Ionization of Solutes:
Electrolytes
Are solutes which dissociate into ions if the dielectric constant of
the solvent is high enough to cause sufficient separation of the
attractive forces between the oppositely charged ions.
Ionization (dissociation) of electrolytes has several consequences
e.g.
Hydrogen ion concentration and pH:
The dissociation of water can be represented by:
H2O -----------H+ + OHIn pure water the concentrations of H+ and OH- ions are equal
and at 25°C both have the values of 1 x 10-7 mol/l.
Acid
Asubstance which donates a proton (or hydrogen ion)
The addition of an acid to water will increase hydrogen ion
concentration (more than 10-7 mol/l)
Base
Asubstance that accepts protons
The addition of a base will decrease the concentration of hydrogen
ions.
 The hydrogen ion concentration range is from 1 mol/l
for a strong acid down to 1 x 10-14 mol/l for a strong
base.
 To avoid the use of such low values  pH has been
introduced as a more convenient measure of hydrogen ion
concentration.
 pH is defined as the negative logarithm of the hydrogen ion
concentration [H+]
pH = -log10 [H+]
 pH of a neutral solution like pure water is 7, why?
because the conc. of H +ions (and OH -) ions = 1 x 10-7 mol/l
 pHs of acidic solutions will be less than 7
pHs of alkaline solutions will be greater than 7
pH has several important applications in pharmaceutical
practice.
- Affect the solubilities of drugs that are weak acids or bases
- Affect the stabilities of many drugs
- Affect the ease of absorption of drugs from the GIT
Dissociation (or ionization) constants and pKa:
In solutions of weak acids or weak bases equilibria exist between
undissociated molecules and their ions.
For a weakly acidic drug HA: The ionization constant
(dissociation constant) Ka of a weak acid can be obtained by
applying the Law of Mass Action: HA -----H+ + A[H+] [A-]
[HA]
pKa = pH + log
Ka=
[HA]
[A-]
pKa = the negative logarithm of Ka
Henderson-Hasselbalch equation:
A general equation that is applicable to any acidic drug with one
ionizable gp : Cu = conc. of the unionized;
Ci = conc. of the ionized species
Cu
pKa = pH + log
Ci
The ionization constant (dissociation constant) Ka of a protonated
weak base is given by B + H+ --------BH+
[H+] [B]
Ka=
[BH+]
Taking the negative log of this equation:
[BH+]
pKa = pH + log
[B]
Henderson-Hasselbalch equation:
A general equation that is applicable to any weak basic drug with
one ionizable group where:
Ci = conc. of protonated ;
Cu = conc. of the unionized species
pKa = pH + log
Ci
Cu
Buffer Capacity:
A buffer counteracts the change in pH of a solution upon the
addition of a strong acid, a strong base, or other agents that
tend to alter the hydrogen ion concentration.
Buffer capacity β: buffer efficiency, buffer index or buffer value
Is the resistance of a buffer to pH changes
upon the addition of a strong acid or base.
Definition
The ratio of amount added of strong base (or acid) to small change
in pH brought about by this addition.
β=
ΔB
Δ pH
ΔB
= the small addition in gram equiv./liter of strong base
added to the buffer solution to produce a pH change
Δ pH = pH change
The buffer capacity of the solution has a value of 1:
when the addition of 1 gram equiv. of strong base (or acid) to 1
liter of the buffer solution results in a change of 1 pH unit.
Example:
Acetate buffer
( acetic acid & sodium acetate)
0.1 mole each in 1 liter of solution.
a) 0.01 mole portions of NaOH is added
HAc +
(0.1 – 0.01)
NaOH ---------NaAc + H2O
(0.01)
(0.1 + 0.01)
b) The conc. of Na acetate (the [salt] in buffer equation)  by 0.01
mol/l & the conc. of acetic acid [acid]  by 0.01 mol/l
because each addition of base converts 0.01 mole of acetic acid
into 0.01 mole of sodium acetate according to the reaction.
