list of engineering chemistry experiments...
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ENGINEERING CHEMISTRY LAB – I B.TECH 1
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
LIST OF ENGINEERING CHEMISTRY EXPERIMENTS
INDEX
S.NO NAME OF THE EXPERIMENT PAGE
NO.
DATE
CONDUCTED
SIGNATURE
1 Preparation of Aspirin
2 Estimation of hardness of water by EDTA
method
3 Determination Of viscosity by Ostwald
viscometer
4 Conductometric titrations of a strong acid vs
strong Base
5 Conductometric titrations of a mixture of acids
vs strong base
6 Determination of percentage purity of
Pyrolusite
7 Potentiometric titrations of a strong acid vs
strong base
8 Potentiometric titrations of a weak acid vs
strong base
9 Determination of % of copper in brass
10 Preparation of Thiokol rubber
11 Adsorption of Oxalic acid on charcoal
12 Estimation of ferrous iron by Dichrometry.
13 Determination of surface tension using
Stalagmometer
ENGINEERING CHEMISTRY LAB – I B.TECH 2
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
SAFETY NORMS TO BE FOLLOWED.
DO’S AND DONT’S IN CHEMISTRY LABORATORY.
NEVER SNIFF, TASTE OR TOUCH THE CHEMICALS.
ALWAYS POUR ACIDS INTO WATER. NEVER ADD WATER TO ACID. IF POUR WATER
INTO ACID , THE HEAT OF REACTION WILL CAUSE WATER TO EXPLODE INTO STEAM,
SOME TIMES VIOLENTLY, AND THE ACID WILL SPLATTER.
KNOW WHAT CHEMICALS YOU ARE USING. READ THE LABEL TWICE BEFORE YOU USE
THEM.
LEARN WHERE THE SAFETY AND FIRST AID EQUIPMENT IS LOCATED, LIKE FIRE
EXTINGUISHERS, EYE WASH STATIONS.
CONSIDER ALL CHEMICALS HAZARDOUS UNLESS YOU ARE INSTRUCTED OTHERWISE.
NEVER EXPERIMENT ON YOUR OWN.
DO NOT EAT /DRINK IN LAB AT ANY TIME.
DO NOT WEAR CONTACT LENS IN THE LAB.
NEVER LEAVE THE LAB WITHOUT WASHING YOUR HANDS.
DO NOT PIPETTE BY MOUTH.
DON’T CASUALLY DISPOSE OF CHEMICALS DOWN THE DRAIN.
MANY COMMON REAGENTS FOR EXAMPLE ALCOHOL AND ACETONE ARE HIGHLY
INFLAMMABLE. DO NOT USE THEM ANYWHERE NEAR OPEN FLAMES.
IF CHEMICALS COME IN CONTACT WITH YOUR SKIN OR EYES, FLUSH IMMEDIATELY
WITH WATER FOR ATLEAST 10 MIN. AND CONSULT YOUR INSTRUCTOR.
ENGINEERING CHEMISTRY LAB – I B.TECH 3
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Experiment No. 1
PREPARATION OF ASPIRIN
AIM:To prepare aspirin from salicylic acid and acetic anhydride in presence of conc.H2SO4
CHEMICALS REQUIRED:
Salicylic acid = 2.5 gm or 5 gm
Acetic anhydride
Con. H2SO4
APPARATUS :
Conical flask
Beaker
Funnel
Glass rod
PROCEDURE:
Take 2.5 g of salicylic acid 5 ml of acetic anhydride in 100 ml conical flask.
Add 3 drop conc. H2SO4 and shake well. Heat the flask on water bath keeping temp below
60-70oC for about 20 min Cool the mixture in an ice bath with stirring then add about 20 ml
of ice cold water to decompose excess of acetic anhydride. Pour the content of the conical
flask into a 250 ml beaker containing 50 g of crushed ice. Stir it until white solid appears.
Scratch the side of the flask with a glass rod or stainless steel spatula. If white solid does not
appear, filter through funnel and dry by pressing into the folds of filter paper. Recrystallise
the crude acetylsalicylic acid.
RECRYSTALLISATION:
Dissolve the solid in minimum amount of ethanol and dissolve it
until clear solution is obtained. Clear solution is allowed to cool slowly to get needle shaped
crystals of Aspirin.
Calculations:
Theoretical yield = 5 gm X M .Wt of product
M .Wt of starting material
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% yield of Aspirin = Experimental value X 100
Theoretical value
RESULT: The amount of crude Aspirin found =
The amount of pure Aspirin found =
VIVA QUESTIONS :
1) What is the chemical name of aspirin?
2) What is the medicinal importance of aspirin?
3) What are the chemicals required for the preparation of aspirin?
4) what is the process involved in the preparation of aspirin?
5) what is the function of acetic anhydride and sulphuric acid?
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Experiment No. 2
HARDNESS OF WATER BY EDTA METHOD
AIM:To determine the total, permanent and temporary hardness of water by EDTA method.
CHEMICALS REQUIRED:
Standard ZnSO4 solution,
EDTA Solution,
Ammonia– Ammonium chloride buffer
water sample
EBT
APPARATUS:
Burette
Conical flask
Pipette
INDICATOR: Eriochrome black –T
END POINT: The end point is noted when the colour of the solution changes from wine
red to blue
PRINCIPLE:
The hardness of water is due to the presence of salts of Ca2+
and Mg2+
. The
bicarbonates of Ca2+
and Mg2+
impart temporary hardness to water which can be removed by
boiling. The amount of Ca2+
and Mg2+
is estimated by complexometric method using EDTA.
The obtained value gives the total hardness. The permanent hardness is determined first by
precipitating the bicarbonates of Ca2+
and Mg2+
by heating and filtering off and the filtrate is
titrated with EDTA. The temporary hardness is obtained by subtracting permanent hardness
from total hardness.
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EDTA Titrations (or complexometric titrations): The titrations involving EDTA as a
complexing agent are known as “EDTA Titrations” (or) “Complex Formation Titrations”.
EDTA stands for ‘Ethylene Di-amineTetra Acetic acid”.
