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1
CYCLE I
2
Known Iron(III):
S.NO Volume of Iron(III)
sample in ml Absorbance /
Optical density
1 5
2 10
3 15
4 20
5 25
6 30
3
EXPERIMENT – 1
COLORIMETRIC ESTIMATION OF IRON (III) WITH THIOCYANATE
Aim: To estimate ferric ion using thiocyanate as complexing agent.
Reagents:
1. Ferric ion solution: In a 250 ml volumetric flask 0.432 gm of Ammonium ferric sulphate
is weight. To this 5 ml conc. HCl is add to produce clear solution. The solution make up to
250 ml with distilled water.
2. 10% Potassium thiocyanate (KSCN): 10 gm of KSCN crystals are dissolving in 100 ml
of distilled water.
3. 4N Hydrochloric acid: 18.5 ml of conc. HCl is dilute to 100 ml with distilled water.
4. Working solution: 5 ml of ferric ion solution is dilute to 250 ml in a volumetric flask with
distilled water.
Theory: Ferric ion reacts with thiocyanate to give a series of intensified red color compounds.
These complexes are red and can be formulated as [Fe(SCN)n]2+ where n = 1,2----6. At low
thiocyanate concentration the predominant colour complex is [Fe(SCN)]2+. At very high SCN-
concentration, it is [Fe(SCN)6]3-
Fe3+ + SCN- [Fe(SCN)]2+
Principle:
Iron is one of the many minerals required by the human body. It is used in the
manufacture of the oxygen-carrying proteins, haemoglobin and myoglobin. A deficiency of iron
in the body can leave a person feeling tired and listless, and can lead to a disorder called anemia.
Many of the foods we eat contain small quantities of iron. In this analysis the iron present in an
iron tablet (dietary supplement) or a sample of food is extracted to form a solution containing
Fe3+ (ferric) ions. To make the presence of these ions in solution visible, thiocyanate ions
(SCN−) are added. These react with the Fe3+ ions to form a blood-red coloured complex.
By comparing the intensity of the colour of this solution with the colours of a series of standard
solutions, with known Fe3+ concentrations, the concentration of iron in the tablet or food sample
may be determined. This technique is called colorimetry.
Colorimeter measures the optical density of an absorbing substance where optical density (O.D)
is defined as O.D = log 1oI
I ; Where oI = Intensity of incident light; I = Intensity of
transmitted light.
As per beers law, optical density of an absorbing substance is related to the concentration by the
equation. . . .O D E C l (or) . ( . ). 2O D E l C
Where ‘C’ is the concentration of the substance, l is the path length, which represents the width
of the cell used and is constant for a given cell used, E is the molar absorption coefficient and is
a constant for given substance. Equation 2 may be written as
O.D. C 3
Equation 3 represents the quantitative form of Beer’s law. If the optical density of a substance is
determined at varying concentration, a plot of O.D. Vs C gives a straight line
4
Unknown Iron(III):
S.NO Volume of Iron(III)
sample in ml Absorbance /
Optical density
1 Unknown-I
2 Unknown-II
5
Procedure:
1. About six 50 ml volumetric flasks are take and into each flask x ml (i.e, x= 5 to 30)of
iron(III) ion solution, 5 ml of 10% thiocyanate solution and 4 ml of 4N HCL are run down
the solution is make up to the mark instantaneously well shake the solution.
2. Add same quantities of 10% thiocyanate solution and 4N HCL to unknown iron (III)
samples and make up to the mark instantaneously well shake the solutions.
3. The optical density (OD) of each sample is measure with 490 nm visible light immediately.
This cyanate complex is unstable, so the readings are taking quickly.
4. A graph is draw with OD along y-axis and volume of the iron (III) along x-axis for known
solutions. The graph is obtained is a straight line.
5. Using optical density values of unknown samples determine the volume of unknown iron
(III) from graph.
Report:
S.NO. Volume of Iron(III)
given (ml)
Volume of Iron(III)
reported (ml)
% Error Marks Signature of the
faculty
1
2
6
Observation and Calculations:
S. No. Temperature
Time required to
flow 50 ml of oil (in
Seconds)
Kinematic Viscosity
(Centistokes)
V = At – B/t
Average
Kinematic
Viscosity
(Centistokes)
The kinematic viscosity of the liquid is given by the formula
V = At - B/t
V = Kinematic viscosity of oil in centistokes
t = Time of flow for 50 ml of oil in seconds
A and B are instrument constants
S. No Type of
equipment Time of flow A value B value
1 Redwood 1 40 to 85 secs 0.264 190
2 Redwood 1 85 to 2000 secs 0.247 65
3 Redwood 2 --- 0.027 20
7
EXPERIMENT – 2
ESTIMATION OF KINEMATIC VISCOSITY OF THE LUBRICATING
OIL
Aim: To determine the kinematic viscosity of given lubricating oil at a given temperature by
using Redwood Viscometer
Principle: The internal drag arises between two successive layers of the liquid is known as
viscosity. Further, the force per unit area required to maintain the velocity gradient by one unit
between two successive layers of one unit length apart is known as viscosity coefficient. High
viscous liquids move slowly while low viscous liquids move fast through a given capillary.