Before the addition of the first portion of NaOH,
the pH of the buffer solution is:
pKa = pH + log C u(Acid )
Ci(Salt)
pH = pKa_ log(Acid)
(Salt)
pH = pKa+ log(Salt)
(Acid)
pH = 4.76 + log (0.1) = 4.76
(0.1)
pH = pKa
The changes in concentration of the salt and the acid by the
addition of a base are represented by
pH = pKa + log (Salt)+(Base)
(Acid) – (Base)
pH = 4.76 + log (0.1) + 0.01
(0.1) – 0.01
Buffer
β ,Capacity
pH of
solution
0.11
Moles of
NaOH
added
0
0.11
0.01
4.85
0.11
0.02
4.94
0.10
0.03
5.03
0.09
0.04
5.13
0.08
0.05
5.24
0.07
0.06
5.36
4.76
 The buffer capacity is not a fixed value for a given buffer
system, but depends on the amount of base added.
 With the addition of more NaOH, the buffer capacity
decreases rapidly, and, when sufficient base has been
added the acid convert completely into sodium ions and
acetate ions
 The buffer has it’s greatest capacity before any base is
added where [salt] / [acid] = 1, and according to
equation, pH = pKa.
 The buffer capacity is influenced by an increase in the
total conc. of the buffer constituents since a greater
conc. of salt and acid provides a greater alkaline and
acid reserve.
Van Slyke developed a more exact equation for
calculation of buffer capacity β
β= 2.3 C Ka [H3O+]
(Ka +[H3O+])2
C = The total buffer concentration (the sum of the molar
concentrations of the acid and the salt).
Ka = dissociation constant
H3O+ = hydrogen ion concentration
The equation permits the calculation of the buffer capacity at
any hydrogen ion concentration, i.e. when no acid or base has
been added to the buffer
Example:
If hydrogen ion concentration is 1.75 x 10-5, pH = 4.76
what is the capacity of the buffer containing 0.10 mole of each
of acetic acid and sodium acetate per liter of solution ?
The total concentration , C = [acid] + [salt], is 0.20 mol/l and
the dissociation constant Ka is 1.75 x 10-5
β= 2.3 C Ka [H3O+]
(Ka +[H3O+])2
β = 2.3 x 0.20 x (1.75x10-5) x (1.75 X 10-5) = 0.115
[(1.75x10-5) +(1.75 X 10-5)]2
Maximum buffer capacity .
The maximum buffer capacity occurs when pH = pKa or
when (H3O+) = Ka
β max = 2.303 C
(H3O+) 2 = 0.576 C
(2 H3O+) 2
β max = 0.576 C
Where C is the total buffer concentration
Example:
What is the maximum buffer capacity of an acetate buffer with
a total concentration of 0.20 mol/l?
β max = 0.576 C
= 0.01152 = 0.01
Buffer in biological & pharmaceutical systems
I. In vivo biological buffer systems
a) Blood
 Blood is maintained at a pH of about 7.4 by:
the 1° buffers in the plasma &
the 2° buffers in the erythrocytes.
 The buffer capacity of blood = 0.039 gram equiv. per
liter/pH unit for whole blood of which: 0.031 by the
cells
0.008 by the plasma
* When the pH of the blood goes below 7.0 or
above 7.8, life is in serious danger.
* The pH of the blood in diabetic coma is dropped to about
6.8
b) Lacrimal fluid
Tears have a great degree of buffer capacity, allowing a
dilution of 1:15 with neutral distilled water before an
alteration of pH is noticed.
The pH of tears is about 7.4 with a range of 7 to 8
II. Pharmaceutical Buffers
Buffer solutions are used in pharmaceutical formulation
particularly in ophthalmic preparations
Gifford suggested two stock solutions of:
- boric acid and monohydrated sodium carbonate
- mixed in various proportions to yield buffer solutions
of pH values from about 5 - 9.
Sorensen proposed a mixture of the salts of:
- sodium phosphate for buffer solutions of pH 6 to 8.