The structure of EDTA is given by
CH2COOH CH2COOH
N - CH2 – CH2 – N
CH2COOH CH2COOH
EDTA forms complexes with metal ions like Ca2+
, Mg2+
, Ba2+
, and
Sr2+
in aqueous solution. Complexo metric titration is the titration of metal ion with a
reagent, usually EDTA, which forms chelated complex with a metal. The common
indicator used is Erichrome Black-T Indicator (EBT), at the end point the colour
generally changes from red to blue.
HOOCH2C CH2COO-Na
+
N - CH2 – CH2 – N
+Na
-OOCH2C CH2COOH
Disodium salt of EDTA
Ethylene di amine tetra acetic acid [EDTA] forms colorless stable
complex with Ca+2
and Mg+2
ions present in water at pH
9-10. To maintain pH
of the
solution at 9-10 NH3-NH4Cl buffer solution is used. Erio chrome black –T (EBT) is
used as indicator. The hard water when treated with (EBT) in presence of buffer
solution forms unstable wine red color complex with Ca+2
and Mg+2
ions present in
water.
The stability of a metal indicator complex is less than that of metal EDTA
complex. Hence during the titration EDTA extracts the metal ions from the metal ion
EBT indicator complex and forms stable colorless metal EDTA complex releasing free
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indicator. Hence the end point of the titration is the color change from wine red to blue
color.
PROCEDURE:
PART A: STANDARDISATION OF EDTA SOLUTION:
20ml of standard Zn2+
solution is pipette out into a well cleaned conical
flask. To it 2- 3 drops of EBT indicator and 5 ml of buffer are added. Then the solution is
titrated against EDTA until wine red color changes from red to blue. This is taken as end
point. Titrations are repeated until two successively concurrent values are obtained. The
morality of EDTA is calculated by using the formula
V1M1=V2M2
PART B: DETERMINATION OF TOTAL HARDNESS OF TAP WATER:
20ml of tap water is taken into a well cleaned conical flask. To it 5ml of
buffer and 2-3 drops of EBT indicator are added. The color of resulting solution is wine
red. The contents of the conical flask are titrated against EDTA which is taken in the
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burette until the colour of the solution changes from wine red to blue. This is the end point.
Repeat the titration until two successively concurrent values are obtained.
The molarity of tap water is calculated by using the formula M2V2=M3V3 and
M1V1=M2V2
M1 = Molarity of standard Zn2 +
solution
V1= Volume of standard Zn2 +
solution
V2=Volume of EDTA rundown
M2 = Molarity of DTA Solution
M2 = M1V1/V2
OBSERVATION: PART A: STANDARDISATION OF EDTA SOLUTION:
S.NO
Volume of
standard Zn+2
solution (ml)
Burette readings
Initial Final Volume of EDTA run
down ‘x’ ml
1. 20 0
2. 20 0
3. 20 0
V1M1=V2M2
M1= Molarity of tap water
V1= Volume of tap water
M2= Molarity of EDTA solution
V2= Volume of EDTA rundown
M2=M1V1/V2.
ENGINEERING CHEMISTRY LAB – I B.TECH 11
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PART B:
DETERMINATION OF TOTAL HARDNESS OF TAP WATER:
S.NO Volume of tap
water (ml)
Burette readings (ml)
Initial Final Volume of EDTA run
down ‘x’ ml
1. 20 0
2. 20 0
3. 20 0
M2V2=M3V3
M2= Molarity of EDTA solution
V2= Volume of EDTA rundown
M3= Molarity of tap water
V3= Volume of tap water
M3=M2V2/V3.
PART C:
DETERMINATION OF PERMANENT HARDNESS OF WATER:
Place 100ml of the sample of water in a beaker and boil gently for 10-
20 minutes. Cool and filter, collecting the filtrate into a 100ml standerd flask. Make up the
solution to the mark by adding distilled water and shake the solution well. Pipette out 50ml of
this made up solution into a clean conical flask, which has been rinsed wth distilled water.
Add 10ml of pH = 10 buffer solution and 2 to 3 drops of EBT indicator. Titrate this solution
against the standard EDTA solution until the colour changes from wine red to blue. Note
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down the reading, repeat the process to get at least two equal titer values. Calculate the
permanent and then temporary hardness as parts per million of CaCO3.
EDTA BOILED WATER
M2 = calculated in 2nd
step M4 =?
V2 = burette reading V4 = 50 ml
n2 = 1 n4 = 1
M2V2 M4V4
n2 = n4
M4 = M2V2
V4
CALCULATIONS:
Hardness = conc.x 105 ppm.
Total hardness = M3 x 105 ppm
Permanent hardness = M4 x 10
5 ppm
RESULT: Amount of total hardness of the given sample water = ----------ppm
Amount of permanent hardness of the given sample water = --------ppm
Amount of temporary hardness of the given sample water = --------ppm
ENGINEERING CHEMISTRY LAB – I B.TECH 13
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VIVA QUESTIONS :
1) what is EDTA and write its structure?
2) what is buffer solution and mention its function?
3) which hardness can be removed by boiling?
4) what is hardness?
5) what is the indicator used and what is the colour change in this experiment?
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Experiment No . 3
DETERMINATION OF VISCOSITY OF OIL BY OSTWALD VISCOMETER
AIM: To determine the viscosity of given oil.
APPARATUS:
Ostwald Viscometer
stop watch.
Rubber bulb
Specific gravity bottle
Beaker
CHEMICALS:
Given oil (test liquid)
water.(std.liquid)
THEORY:
In the laboratory the viscometer commonly used for comparing viscosities of
liquids is the Ostwald viscometer. In its operation different liquids are taken in exactly the
same volume. This is essential so that the height of the liquid columns creating the pressure
heads are equal and then the pressure heads will be directly proportional to the densities of
the respective liquids. The liquids are sucked up in to the upper bulb A. initially enough
liquid volume is taken in bulb B such that when the liquid stands above the mark X in bulb A,
a little is still left in the bend and the bulb B. the time of flow (t) is measured for each liquid
for its flow from X to Y. the driving pressure P at all stages of the flow of the liquid is given
by h*d*g, where g is the acceleration due to gravity; d is density of liquids; h is the difference
between the heights of the liquids in the upper and lower bulbs (same for all liquids between
the same points x and y)
For any given capillary, radius, length being constant and volume fixed, kPt
where k includes all constant terms.
