Добавил:
Sekretar
kiopkiopkiop18@yandex.ru
t.me/Prokururor I Вовсе не секретарь, но почту проверяю
Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз:
Предмет:
Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_5708_Библиотеки_им_академика_М_И_Перельмана.pdf
227
Solution
Table 8.3
Critical solution temperature of some binary systems and their composition at CST
Components Upper CST
(˚C)
Lower CST
(˚C)
Weight percent of
B at CST
AB
Water Isobutyl alcohol 134.0
Water Phenol 68.1
36.1
Cyclohexane Methanol 49.0
29.0
Methanol Carbon disulphide 40.5
80.0
Aniline Hexane 59.6
48.0
Water Aniline 167.0
58.0
Isopentane Phenol 63.5
49.0
Water Triethyl amine
18.5
Glycerine
mToludine
120.0 6.7
Water Nicotine 208.0 60.8 32 and 22
Experiment 6 Determine CST and composition of solution at the CST
for the phenolwater system.
Requirement
Test tubes, cork carrying thermometer, wire stirrers, 50 mL beaker for bath
liquid (water), graduated pipette, weighing bottel, burner, liquid phenol (80%)
phenol), and distilled water.
Caution
Phenol is highly corrosive which produces painful burns if brought in contact
with the skin. In case of accidental contact with phenol or its solution, wash
the exposed part throughly with water and then apply glycerine or vaseline.
Procedure
1. Prepare different mixture of phenol and water by mixing different volumes
of phenol (80%) and water as given in table here and find the micibilty
temperature of each composition by following the procedure given below.
Test tube no 12345678910
Phenol solution (mL) 10987654321
Water (mL) 0123456789
Phenol % (wt/v) 80 72 64 56 48 40 36 32 28 24
2. Place enough of water in a 250 -mL beaker and place it on tripot stand
and heat it slowly.
228
Practical Pharmaceutical Analytical Chemistry
3. Take test tube no. 1 and check whether the two componets are already
completely micible or not, if they are completely micible then note the
temperature of mixture by placing thermometer in the solution.
4. If the two components are not already completely miscible, stir them by
the stirring wire or a glass rod.
5. A cloudiness will appear in such mixtures. Place the mixture in beaker
containing water and raise the temperature of water bath gradually while
stirring the mixture.
6. Note down the temperature at which last cloudiness disappears.
7. Remove the flame from under the beaker while continuing the stirring.
Allow the temperature of water bath to decrease gradually.
8. Note the temperature at which cloudiness just reappears.
9. The mean of two temperatures is considered as miscibility temperature of
the mixture.
10. Repeat the experiment to record the concordant reading of the micibility
temperature of a mixture test tube no. 1.
11. Take test tube no. 2 and repeat steps 4–10 and record the micibility tem-
perature of the mixture.
12. Repeat the experiment (steps 4–10) for all other test tubes to record the
concordant reading of the micibility temperature of all mixtures in rest of
the test tubes.
13. After the experiment, transfer all the mixtures to a used phenol bottle kept
in the laboratory for recovery of phenol by fractional distillation.
14. Plot a graph between percentage composition of mixture along the X-axis
and the micibility temperature along the Y-axis.
15. Draw a smooth curve through the points and mark the highest point
on the curve visually and by joining the line of mid-points of the tie
lines.
16. Read the temperature and composition of the mixture at the maximum
point from the graph.
17. The temperature for the maximum point is the CST for the phenol-water
system.
Observation and Calculations
Room temperature _________ °C
Density of water at room temperature _________ g/mL
Weight of phenol taken in test tube A _________ g
Weight of phenol taken in test tube B _________ g
229
Solution
Observation
S. No. Weight of
phenol
Weight of water
(Volume u density)
Percentage
of Phenol
Mean Micibility
temperature of
mixture
1
2
3
4
5
6
7
Result The CST of phenol water system was found to be ______˚C and the
concentration of phenol at CST was found to be ______ %
Experiment 7 Study of CST and composition of solution at the CST for the
aniline and water systems.
Requirment
Test tubes, cork carrying thermometer, wire stirreres, 50 mL beaker for bath liq-
uid (water), graduated pipe, weighing bottle, burner, aniline, and distilled water
Procedure
The experiment is carried out in a similar fashion as Experiment 6.