The Clark-Lubs mixtures and their pH ranges
a. pH 1.2 to 2.2: HCI and KCI
b. pH 2.2 to 4.0: HCI and potassium hydrogen phthalate
c. pH 4.0 to 6.2: NaOH and potassium hydrogen phthalate
d. pH 5.8 to 8.0: NaOH and KH2PO4
e. pH 7.8 to 10 : H3BO3, NaOH and KCl
Sodium chloride is added to buffer mixture to make it isotonic
with body.
Preparation
Factors of some importance in the choice of pharmaceutical
buffer include:
 Availability and cost of chemicals
 Sterility of the final solution.
 Stability of the drug and buffer on aging.
 Freedom from toxicity.
For example, a borate buffer, because of its toxic effects,
cannot be used for a solution to be administrated orally or
parenterally.
The following steps should be used in preparing buffer
systems
a. Select a weak acid having a pKa approximately equal to
the pH wanted to insure maximum buffer capacity.
b. From the buffer equation, calculate the ratio of salt and
weak acid required to obtain the desired pH.
log Cu = pKa - pH
Ci
c. Consider the individual concentrations of the buffer salt
and acid needed to obtain a suitable buffer capacity.
β = 2.3 C Ka [H3O+]
)Ka + [H3O+])2
A concentration of 0.05 to 0.5 molar is sufficient and
a buffer capacity of 0.01 to 0.1 is sufficient.
d. Finally, determine the pH and buffer capacity of the
completed
buffered solution using a pH meter.
Summary of Buffer Systems
System pKa
Range
acetic acid/acetate
4.76
malic acid/malate
1.96, 6.23
citric acid/citrate
4.74 (pKa2)
tartaric acid/tartrate
2.93, 4.23
pH
3.5-5.7
2.5-5.0
2.5-6.0
2.5-5.0
lactic acid/lactate
phosphoric acid/phosphate
glycine/glycinate
glutamic/glutamate
3.8
7.2 pKa2
2.4, 9.8
9.67 pKa3
3.0-5.0
6.0-8.2
6.5-7.5
8.2-10.2
III. Influence of Buffer Capacity and pH
on Tissue Irritation
 Solutions to be applied to tissues or administered
parenterally are liable to cause irritation, if their pH is
greatly away from the normal pH of the body fluid.
  must be considered when formulating:
- ophthalmic solutions
- parenteral products
- fluids to be applied to abraded surfaces.
 Factors affecting: (i &ii are of greater significance)
i) The buffer capacity of the solution
ii) The volume to be used in relation to that of body fluid
with which the buffered solution will come in contact
iii) Actual pH of the solution
iv) The buffer capacity of the body fluid
Tissue irritation due to large pH differences between:
the solution administered & the physiological fluid
is minimized:
(a) The lower the buffer capacity of the solution
(b) The smaller the volume used for a given concentration.
(c) The larger the volume and buffer capacity of the
physiological fluid
 The pH of solutions for introduction into the eye may vary
from 4.5 to 11.5 without marked pain or damage.
This is true only if the buffer capacity was kept low.
 Sorensen phosphate buffer produced irritation in the eyes of
a number of subjects when used outside the pH range of 6.5
to 8
 Boric acid solution of pH 5 produced no discomfort in the
eyes of the same subjects. Why?
Because of the very low buffer capacity of boric acid compared
to that of the phosphate buffer.
Parenteral solutions for injection into the blood stream are usually
not buffered or they are buffered to a very low capacity so that the
buffers of the blood may bring them
within the physiological pH range.
Influence of Buffer Capacity and pH
on Optimum Therapeutic Response
• The undissociated form of a weakly acidic or basic drug
has a higher therapeutic activity than the dissociated
salt form WHY?
• Because: the undissociated form is lipid soluble and can
penetrate body membranes, whereas the ionic form is
not lipid-soluble and can only penetrate membranes with
great difficulty.
• Thus, the therapeutic response of weakly basic alkaloids
(used as ophthalmic drugs) increases as the pH of the
solution increases, and hence-concentration of the
undissociated base, was increased.
