Transcript Pharmaceutical Instrumental Analysis 501413 Dr. Wael Abu

Pharmaceutical Analysis
501722
Dr. Wael Abu Dayyih
Group (3)
2011
1
Physical and Chemical Properties
of Drug Molecules
2
Introduction
• The physical properties of organic molecules, such as pKa and
partition coefficient, are dealt with extensively in pharmacy
courses but do not feature greatly in analytical chemistry
courses. It is often surprising that analytical chemists cannot
distinguish between, for instance, basic, weakly basic, acidic,
weakly acidic and natural nitrogen functions. The physical
properties of drug molecules, along with simple chemical
dramatization and degradation reactions, play an important
part in the design of analytical methods. Drug molecules can be
complex, containing multiple functional groups that in
combination produce the overall properties of the molecules.
The chapter will serve as a starting point for understanding the
chemical and physio-chemical behavior of drug molecules,
which influence the development of analytical methods. The
latter part of the chapter focuses on some typical drugs that are
representative of a class of drug molecules and lists their
physical properties and the properties of their functional groups
in so far as they are known.
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Calculation of
pH value of aqueous solutions of strong and weak acids and
bases
• Dissociation of water
• The pH of a solution is defined as –log [H+],
where [H+] is the concentration of hydrogen ions
in solution.
• In pure water the concentration of hydrogen ions
is governed by the equilibrium


H 2O 
H

HO
Ka
• Ka is the dissociation constant for the
equilibrium, is known as Kw in the case of the
dissociation of water and is determined by the
following expression:
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Dissociation of water


[ H ][ HO ]


14
Kw 
[ H ][ HO ] 10
[ H 2O]
Since the concentration of water does not change
appreciably as a result of ionization, its concentration
can be regarded as not having an effect on the
equilibrium and it can be omitted from the equation,
and this means that in pure water:
[H+] = [HO-] = 10-7
The pH of water is thus given by –log 10-7 = 7.00
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Strong Acids and Bases
• If an acid is introduced into an aqueous solution the [H+] increases.
• If the pH of an aqueous solution is known, the [H+] is given by the
expression 10-pH, e.g. [H+] in pH 4 solution = 10-4 M = 0.0001 M.
Since [H+] [OH-] = 10-14 for water, the concentration of [OH-] in this
solution is 10-6 M.
• A strong acid is completely ionized in water and [H+] is equal to its
molarity, e.g. 0.1M HCl contains 0.1 M H+ (10-1 H+) and has a pH of –
log 0.1 = 1. For a solution of a strong base such as 0.1 M NaOH, [OH] = 0.1M and [0.1] [H+] = 10-14, therefore, [H+] = 10-13 M and the pH
of the solution = 13. Although the pH range is regarded as being
between 0 and 14, it does extend above and below these values,
e.g. 10 M HCl, in theory, has a pH of -1.
• Self-test 2.1
• Calculate the pH of the following solution:
• 0.05 M HCl
• 0.05 M NaOH
• 0.05 M H2SO4
• Answers: (i) 1.3, (ii) 1.0, since 0.05 M H2SO4contains 0.1 M H+.
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Weak acids and bases
• Weak acids are not completely ionized in
aqueous solution and are in equilibrium with
the dissociated acid, as is the case for water,
which is a very weal acid. The dissociation
constant Ka is given by the expression below:

HA 
A
 H
Ka
[ A  ][ H
Ka 
[ HA]


]
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Weak acids and bases
• For instance, in a 0.1 M solution of acetic acid
(Ka = 1.75 10-5), the equilibrium can be
written as follows:
Ka
CH 3COOH 
CH 3COO   H 
(0.1  x)
x
x
[CH 3COO ][ H  ]
Ka 
[CH 3COOH ]
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In comparison, the pH of 0.1 M HCl is 1.
Weak acids and bases
• The pH can be calculated as follows:
1.75105
x2