This equation leads to an easy comparison between the viscosities of different liquids.
22
11
22
11
2
1
tP
tP
tkP
tkP
ENGINEERING CHEMISTRY LAB – I B.TECH 18
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This equation can be simplified into
22
11
22
11
2
1
td
td
gthdgthd
PROCEDURE:
Take a clean and dry Ostwald viscometer and set it vertically on a stand
Fill the water through the larger bulb. Suck the liquid up into the bulb A through a rubber
cock attached to the end D to a level the mark X Allow the liquid to flow freely through the
capillary and note the time t1 for the liquid to flow from X to Y. Repeat steps 3 to 4 to obtain
the average time for the liquid to flow from X to Y.
OBSERVATION TABLE:
S.No Liquid taken Density (gm/cc) Time (Sec) Viscosity of
liquid(milli poise)
1 Water d1 = t1 = 1 =8.90
2 Test liq 1 d2 = t2 = 2 =
3 Test liq 2 d3 = t3 = η3=
CALCULATIONS:
2
1
22
11
td
td
ENGINEERING CHEMISTRY LAB – I B.TECH 19
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Y
X
Temp. 0C
RESULT: The viscosity of the given liquid = ………….. milli poise
VIVA QUESTIONS :
1) what is viscosity of a liquid?
2) what are the units of viscosity?
3) what are the effects of temperature and density on viscosity?
4) what are the apparatus used to determine the viscosity of a liquid?
5) what is the viscosity of water?
v
i
s
c
o
s
i
t
y
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Experiment No : 4
CONDUCTOMETRIC TITRATION
AIM:To determine the neutralization point in an acid base titration using conductivity meter.
APPARATUS :
Conductivity bridge
Conductivity Cell
Distilled water
250 ml beaker
Stirrer
CHEMICALS REQUIRED:
0.1M HCl
1.0 M NaOH.
PRINCIPLE:
As the alkali is added from the burette into the cell containing acid, the
concentration of H+ ions change in a graphical manner, which leads to a considerable change
in the electrical conductance of the solution, which is measured using a conductivity meter.
Then from the plot of conductance versus volume of alkali, the precise neutralization point is
determined. The titration of strong acid vz strong base involves the following equation.
HCl + NaOH ― NaCl + H2O
In the titration of HCl against NaOH, initially the conductance of HCl solution is maximum
due to complete ionization. As the alkali is added, the conductance of solution decreases and
after the neutralization point the conductance starts increasing. This is because of the addition
of the alkali. The fast moving H+ ions are replaced by Na
+ ions (slow moving) once the
neutralization point is reached, addition of alkali introduces fast moving OH- ions, there by
increasing the conductivity of the solution.
ENGINEERING CHEMISTRY LAB – I B.TECH 24
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PROCEDURE:
Pipette out 20ml of 0.1M HCl into a clean 250ml beaker. Fill the burette with
1.0M NaOH solution. Dip a conductivity cell in the 0.1 M HCl solution and connect it to the
conductivity meter and note the initial conductance of HCl solution. Run down the NaOH
solution from the burette in small volumes of 1ml with stirring till the end point to be
expected in to the cell and note the conductance.
OBSERVATION:
S.NO. Volume of NaOH (ml)
Conductance
(Ohm-1
or mhos)
1 0 ml
2. 1ml
3. 2ml
4. 3ml
5. 4ml
6. 5ml
7. 6ml
8. 7ml
9. 8ml
10. 9ml
11 10ml
12 11ml
13 12ml
14 13ml
15 14ml
ENGINEERING CHEMISTRY LAB – I B.TECH 25
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
Y
EP
X
M1V1=M2V2
RESULT: From the graph the volume of 1.0 M NaOH required for the neutralization (Point
of the graph) of 0.1M HCl is ------------
Conductance
Volume of NaOH
ENGINEERING CHEMISTRY LAB – I B.TECH 26
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VIVA QUESTIONS :
1)what is neutralization point?
2)why we don’t require an indicator in this experiment?
3)why the conductance decreases when we add NaOH to HCl solution?
4)After neutralization point why there is a rapid increase in the conductance?
5)what are the units of conductance?
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Experiment No : 5
CONDUCTOMETRIC TITRATION OF MIXTURE OF
ACIDS VS STRONG BASE
AIM: To determine the neutralization point in an mixture of acids-base titration using
conductivity meter.
APPARATUS :
Conductivity bridge
Conductivity Cell
Distilled water
250 ml beaker
Stirrer
CHEMICALS REQUIRED:
0.1M HCl
0.1M CH3COOH
1.0 M NaOH.
PRINCIPLE:
when a mixture containing CH3COOH & HCl is titrated against an alkali
,strong acid will b neutralized first, the neutralization of week acid [CH3COOH] commences
only after the complete neutralization of strong acid[HCl]. Then conductance titration curve
will be marked by two breaks. The first one corresponds to neutralization of HCl and second
one two that of CH3COOH.Let V1 & V2 ml be volume of alkali corresponding to first and
second breaks, respectively.
V1 ml of NaOH = HCl
(V1-V2) ml of NaOH=CH3COOH
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PROCEDURE:
Take a beaker and add 20ml of CH3COOH(0.1M) and 20ml of HCl(0.1M) . Fill
the burette with 0.1M NaOH solution. Place the conductivity cell in distilled water.. Now
measure initial conductance of mixture of solution. Then 1 ml of NaOH is added every time
from burette into the solution and stirred well each time. Note down the conductance values
till the conductance values decreases and increase considerably.
OBSERVATION:
S.NO. Volume of NaOH (ml) Conductance (µ or
mhos)
1 0 ml
2. 1ml
3. 2ml
4. 3ml
5 4ml
6 5ml
7 6ml
8 7ml
9 8ml
10 9ml
11 10ml
12 11ml
13 12ml
14 13ml
15 14ml
ENGINEERING CHEMISTRY LAB – I B.TECH 31
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GRAPH:
Y
Conductance v2
v1
X
Vol. of NaOH
M1V1=M2V2
RESULT: From the graph the volume of 1.0 M NaOH required for the neutralization (Point
of the graph) of 0.1M HCl and 0.1M CH3COOH is---------
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VIVA QUESTIONS :
1) How many neutralization points are there in this experiment?