Mandelic acid, benzoic acid and salicylic acid have
pronounced antibacterial activity in non ionized form but have
no activity in ionized form. Accordingly, these substances
require an acidic pH to function as antibacterial agents. Thus
sodium benzoate is effective as a preservative:
4% concentration at pH 7
0.06 to 0.1% concentration at pH 3.5 to 4
0.02 to 0.03% concentration at pH 2.3 to 2.4
. Influence of Buffer Capacity and pH on
Drug Stability
 Buffer is used to prevent changes in pH due to the
alkalinity of the glass or acidity of CO2 from dissolved
air
 Solutions as Thiamine hydrochloride may be sterilized
by autoclaving without decomposition if the pH is below
5 above this pH thiamine hydrochloride is unstable.
 The stability of emulsions is pH dependent.
PH and solubility.
The influence of buffering on the solubility of the alkaloidal
base:
 At a low pH a base is predominantly in the ionic form
which is usually very soluble in aqueous media
as the pH is raised more undissociated base is formed.
 Therefore, solution should be buffered at sufficiently low
pH.
 Yet, when the solution is instilled in the eye, the tears
participate in the gradual neutralization of the solution
and the conversion of the drug from the physiologically
inactive form to the undissociated base the base can
then readily penetrate the lipoidal membrane.
 As the base is absorbed at the pH of the eye, more of the
salt is converted into base to preserve the equilibrium,
hence the alkaloidal drug is gradually absorbed.
Isotonic solution
is that solution having the same salt concentration and
hence the same osmotic pressure as the red blood cell
contents 
Pharmaceutical solutions that are applied to
delicate membranes of the body should be:
adjusted to approximately the same osmotic
pressure of the body fluids.
Isotonic solution
Cause no swelling or contraction of the tissues with
which they come into contact
produce no discomfort when instilled into the eye, nasal
tract, blood stream or other body tissues.
Method of adjusting Tonicity
Class I Method
1. The cryoscopic method
2. The Sodium Chloride Equivalent method
1. The cryoscopic method
■ NaCl or some other substance is added to the solution
of the drug to lower the freezing point of the solution to
0.52 ° C and thus make it isotonic with body fluids.
■ The freezing point depressions of a number of drug
solutions are found in tables
Substance
M
E
V
Tf1%
Liso
Apomorphine
hydrochloride
312.79
0.14
4.7
0.08
2.6
M is the molecular weight of the drug.
E is the sodium chloride equivalent of the drug.
V is the volume in ml of isotonic solution prepared by
adding water to 0.3 gm of the drug
Tf1% is the freezing point lowering of a 1 % solution of the
drug.
Liso is the molar freezing point depression of the drug at a
concentration approximately isotonic with blood and lacrimal
fluid.
Example: How much NaCl is required to render 100 ml of a 1
% solution of apomorphine HCl isotonic with blood serum?
■ From the Table it is found that a 1 % solution of drug has
a freezing point lowering of 0.08 .
■ To make this solution isotonic with blood, sufficient
NaCl must be added to reduce the freezing point
lowering additional 0.44 to be = to 0.52 (blood serum)
■ 0.52 - 0.08= 0.44
By the method of proportion: (Tf1% of NaCl = 0.58)
1%
= 0.58
X= 0.76%
X
0.44
■ 0.76 % NaCl will lower the freezing point to the required
and will render the solution isotonic.
■ The solution is prepared by dissolving 1.0 g of apomorphine HCl and 0.76 g of NaCl in sufficient water to
make 100 ml of isotonic solution.
2. Sodium Chloride Equivalent Method
The Sodium Chloride Equivalent E of a drug is:
the amount of NaCl which has the same osmotic effect as 1 g
of the drug.
