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Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_5986_Библиотеки_им_академика_М_И_Перельмана.pdf
217
Solution
Result Solubility of given salt at various temperatures, viz. 90, 80, 70, 60, 50,
40, 30˚C and room temperature _____, _____, _____, _____, _____, _____,
_____, and _____ g/10 g.
Experiment 2 To determine the solubility of benzoic acid at room temperature
and a temperature above and below the room temperature by volumetric method.
Requirements
Benzoic acid, sodium hydroxide, hydrochloric acid, phenolphthalein, distilled
water, thermostat, thermometer, conical flask, beaker (50, 100, 250 mL), and pipette.
Procedure
1. Clean all glasswares using chromic acid, detergent and finally rinse
thoroughly with water and dry.
2. Take 5 mL of distilled water in a 100 mL beaker and heat to 50˚C.
3. Add enough of benzoic acid and stir with a glass rod to make saturated
solution (some acid should remain undissolved).
4. Maintain temperature for few minutes and then take the beaker containing
benzoic acid solution off from the heat.
5. Allow the solution to cool to 40˚C and pipette out 1 mL of the acid solu-
tion taking care that no solid enters the pipette (this can be done by using
pipette attached to a cotton guard plug for preventing any solid particles
of acid entering the pipette and removing the tip before discharging the
contents in conical flask).
6. Meanwhile standardize sodium hydroxide solution using phenolphthalein
as an indicator. Record concordant reading.
7. Collect the acid solution in 10 mL conical flask and add a few drops of
phenolphthalein and titrate with 0.0 N NaOH.
8. Wash the pipette thoroughly with warm water and repeat the titration for
concordant reading.
9. Cool down the acid solution to room temperature and again pipette out
1 mL of the acid solution as in step 5.
10. Repeat steps 6 and 7 and record the readings.
11. Cool down the acid solution to 15˚C by placing the solution in ice bath
and again pipette out 1 mL of the acid solution as in step 5.
12. Repeat steps 6 and 7 and record the readings.
Observations and Calculations
Equivalent weight of benzoic acid 122
Room temperature ______˚C
218
Practical Pharmaceutical Analytical Chemistry
Temperature Initial reading Final reading
Volume of 0.02 N
NaOH consumed
Solubility of
benzoic acid
g/100 mL
Temperature
above room
temperature
(
_____˚C)
Room
temperature
(
_____˚C)
Temperature
below room
temperature
(
_____˚C)
Observations
Normality of benzoic acid
Benzoic acid solution NaOH
11
2
3
N
1
N
2
V
2
/V
1
(4)
where N
1
is the normality of benzoic acid solution, V
1
is the volume of benzoic
acid solution used, N
2
is the normality of sodium hydroxide solution and V
1
is
the volume of sodium hydroxide solution consumed. Solubility of benzoic acid
per 100 mL of solution (S ) at temperature above room temperature is deter-
mined by the following formula.
S
NE
=
×
1
100
1000
g/100m
l
(5)
Similarly, calculate the solubility of benzoic acid at room temperature and
temperature below room temperature.
Result Solubility of benzoic acid at room temperature ( _____˚C) was found
to be _____ while solubility at a temperature above ( _____˚C) and below
room temperature ( _____˚C) was found to be _____ and _____, respectively.
Experiment 3 Determine the heat of the solution of a substance by solubil-
ity method.
219
Solution
Requirements
Same as Experiment 2.
Procedure and Observations
Perform the experiment at two different temperatures as in Experiment 2 and
calculate the solubility at two different temperatures.
Calculations
Suppose solubility of benzoic acid at two different temperatures T
1
and T
2
is S
1
and
S
2
, respectively. The heat of the solution is determined by the following equation:
log log
.
10 2 10 1
1
2 303
SS
H
T
−=
+
−
Δ
R[(1/ ) (1/ )
12
where 'H is heat of the solution.
Result The heat of solution of benzoic acid __________ cals/mol
Experiment 4 Study the effect of temperature on solubility of paracetamol
in distilled water.
Requirements
Paracetamol, distilled water, dissolution apparatus or mechanical or magnetic
stirrer, thermostat, and UV–visible spectrophotometer
Procedure
1. Clean the dissolution apparatus thoroughly with water.
2. Place about 250 mL of distilled water in the container and add excess of
paracetmol.
3. Increase the temperature with constant stirring and pipette out 10 mL of
the sample solution.
4. After each withdrawal of sample add 10 mL distilled water to keep the
volume constant.