(0.1  x)
• Since the dissolution of the acetic acid does not
greatly change the concentration of the un-ionised
acid, the above expression can be approximated to:
5
1 .7 5  1 0
x2

0 .1
x  [ H ] 
p H  2 .9 0
1.7 5  1 0 6  0.0 0 1 3 2
M
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Weak acids and bases
• The calculation of the pH of a weak base can be
considered in the same way. For instance, in a 0.1 M
solution of ammonia (Kb = 1.810-5), the equilibrium
can be written as follows:

Kb
NH 3  H 2 O 
NH 4  HO 
(0.1  x)
x
x
• If the concentration / activity of water is regarded as
being 1, then the equilibrium constant is given by the
following expression:
[ NH ][ HO ]
Kb 


4
[ NH 3]
5
1.8  10
x2

(0.1  x)
10
Weak acids and bases
• The concentration of NH3 can be regarded as
unchanged by a small amount of ionization
and the expression can be written as:
x  [ HO  ]  1.8  106  0.0013M
14
K
10
w
[H  ] 

 7.7  1012 M
0.0013 0.0013
pH  11.1
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Self-test
•
•
•
•
Calculate the pH of the following solutions:
0.1 M formic acid (Ka = 1.7710-4)
0.05 M phenol (Ka = 1.310-10)
0.15 M ethylamine (Kb = 5.610-4)
•
Answers: (i) 2.4, (ii) 5.6, (iii) 11.9
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Acidic and basic strength and pKa
• The pKa value of a compound is defined as: pKa = -log Ka.
• A pKa value can be assigned to both acids and bases.
• For an acid, the higher the [H+] the stronger the acid,
e.g.:
Ka


CH 3 COOH CH 3COO  H
• In the case of a base, it is the protonated form of the
base that acts as a proton donor, e.g.:
Ka
NH  
NH 3  H 
• In this case, the lower the [H+] the stronger the base.
• If pKa is used as a measure of acidic or basic strength, for
an acid, the smaller the pKa value the stronger the acid;
for a base the larger the pKa value the stronger the base.
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Henderson-Hasselbach equation