The sodium chloride equivalents E for a number of drugs are listed
in the following table:
ISOTONIC SOLUTIONS
TABLE OF SODIUM CHLORIDE EQUIVALENTS
Substance
Boric acid
Molecular weight
Ions Sodium chloride
equivalent E
61.8
1
0.52
cocaine
hydrochloride
340
2
0.17
ephedrine sulfate
429
3
0.2
rmining the amount of NaCl used to cause a solution isotonic:
ltiply the quantity of each drug in the prescription by it’s sodium chloride
uivalent E
btract this value from the concentration of NaCl which is isotonic with body
ds (0.9 gm per 100 ml)
ample :
olution contains 1 g of ephedrine sulfate in 100 ml.
w much NaCl must be added to make solution isotonic ?
w much dextrose would be required for this purpose?
dium chloride
Cl equivalent (E) of Ephedrine sulfate is 0.2 i.e.
of Ephedrine sulfate is osmotically equiv. to 0.2 g NaCl
g X 0.2 = 0.2 g
g of NaCl is required for isotonicity of 100 ml solution
– 0.2 = 0.7 g of NaCl must be added.
xtrose
use dextrose instead of NaCl for adjusting the tonicity.
Cl equivalent of dextrose = 0.16
1 gm dextrose =
X
0.16 g NaCl
0.7 g NaCl
4.37 g of dextrose
ss I I Method
ite-Vincent Method.
It involves the addition of water to the drugs to make an isotonic solution.
ample :
make 30 ml of a 1 % solution of cocaine HCl isotonic with body fluid.
The weight of the drug is multiplied by the NaCl
equivalent E
0.3 g X 0.17 =0.051 g
This is the quantity of NaCl osmotically equivalent to
0.3 gm of cocaine HCl
is known that 0.9 gm of sodium chloride in 100 ml water yields isotonic
ution. The volume V of isotonic solution which can be prepared from 0.051
of sodium chloride
equivalent to 0.3 gm of cocaine hydrochloride)
w X E X 111.1
0/0.9 = 111.1
the volume in ml of isotonic solution which may be
prepared by mixing the drug with water
= the weight in grams of the drug given in the problem,
= the sodium chloride equivalent obtained from Table
1.1= the volume in ml of isotonic solution obtained by
dissolving 1 gm of Sodium Chloride in water.
0.3X 0.17 X 111.1
5.66 ml
In order to complete the isotonic solution to make
ml of the finished product, enough isotonic sodium chloride
ution, or an isotonicbuffered solution is added.
ample :
Make the following solution isotonic with respect to an
al membrane.
/
Phenacaine hydrochloride
0.06 gm
oric acid
0.30 gm
erilized distilled water to
100 ml
W X E X 111.1
W X E X 111.1 = [(0.06 X 0.17) +(0.3 X 0.52)] X 111.1
18.4 ml
The drugs are mixed with water to make 18.4 ml
an isotonic solution and the preparation is completed with isotonic solution
make 100 ml of the finished product. Isotonic Formulation example:
ke the following solution isotonic assuming an ideal membrane:
enacaine hydrochloride
0.06 g
ric acid
0.30 g
uting buffer to make
100 ml
w x E x 111.1, where wis weight in grams, and E is data from the table.
[(0.06 x 0.20) + (0.30 x 0.50)] x 111.1
18 ml
mponents are mixed with WFI to make 18 ml, diluting buffer is then added
make 100ml
epare the following sterile eye drop:
opine Sulfate
1.1%
osphate Buffered isotonic solution (Ph 7), ad
30.0 ml
e equivalence value of Atropine Sulfate to NaCl is 0.12
ng the sodium chloride equivalent method
yoscopic method
ezing point depression of the drug is found in table
e value is subtracted from 0.52 (of blood)
culate how much NaCl is needed to make it 0.52
Cl equivalent method
antity of drug is multiplied by NaCl equivalent from table
s is subtracted from 0.9g.
e result is the amount of NaCl added to make it isotonic
ite-Vincent
e Volume of isotonic solution which is prepared by the given quantity of
ug when mixed water is calculated by:
V= w x E x 111.1
e volume required is then completed by isotonic solution
For determining the amount of NaCl used to cause a solution
isotonic:
■ multiply the quantity of each drug in the prescription by
it’s sodium chloride equivalent E
■ subtract this value from the concentration of NaCl which
is isotonic with body fluids (0.9 gm per 100 ml)
Example :
A solution contains 1 g of ephedrine sulfate in 100 ml.
