Faculty of Science
Chemistry Department
1
Content
Experiment
No.
1
Determination of absolute and relative density of ethanol.
2
Determination of absolute and relative viscosity of ethanol.
3
Determination of order of the reaction and rate constant for
hydrolysis of methyl acetate.
4
Determination of the saponification rate constant of ethyl acetate in
alkaline medium.
5
Clock reaction: Determination of the rate constant and order of the
sulphite-iodate reaction.
6
Determination of the rate constant and the energy of activation of the
reaction between potassium persulphate and potassium iodide.
7
Three components system.
8
Determination of the adsorption isotherm of oxalic acid on charcoal.
9
Distribution of acetic acid between benzene and water (strength of
hydrogen bond).
10
Determination of heat solution from solubility measurements (oxalic
acid in water)
11
Molecular weight of a polymer from viscosity measurements
2
Experiment (1)
Determination of Absolute and Relative Density of Ethanol
Density:
The density of a liquid is the mass per unit volume of the liquid (w/v),
the generally accepted unit of volume is milliliter (ml), which is defined as the
volume occupied by 1gram of water at the temperature of maximum density
(4°C). The density of water at this temperature in gm/ml is unity and the
density at any other temperature is expressed relatively to that water at 4°C as
standard and represented by dt
Liq
.
The relative density is the ratio of the weight of a given volume of the
substance to the weight of an equal volume of water at the same temperature
dt Liq and the density of the substance at temperature t ° is equal to the relative
density multiplied by the density of water at that temperature.
Apparatus and Chemicals:
Pyknometer or (density bottle), support, distilled water, ethyl alcohol
and thermostat water bath.
Method:
1- Wash the pyknometer or (density bottle) with chromic acid, water and
distilled water.
2- For drying the pyknometer, wash it with alcohol followed by ether,
then pass a current of dry air through it by attaching one of its ends to a
water suction pump.
3- Weight the pyknometer or (density bottle) empty with its caps on.
3
4- Fill the pyknometer or the density bottle with freshly prepared boiled
and cooled distilled water.
5- Adjust the thermostat at the required temperature (30 °C).
6- Suspend the pyknometer in the thermostat and leave it to acquire the
temperature of the bath (10-15 mins).
7- Remove the pyknometer from the bath, dry it well with a filter paper
and weight it.
8- Repeat all the stages at 25, 35, 40, 45°C.
9- Remove the water from the pyknometer and rinse with the liquid or
solution then repeat the stages 4,5,6,7 using the given unknown.
Calculations:
Let w1 be the weight of liquid that fills the pyknometer d1 its density
and let w2 and d2 be the weight and density of water.
If v is the volume of pyknometer:
d1 =
W1
V
d2 =
W2
V
d1
W
= 1
d2
W2
The ratio
d1
is known as the relative density.
d2
If d2 is known (from any physical constants book), the absolute density d1 can
be calculated
4
Density of water at different temperatures
Temperature °C
Density
25
0.99707
30
0.99567
35
0.99406
40
0.99224
45
0.99025
50
0.98807
5
Experiment (2)
Determination of the Relative and Absolute Viscosity of a Liquid
Viscosity:
A liquid moving through a tube may be considered as composed of
concentric layers moving with different velocities layers adhering to the sides
of the tube are considered to be stationary, and the velocity increases as the
middle of the tube is approached. Hence there will be a velocity gradient
between the layers due to frictional forces acting between the different layers
of the liquid. These forces are responsible for the so-called viscosity of the
liquid.
It has been experimentally shown that the force f required to maintain a
constant difference between the velocities of the parallel layers of liquid
moving in the same direction varies directly with the difference in velocity V
and the area A, of the surface of contact of the two layers, and inversely as the
distance d, between the layers.
f = ηAv/d
Where η is a proportionality factor known as the coefficient of
viscosity, the unit of viscosity is the poise. This is defined as the force
necessary to move a layer of liquid of area 1 cm2 with a velocity of 1 cm/sec
past another layer at distance of one centimeter. The velocity of a liquid is
generally measured by observing the time required for a definite volume of
liquid to flow through a standardized capillary tube under a known difference
of pressure. The apparatus commonly used in the laboratory is the Ostwald’s
6
viscometer the law governing the flow of liquids through capillary tubes was
discovered by Poisenille and is given by:
η=
 Pr 4 t
8VL
In which V denotes the volume of a liquid of viscosity η flowing
through a capillary tube of length L and radius r in time t and under the
pressure p.