2) which acid reacts first with NaOH?
3) write the reaction between CH3COOH and NaOH?
4) what is the basic principle involved in this experiment?
5) How to determine the end points?
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Experiment No . 6
ESTIMATION OF MANGANESE DIOXIDE IN PYROLUSITE
AIM: To estimate the amount of MnO2 present in the given sample of pyrolusite
APPARATUS:
Standard flask
Pipette
Burette
Conical flask.
CHEMICALS REQUIRED:
Pyrolusite
Oxalic acid
Dil H2SO4
KMnO4
PRINCIPLE :
Manganese dioxide occurs in nature in the form of pyrolusite. The percentage
of MnO2 is determined by treatment with excess of acidified solution of a reducing agent
thus the MnO2present in pyrolusite sample is reduced to a known excess of standard
sodium oxalate as acid medium and the unreacted sodium oxalate is titrates against
standard solution of KMnO4.
MnO2 + H2SO4 + H2C2O4 2CO2+2H2O+MnSO4
The ore is graded on the basis of its available oxygen content rather than on its percentage
of Manganese.
PROCEDURE :
STEP I : STANDARDIZATION OF KMnO4 SOLUTION :
Pipette out 20 ml of oxalic acid in a clean conical flask add 20 ml of 3M H2SO4
and heat the conical flask to 700c . Fill the burette with KMnO4 solution and place the hot
conical flask under the burette and rundown KMnO4 in drop wise manner .At the end
point solution turns from colorless to pale pink . Note the burette reading and repeat the
experiment for concurrent values .
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STEP II :
ESTIMATION OF MnO2 IN PYROLUSITE :
Weigh accurately 0.2 gm of finely powdered dry pyrolusite in to 250 ml conical flask and
add 50 ml of N/10 oxalic acid .Add 50 ml of diluteH2SO4 (10%) and place a short funnel
over the conical flask .Boil the contents of the flask gently until no black particles are
visible in the flask . Allow it to cool about 700c and titrate the excess oxalate with standard
KMnO4 solution until a permanent pink color appears .Note the burette reading and repeat
the experiment for concurrent values .
CALCULATIONS :
S.No Volume of oxalic
acid
Burette reading
(ml)
Volume of
KMnO4(ml)
1 20ml Initial Final
2 20ml
3 20ml
M1V1 M2V2
-------- = --------
n1 n2
oxalic acid (n1)=5 KMnO4 (n2)=2
M2= M1 * 20 * 2
V2 * 5
Molarity of KMnO4 = -----------------------M
Molarity * M.Wt=Normality * equivalent wt
. Normality= Molarity * 158
31.6
=-------------------------N
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STEP 3:
Weight ore taken =0.2 gm
Volume of KMnO4 required for titration(V)= -----------------
Volume of KMnO4 consumed = 50-V
1 ml of 1 N KMnO4 = 0.04346gm of MnO2
1 ml of --------------- N KMnO4 = ?
0.04346 * N of KMnO4
--------------------------- = Y gm of MnO2
1 N
CALCULATIONS:
1 ml of ----------------N KMnO4 = Y gm of MnO2
(50-V) ml of ---------- N KMnO4 = ?
(50-V) * Y = A gm of MnO2
0.2 gm of Pyrolusite ore contains A gm of MnO2
100 gm of Pyrolusite ore contains ---------------------
100 * A = B gm
0. 2
Therefore % of MnO2 in pyrolusite = B %
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RESULT:
Molarity of oxalic acid = -----------------
Molarity of KMnO4 = -----------------
The percentage of MnO2 present in the given Pyrolusite Sample = B %
VIVA QUESTIONS :
1) what is pyrolusite?
2) what is self indicator in this experiment?
3) what is the equivalent weight of KMnO4?
4) what is molarity and normality?
5) what is the basic principle of pyrolusite experiment?
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Experiment No . 7
POTENTIOMETRIC TITRATION OF STRONG ACID VS STRONG BASE
AIM: To determine the neutralization point in the potentiometric titration of strong acid
vs strong base.
APPARATUS:
Potentiometer bridge
saturated calomel electrode
platinum electrode
Beaker
Burette
Stirrer
CHEMICALS REQUIRED:
N/10HCl
N/10 NaOH
Distilled water.
PRINCIPLE:
In the titration of 0.1N HCl with 0.1N NaOH on addition of the alkali, there is
variation in the concentration of H+ ions, there variations in the concentration of H
+ ions i.e.
PH is followed potentiometrically using Quinhydrone (reversible to hydrogen ion) as, the
indicator electrode in the HCl solution and coupling it with saturated calomel electrode
(reference electrode) since the potential of the lather remains constant, the emf of the cell will
vary only with PH of HCl solution. Therefore by measuring this emf act each stage of the
titration and plotting it against the volume of base, we can deduce the equivalence point from
the plot. At the end point, the emf increases at once which is clearly detected in the of graph.
Cell reaction is:
Hg / Hg2Cl2 (s).KCl(saturated solution)(s) // H+, QH2/Pt.
ENGINEERING CHEMISTRY LAB – I B.TECH 42
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
PROCEDURE:
Pipette out 20ml of 0.1N HCl solution into beaker and saturate it with
quinhydrone and dip the indicator electrode (platinum electrode) connect the indicator
electrode and saturated calomel electrode (reference electrode) to the potentiometer. The two
half cells are connected by means of a salt bridge. The potentiometer is standardized and
used for measuring the emf directly.
Take 20 ml of HCl solution in the beaker and add 1 ml of 0.1N NaOH
from the burette every time. Shake well after each addition and measure the cell emf. From
this rough titration find out the approximate volume needed for the end points. Now a fair
titration is repeated by adding volumes of 1ml alkali. Subsequent additions are made in
steps of 1or 2 ml of alkali. Then plot a graph between the measured emf on Y-axis and
volume of alkali on X-axis.