Further, the time required to flow a given volume of liquid through a capillary depends on its
viscosity. Therefore, the viscosity of liquid can be determined by determining the time required
to flow the known volume of liquid through a standard capillary. Viscosity is expressed in poise.
Procedure:
The Redwood viscometer consists of oil cup which is opened at the upper end and it is
fitted with an orifice
It is cleaned thoroughly with suitable solvent and then dried
The orifice is covered with brass ball to stop the flow of oil
The oil cup is placed in the cylindrical copper vessel which serves as water bath
The bath is filled with suitable liquid which has the boiling point higher than the
temperature at which the viscosity of oil to be determined
If the viscosity of the oil is to be determined at 800C or below, the bath is filled with water
The instrument level is adjusted on the tripod stand with the help of the leveling screws
Now the oil cup is filled with oil to be tested carefully up to the level indicated and the
covered with lid
Two thermometers, one is in the oil and the other one is in the liquid (water) are immersed
Similarly two stirrers also placed in the oil and the liquid
One 50 ml flask is kept in position below the jet
Now the oil is heated slowly with constant stirring of oil and the water until it reaches to the
required temperature at which the viscosity of the oil is to be determined
When the temperature of the oil has quite steady and reaches the required temperature, the
brass ball is lifted and simultaneously the stop watch is started.
The oil is allowed to flow through the orifice and collected in the flask
Stop watch is stopped when 50 ml of oil is collected in the flask up to the mark and
immediately the orifice is covered with brass ball to stop the over flow of the oil
The time required to flow the 50 ml of oil is noted
The oil cup is refilled again with oil and same procedure is repeated for five to six times
The viscosity of oil is calculated at given temperature
8
9
Precautions
The oil should be filtered through a 100 mesh wire sieve before testing for its viscosity
Receiving flask should be placed in such a way that the oil jet touches inside layer of the
flask and does not form foaming
Same receiving flask should be used for all readings
After each reading, oil should be completely drained out of the receiving flask and it
should be thoroughly cleaned and dried
Report:
Name of the
Lubricating Oil
Average of kinematic viscosity
(Centistokes)
Marks awarded Signature of the faculty
10
Step-1: Observations & Calculations:
Burette : NaOH Solution
Conical Flask : 20 ml Oxalic Acid
Indicator : Phenolphthalein
End Point : Colour less to Pale pink.
S.No. Volume of Oxalic
acid (v1 ml)
Burette readings (ml) Volume of NaOH Rundown
(v2 ml) Initial Final
1
2
3
4
Calculations:
Oxalic Acid:
M1 = Molarity of Oxalic Acid = 0.05M
V1 = Volume of Oxalic Acid = 20 ml
n1 = Moles of Oxalic Acid = 1 mole
Sodium Hydroxide:
M2 = Molarity of NaOH =?
V2 = Volume of NaOH =
n2 = Moles of NaOH = 2 mole
Molarity of NaOH = ----------------- M
11
EXPERIMENT-3
CONDUCTOMETRIC TITRATION OF ACID BY BASE
Aim: To determine the amount of unknown acid solution with standard base solution by
conductometric method.
Apparatus: Conductivity meter (with cell), burette (10ml), volumetric flask (100 ml), beakers
(100 ml), stirrer / glass rod.
Chemicals: Stock acid solution, 0.05 M oxalic acid in100ml volumetric flask and Stock base
solition.
PRINCIPLE: Conductometric titrations works on the principle of Ohm's law. As current is
inversely proportional to Resistance (R) and the reciprocal of resistance is termed as
Conductance, and its unit is Siemen (mho) cm-1. The electrical conductivity of a solution
depends on the number of ions and their mobility. In Conductometric titrations, the titrant is
added from the burette, and the conductivity readings are plotted against the volume of the
titrant. Upon adding a strong base to the strong acids, the conductance falls until the strong acid
is neutralized then raised. Such a titration curve consists of 2 lines which intersect at a particular
point, known as the End point or Equivalence point. The method can be used for titrating
coloured solutions or homogeneous, which cannot be used with normal indicators.
Strong Acid with a Strong Base:
For example, in the titration of HCl versus NaOH, the addition of a strong base (NaOH)
to a strong acid (Hcl). Before NaOH is added, the conductance is high due to the presence of
highly mobile hydrogen ions. When the base is added, the conductance falls due to the
replacement of hydrogen ions by the added cation as H+ ions react with OH- ions to form
undissociated water. This decrease in the conductance continues till the equivalence point. At the
equivalence point, the solution contains only NaCl. After the equivalence point, the conductance
increases due to the large conductivity of OH- ions.
