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207
Properties and Molecular Constitution of Substances
Refractive index of the liquid, h
D
____
Density of the liquid at room temperature, r ____ gm/cm
2
Specific refractivity R, by using the formula:
R =
−
+
×
η
ηρ
2
2
1
2
1
where h is the refractive index of liquid observed and r is the density of the
liquid in gm/cm
2
.
Result The refractive index of given liquid samples (a and b) was found to
be ____ and ____cm
2
/g, respectively.
Experiment 5 Determine the molar refractivity of methyl alcohol, ethyl
alcohol, n-hexane and chloroform and calculate the atomic refractivities of C,
H and Cl.
Requirement
Pure sample of methyl alcohol, ethyl alcohol, n-hexane and chloroform, Abbe’s
refractometer, dropper, specific gravity bottle, cotton, alcohol, acetone, filter paper.
Procedure
1. Determine the refractive index of the samples as in Experiment 1.
2. Calculate the specific refractivities of given liquids as mentioned.
3. Calculate the molecular refractivities of the liquids by multiplying their
specific refractions by the respective molar masses.
Specific refraction R may be calculated from
R −
−
+
×
η
ηρ
2
2
1
1
1
and molar refractivity from
R
M
R u M
where M is the molecular mass of the liquid, r is the density and n is the observed
refractive index of liquid.
Observations
Liquid Observed
refractive index (n)
Density (
r)
RM R
M
Methyl alcohol
Ethyl alcohol
Chloroform
n-Hexane
208
Practical Pharmaceutical Analytical Chemistry
Calculations
The atomic refractivities can be calculated from molar refractivities of the
compounds as:
[R]CH
3
OH [R] CH
3
CH
2
OH Refraction of CH
2
[R]C
6
H
14
[6 u Refraction of CH
2
] Refraction of 2H
Refraction of CH
2
[2 u Refraction of H] Refraction of C
[R]CHCl
3
Refraction of CH Refraction of 3Cl
1. Molecular refractivity of CH
2
group
2. Atomic refractivity of C
3. Atomic refractivity of H
4. Atomic refractivity of Cl
Result The molar refractivity of methyl alcohol, ethyl alcohol, n-hexane and
chloroform and was found to be ____, ____ and ____, respectively and the
atomic refractivities of C, H and Cl were calculated as ____, ____, and ____,
respectively.
Experiment 6 Determine the composition of mixture of sample liquids
(A and B) by using Abbe’s refractometer.
Requirement
Pure sample liquids A and B, Abbe’s refractometer, test tube, pipettes, dropper,
specific gravity bottle, cotton, alcohol, acetone, filter paper.
Procedure
1. Prepare five mixture of A and B between the pure liquids by taking different
volumes of A and B as given in the table.
2. Record their refractive indices as in Experiment 4.
3. Record the refractive indices of sample mixture provided by following same
procedure.
4. Plot a graph between compositions of liquids A and B on X-axis and
refractive index on Y-axis.
209
Properties and Molecular Constitution of Substances
Observations
Room temperature: ____˚C
Entry Components
A (mL) B (mL)
Refractive indices
A 10 0
182
264
355
446
528
B010
Sample ? ?
Read the compositions of A and B from the graph.
Result The composition of the given mixture of A and B was found to
contain ____% of A and ____% of B.
Experiment 7 Find the composition of two solutions of sugarcane in water (A
and B) with less than 20% sugar using a refractometer. Assuming that the interac-
tion between sugar and water molecules can be ignored, calculate the refractive
index of cane sugar from refractive indices of 5, 10 and 20% sugar solution.
Requirement
Sugar solutions between 0% and 20%, Abbe’s refractometer, dropper, specific
gravity bottle, cotton, alcohol, acetone, filter paper, 25 mL graduated flasks.
Procedure
1. Prepare 20, 18, 16, 14 and 12% sugar solutions by directly weighing 10, 8, 6,
5, 4 and 3% by dilution in 25 mL graduated flasks and make up volume by
distilled water.
2. Find the refractive indicies of the solutions as measured in the previous
exercises and determine the densities of 20, 10 and 5% solutions.
3. For densities, weigh the 25 mL volumetric flask empty and with enough
sugar to make 20, 10 and 5% solutions and then after making up the solution
volumes upto the marks, weigh them again.
4. This will give the densities of solutions, weights of water in them and the
mole fraction of water and sugar in them.
R
M
value of a mixture can be calculated from the refractive index, density and mole
fraction of components using Equation (19). We can calculate the R
M
value of
210
Practical Pharmaceutical Analytical Chemistry
cane sugar. From this R
M
value, refractive index of sugar crystals can be calculated
from a graph of refractive indices verses sugar percentage in solutions. Percentages
of sugar in solution X and Y can be calculated using their refractive indices.