If the times of flow of equal volumes of two liquids through the same
capillary tube are measured under the same experimental conditions, it
follows that:
1
Pt
dt
= 11 = 11
2
P2 t 2
d 2t2
η1 and η2 are the viscosity coefficient of the two liquids. d1 and d2 their
densities, and t1 and t2 their times of flow. This equation is used to calculate
the relative viscosity.
Method:
1- The viscometer is cleaned and dried in the same way as pyknometer.
2- The viscometer is clamped vertically in a water thermostat. It must be
dipped to the mark (a) at temperature 30°C.
3- A definite volume of the liquid is introduced in the wide tube (limb c)
of the viscometer and is left until it acquires the required temperature of
the bath, the liquid is then forced up through the capillary tube by
suction through a rubber tube attached to the end (a) until the liquid
fills the bulb (E) and rises slightly above the mark a. The volume of the
liquid introduced must be sufficient to fill the bulb (E), the lower bend
(F) and extends up slightly into the wide tube (c); otherwise air bubbles
will form in the capillary and affect the time of flow.
4- The liquid is then allowed to flow back through the capillary tube and a
stopwatch is started when the meniscus passes by the upper mark (a)
and it is stopped when the meniscus passes by the lower mark (b).
Repeat twice.
7
5- Repeat the steps at 25°C, 35°C, 40°C and 50°C.
6- Repeat the steps on freshly boiled then cooled distilled water using the
same volume as taken of the liquid.
Calculations:
Let the time of flow of the liquid be t 1, its density d1 and its absolute
viscosity η1, the time of flow of water be t2, density d2 and viscosity η2
1
dt
= 11
2
d 2t2
And the viscosity of the liquid η1 can be known if η2 is known (from any
physical constants book).
Viscosity of water at different temperatures
Temperature °C
Viscosity
25
0.8937
30
0.8007
35
0.7225
40
0.6540
50
0.5494
8
Experiment (3)
Determination of Order of the Reaction and Rate Constant for the
Hydrolysis of Methyl Acetate
Hydrolysis of Methyl Acetate:
Methyl acetate hydrolyses in water to give methanol and acetic acid in
accordance with:
H+
CH3COOCH3 + H2O
CH3COOH + CH3OH
The reaction does not proceed at any measurable rate in pure water but
is highly catalyzed by hydrogen ions. Although more than one molecular
species are involved, the reaction is kinetically of the first order since water is
usually present in large excess. For a first order reaction, the rate constant k
may be expressed as:
k=
1
a
1n
ax
t
Or
log (a-x) = -
kt
+ log a
2.303
Where (a) is the original concentration, (a-x) the concentration after time t
Procedure:
1- Prepare 0.1 N HC, 0.1 N NaOH.
2- Transfer 100 ml of the acid into a flask, and put 10 ml methyl acetate
into another flask. Thermostat both flasks at 25°C (or at room
temperature).
3- By means of a pipette, transfer 5 ml methyl acetate into the acidcontaining flask. Shake thoroughly and record the time of mixing as the
start time of reaction.
4- After about 5 minutes, withdraw 10 ml of the solution, run into about
20 ml ice-cooled water (to arrest the reaction) and record the time.
Titrate the solution rapidly with 0.1 N NaOH using phenolphthalein as
9
indicator. Repeat at increasing time intervals making a total of about
ten determinations over a period extending for about two hours.
5- Carry about one similar determination after sufficiently long time, say
24 hrs; when hydrolysis is complete, or alternatively run out 10 ml of
the ester solution into a small flask containing some water and fitted
with an air condenser. Boil on a water bath for about one hour to affect
complete hydrolysis. Wash the condenser into the flask and titrate with
0.1 N NaOH.
Calculations:
1- In any titration, the amount of alkali consumed (X ml) is equivalent to
the amount of HCl in 10 ml plus the amount of acetic acid liberated.
After complete hydrolysis, the amount of alkali consumed (a ml) is
equivalent to amount of HCl in 10 ml plus the amount of acetic acid
equivalent to all ester. The remaining ester concentration after time t
min. is therefore a-x.
2- Plot log (a-x) against t and identify the order of the reaction.
3- From the graph calculate the rate constant k and the half-life period (t1/2
=
0.69327
).
k
log (a-x)
slope = -
log a
t
10
k
2.303
Experiment (4)
Determination of the Saponification Rate Constant of Ethyl Acetate in
Alkaline Medium
Saponification of Ethyl Acetate:
In presence of alkali, ethyl acetate undergoes saponification in
accordance with:
CH3COOC2H5 + OH-
CH3COO- + C2H5OH
The rate of saponification is directly proportional to both
concentrations of ester and alkali, and the reaction therefore is a second order
one.
dx
= k (a-x) (b-x)
dt
a and b are the initial molar concentrations of ester and alkali
respectively and k the reaction rate constant. The integrated form of the above
rate equation is:
 2.303   b(a  x) 
k
 log a(b  x) 
 t ( a  b)  