[ A ][ H ]
Ka 
[ HA]
• Can be rearranged substituting pH for –log
[H+] and pKa for –log Ka to give:
[ A ]
pH  pKa  log
[ HA]
• For example, when acetic acid (pKa 4.76) is in solution at pH
4.76. the Henderson-Hsselbalch equation can be written as
follows: pH=4.76+log[CH3COO-]/[CH3COOH]
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Self-test
• Determine the percentage of ionization of
acetic acid at (i) pH 3.76 and (ii) pH 5.76.
• Acetic acid is 50% ionized at pH 4.76. in the
case of a weak acid, it is the protonated form
of the acid that is un-ionised, and, ad the pH
falls, the acid becomes less ionized.
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Self-test
• Calculate the percentage of ionization of
ammonia at (i) pH 8.25 and (II) pH 10.25
•
Answers: (i) 90.9%, (ii) 9.09)
• An alternative way of writing the expression giving the
percentage of ionization for an acid or base of a particular pka
value at a particular pH value is:
• Acid: % ionisation = { 10(pH-pKa)/1+10(pH-pKa)} × 100
• Base: % ionisation = {10(pka-pH)/1+10(pKa-pH) } × 100
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Ionization of drug molecules
• The ionization of drug molecules is important with regard
to their absorption into the circulation and their
distribution to different tissues within the body. The pKa
value of a drug is also important with regard to its
formulation into a medicine and to the design of analytical
methods for its determination.
• Calculation example 2.1
• Calculate the percentage of ionization of the drugs at pH
7.0 and pka as shown
• Diphenhydramine
• This drug contains one basic nitrogen andpka 9.0, at pH7.0,
its percentage of ionisation can be calculated as follows:
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Ionization of drug molecules
• Ibuprofen
• This drug contains one acidic group
andpKa4.40, at pH 7.0, its percentage of
ionization can be calculated as follows:
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Self-test
• Calculate the percentage of ionisation of the
following drugs which contain 1 group that
ionizes in the pH range 0-14, at the pH values
of (i) 4 and (ii) 9.
•
• Answers: Bupivacaine (i) 99.99%, (ii) 11.2%,
(iii) 79.9%; Indomethacin (i) 24.0%m (ii)
99.99%; 5-fluorouracil (i) 0.01%, (ii) 90.9%.
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Buffers Solution
• Buffers can be prepared from any weak acid or base and are used to
maintain the pH of a solution in a narrow range. This is important in living
systems; for example human plasma is buffered at pH 7.4 by a carbonic
acid/ bicarbonate buffer system.
• Buffers are used in a number of areas of analytical chemistry, such as the
preparation of mobile phases for chromatography and the extraction of
drugs from aqueous solution. The simplest type of buffer is composed of
a weak acid or base in combination with a strong base or acid. A common
buffer system is the sodium acetate/ acetic acid buffer system. The most
direct way of preparing this buffer is by the addition of sodium hydroxide
to a solution of acetic acid until the required pH is reached. The most
effective range for a buffer is 1 pH unit either side of he pKa value of the
weak acid or base used in the buffer. He pKa value of acetic acid is 4.76;
thus its effective buffer range is 3.76-5.76.
1
pH  ( pKa  Pkw  log[ salt ]
2
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Salt Hydrolysis
• When the salt of a strong acid and a strong base is dissolved in
water it produces a pH of ca 7.0. when salts of a weak acid and a
strong base or of a strong acid and a weak base are dissolved in
water, they will produce, respectively, alkaline and acidic solutions.
• When sodium acetate is dissolved in water, the acetate ion behaves
as a base removing protons from solution. For a weak electrolyte in
Kb × Ka = Kw. If a 0.1 M solution of sodium acetate in water is
considered.
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Activity, ionic strength an dielectric
constant
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Partition coefficient
•
An understanding of partition coefficient and the effect of pH on partition
coefficient is useful in relation to the extraction and chromatography of drugs. The
partition coefficient for a compound (P) can be simply defined as follows:
• P = Co/Cw
•
•
Where Co is the concentration of the substance in an organic phase and Cw is the
concentration of the substance in water.
The greater P the more a substance has an affinity for organic media. The value of
P for a given substance of course depends on the particular organic solvent used
to make the measurement. Many measurements have been made of partition
between n-octanol and water, since n-octanol, to some extent, resembles
biological membranes and is also quite a good model for reverse-phase
chromatographic partitioning. P is often quoted as a log P-value, e.g. a log P of 1 is
equivalent to P=10. Where P=10 for a particular compound partitioning into a
particular organic solvent, and partitioning is carried out between equal volumes
of the organic solvent and water, then ten parts of the compound will be present
in the organic layer for each part present in the water layer.
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Effect of pH on partitioning
•
•
Many drugs contains ionisable groups and their partition coefficient at a given pH
may be difficult to predict if more than one ionized group is involved. However,
often one group in a molecule may be much more ionized than another at a
particular pH, thus governing its portioning 0f 0rganic acids and bases into organic
solvent with represent to the pH of the solution that they are dissolved in.
From the Henderson – Hasselbalch equation
• For acid : Papp = P/1+10pH-pKa
• For bases : Papp = P/1+10pka-pH
•
Papp is the apparent partition coefficient which varies with pH. Thus it can be
seen that, when a compound, acid or base, is 50% ionized (i.e. pH = pKa), its
partition coefficient is half that of the drug in the un-ionised state.
•
•
Papp =P/1+10 0 = P/2
AS a general rule, for the efficient extraction of a base into an organic medium
from an aqueous medium, the pKa of the aqueous medium should be at least 1 pH
unit higher than the pKa value of the base, and, in the same situation for an acid,
the pH should be 1 .pH until lower than the pka value of the acid.
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