OBSERVATION:
Vol. of alkali
(ml) EMF (mv)
1ml
2ml
3ml
4ml
5ml
6ml
7ml
8ml
9ml
10ml
ENGINEERING CHEMISTRY LAB – I B.TECH 43
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
CALCULATIONS: N1V1 = N2V2
Where N1 = Strength of acid(0.1N)
V1 = Volume of acid taken in the beaker
N2 = Strength of the Base(0.1N)
V2 = ? (Volume of alkali from the graph)
RESULT:
From the graph the equivalence point for the acid base titration, is
potentiometrically determined to be ……ml
ENGINEERING CHEMISTRY LAB – I B.TECH 44
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
VIVA QUESTIONS :
1) what is a potentiometric titration?
2) what are the electrodes used in the experiment?
3) what is the indicator electrode?
4) what is neutralization point?
5) How can you determine the strength of acid from neutralization point?
ENGINEERING CHEMISTRY LAB – I B.TECH 45
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
SPACE FOR OBSERVATIONS AND CALCULATIONS
ENGINEERING CHEMISTRY LAB – I B.TECH 46
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
SPACE FOR OBSERVATIONS AND CALCULATIONS
ENGINEERING CHEMISTRY LAB – I B.TECH 47
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
Experiment No . 8
POTENTIOMETRIC TITRATION OF WEAK ACID VS STRONG BASE
AIM: To determine the neutralization point in the potentiometric titration of weak acid
versus strong base.
APPARATUS:
Potentiometer bridge
saturated calomel electrode
platinum electrode
Beaker
Burette
Stirrer
CHEMICALS REQUIRED:
N/10 CH3COOH
N/10 NaOH
Distilled water.
PRINCIPLE:
In the titration of 0.1N CH3COOH with 0.1N NaOH on addition of the alkali,
there is variation in the concentration of H+ ions, there variations in the concentration of H
+
ions i.e. PH is followed potentiometrically using Quinhydrone (reversible to hydrogen ion)
as, the indicator electrode in the CH3COOH solution and coupling it with saturated calomel
electrode (reference electrode) since the potential of the lather remains constant, the emf of
the cell will vary only with PH of CH3COOH solution. Therefore by measuring this emf act
each stage of the titration and plotting it against the volume of base, we can deduce the
equivalence point from the plot. At the end point, the emf increases at once which is clearly
detected in the of graph.
ENGINEERING CHEMISTRY LAB – I B.TECH 48
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
Cell reaction is
Hg/Hg2Cl2 (s).KCl(saturated solution)(s) // H+, QH2/Pt.
PROCEDURE:
Pipette out 20ml of 0.1N CH3COOH solution into beaker and saturate it
with quinhydrone and dip the indicator electrode (platinum electrode) connect the indicator
electrode and saturated calomel electrode (reference electrode) to the potentiometer. The two
half cells are connected by means of a salt bridge. The potentiometer is standardized and
used for measuring the emf directly.
Take 0.1N NaOH in the burette and add 1 ml of CH3COOH solution in the
beaker everytime. Shake well after each addition and measure the cell emf. From this rough
titration find out the approximate volume needed for the end points. Now a fair titration is
repeated by adding volumes of 1ml alkali. Subsequent additions are made in steps of 1or 2
ml of alkali. Then plot a graph between the measured emf on Y-axis and volume of alkali on
X-axis.
OBSERVATION:
Vol. of alkali
(ml) EMF (MV)
1ml
2ml
3ml
4ml
5ml
6ml
7ml
8ml
ENGINEERING CHEMISTRY LAB – I B.TECH 49
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
GRAPH:
Y
EP
EMF
X
Vol.of NaOH
CALCULATIONS:
N1V1 = N2V2
Where N1 = Strength of the Acid
V1 = Volume of acid taken in the beaker
V2 = ? (Volume of alkali from the graph)
N2 = Strength of the base
RESULT:
From the graph the equivalence point for the acid base titration, is
potentiometrically determined to be ……ml
ENGINEERING CHEMISTRY LAB – I B.TECH 50
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
VIVA QUESTIONS :
1) what is a potentiometric titration?
2) what are the electrodes used in the experiment?
3) what is the indicator electrode?
4) what is neutralization point?
5) How can you determine the strength of acid from neutralization point?
ENGINEERING CHEMISTRY LAB – I B.TECH 51
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
SPACE FOR OBSERVATIONS AND CALCULATIONS
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MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
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MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
Experiment no . 9
DETERMINATION OF PERCENTAGE OF COPPER IN BRASS
AIM: To determine the Percentage of Copper in the given sample of brass
APPARATUS: Conical flask
250 ml standard flask
Test tube
Burette
Pipette
Beaker
CHEMICALS:
Brass alloy
Nitric acid
Urea
Ammonium hydroxide
Hypo
Starch
PRINCIPLE:
Brass is an alloy of copper and zinc. When dissolved in concentrated nitric
acid, both the metals get converted to their nitrates.
Cu + 4HNO3 Cu(NO3)2 + 2H2O +2NO2
When brass solution is treated with excess of KI solution, Cu+2
ions oxidize to
liberate equivalent quantity of I2.
2 Cu(NO3)2 + 4KI Cu2I2 + 4KNO3 + I2
The liberated I2 is titrated with hypo solution using starch as indicator.
Na2S2O3 + I2 Na2S4O6 + 2NaI
The oxides of nitrogen present in brass solution are destroyed by adding urea.
The presence of nitrogen oxides will be responsible for the liberation of extra I2 from KI, as
they are also good oxidizing agents. The nitric acid present in brass solution is neutralized by
adding NH4OH till a pale blue ppt of Cu(OH)2 is obtained. Otherwise HNO3 may also liberate
ENGINEERING CHEMISTRY LAB – I B.TECH 54
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
I2 from KI. The Cu(OH)2 ppt is dissolved in dilute acetic acid. Other mineral acids are not
preferable as they lower the PH to a very low value and at which liberation of I2 from KI by
Cu+2
is not quantitative .Starch reacts with I2 to form a blue colored complex. At the end
point, when free I2 is exhausted in the solution, added quantity of hypo dissociates starch-I2
and liberates starch, thereby discharging the blue color. End point is the color change from
BLUE to COLORLESS. During the liberation of I2 from KI, Cu+2
gets reduced to
Cu+1
.Hence the equivalent weight of Cu = atomic weight of Cu= 63.54
PROCEDURE:
Weigh accurately about 1 gm brass alloy. Transfer into a clean beaker. Dissolve the
alloy in ¼ test tube conc. HNO3. Then add about 1 gm of Urea and 1 test tube of distilled
water. Boil for 2 minutes. Cool the products and transfer quantitatively into a 250 ml standard
flask. Make up the solution to the mark and shake well .Pipette out 25 ml of Brass solution
into a conical flask. Add NH4OH solution drop wise until a bluish white ppt persists. Add
dilute acetic acid solution drop wise just to dissolve the ppt and about 5 ml excess, followed
by the addition of 10 ml of 10% KI solution .Titrate this solution against hypo solution taken
in the burette until the color changes to pale yellow. At this stage add starch solution (1 ml)
and continue the titration till the blue color changes to white .Repeat the titration for
concordant values.