If the initial concentration a are the same for ester and alkali, the above
equation reduces simply to:

1
x

t  a (a  x) 
k = 
Or
1
1
=kt+
ax
a
Procedure:
1- Prepare exactly 0.1 N Na2CO3, 0.1 N HCl and 0.1 N NaOH.
Standardize the acid against carbonate and hydroxide against the acid.
2- By appropriate dilution, prepare
3- Prepare
N
N N
HCl, ,
NaOH.
40
40 20
N
ethyl acetate (density of ethyl acetate is 0.901).
20
11
4- By means of a pipette, transfer 50 ml
flask, and 50ml
N
ethyl acetate into a clean dry
20
N
NaOH into another flask, stopper the two flasks.
20
5- Add quickly the alkali to the ester, mix thoroughly and meanwhile
record the time. The mixture is
N
with respect each of alkali and ester.
40
6- Withdraw 10 ml portion of the reacting mixture, record the time (about
2 mins. from the start) and run immediately into a flask containing
about 50 ml distilled water and exactly 10 ml
excess HCl with
N
HCl. Titrate back the
40
N
NaOH using phenolphthalein indicator.
40
7- Repeat step 6 at time intervals extending for about 100 mins. A total of
eight titrations would be sufficient.
Calculations:
If (a) is the amount of HCl equivalent to the original concentration of
alkali and ester, and x the amount of alkali equivalent to the excess HCl after
time t, a-x lives hence the amount of each of alkali and ester remaining.
Tabulate the results as follows:
t,
Plot
1
ax
1
against t to identify the order of reaction, from the slope of the
ax
curve, the reaction rate constant k may be calculated.
12
Experiment (5)
Determination of the Rate Constant and Order of the Sulphite- Iodate
Reaction
Introduction:
A slow reaction is sometimes followed by much rapid one which
however, does not occur except when the first is complete. The time elapsed
from the start of the reaction until the second step of reaction occurs is termed
induction or incubation period. As an example, is the interesting reaction of
H. Landolt in which sulphite is oxidized by iodate in acid medium. The first
slow reaction is:
3 H2SO3 + HIO3
3 H2SO4 + HI
When all sulphite has been oxidized to sulfate, the hydroiodic acid
produced reacts immediately with iodic acid in accordance with:
5 HI + HIO3
3 I2 + 3 H2O
Consequently, if starch is present in the reacting mixture, a dark violet
color suddenly develops at the end of the first slow reaction.
Apparatus and Chemical:
8 flat bottom flasks, stop watch, starch, sodium sulphite and potassium
iodate.
Procedure:
1- An acid sulphite solution is prepared as follows:
About 2g starch are mixed thoroughly with 10 ml water, and the
suspension is added drop wise to about 100 ml boiling water. Cool and
transfer into a 250 ml measuring flask containing about 25 ml water to
which 1 ml concentrated sulfuric acid is added. Accurately weighed
0.315 g sodium sulphite are dissolved in the least amount of water, and
then carefully introduced into the flask which is then completed to the
13
mark with distilled water. Prepare also 100 ml of exactly 0.0211
potassium iodate solution.
2- In dry small flat bottom flasks make up the following two series of
sulphite concentrations.
Series 1
Series 2
ml sulphite
ml water
ml sulphite
ml water
25
70
25
68
20
75
20
73
15
80
15
78
10
85
10
83
The volume of each mixture of (series 1) is 95 ml and of (series 2) is 93 ml.
3- To each mixture of series 1 in turn adds exactly 5 ml iodate (making a
total volume of 100 ml), mix thoroughly and meanwhile start a clock
on. Determine the induction period indicated by the sudden appearance
of the blue color.
4- To each mixture of series 2 in turn add exactly 7 ml iodate (making a
total volume of 100 ml), proceed then in the same way as in step (3).
5- For a constant iodate concentration, the rate equation may be written
as:
Rate = R = k cn
Where k is the rate constant, c sulphite concentration in mixture
expressed as gram mole per liter, and n the order of reaction with respect
to sulphite.
6- From the fact that all sulphite should have been consumed at the
moment the blue color appears, the average rate R is calculated from,
c
t
R= ; c is the sulphite concentration and t the induction period.
7- Plot log R against log c for each series of mixtures, and from the slope
of the line obtained, find out the order of reaction n. from the intercept
using the relation below, and calculate the rate constant k.
log R = log k + n log c
14
Experiment (6)
Determination of the Rate Constant and the Energy of Activation of the
Reaction between Potassium Persulphate and Potassium Iodide
Consider the reaction:
K2S2O8 + 2KI
I2 + 2K2SO4
S2O8-2 + I -
(S2O8I)-3
slow
(S2O8I)-3 + I-
I2 + 2SO4-2
rapid
I2 + 2Na2S2O3
Na2S4O6 + 2NaI
Before reaction occurs, molecules must be activated, they posses
energy in excess of a certain amount. These activated molecules will then
collide and lead to the reaction. Collisions between molecules that are not
activated will be of no use and no reaction will take place. The minimum
energy which the molecules must absorb before the reaction can take place is
known as the energy of activation Ea.
The rate constant k varies with temperature in a manner described by
the following equation:
Ea
k=Ae
– RT
Where A is the pre-exponential factor (or frequency factor), Ea is the
energy of activation for the reaction in (J/mole), R is the molar gas
constant (8.3143 J/k.mole), and T is the absolute temperature (Kelvin).
The equation is known as the Arrhenius Equation.
Log k = log A –
15
Ea
2.303RT
According to this equation, a plot of log k against
line with a slope –
1
should be straight
T
Ea
and intercept log A as shown is the figure.
2.303R
Knowing the slope, the value of Ea can be calculated.
Ea can also be determined if the rate constant is known at two different
temperatures. If k1 is the rate constant at temperature T1 and k2 at another
temperature T2, then from the equation:
log (
Ea  1 1 
k1
  