5. Record temperature and measure the absorbance by spectrophotometer
at 245 nm.
6. Note down the absorbance against the temperature at which sample was
withdrawn.
7. Check the increasing temperature and pipette out 10 mL of the sample
solution after every 10˚C rise in temperature.
8. Repeat steps 6 and 7, till the temperature reaches 105˚C.
9. Plot a graph between temperature and solubility of the paracetamol and
observe the effect of temperature on solubility of paracetamol in distilled water.
220
Practical Pharmaceutical Analytical Chemistry
Observations and Calculations
Observation
Temperature (˚C) Volume of sample used Absorbance
RT (25)
35
45
55
65
75
85
95
105
115
125
Result As indicated by absorbance the solubility of paracetamol _____
______________ with increase in temperature
8.1.3 Solubility of Sparingly Soluble Solids in Liquids
The equilibrium between solids and ions is different from the equilibrium
between gases. The equilibrium between solids and ions is known as phase
equilibrium [e.g. NaCl (aq)].
8.1.4 Solubility Rules of Ionic Solids in Water
Solubility is a result of an interaction between polar water molecules and the
ions. The extent of formation of a solution is determined by the following two
forces:
1. Force of attraction between H
2
O molecules and the ions of the solid: This force
tends to bring ions into solution. If this is the predominant factor, then the
compound may be highly soluble in water.
2. Force of attraction between oppositely charged ions: This force tends to keep the
ions in the solid state. When it is a major factor, then water solubility may
be very low.
221
Solution
But it is not easy to estimate the relative magnitudes of these two forces or to
quantitatively predict water solubilities of electrolytes. Therefore, it is easier to
refer to a set of generalizations, sometimes called “solubility rules” that are based
upon experimentation.
Solubility Rules
NO
3
− All nitrates are soluble.
Cl
- All chlorides are soluble except AgCl, Hg
2
Cl
2
, and PbCl
2
.
SO
4
2
- Most sulphates are soluble, exceptions include BaSO
4
, PbSO
4
, and
SrSO
4
.
CO
3
2
- All carbonates are insoluble except NH
4
and those of the Group 1
elements
OH
- All hydroxides are insoluble except those of the Group 1 elements,
Ba(OH)
2
, and Sr(OH)
2
. Ca(OH)
2
is slightly soluble.
S
2
- All sulfides are insoluble except those of Groups 1 and 2 elements and
NH
4
.
These rules give us a qualitative sense of whether a compound will have a high or
low solubility in water. By considering solubility equilibria, quantitative predic-
tions about the amount of a given compound that will dissolve can be made. Also,
the factors that affect solubility can be analyzed by the use of these equilibria.
A saturated solution of a slightly soluble salt in contact with undissolved salt
involves an equilibrium as
CaF s Ca aq 2F aq
2
2+
()
←→⎯⎯⎯
() ()
−
(7)
Ca OH s Ca aq 2OH aq
2
2
()()
←→⎯⎯⎯
() ()
−
(8)
In the equilibrium constant expression for a heterogeneous equilibria, the
concentrations of pure liquids and solids is generally ignored and can be written as
K
sp
2+
2
Ca F or
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
−
(9)
K
sp
2
2
Ca OH a
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
−
(10)
This equilibrium constant is called a solubility–product constant. Even though
[CaF
2
] is excluded from the equilibrium constant expression, some undissolved
CaF
2
(s) must be present in order for the system to be at equilibrium. In general,
the solubility product constant (K
sp
) is the equilibrium constant for the equilib-
rium that exists between a solid ionic solute and its ions in a saturated aqueous
solution. The solubility product is equal to the product of the concentrations of
the ions involved in the equilibrium, each raised to the power of its coefficient in
the equilibrium equation or the solubility product is equal to the product of the
concentrations of the ions involved in the equilibrium, each raised to the power
the number times they appear in equation.
222
Practical Pharmaceutical Analytical Chemistry
The solubility constant, K
sp
, has a fixed value at a given temperature and is
independent of the concentration of the individual ions. The values of K
sp
for
several slightly soluble ionic solids are given in Table 8.2.