OBSERVATIONS AND CALCULATIONS:
Weight of brass = …………gm
In burette = hypo solution
In conical flask = 25 ml of brass solution, NH4OH, Acetic acid,10 ml of 10% KI
Indicator = starch solution
Color change = Blue To Colourless
S.No Volume of
Brass
solution
Initial
reading on
the burette
Final reading Volume of hypo used
V1 ml
1 25 ml
2 25 ml
3 25 ml
ENGINEERING CHEMISTRY LAB – I B.TECH 55
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
N1= Normality of Hypo
V1 = Volume of Hypo
N1 × V1 = N2 × V2
N2 = N1 X V1 = N1 X V1
V2 25
Weight of Copper in 1litre of solution = N2 X Equivalent weight of copper
= N2 X 63.54 = …………gm
N2 X 63.54
Wt of Copper in 250ml is =
4
N2 X 63.54 x 100
% of Copper in Brass =
4 X Wt of Brass
RESULT:
Percentage of Copper in the given sample of brass is ………………….
ENGINEERING CHEMISTRY LAB – I B.TECH 56
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
VIVA QUESTIONS :
1) What is brass?
2) What is the colour formed when starch reacts with iodine?
3) What is the role of KI in the experiment?
4) What is the indicator used?
5) What is the basic principal involved in this experiment?
ENGINEERING CHEMISTRY LAB – I B.TECH 57
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
SPACE FOR OBSERVATIONS AND CALCULATIONS
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MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
Experiment No . 10
PREPARATION OF THIOKOL RUBBER
AIM:To Synthesize Thiokol rubber using sodium poly sulphide with 1,2 dichloro ethane.
APPARATUS:
Beakers
Glass rod.
CHEMICALS:
NaOH,
Powdered Sulphur
Ethylene chloride (1,2 dichloride ethane )
Benzene
5% Sulphuric acid
Nitric acid.
THEORY:
It is a rubbery white substance and is obtained by treating sodium polysulfide with
1, 2 – dichloro ethane.
S S S S
|| || || ||
nCl – CH2 – CH2 – Cl + n Na+
- S- - S
- - Na
+ → (- CH2 – CH2 - S – S - )n + (2n – 1 ) NaCl
PROCEDURE:
In a 100 ml beaker dissolve 2 g NaOH in 50 -60 ml warm water.Boil the
solution and to this add in small lots with constant stirring 4 g of powdered sulfur. During
addition and stirring the yellow solution turns deep – red.Cool it to 60 – 700C and add 10 ml
of 1,2 – dichloromethane (ethylene chloride)with stirring . Stir for an additional period of 20
minutes while rubber polymer separated out as a lump.Pour out the liquid from the beaker in
the sink to obtain Thiokol rubber. Wash it with water under the tap.Dry in the fold of filter
papers. The yield is about 1.5 g. Determine the solubility of the polymer in benzene, acetone,
5% H2SO4 and nitric acid.
ENGINEERING CHEMISTRY LAB – I B.TECH 60
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
NOTE: If some sulfur remains un dissolved filter the solution.
RESULT:
Yield obtained = --------------g.
VIVA QUESTIONS :
1) what are the chemicals used in the preparation of Thiokol?
2) what are the different monomers are used?
3) what is the End point colour?
4) Is it natural rubber or synthetic rubber?
5) which rubber is not vulcanized with sulphur?
ENGINEERING CHEMISTRY LAB – I B.TECH 61
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
SPACE FOR OBSERVATIONS AND CALCULATIONS
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ENGINEERING CHEMISTRY LAB – I B.TECH 63
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
Experiment No . 11
STUDY OF ADSORPTION OF OXALIC ACID
ON ACTIVATED CHARCOAL
AIM: Study of adsorption of oxalic acid from solution on activated charcoal
CHEMICALS :
Oxalic acid
NaOH
Phenolphthalein indicator
Charcoal
APPARATUS: Reagent bottles
Funnel
Filter paper
Glass rod
BASIC PRINCIPLE:
Physical adsorption ,chemical adsorption and adsorption isotherms.
Adsorption :- This is a surface phenomena
Absorption :- This is a bulk phenomena
Adsorbent :- The Solid substance on which adsorption takes place.
Adsorbate :- The substance which undergo adsorption .
PROCEDURE:
1. PREAPARATION OF 0.25M OXALIC ACID: Weigh out exactly
3.15gm of oxalic acid into a 1000 ml standard flask and make up the solution to the mark
with distilled water after dissolving the salt in little distilled water. Shake the flask well for
uniform concentration.
ENGINEERING CHEMISTRY LAB – I B.TECH 64
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
2. PREPARATION OF 0.1M NaOH SOLUTION: Dissolve 4 gms of
NaOH in 1000 ml of water and stir the solution well for uniform concentration.
3. PHENOLPHTHALEIN INDICATOR: Dissolve 1 gm of
phenolphthalein indicator in100ml of ethanol.
Take well cleaned and dried 6 stoppered reagent bottles and label them. With the
help of two burettes transfer and oxalic acid and distilled water into these 6 bottles as shown
below:
Reagent bottle no. 1 2 3 4 5 6
Oxalic acid
(0.25M)(ml)
50 40 30 20 10 00
Distilled water(ml) 50 60 70 80 90 100
Mix the solutions well and keep all the six bottles in a water bath for some time
to acquire the temperature of the water bath. Weigh exactly 2 gms of activated charcoal on 6
glazed papers and add the same to each bottle. Shake the bottle well and keep in the water
bath for half an hour time while shaking from time to time. Filter the contents of all the 6
bottles into 6 different labeled dried conical flasks. Titrate the 6 samples with M/10 NaOH
taken in the burette using phenolphthalein indicator till pale pink colored end point is
obtained. Let the titre values be V3,V4,V5,V6,V7,V8respectivelyfor bottle no.s 1,2,3,4,5,6.