)=
2.303R  T2 T1 
k2
log (
Ea  T2  T1 
k2
)=


2.303R  T2  T1 
k1
Or
If this equation is solved for the energy of activation Ea, the following
relationship is obtained:
 T2T1 
k2
 log ( )
k1
 T2  T1 
Ea = 2.303R 
log k
slope = -
Ea
2.303R
1/T
Apparatus and Chemicals:
Thermostat, burette, pipettes, conical flasks, measuring flasks (cylinders),
0.01 N Na2S2O3, 0.4 N KI, 0.04 N K2S2O8 and starch
Procedure:
1- Pipette 50 ml of each of 0.4 N KI and 0.4 N K2S2O8 in two Stoppard
bottles. Allow the solution to reach thermal equilibrium at 25°C.
16
2- Add the KI solution to the K2S2O8 solution and report the time of
starting reaction.
3- Three minutes after the start, remove 10 ml of the reaction mixture, run
it into 500 ml conical flask containing 200 ml cooled distilled water.
Titrate the liberated iodine against 0.01 N sodium thiosulfate solution
using starch as indicator.
4- Repeat step 3 at suitable time intervals, say (6, 10, 15, 20, 27, 35,
…..minutes). The volume of sodium sulfate at any time t, (which is
equivalent to the amount of K2S2O8 decomposed) is equivalent to (x).
5- For reaction completion, heat the remaining reaction mixture in a water
bath for one hour at 45 - 50°C. (2 g of KI).
6- Withdraw 10 ml the reaction mixture, run it into 250 ml conical flask,
titrate against the standard sodium thiosulfate solution using starch as
indicator. In this case the volume of sodium thiosulfate which is
equivalent to the iodine liberated, corresponds to (a), the initial
concentration of K2S2O8.
7- Carry out the entire experiment at another two different temperatures.
8- Decomposition of potassium persulfate (K2S2O8) with potassium iodide
is a second order reaction (first order with respect to both KI and
K2S2O8). By taking the concentration of KI ten times greater than
K2S2O8, you can treat the same reaction as pseudo-first-order. Plot log
a
against t, and calculate the values of k from the slope of the
ax
k
obtained straight lines. The slope is equal to; slope =
.
2.303
1
9- Plot log k against , and calculate Ea from the obtained straight line.
T
The slope is equal to; slope = 
Ea
.
2.303R
10- Calculate the thermodynamic parameters of the activated complex
H*, G*, & S*.
17
Experiment (7)
Three Component Systems
The mutual solubility of a pair of partially miscible liquids may be
altered by the addition of a third component. Thus, if the third component is
soluble in only one of the two other components, the mutual solubility of the
two liquids decreases, if the third component dissolves readily in each of the
two other components, the mutual solubility of the latter two components
increases, until a point is reached at which the mixture becomes
homogeneous. This behavior is illustrated by the system [ethyl acetate - ethyl
alcohol – water].
If ethyl alcohol is added to a heterogeneous mixture of ethyl acetate
and water, the mutual solubility of the latter two liquids is increased giving a
homogeneous system at a definite temperature.
If ethyl acetate is added to a homogeneous mixture of ethyl alcohol and
water, the mutual solubility is decreased (because ethyl acetate dissolves in
ethyl alcohol only).
The phase properties are best defined by using the triangular diagram.
On an equilateral triangle, the apexes A, B, and C represent pure components
(100%). A point on any one side represents a mixture of two components
only. A point p inside the triangle represents the composition of a mixture of
A, B and C of percentages respectively proportional to the lengths of lines p a,
pb, pc drawn parallel to the sides of the triangle. This contention is borne out
by the fact that in an equilateral triangle the sum of pa + pb + pc (also for any
point other than p) is always equal to the length of any one side
(corresponding to a total of 100%).
18
A
p
C
B
The triangular diagram
The triangular diagram usually consists of an equilateral triangle. The
length of the side is taken as 100, and represents therefore the percentage
amount of the three components; each corner of the triangle represents 100%
of one component. In plotting the composition of a tertiary mixture, 2 points
are marked on two sides of the triangle representing the % amount of two
components and from these points line is drawn paralleled to the other two
sides of the triangle. The point of intersection gives the composition of the
tertiary mixture. The line connecting the various points is the binodal curve.
Within this curve all the lines show immiscible region and outside the curve,
the region of complete miscibility
20 % A, 60 % B, 20 % C
20 % A, 15 % B, 65 % C
19
Procedure:
1- Into five dry stopper bottles, introduce the following volumes of ethyl
acetate and water.
Bottle
Ethyl acetate (ml)
Water (ml)
1
10
2
2
8
4
3
6
6
4
4
8
5
2
10
The mixture will appear turbid.
2- From a burette, run in ethyl alcohol in small quantities at time into each
of the bottles in turn. Shake well after each addition.
3- Continue addition of ethyl alcohol till turbidity just disappears on
shaking (the final addition should be made drop by drop).
Calculations:
1- Calculate the percentage composition by weight of each mixture at the
stage when turbidity just disappears.
Density of ethyl acetate = 0.894 g/c.c
Ethyl alcohol = 0.789 g/c.c
Water
= 0.996 g/c.c
For example:
v1 x d1 (ethyl acetate)
% by weight for =
Ethyl acetate
x 100
(v1 x d1) + (v2 x d2) + (v3 x d3)
Ethyl acetate water ethyl alcohol
2- Tabulate the results and plot then on a triangular diagram. Join the
points by smooth curve to obtain the binodal curve.
20
Experiment (8)
SURFACE CHEMESTRY
Adsorption by Solids from Solution
Introduction:
At the surface of a solid or a liquid molecular force are usually