Table 8.2 Solubility product constants of some sparingly soluble salts at 25 °C
Bromides Chromates Iodides
AgBr 5 10
13
PbBr
2
5 u 10
6
Ag
2
CrO
4
2 u 10
12
BaCrO
4
2 u 10
10
PbCrO
4
1 u 10
16
AgI 1 u 10
16
PbI
2
1 u 10
8
Carbonates Fluorides Sulphates
BaCO
3
2 u 10
9
CaCO
3
5 u 10
9
MgCO
3
2 u 10
8
BaF
2
2 u 10
6
CaF
2
2 u 10
10
PbF
2
4 u 10
8
BaSO
4
1.4 u 10
9
CaSO
4
3 u 10
5
PbSO
4
1 u 10
8
Chlorides Hydroxides Sulphides
AgCl 1.6 u 10
10
Hg
2
Cl
2
1 u 10
18
PbCl
2
1.7 u 10
5
Al(OH)
3
5 u 10
33
Cr(OH)
3
4 u 10
38
Fe(OH)
2
1 u 10
15
Fe(OH)
3
5 u 10
38
Mg(OH)
2
1 u 10
11
Zn(OH)
2
5 u 10
17
Ag
2
S 1 u 10
49
CdS 1 u 10
26
CuS 1 u 10
35
FeS 1 u 10
17
PbS 1 u 10
27
ZnS 1 u 10
20
Experiment 5 Determine solubility product of calcium hydroxide at room
temperature.
Requirement
Conical flask 250 mL, sodium hydroxide, concentrated HCl, sodium carbonate,
phenolphthalein, methyl orange, graduated cylinder, pipette, and distilled water.
Procedure
1. Prepare 0.05 N HCl solution by measuring 4.5 mL of conc. HCl and mak-
ing up volume to 1000 mL.
2. Standardize the 0.05 N HCl solution by using sodium carbonate using
methyl orange as an indicator.
3. Prepare a solution to get 0.1 N NaOH by dissolving 0.8 g of NaOH into
200 mL of distilled water.
4. Standardize 0.1 N NaOH solution by standardized 0.05 N HCl, using
phenophthalein as an indicator and record concordant reading.
5. Take three conical flasks of 250 mL capacity and label them as 1, 2 and 3.
6. Place accurately measured 50 mL of 0.1 N NaOH solution in flask 1.
7. Place accurately measured 50 and 25 mL of 0.1 N NaOH in two volumetric
flask (100 mL) and make up the volume with distilled water to give 0.05 N
and 0.025 N NaOH and transfer 50 mL of each to flasks 2 and 3, respectively
223
Solution
8. Pipette out 10 mL of 0.05 N NaOH and titrate against 0.05 N NHCl
to measure hydroxyl ion content using phenophthalein as an indicator.
Repeat the experiment and obtain the readings for concordance.
9. Repeat step 8 with 0.025 N NaOH.
10. Place about 2 g of Ca(OH)
2
into each conical flask labelled 1, 2 and 3.
11. Cork and shake the flask vigorously; continue shaking for another 1 h.
12. Allow the solutions to stand and filter solution from flask 1 through a dry
filter in another conical flask, discard first few millilitre of filtrate, while
keeping the flask covered to avoid absorption of carbon dioxide.
13. Pipette out 2 mL aliquot of solution filtrate and add few drops of phenol-
phthalein and titrate against standardized 0.0 N HCl. Note the reading and
repeat for concordant reading.
14. Repeat the procedure (steps 10 and 11) for flasks 2 and 3 and note the
reading and repeat for concordant reading.
Observation and Calculations
Observation
Flask
no
Initial
concentration
of [OH
] due
to [NaOH] (a
mol/dm
3
)
Total
concentration
of [OH
] due
to [NaOH]
and Ca(OH)
2
(b mol/dm
3
)
Concentration
of [OH
] due
to Ca(OH)
2
(ba) mol/
dm
3
Concentration
of [Ca
]
½(ba) mol/
dm
3
Solubility
of
Ca(OH)
2
½(ba)
Solubility
Product
½ (ba)
x b
2
mol/dm
3
1.
a.
2.
b.
3.
c.
Obtain the concentration of OH
(a, mol/dm
3
) due to NaOH and total OH
(b, mol/dm
3
) due to NaOH Ca(OH)
2
from titration of 0.1 N, 0.05 N, 0.025 N
NaOH and flasks 1, 2 and 3, respectively
Obtain the concentration of OH
(b a, mol/dm
3
) due to Ca(OH)
2
only by
subtracting the value of OH (a, mol/dm
3
) ions concentration from total OH
concentration (b, mol/dm
3
)
Form each Ca(OH)
2
molecule, one calcium ion [Ca
] is produced and two
OH
are produced, i.e calcium ion concentration is half the concentration of
hydroxyl ions produced in the solution. Therefore, the concentration of calcium
ion [Ca
] can be taken as half the concentration of OH
ion, i.e. concentration
of calcium ion [Ca
] is given by ½(b a) mol/dm
3
224
Practical Pharmaceutical Analytical Chemistry
Solubility product (K
sp
) of Ca(OH)
2
[Ca
2
][OH
]
2
½(b a) u (b)
2
Since, solubility product is a principle which states that solubility product
remains constant irrespective of the source of ions, the result from all the three
flasks should lie within narrow range.