Result The CST of anilinewater system was found to be _____˚C and the
concentration of aniline at CST was found to be _____ %
Experiment 8 Study the effect of presence of impurity in water on CST of
the phenolwater system.
Requirment
Test tubes, cork carrying thermometer, wire stirrers, 50 mL beaker for bath liq-
uid (water), graduated pipe, weighing bottle, burner, phenol (80%), and distilled
water.
Theory
The effect of impurity present in a component on CST of the system study
is studied by adding impurity like any salt in one phase and then determining
the CST of the system. Critical solution temperatures are very sensitive to the
230
Practical Pharmaceutical Analytical Chemistry
presence of impurities soluble in one or both the components of the system.
If an impurity is soluble in one component only, the upper CST of the system
increases while the lower CST decreases. If the impurities are soluble in both the
components of the system, the upper CST of the system decreases whereas the
lower CST rises. The change in the CST of system has been found to be linearly
proportional to the percentage of impurity.
In similar fashion the micibility temperature of fixed proportions of compo-
nents is affected by the presence of impurity. This property has been utilized to
determine the amount of impurity in the system.
Procedure
1. Take seven test tubes and add 5 mL of 80% phenol solution in each.
2. Prepare a stock solution of 1% NaCl by dissolving g in 10 mL of distilled
water.
3. Prepare 1 mL each of 0.2, 0.4, 0.6, and 0.8% NaCl solution by mixing
2, 4, 6, mL of 1% NaCl solution from a graduate pipette or burette with 8,
6, 4 and mL, respectively of distilled water.
4. Place 5 mL of 0.2, 0.4, 0.6, 0.8, 1% NaCl solution and pure water, respectively
into six of test tubes and 5 mL of the test solution in the seventh test tube.
5. Determine the micibility temperature of all the seven mixtures and plot a
graph between the percentages of salt in aqueous solutions used (X-axis)
and the complete micibility temperatures (Y-axis).
6. Draw a line through the points and from the graph read the composition
corresponding to the complete miscibility temperature observed with the
unknown solution.
Observation
S. No. Percentage of NaCl Miscibility temperature (˚C)
1 0.0
2 0.2
3 0.4
4 0.6
5 0.8
61
7 Unknown
Result The presence of impurity affects the CST of phenol water system.
Percentage of salt in the solution of unknown percentage ______%
231
Solution
8.1.6 Distribution of Solute in Two Adjacent Phases
Partition coefficient, unlike solubility, measures the differential ability of a solute to
dissolve in two insoluble phases. Most commonly the partition coefficient is meas-
ured in octanol and water—the so-called K
OW
. Almost all molecules will dissolve
at least marginally in both phases. The use of a separatory funnel with diethyl ether
and water is a very well-known example of using partition coefficient to purify
an organic compound. Partition coefficient is measured as the logarithmic ratio of
the concentration of the compound in the two phases. The value is a measure of
hydrophobicity and hydrophilicity of the solute. The value is commonly used in
medicinal applications, where it is a predictor of the ability to penetrate cells and
the skin. When an excess of solute is added to a mixture of two immiscible liq-
uids, the solute which is slightly soluble in the liquids distributes itself in both the
liquids/phases according to its solubility in each liquid till the two liquids/phases
become saturated. If the solute added is not sufficient to saturate the two liquids/
phases it will still distribute itself to the two phases in a definite concentration
ratio. This was first observed by Berthelot and Jangfleish in 1872 while performing
the solubility study of iodine in water and carbon disulphide. They observed that
the iodine distributed itself between the two phases and the concentration in two
layers was constant irrespective of amount of iodine added at given temperature.
If C
1
and C
2
are the concentrations of solute in two phase (solvents 1 and 2)
in contact with each other at constant temperature, the equilibrium expression
can be written as
CC K
12
/
D
11
The equilibrium constant K is termed as the distribution ratio, distribution
coefficient or partition coefficient. Equation (1) is known as distribution law.
Values of K are unitless.
The partition coefficient is a ratio of concentrations of compound in similar
molecular species between the two solutions. To measure the partition coefficient
of ionizable solutes, the pH of the aqueous phase is adjusted such that the pre-
dominant form of the compound is unionized. The logarithm of the ratio of the
concentrations of the solute in the solvents is called log P.