unbalanced or unsaturated. As a result of this unsaturation, exposed surface
tend to satisfy their residual forces by attracting and retaining onto them other
substances with which they come in contact. The phenomenon is known as
adsorption.
Solids may adsorb dissolved substances from solutions as well as gases.
In sugar refining, for example, colored materials and impurities may be
removed by filtering through adsorbents such as charcoal. Adsorption of
solutes from solution involves the establishment of equilibrium between the
amount adsorbed and the concentration of substance in solution. The variation
of the amount adsorbed with concentration may be represented by an isotherm
of the Freundliche type:
x
= KCn
m
Where x is the amount of solute adsorbed per m grams adsorbent, C the
equilibrium concentration of solute in solution, K and n are constants. The
relation may be written in the form:
log
x
= n log C + log K
m
A plot of log
x
against log C should therefore be straight line of slope
m
n and intercept log K.
Types of Adsorption:
There are two types of adsorption: Physical adsorption (physisorption)
and chemical adsorption (chemisorption).
21
Chemisorption is distinguished qualitatively from physisorption in the
following ways:
Physisorption
Chemisorption
1-The forces causing physisorption
are generally referred to as van der –
Waals forces.
It involves the formation of chemical
bond.
2-The heat (enthalpy) involved in the
physisorption process is usually near
that of the heat of condensation (10 –
20 kJ/mol).
The heat of adsorption for chemisorption is greater than that for physisorption, lying in the range (40 – 200
kJ/mol).
3-It often results in the formation of
multilayers of adsorbed molecules.
Chemisorption leads only to a monolayer.
4- Physisorption is none-specific,
occurs very rapidly and readily
reversible (removed by lowering the
gas pressure or the concentration of
the solute).
It is more specific than physisorption,
can occur either rapidly or slowly and
irreversible (often accompanied by
chemical change).
5-The extent of physisorption is
smaller at higher temperatures.
May not occur at an appreciable rate
at low temperatures because it has an
activation energy.
Many substances, both liquid and gas, adsorb to solid surface. The amount
of substance adsorbed depends on:
1. The specific nature of the solid and of the molecules being adsorbed.
2. The temperature, the amount of substance adsorbed at any surface
decreases with rise of temperature, since all adsorption processes
(physisorption) are exothermic.
3. The concentration (or pressure). At constant (isotherm) the amount
adsorbed increases with the concentration of the adsorbate.
22
Experiment:
Determination of the Adsorption Isotherm of Oxalic Acid on Charcoal
Procedure:
1- Prepare exactly 0.3N KMnO4, 0.5N oxalic acid and 2NH2SO4.
2- Standardize the permanganate against the acid and determine its exact
normality (using 2NH2SO4).
3- Into 5 small flasks (bottles) introduce the following solutions:
a- 100 ml 0.5 N oxalic acid.
b- 80 ml 0.5 N oxalic acid + 20ml distilled H2O.
c- 60 ml 0.5 N oxalic acid + 40ml distilled H2O.
d- 40 ml 0.5 N oxalic acid + 60ml distilled H2O.
e- 20 ml 0.5 N oxalic acid + 80ml distilled H2O.
4- Take 10ml of each bottle and titrate against potassium permanganate
(in presence of 10 ml of 2 N H2SO4 with heating).
5- To each bottle, a known weight of charcoal (about 1 gm) is added;
leave the solutions for about 30 mins. with occasional shaking.
6- Filter each solution through a dry small filter paper into a dry receiver,
rejecting the first few mls of filtrate.
7- Titrate 10 ml of each filtrate with 0.3 N KMnO4 with heating in the
presence of 10 ml of 2 N H2SO4.
23
Calculations:
1. Tabulate the results as follows:
V1 ml of KMnO4= 10 ml acid before adsorption.
V2 ml of KMnO4= 10 ml acid after adsorption.
Wt of oxalic acid before adsorption (w1) = N. V1. eq.wt = mg/10ml
Wt of oxalic acid after adsorption (w2)= N. V2. eq.wt
= mg/10ml
w1 - w2 = wt of oxalic acid adsorbed = x
= mg/10ml
Amount of oxalic adsorbed in 1L = x.
1000
10
Amount in gms. of oxalic adsorbed in 1L = x .
= mg/L
1000
x
x 1000 =
g/L
10
10
m = weight of charcoal
= 1 gm.
C = equilibrium concentration in gm/L
C = w2 x
2. Plot
100
1000
x
x
against C and log
against log C.
m
m
3. from the curve find out the value each of the constants n and K.
24
Experiment (9)
Distribution of Acetic Acid between Benzene and Water
Strength of Hydrogen Bond
The variation with concentration of the partition coefficient of acetic
acid between benzene and water at different temperatures may be used to
determine the strength of the hydrogen bridges responsible for double
molecule formation in benzene according to:
The dotted lines represent the two hydrogen bridges, which join the two
molecules.
In strong aqueous solutions, acetic acid exists almost as single
unionized molecules. In benzene, a considerable amount of dimerization
occurs giving:
Water
HAC
Cw
Benzene
HAC
C1
1/2 (HAC)2 C2
Cw and C1 are the monomer concentration in water and benzene
respectively, C2 that of dimmer in benzene layer. The total concentration in
benzene expressed as monomer is:
CB = C1+ 2 C2
Setting up the equilibrium equation for the system, it follows
K1 =
C1
Cw
K2 =
C12
C2
K1 is the partition coefficient of acetic acid monomer between benzene
and water
K2 the equilibrium constant for acetic acid association in benzene
Hence,
CB = C1+ 2 C2
25
CB = K1Cw +
2C12
K2
CB = K1Cw +
2 K 12 C w2
K2
Dividing by Cw
CB / Cw = K1 +
2 K 12 C w
K2
The energy H required to break the two bonds joining the two
molecules in the dimmer is obtained using the Van’t Hoff isochore:
 TT 
 K (atT2 ) 
H  R 1 2 2.303 log  2