R
M
d
M
=
−
+
×
I
I
2
2
1
2
R
XM XM
d
M ()
()
()
mix
mix
mix
AA BB
mix
=
−
+
×
+I
I
2
2
1
2
RXRXR
MMM() () ()mix A A B B
=+
Observations
Entry Percentage of sugar Refractive indices
120
218
316
414
512
610
78
86
95
10 4
11 3
12 0
X ?
Y ?
Densities
Percentage of
sugar
Mass of
sugar (g)
Mass of
water (g)
Mass of 25 mL
solution (g)
Density
00
5 1.25
10 2.50
20 5.00
Plot a graph of refractive indices verses percentages and read the percentage of
sugar in X and Y.
211
Properties and Molecular Constitution of Substances
Observations
Room temperature _____ °C
Refractive index of water at room temperature ______
S. no Concentration (%) Density Refractive index Mean
ABC
1 0.1
2 0.2
3 0.3
4 0.4
5 0.5
6 0.6
7 0.7
8 0.8
9 0.9
10 1.0
Calculate the mole fraction of sugar and water in 20, 10 and 5% solutions.
Result The refractive index of cane sugar was found to be ____ and
percentage composition.
Experiment 8 Determine the effect of concentration on refractive index
using Abbe refractometer.
Requirement
Sodium benzoate, equipments as in Experiment 4.
Prodcedure
1. Prepare a stock solution of sodium benzoate of 1% by dissolving 1 gm of
sodium benzoate in distilled water and making up the volume upto 100 mL
in a 100 mL volumetric flask.
2. Prepare different dilutions (0.1, 0.2, …, 0.9%) by aligation method from the
stock solution.
3. Determine the refractive index of all the solutions and stock solution as
Experiment 4.
4. Determine the viscosity of substance at different dilutions.
Plot a graph between concentrations and refractive index of the dilutions.
Result The effect of concentration was studied and ____ is the concentration
at which highest refractive index was observed.
212
Practical Pharmaceutical Analytical Chemistry
Judge yourself
1. Define the angle of rotation, specific, molecular and intrinsic rotations.
2. Write the equation for calculation of specific and molecular rotations.
3. Which of the following molecules cannot be optically active:
(i) H
2
(ii) CH
4,
CHClBr
2
(iii) CHClBrF (iv) C
3
H
6
Br
2
(v) CH
3
CH-
ClCH
2
Cl
4. What do you understand by chiral centre?
5. When does a molecule show optical activity?
6. What is the difference between dextrorotatory and levorotatory
substances?
7. How can you follow the rate of reaction by optical activity measurements?
8. What are the factors that affect optical rotation measurement?
9. What is the importance of intrinsic rotation calculation?
10. Define the refractive index and molar refractivity.
11. Which method is used to determine the refractive index of a compound?
12. Explain Snell’s law of refraction.
13. What are the factors that affect refractive index of a substance?
14. What do you understand by the critical angle?
15. What is the source of incident light in Abbes refractometer?
16. How is refractive index of solutions determined?
17. Describe the principle of Abbes refractometer.
18. Name various instruments used to measure refractive index.
Chapter
8
Chapter
Solution
Solution is defined as a homogenous mixture of two or more substances within
limits. A solution has two components, solvent and solute. Solvent is generally that
part of a solution which is in larger proportion and solute is the one in lesser
proportion.
Saturated solution is a solution in which the substance in solution is in equi-
librium with the solid phase at a particular temperature; that is, a soluble solute
is in contact with the excess of solute. A saturated solution is generally prepared
by allowing a supersaturated solution to cool down to the required temperature.
Supersaturated solution Supersaturated solution contains more solute than
required for the preparation of the saturated solution at a definite temperature.
A supersaturated solution is prepared by heating the solution.
Expression of concentrations of solution: Concentrations of solution are generally
expressed as:
Molarity
: Number of moles of the solute present in 1 L of the solution
Molality: Number of moles of the solute present in 1000 g of solvent.
Normality: Number of gram equivalents of the solute present in 1 L of the
solution.
Mole fraction: Ratio of number of moles of solute present in solution to the
total number of moles of solute and solvent present in the solution.
8.1 Solubility
Solubility is an important phenomenon in chemistry and pharmaceutical sci-
ence. It plays an effective and prominent role in chemical reactions and in the
formulation of dosage forms. Solubility is defined as the amount of solute dis-
solved in a specified amount of a solvent to form a saturated solution at a given
temperature. In other words solubility is the concentration of the solute in a
saturated solution at a certain temperature. Also, solubility is defined a measure
214
Practical Pharmaceutical Analytical Chemistry
of the maximum amount of solute that can be clearly dissolved into a solvent at a
given temperature and pressure. A solute will have different degree of solubilities
in different solvents depending upon many factors like nature of the solute and
solvent, temperature, etc.
8.1.1 Factors Affecting Solubility
Certain factors affect solubility of a solute in a particular solvent. Also, polarity
and temperature play a major role among other factors.