Take 20 ml of the stock solution into a clean 250 ml conical flask, add 2 drops of
phenolphthalein indicator and titrate against 0.1M NaOH solution till pale yellow color is
obtained as end point. Repeat the titration concurrent values. Let the titre value be x ml.
CALCULATIONS:
Molarity of oxalic acid solution prepared (M1) = Wt. of oxalic acid X 10 /126
Molarity of NaOH solution (M1) = V1M1/n1 = V2M2/n2
V1 = Volume of oxalic acid (20ml)
M1 = Molarity of oxalic acid
n1 = no. of moles of oxalic acid = 1
ENGINEERING CHEMISTRY LAB – I B.TECH 65
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
V2 = Volume of NaOH (titre value x ml)
n2 = no. of moles of NaOH reacted = 2
M2 = V1M1/n1 X n2/ V2
The initial concentration of the oxalic acid solutions in 1 to 6 conical flasks
Bottle no. Initial concentration(C)
1 50 X M2 /100 = a
2 40 X M2 /100 = b
3 30 X M2 /100 = c
4 20 X M2 /100 = d
5 10 X M2 /100 = e
6 00 X M2 /100 = zero
Concentration of oxalic acid solutions after adsorption:
Bottle no. 1 = M (oxalic acid) = 2 X M (NaOH) X V3 / 50 = p
Bottle no. 2 = M (oxalic acid) = 2 X M (NaOH) X V3 / 40 = q
Bottle no. 3 = M (oxalic acid) = 2 X M (NaOH) X V3 / 30 = r
Bottle no. 4 = M (oxalic acid) = 2 X M (NaOH) X V3 / 20 = s
Bottle no. 5 = M (oxalic acid) = 2 X M (NaOH) X V3 / 10 = t
Amount of Oxalic acid adsorbed x = (C1-C2) X100/1000 = (C1-C2)/10
Where, C1 = initial concentration
C2 = conc. after adsorption
ENGINEERING CHEMISTRY LAB – I B.TECH 66
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
Bottle no. (C1) initial
concentration
(C2) concen.
after adsorption
Amount of oxalic acid
adsorbed (C1-C2)/10 =
x
1
a p a-p/10
2
b q b-q/10
3
c r c-r/10
4
d s d-s/10
5
e t e-t/10
RESULT:
Amount of Oxalic acid adsorbed x = ………….
ENGINEERING CHEMISTRY LAB – I B.TECH 67
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
VIVA QUESTIONS :
1) what is adsorption?
2) what is adsorbate in the experiment?
3) what is adsorbent in the experiment?
4) what is the indicator used and what is the colour change?
5) Is adsorption bulk phenomena or surface phenomena?
ENGINEERING CHEMISTRY LAB – I B.TECH 68
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
SPACE FOR OBSERVATIONS AND CALCULATIONS
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Experiment : 12
ESTIMATION OF FERROUS IRON BY DICHROMETRY.
AIM: To estimate the amount of ferrous iron present in the given solution using standard
Potassium Dichromate.
APPARATUS:
Burette
Pipette
Conical flask
Burette stand
standard flask
weighing box
weighing bottle.
CHEMICALS:
Potassium Dichromate
Iron solution
sulphuric acid
phosphoric acid
Diphenylamine indicator.
PRINCIPLE:
Potassium Dichromate is an oxidizing agent. It liberates nascent oxygen
atoms in presence of dilute sulphuric acid.
K2Cr2O7 + 4H2SO4 K2SO4 + Cr2 (SO4)3 + 4H2O + 3(O)
The liberated nascent oxygen oxidizes ferrous salt to ferric salt in cold and in
presence of acid.
6FeSO4 + 3H2SO4 + 3(O) 3Fe2(SO4)3 + 3H2O
K2Cr2O7 +7H2SO4 + 6FeSO4 K2SO4 + Cr2(SO4)3 + 3Fe2(SO4)3 + 7H2O
ENGINEERING CHEMISTRY LAB – I B.TECH 72
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
PROCEDURE:
Estimation of ferrous iron involves two steps:
1) Preparation of standard solution of potassium dichromate,
2)Estimation of ferrous iron.
STEP-I
PREPARATION OF STANDARD POTASSIUM DICHROMATE
Weight of empty weighing bottle (w1) =
Weight of the weighing bottle + salt (w2) =
Weight of K2Cr2O7 (W3) = w2-w1
= gm
Normality of potassium dichromate= Weight of K2Cr2O7 × 1000
Gram equivalent weight V ml
STEP-I
PREPARATION OF STANDARD SOLUTION OF POTASSIUM
DICHROMATE:
About 0.5 g of potassium dichromate crystals are taken in a
clean dry weighing bottle and its weight is found accurately in analytical balance .Then it is
transferred into 100ml standard flask through the funnel. The weight of the weighing bottle is
again found out accurately. The difference between two weights gives the weight of
potassium dichromate. Then the substance is dissolved and made up to the mark with distilled
water, shaken well to get uniform concentration and kept aside for use. The Normality of
potassium dichromate is calculated as follows:
Normality of potassium dichromate=Weight of K2Cr2O7 × 1000
Gram equivalent weight v ml
ENGINEERING CHEMISTRY LAB – I B.TECH 73
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
STEP-II
ESTIMATION OF FERROUS IRON:
The given ferrous solution is made up to the mark with
distilled water and shaken well for uniform concentration. The burette is rinsed with distilled
water and then it is filled with potassium dichromate and initial reading is noted. Pipette is
cleaned and rinsed with given iron solution. A 20ml of iron solution is pipette out in a clean
conical flask and 5ml of acid mixture and few drops of diphenylamine indicator are added. It
is titrated by adding potassium dichromate from burette until violet blue is observed. This is
the end point. Then final burette reading is noted and the experiment is repeated to get
concurrent readings. Then estimate the amount of ferrous iron present in the whole of given
solution.