 T2  T1 
 K1 (atT1 ) 
The strength of a hydrogen bond is one half H.
Determination of the strength of the hydrogen bridge responsible for the
dimerization of acetic acid in benzene
Procedure:
1. Prepare 100 ml of each exactly 0.1N Na2CO3, 0.1N HCl, 0.1N NaOH and
0.1N CH3COOH. Standardize HCl against Na2CO3 and NaOH against HCl.
By appropriate dilution prepare 250 ml of each exactly 0.05N and 0.02N
NaOH solutions.
2. In a dry stopped tube, shake up 30 ml benzene with 25ml normal acetic
acid, in another tube shake up 30 ml benzene with 15 ml normal acetic acid
and 10 ml distilled water, in the third tube shake up 30 ml benzene with 10
ml normal acetic acid and 15 ml distilled water. Keep in a thermostat at
10°C for at least 30 mins. with occasional shaking and leave until the two
layers separate.
3. The concentration of acetic acid in the benzene layer CB is determined by
pipetting 10 ml into a conical flask containing some distilled water and
titrating with 0.02 N NaOH using phenolphthalein as indicator.
4. Pipette out 10 ml of the aqueous layer and dilute to 100 ml in a measuring
flask. Use 10 ml of this solution to determine the concentration of acetic
26
acid in the aqueous layer, Cw, by titrating with 0.05N NaOH. Express
concentrations in moles / liter.
5. Repeat step (2) at 40°C and determine CB and Cw at each temperature as
described in steps (3), (4).
Calculation:
1. Plot
CB
against Cw. The intercept of the line obtained is K 1 and the slope
Cw
2 K12
is
.
K2
2. Knowing the value of K1, the value of K2 obtained.
3. Substitute in the Van’t Hoff isochore to obtain H.
27
Experiment (10)
Determination of heat of solution from solubility measurements
(Oxalic acid in water)
Heat of Solution:
The equilibrium between a solid and its saturated solution may be
represented as:
Solid solute
dissolved solute
An equilibrium constant KS may be used to define this equilibrium,
such as:
KS =
[dissolved solute]
 [dissolved solute]  CS
[Solid solute]
CS is the molal concentration of dissolved solute (mol/1000gm solvent).
If H is the change in heat content when one mole of solute is dissolved in a
large volume of the nearly saturated solution (heat of solution), then
according to Van’t Hoff isochore:
d 1n
Ks
H
=
RT 2
dT
or
d 1n
Cs
H
=
RT 2
dT
If H is independent of T, then by integration
d 1n CS =
H
R