Result The solubility product (K
sp
) of Ca(OH)
2
is found to be ________
mol/dm
3
8.1.5 Solubility of Partially Miscible Liquid Pairs
There are a number of liquid pairs which are only partially miscible with one
another. The miscibility varies with temperature in a characteristic manner.
When two partially miscible liquids are mixed together properly, two differ-
ent types of composition of the liquid are obtained, e.g. phenol-water system.
Phenol and water are partially miscible at room temperature. On shaking these
two liquids with each other, two layers of two saturated solution of different
compositions are obtained, one of phenol in water (lower layer) and other water
in phenol (upper layer). Such solutions are termed as conjugate solutions. At a
particular temperature the composition of two layers is fixed and both exist in
equilibrium but as the temperature of such a system is increased, the solubility
of phenol in water and water in phenol is increased till a point is reached where
two conjugate solution change into a homogenous solution.
The temperature at which and above which the two partially miscible
liquids form homogenous solution is termed as critical solution temperature
(CST) or the consolute temperature. The CST of phenol-water system is
68.1˚C and the composition of the solution at CST is 36.1% phenol and
63.9% water. Such type of CST is known as upper CST, i.e. CST obtained by
increase in temperature. Another example showing upper CST is aniline and
hexane. The consolute temperature of this system is 59.6˚C and composition
of solution is 52% aniline and 48% hexane. Water-aniline, aniline-benzene,
CS
2
-methyl alcohol systems also show upper CST at 167˚C, 59.6˚C and
40.5˚C, respectively.
In some cases like triethyl amine and water system mutual solubility increases
by decreasing the temperature, i.e. CST is attained by lowering the tempera-
ture. It is said that this system shows lower CST (CST obtained by decrease
in temperature). Other examples of systems showing lower CST are 1-meth-
ylpiperidine-water systems. Another example is of a system whose solubility
increases with both increase or decrease of temperature, e.g. for nicotine and
water system. CST can be obtained either by increasing the temperature or by
225
Solution
decreasing the temperature. That is, this system shows upper as well as lower
CST (CST obtained by increase as well as decrease in temperature). In these
cases CST is affected by pressure. With increase in pressure, the two CST’s
approach each other and composition curve decreases gradually. At very high
pressure the two partially immicible liquids becomes miscible at all tempera-
tures (Table 8.3).
Fig. 8.1
Phenol–water system
showing CST and %
phenol at CST
Phase Diagrams of Various Binary Systems
C
0 36.1 100
A
z
y
x
ab
y
r
68.1 C
11
63
phen
e perat re C
The curve ABC shows the limits of temperature and concentration within
which two liquid phases exist in equilibrium. The region outside this curve
contains system having one liquid phase. Starting from point y equivalent
to system containing 100% water at 50˚C the addition of known amounts
of phenol to a fixed quantity of water at the same temperature will result in
formation of single liquid phase until the point a is reached. At this point
addition of further amount of phenol will result in the formation second
phase. The concentration of phenol at which this occurs is 11% by weight of
phenol in water. As we keep on increasing the phenol content, we proceed
across the diagram from point a to point b and form a system which is rich in
phenol. As the concentration of phenol exceeds 63% at 50˚C, a single phenol
rich phase is formed.
The line ab drawn across the region containing two phases is termed as tie
line and is always parallel to base line in the two components.
An important feature of this type of phase diagrams is that all systems
prepared on tie line at equilibrium will separate into two phases of constant
composition. Such phases are termed as conjugate phases.
226
Practical Pharmaceutical Analytical Chemistry
pha e
18.5 C
ne pha e
rieth a ine in ater ei ht
0 100
100
e perat re
pha e
60.8
208
ne pha e
i tine in ater ei ht
0 100
300
e perat re
Fig. 8.2
Phase diagram for the
system triethylanime-
water showing lower
CST
Fig. 8.3
Phase diagram for the
system nicotine-water
showing upper and
lower CST