Type of solute and solvent: Chemically, similar type of compounds will
have more solubility than dissimilar compounds. Commonly, it is stated
that 'like dissolve like'. Therefore, polar compounds will be soluble in
polar solvents and nonpolar in nonpolar solvents. The solubility can be
expressed in percent by weight at a specific temperature.
s Temperature affects solubility of almost all compounds: Dependency of solubil-
ity has been referred to as temperature coefficient. Some solvents and solutes
show little temperature dependence, such as NaCl, which has solubility
in water of 26.3% by weight at 25°C and 27.7% by weight at 100°C
whereas others like glucose show more temperature dependence, which
has a solubility of 47.6% by weight at 25°C and 84% by weight at 100°C.
Some substances are infinitely soluble, like water and ethanol; still others
have vanishing small solubility like silver chloride in water. There really
is no completely insoluble material since that would require zero dis-
solution.
s
pH: Certain compounds are soluble at a particular pH like complex of
various metal are soluble at basic pH. The solubilities of many substances
depend on the pH of the solution. For example, the solubility equilib-
rium of calcium hydroxide, Ca(OH)
2
:
Ca(OH) (s) Ca (aq)+2OH (aq)
2
2+
←→⎯⎯⎯
−
(1)
By increasing the pH, i.e., adding OH
-
shifts the equilibrium from right
to left, decreasing the solubility of Ca(OH)
2
or by increasing H
ions
or decreasing the pH will shift the equilibrium to the right, and the
solubility of Ca(OH)
2
increases. This is because the H
ions remove
OH
ions from the right-hand side of the equilibrium expression to
form water, and according to Le Chatelier’s principle the equilibrium
will shift toward the side where something is removed.
s Common ion effect: Addition of common ion will result in precipitation of
less soluble ion. This is explained earlier under pH.
215
Solution
In general, solubility of solids in liquids is given as soluble and sparingly soluble
while the solubility of liquids in liquids is given as three types, viz.
1. completely miscible liquids,
2. partially miscible liquids,
3. Immiscible liquids.
8.1.2 Solubility of Solids in Liquids
Experiment 1 Determine the solubility of an inorganic salt at different
temperatures.
Requirement
Potassium chloride, sodium chloride, potassium sulphate, thermometer, porce-
lain dishes, beaker (50, 100, 250 mL), pipette, hot plate or water bath.
Procedure
1. Clean all glasswares using chromic acid, detergent and finally rinse thor-
oughly with water.
2. Take 50 mL of distilled water in 100 mL beaker.
Expression of Solubility
1. %w/w (weight by weight)
2. %w/v (weight by volume)
3. %v/v (volume by volume)
Solubility of drugs is expressed in various units in the Merck Index. From experi-
ence, we are aware that ethyl alcohol and water are miscible in all proportions
whereas water and mercury are completely immiscible for all practical purposes.
Between these two extremes lies a whole range of systems that exhibit partial
miscibility (Table 8.1).
Table 8.1 Terms used for solubility
Term
Parts of solvent required for one part
of solute
Very soluble Less than 1 part
Freely soluble
110 parts
Soluble
1030 parts
Sparingly soluble
30100 parts
Slightly soluble
1001000 parts
Very slightly soluble
100010,000 parts
Practically insoluble More than 10,000 parts
216
Practical Pharmaceutical Analytical Chemistry
3. Add some amount of any one salt, e.g. potassium chloride and stir with a
glass rod.
4. Place the beaker on hot plate on water bath and heat upto 90˚C with con-
tinuous stirring.
5. Maintain temperature for a few minutes and then take the beaker contain-
ing salt solution off from the heat.
6. Weigh empty porcelain dish and note the weight of porecelain dish. Pipette
out 1 mL of the salt solution taking care that no solid enters the pipette (this
can be done by tieing cotton on the tip of pipette and removing the tip
before discharging the contents in porcelain dish).
7. Allow the solution to cool to 80˚C and again pipette out 1 mL of the salt
solution as in step 6.
8. Pipette out 10 mL of the salt solution at temperature 70, 60, 50, 40, 30˚C
and room temperature in porcelain dish.
9. Evaporate the solution of each porcelain dish on flame or hot plate.
10. Dry the solution till constant weight.
11. Take weight of the dish and calculate weight of solid.
Observations and Calculation
Observation
S.
No.
Temperature
(˚C)
Weight
of
empty
dish
W
1
g
Weight
of
dish
Solution
W
2
g
Weight
of
dish
residue
W
3
g
Weight
of
residue
W
3
W
1
g
Weight
of
solvent
W
2
W
3
g
Solubility
in g/100 g
190
280
370
460
550
640
730
8RT
Solubility of solvent at a temperature (80˚C) in g/100 g
WW
WW
3
−
−
×
1
23
100
(2)