How much NaCl must be added to make solution isotonic ?
How much dextrose would be required for this purpose?
Sodium chloride
NaCl equivalent (E) of Ephedrine sulfate is 0.2 i.e.
1 g of Ephedrine sulfate is osmotically equiv. to 0.2 g NaCl
1.0 g X 0.2 = 0.2 g
0.9 g of NaCl is required for isotonicity of 100 ml solution
0.9 – 0.2 = 0.7 g of NaCl must be added.
Dextrose
To use dextrose instead of NaCl for adjusting the tonicity.
NaCl equivalent of dextrose = 0.16
1 gm dextrose =
X
0.16 g NaCl
0.7 g NaCl
X = 4.37 g of dextrose
Class I I Method
White-Vincent Method.
It involves the addition of water to the drugs to make an
isotonic solution.
Example :
To make 30 ml of a 1 % solution of cocaine HCl isotonic with
body fluid.
1. The weight of the drug is multiplied by the NaCl
equivalent E
0.3 g X 0.17 =0.051 g
This is the quantity of NaCl osmotically equivalent to
0.3 gm of cocaine HCl
2. It is known that 0.9 gm of sodium chloride in 100 ml water
yields isotonic solution. The volume V of isotonic solution
which can be prepared from 0.051 gm of sodium chloride
(equivalent to 0.3 gm of cocaine hydrochloride)
V = w X E X 111.1
100/0.9 = 111.1
V = the volume in ml of isotonic solution which may be
prepared by mixing the drug with water
w = the weight in grams of the drug given in the problem,
E = the sodium chloride equivalent obtained from Table
111.1= the volume in ml of isotonic solution obtained by
dissolving 1 gm of Sodium Chloride in water.
V = 0.3X 0.17 X 111.1
V = 5.66 ml
In order to complete the isotonic solution to make
30 ml of the finished product, enough isotonic sodium
chloride
solution, or an isotonicbuffered solution is added.
Example :
Make the following solution isotonic with respect to an
ideal membrane.
R/
Phenacaine hydrochloride
0.06 gm
Boric acid
0.30 gm
Sterilized distilled water to
100 ml
V = W X E X 111.1
V = W X E X 111.1 = [(0.06 X 0.17) +(0.3 X 0.52)] X 111.1
V = 18.4 ml
The drugs are mixed with water to make 18.4 ml
of an isotonic solution and the preparation is completed with
isotonic solution to make 100 ml of the finished product.
Isotonic Formulation example:
Make the following solution isotonic assuming an ideal
membrane:
Phenacaine hydrochloride
0.06 g
Boric acid
0.30 g
Diluting buffer to make
100 ml
V = w x E x 111.1, where wis weight in grams, and E is data
from the table.
V = [(0.06 x 0.20) + (0.30 x 0.50)] x 111.1
V = 18 ml
Components are mixed with WFI to make 18 ml, diluting buffer
is then added to make 100ml
Prepare the following sterile eye drop:
RX
Atropine Sulfate
1.1%
Phosphate Buffered isotonic solution (Ph 7), ad
30.0 ml
The equivalence value of Atropine Sulfate to NaCl is 0.12
Using the sodium chloride equivalent method
Cryoscopic method
Freezing point depression of the drug is found in table
The value is subtracted from 0.52 (of blood)
Calculate how much NaCl is needed to make it 0.52
NaCl equivalent method
Quantity of drug is multiplied by NaCl equivalent from table
This is subtracted from 0.9g.
The result is the amount of NaCl added to make it isotonic
White-Vincent
The Volume of isotonic solution which is prepared by the
given quantity of drug when mixed water is calculated by:
V= w x E x 111.1
The volume required is then completed by isotonic solution