dT
T2
H
+ constant
RT
H
log CS = + constant
2.303RT
1
A plot of log CS against
should give a straight line with a slope
T
H
=
, from which H can be calculated in calories/mole.
2.303R
1n CS = -
28
Procedure:
1. Prepare 100 ml of standard oxalic acid solution 0.1N, and 0.1 N Potassium
permanganate. Standardize permanganate solution against oxalic acid
solution.
2. Distilled water in a large test tube is saturated with oxalic acid crystals (20
gms acid per 60 ml distilled water) at about 60°C (notice that some of the
solid oxalic must present in equilibrium with the saturated solution at this
temperature and continuous stirring is necessary to make sure that the
solution is saturated with the solute) and then cooled slowly in a water
thermostat to the desired temperature. The equilibrium is attained rapidly
when approached in this way.
3. Lower slowly the temperature of the thermostat from 60°C to 50°C and
adjust the latter temperature. After equilibrium is reached (about 10 mins.)
transfer by means of a pipette 10 ml of the clear supernatant liquid into
previously weighed weighing bottle. (The tip of the pipette is plugged with
a small roll of cotton wool so as to present withdrawal of fine crystals with
the solution). Determine the weight of the solution then transfer it
quantitatively into 100 ml measuring flask and complete to the mark with
distilled water.
4. Titrate 10 ml of the solution against the standard KMnO4.
5. Repeat the above steps at the temperatures 40°, 30°, 20°, 10° and 0°C.
Calculation:
Let us determine the molal solubility CS at each temperature as follows:
First at 60°C
1. Weight of solution w1 = ---- gm.
2. From the titration we get the normality of the diluted oxalic acid solution:
N VKMnO4 = N` V`oxalic
3. N` (oxalic acid) x eq.wt = y (wt in gm/1)
y
1000 ml.
??
100 ml
yx
100
= wt of oxalic acid in 100 ml = z
1000
29
gms of oxalic acid (z) in x gms saturated solution.
4. Then z gms oxalic acid is soluble in (w1 – z) gms distilled water.
z
w1 – z
??
1000 gms
zx
zx
1000
– z = gms of oxalic acid soluble in 1000 gms.
w1
1000
= gms moles of oxalic acid is soluble in
( w1  z ) M
1000 gms distilled water.
This gives the solubility or molal concentration CS of oxalic acid.
Solubility is the number of gram moles of solute per 1000 grams of solvent.
CS = z x
1000
( w1  z ) M
M is the molecular weight.
5. In a similar manner calculate the solubility at the other temperatures (40°,
30°, 20°, 10° and 0°C).
6. Plot log CS against 1/T (absolute).
7. From the slope of the line obtained, calculate H.
Slope = 
H
and R = 1.987 calories/mole.
2.303R
30
Experiment (11)
Molecular weight of a polymer from viscosity measurements
Introduction:
The ratio of the viscosity η of a solution of non-spherical high polymer
molecules to the viscosity ηο of the solvent is related to the molecular weight
M of the polymer by the expression:
   