The following factors affect distribution law:
1. Temperature: Solubility of any substance in a solvent changes with change
in temperature but the magnitude of change is different for different
solvents. Therefore, while performing partition coefficient determina-
tion experiment temperature should be maintained constant by placing
the system in a thermostat.
2. Existence of similar species: It was observed that distribution law hold true
for only similar kind of species, i.e. if a solute associates or dissociates
232
Practical Pharmaceutical Analytical Chemistry
in any of the solvent it no longer remain in same form as added to the
system. For example, benzoic acid distribution in water and benzene,
in acidified water benzoic acid exists as single molecule C
6
H
5
COOH,
however in benzene it associates and forms dimer (C
6
H
5
COOH)
2
.
3. The solution is dilute.
4. The two liquids are mutually immiscible: The solubility of two layers should
not change the miscibility of two phases.
5. Solute does not interact with any of the solvents.
Distribution law is stated as follows: When a solute is added to two phases
immiscible with each other and in contact with each other, the solute distributes
itself between two phases in such a way that there exists, for similar molecular
species, at constant temperature, a constant ratio of distribution irrespective of
the concentration of solute and presence of any other species.
Knowledge of partition coefficient is important in different phases of drug
design and development. These include absorption and distribution of drugs in
body, preservation of oil–water systems, drug action at non-specific sites, etc.
Experiment 9 Determine the partition coefficient of succinic acid between
ether and water at room temperature.
Requirements
4–5% succinic acid solution in carbon dioxide free water, ether and distilled
water, 0.05 N NaOH solution, phenolphthalein, 250 mL three stoppered conical
flasks, 100 mL conical flasks, burette, pipette, and thermometer.
Procedure
1. Weigh 5 g of succinic acid and dissolve in 12 mL of water.
2. Take three stoppered conical flasks and label them as 1, 2 and 3.
3. Place 30, 40 and 5 mL of succinic acid solution in flasks 1, 2 and 3, respec-
tively.
4. Add 20 and 10 mL of water to flasks 1 and 2, respectively, to make total
volume of 50 mL in each conical flask.
5. Now add 50 mL of ether to each conical flask, cork them and, shake them
vigorously.
6. Place all the three bottles in a thermostat or trough of water at room tem-
perature.
7. Record the temperature of water bath.
8. Let the conical flasks stand in the thermostat for about half an hour while
shaking them after every 4- min in turns.
9. After final shaking, swirl the bottles so that any drops of liquid adhering to
sides of the conical flasks settle down.
233
Solution
10. Take flask 1 and separate the two layers in two conical flasks by using sepa-
rating funnel.
11. Pipette out 1 mL of aqueous layer of flask 1 in a 100 mL conical flask and
titrate with 0.0 N NaOH using phenolphthalein as an indicator.
12. Record concordant reading for flask 1 and repeat the procedure for flasks 2 and 3.
13. Now pipette out 1 mL of ether layer in a 100 mL conical flask containing 2 mL
of water and titrate with 0.0 N NaOH using phenolphthalein as indicator.
14. Record concordant reading for all the three flasks 1, 2 and 3.
Observations and Calculations
Temperature ______˚C
Titration of Upper layer (aqueous layer)
Flask Volume taken
(mL)
Burette reading Concordant
reading
123
Flask 1
Flask 2
Flask 3
Titration of Lower layer (organic layer)
Flask Volume taken
(mL)
Burette reading Concordant
reading
123
Flask 1
Flask 2
Flask 3
Concentration (mol/L) of succinic acid in two layers is calculated as follows:
For aqueous layer (water layer)
Concentration of succinic acid in water in flask 1:
N
1
V
1
N
2
V
2
N
1
0.05 u V
2
10
Concentration of succinic acid in water (C
aq
)
C
aq
(0.01 u V
2
u 118)/10 mol/L
Similarly calculate the concentration of succinic acid in water in flasks 2 and 3.
234
Practical Pharmaceutical Analytical Chemistry
For organic layer (ether layer)
Concentration of succinic acid in ether layer in flask 1
11
22
N
1
0.05 u V
2
10
Concentration of succinic acid in ether layer (C
org
)
C
org
(0.01 u V
2
u 118)/10 mol/L
Similarly, calculate the concentration of succinic acid in water in flasks 2 and 3.
