → x mmol → 1 mL
 Concentration = 3.33 x 10-3 mmol/mL =
3.33 x 10-3 M
 (b) 1 mCi = 2.22 x 109 DPM
9
0.8
x
2.22
x
10
 Total activity =
= 8.9 x 108
2
CPM/mL

848
Problems and Answers
The amount of sucrose in an actively
photosynthesizing canna leaf was
determined by extracting the leaf with
boiling 80% ethanol. To the extract was
add 5 mg 14C-label sucrose, (SA = 80
mCi/mg) and a 3-mg sample showed an
activity of 16 mCi. How much sucrose
was extracted from the leaf?
849
Problems and Answers
Solution
Wa
 SAr = SA
Wa + Wi
5
16
 SAr =
= 80
5 + Wi
3

Wi = 70 mg
850