    1
0  


  KM
c






Equation (1)
In this equation K is a constant for any given type of polymer, solvent and
temperature, α is a function of the geometry of the molecule, and c is the
number of grams of polymer in 100 ml of solution.



  0 


The term    1 is known as the specific viscosity in the form:
 sp
Equation (2)
c
This equation is only valid for very dilute solution (less than 1 percent) and
hence the graph which is drawn of (
 sp
c
) vs. c is extrapolated to zero
concentration. The extrapolated value is known as the intrinsic viscosity [η].
[η] = lim
 sp
c→0
equation (3)
c
If the logarithmic function ln

is expanded as an infinite series, since
0
the second and higher terms can be neglected as the concentration approaches
zero, it will be seen that :
lim
 sp
c→0 c
1

ln( )
0
c→0 c
= lim
equation (4)
31
Hence,
1

ln( )
0
c→0 c
[η] = lim
equation (5)
The intrinsic viscosity is therefore the intercept on the graph of either
1


or ln( ) vs. c. A more reliable value of the intercept is obtained by
c
c 0
drawing both graphs.
The [η] value obtained by this method may then be used to determine the
molecular weight of the polymer by applying the equation:
[η] = K Mα
Apparatus and Chemicals:
Oswald viscometer ( see experiment 2), stop watch, thermostat at 25°C,
25 ml graduated flask, 5 ml and 10 ml pipettes, polystyrene and toluene.
Procedure:
1. Prepare the stock solution of the polystyrene in toluene by weighting
0.25g of the polystyrene dissolved in 50 ml toluene.
2. Make 5 different concentrations by diluting the stock solution to 2, 4, 8,
and 16 times.
3. Measure the average time of flow ( t ) for each of the above five
concentrations using Oswald viscometer at 25°C as explained in
experiment number 2.
4. Measure the average flow time of pour toluene ( t ) at 25°C.


td
t


) from:
0 t0 d 0 t0
0
5. Calculate (
6. Calculate ηsp for each concentration from: ηsp = {

–1}
0
7. Record the results in a tabular form as follows:
c
t

t

 0 t0

– 1 = ηsp
0
32
 sp
c
1

ln( )
c 0
8. Plot
 sp
c
1
c
vs. c and { ln(

) } vs. c.
0
9. Extrapolate the graphs to c→ 0 and obtain the intrinsic viscosity [η].
10.Calculate the molecular weight of polystyrene M from the equation:
[η] = K Mα , where K = 3.7 x 10-4 and α = 0.62.
------------------------------------------------------------------------------------------
GOOD LUCK
33