ESTIMATION OF FERROUS IRON:
S. No Volume of Iron (V2)
ml
Burette Reading(ml) Volume of K2Cr2O7
(V2) ml
Initial Final
1.
2.
3.
NIVI = N2V2
Volume of K2Cr2O7 (V1) =
Normality of K2Cr2O7 (N1) =
Volume of ferrous iron solution (V2) = 20ml
Normality of ferrous iron solution(N2) = ?
ENGINEERING CHEMISTRY LAB – I B.TECH 74
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
Normality of Ferrous solution (N2) = N1V1
V2
= N.
Amount of ferrous solution present in the given solution =
Normality of Iron (N2) ×Eq.Wt of Iron × 100
1000
RESULT:
The amount of Fe2+
present in the given solution is ________gm.
ENGINEERING CHEMISTRY LAB – I B.TECH 75
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
VIVA QUESTIONS :
1) what is an oxidizing agent?
2) what is the indicator used in the experiment?
3) which principal is involved in this experiment?
4) what is the end point colour of this experiment?
5) Define Normality and Molarity?
ENGINEERING CHEMISTRY LAB – I B.TECH 76
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
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ENGINEERING CHEMISTRY LAB – I B.TECH 79
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
Experiment : 13
DETERMINATION OF SURFACE TENSION
AIM: To determine the effect of soap and detergent on the surface tension of water using
a Stalagmometer.
APPARATUS: Stalagmometer
Specific gravity bottle
Rubber cork
Analytical balance
CHEMICALS: Soap solution.
Methanol (or) Ethanol
PRINCIPLE:
Surface tension is the characteristic properly of every liquid and it is due to
intermolecular attraction among molecules of liquid. A molecule in the interior part of the
liquid is attracted by the surrounding molecules on the surface of the liquid are attracted only
towards the interior i.e., sides and the bottom of a constant tension due to the downwards
flow of the molecules in bulk, this tension at the surface is known as surface tension. It is
defined as the force in dynes acting on a surface at right angles to any line of unit length. It is
denoted by ”γ” (gamma) and its units are dynes/cm.
ENGINEERING CHEMISTRY LAB – I B.TECH 80
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
Table given below lists the surface tension of several liquids at 200C.
Liquid Surface tension
(dynes/cm)
Liquid Surface tension
(dynes/cm)
Water 72.8 Ethylene glycol 47.7
Benzene 28.9 Glycerol 63.4
Toluene 28.4 Carbon tetrachloride 27.0
Acetone 23.7 Ethyl iodide 29.9
Methyl alcohol 22.6 Ethyl bromide 24.2
Ethyl alcohol 22.3 Nitrobenzene 41.8
PROCEDURE
The determination of surface tension of a liquid involves the following 2 steps.
1.Determination of density of a liquid
Density of a liquid is mass/unit volume. It is found by using specific gravity bottles.
The specific gravity bottle is first washed with distilled water and finally with alcohol and
dried. The weight of the empty specific gravity bottle is found by using analytical balance.
Let it be W1gm.Then it is filled with the distilled water and its weight is accurately
determined. Let it be W2 gm. Now the given unknown liquid is then filled in the specific
gravity bottles and the weight us found. Let this weight be W3 gm the density of the unknown
liquid (d2) is going to be calculated as:
Density of liquid (d2) = Weight of liquid
X Density of water
Weight of water
ENGINEERING CHEMISTRY LAB – I B.TECH 81
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
2. DETERMINATION OF SURFACE TENSION OF A GIVEN LIQUID:
The stalagmometer is first washed thoroughly with distilled water and finally with
little alcohol and dried. A clean rubber tube is attached to the upper end of the stalagmometer.
A screw pitch cork is fixed on the rubber tube to regulate the flow of liquid by lifting the
influx of air
The stalagmometer is dipped in a beaker of water and suck the water till it rises above
the upper mark (X). In the stalagmometer the water level is carefully brought to the mark (X).
The pitch cork is opened in such a way that it allows the flow of 12-18 drops per minute.
Now the stalagmometer is adjusted.
Then water is sucked above the mark and the no of drops of water that flows from the mark
(X) to the mark (Y) are noted. This is repeated 3 to 4 times with water.
Now the stalagmometer is rinsed with the liquid whose surface tension is to be
determined. The liquid is then filled in the stalagmometer as above and no. of drops that flow
from the mark(X) to mark (Y) is noted. This is repeated 3 or 4 times with the liquid and the
experimental results are noted in the tabular for
CALC ULATIONS:
Calculation of density of the liquid
Room temperature = 27oC
Weight of empty specific gravity bottle (W1) =
Weight of specific gravity bottle + water (W2) =
Weight of specific gravity bottle + given liquid (W3) =
Weight of water = (W2-W1) =
Weight of given liquid = (W3-W1) =
Density of water (d1) = 1 gm/cc
ENGINEERING CHEMISTRY LAB – I B.TECH 82
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
Density of liquid (d2) = Weight of liquid
X Density of water
Weight of water
S. No.
Volume of
liquid
Number of drops
Surface
tension
(dynes/cm) Exp.1 Exp. 2 Average
(N)
1 Std.liquid
2 Test liquid-1
3 Test liquid-2
Therefore the surface tension of given liquid (γ2) is determined busing the following
equation.
n1d2
γ 2 = x γ 1
n2d2
Where
γ 1= Surface tension of water = 72.8 dynes/cm
n1= no. of drops of water
d1= density of water
γ 2= surface tension of given liquid
n2= no. of drops of given liquid
d2= density of given liquid
ENGINEERING CHEMISTRY LAB – I B.TECH 83
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
RESULT:
The surface tension of the given test liquid γ2 = …………… dyne/cm
VIVA QUESTIONS :
1) Define surface tension?
2) what are the units of surface tension?
3) what is the apparatus used to find out surface tension?
4) what are the forces that causes surface tension?
5) what is the value of surface tension of water?
ENGINEERING CHEMISTRY LAB – I B.TECH 84
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
SPACE FOR OBSERVATIONS AND CALCULATIONS
ENGINEERING CHEMISTRY LAB – I B.TECH 85
MARRI LAXMAN REDDY INSTITUTIONS, DUNDIGAL, HYD.
SPACE FOR OBSERVATIONS AND CALCULATIONS