Partition coefficient K C
org
/C
aq.
Since, partition coefficient is the ratio of
concentration of solute in two immicible liquids, therefore the result from all the
three flasks should lie in close range.
Result The partition coefficient of succinic acid between ether and water is
______ irrespective of different concentration in the flask. Suggesting that parti-
tion coefficient is constant irrespective of concentration at a given temperature.
Experiment 10 Determine the partition coefficient of iodine between carbon
tetrachloride and distilled water at room temperature.
Requirements
Iodine, carbon tetrachloride, distilled water, 0.0 N sodium thiosulphate solution,
starch, three 250 mL stoppered conical flasks, 100 conical flasks, burette, pipette,
and, thermometer
Procedure
1. Prepare 50 mL saturated iodine solution in carbon tetrachloride.
2. Take three stoppered conical flasks clean them and label as 1, 2 and 3.
3. Place 25, 15 and 5 mL of iodine solution in flasks 1, 2 and 3, respectively.
4. Add 10 and 20 mL of carbon tetrachloride to flasks 2 and 3, respectively, to
make total volume of 25 mL in each conical flask.
5. Now add 100 mL of distilled water to each conical flask and cork them,
shake them vigorously.
6. Place all the three bottles in a thermostat or trough of water at room tem-
perature.
7. Record the temperature of water bath.
8. Let the conical flasks stand in the thermostat for about half an hour while
shaking them after every 4- min in turns.
9. After final shaking, swirl the bottles so that any drops of liquid adhering to
sides of the conical flasks settle down.
10. Take the flasks one by one and separate the two layers in two conical flasks
by using separating funnel.
235
Solution
11. Pipette out 10 mL of aqueous layer in flask 1 of a 100 mL conical flask
and titrate with 0.01 N sodium thiosulphate using starch as indicator (add
starch 2–3 drops towards end point of the titration).
12. Record the titration value and take concordant reading by repeating step
and 11 for flask 1.
13. Similarly, titrate the aqueous layer from flasks 2 and 3.
14. Now pipette out 1 mL of organic layer in a 100 mL conical flask containing
2 mL of water and titrate with 0.0 N sodium thiosulphate using starch as
indicator (add starch 23 drops towards end point of the titration).
15. Record the titration value and take concordant reading by repeating step
14 for flask 1.
16. Similarly, titrate the organic layer from flasks 2 and 3.
1 7. Calculate the concentration of iodine in both phases, i.e. aqueous and
organic phase.
Observations and Calculations
Temperature _____˚C
Titration of Upper layer (aqueous layer)
Flask Volume taken
(mL)
Burette reading Concordant
reading
123
Flask 1
Flask 2
Flask 3
Lower layer (organic layer)
Flask Volume taken
(mL)
Burette reading Concordant
reading
123
Flask 1
Flask 2
Flask 3
Gram equivalent weight of iodine is 12 g.
Calculate concentration (mol/L) of iodine in two layers as in Experiment 9.
Partition coefficient K C
org
/ C
aq
Result The partition coefficient of iodine between carbon tetrachloride and
water is ______ and is constant irrespective of different concentrations in the flask.
236
Practical Pharmaceutical Analytical Chemistry
Experiment 11 Study distribution of succinic acid between distilled water
and benzene.
Requirements
Succinic acid, benzene, distilled water, 0.0 N sodium hydroxide solution, phenol-
phthalein, and apparatus same as in Experiment 10.
Procedure Same as in Experiment 10.
Observations and Calculations
Temperature ______˚C
Titration of Upper layer (aqueous layer)
Flask
Volume taken
(mL)
Burette reading
Concordant
reading
123
Flask 1
Flask 2
Flask 3
Titration of Lower layer (organic layer)
Flask Volume taken
(mL)
Burette reading Concordant
reading
123
Flask 1
Flask 2
Flask 3
Concentration (mol/L) of succinic acid in two layers is calculated as in
Experiment 10.
Result The partition coefficient of succinic acid between benzene and dis-
tilled water is ________.
8.1.7 Effect on Partition Coefficient due to Association
or Ionization of Molecules
The distribution law is applicable to only similar kind of species in both phases.
But a solute may associate or dissociate in any of the solvents (Fig. 8.4).