H2C2O4 + 2 NaOH Na2C2O4 + 2H2O
HCl + NaOH NaCl + H2O
Formula:
Procedure:
Step 1: Standardization of sodium hydroxide by using oxalic acid
1. Rinse and fill the burette with the given NaOH solution
2. Pipette out 20 ml of 0.05 M oxalic acid solution into a clean conical flask
3. Add 1 or 2 drops of phenolphthalein indicator to oxalic acid solution.
4. Titrate the solution against sodium hydroxide solution drop wise with shaking till the
solution changes to pale pin
5. Note the volume of NaOH used. It is the end point.
6. Repeat the titration until the concordant readings are obtained
7. Calculate the molarity of NaOH by using the formula mentioned above
12
Observations and Calculations: Conductometric titration in between HCl and NaOH
Volume of base added Conductance Corrected conductance
C1= C[(v+V)/V]
Calculation of Unknown molarity of HCl solution:
Sodium Hydroxide:
M2 = Molarity of NaOH =
V2 = Volume of NaOH =
n2 = Moles of NaOH = 1 mole
HCl:
M3= Molarity of HCl =?
V3= Volume of HCl = 25 ml
n3 = Moles of HCl = 1 mole
Molarity of HCl = ----------------- M
Amount of HCl = ---------------- grs/l
13
Step 2: Determination of molarity of unknown HCl by using standard NaOH through
conductometric titration
1. In 100 ml beaker take 25ml of given unknown HCl solution and add 25ml of distilled
water. The contents are shaken thoroughly.
2. Now, the conductivity cell is immersed in the beaker and the initial conductance of the
solution is taken by stirring the solution and keeping it constant.
3. Then, 0.5 ml portions of base is added from the burette and stirred well. The conductance
of the solution for each addition is to be noted.
4. The conductivity is corrected by multiplying with the factor [(v+V)/V], where 'v' is the
volume of base added and 'V' is the volume of solution initially taken in the beake
5. Plot the graph with respect to the volume of base consumed versus corrected
conductance. From the intersection point on the graph which gives value represents the
equivalence points of acid and base.
Report:
S.No Given Amount of
unknown Acid
Reported Amount of
unknown Acid
% Error Marks Signature of the
Faculty
1
14
Observations & Calculations:
Step-1: Standardization of EDTA
Burette : EDTA Solution
Concial Flask : 20 ml CaCl2 + 1 ml. of Ammonia Buffer
Indicator : Erichrome Black-T Indicator
End Point : Wine red to Sky Blue.
S.No. Volume of CaCl2
(v1 ml)
Burette readings (ml) Volume of EDTA Rundown
(v2 ml) Initial Final
1
2
3
N1V1 = N2V2
Normality of calcium chloride (N1) = 0.02N
Volume of calcium chloride (V=) = 20 ml
Normality of EDTA (N2) = ?
Volume of EDTA (V2) = volume of EDTA rundown in ml
N2 = N1V1/V2
N2=---------------- N
15
EXPERIMENT NO- 4
DETERMINATION OF HARDNESS OF WATER WITH EDTA
Sample Details: Area: The water sample was collected from ______________(Write the source)
Source: Aim: To estimate the total hardness present in the given water sample.
Apparatus: Burette, Burette Stand, Pipette, Conical Flask, Beaker, Wash Bottle, Glazed tile,
Glass Funnel.
Reagents: EDTA, CaCl2 (0.02N), Ammonia Buffer (pH=10), Erichrome Black-T (EBT)
indicator, Distilled Water.
Principle: In alkaline condition, EDTA reacts with Calcium and Magnesium to form chelated
complex. Ca and Mg develop winered colour with EBT indicator under alkaline condition.
When EDTA is added as titrant, Ca & Mg divalent ions get complexed resulting in a sharp
change from wine red to sky blue colour which indicates end point of the titration. The pH for
the titration has to be maintained at 10.0
complexEBTMg
CaEBT
Mg
Ca pH
10
2
2
Wine red (unstable)
EBTcomplexEDTAMg
CaEDTAcomplexEBT
Mg
CapH
10
Wine red (unstable) Stable Sky Blue
Procedure:
Step-1: Standardization of EDTA
1) Take 20ml. CaCl2 (0.02N) solution in a clean conical flask.
2) To this add 1 ml. of Ammonia Buffer.
3) Add a pinch / 1 (or) 2 drops of EBT indicator.
4) The solution turns to Wine red colour.
5) Titrate with EDTA till the colour changes from Wine red to sky Blue.
6) Note down the volume of EDTA rundown (v2 ml).
7) Repeat the procedure till the concordant readings are obtained.
Step-2: Determination Total Hardness of water sample
1) Take 20ml. water sample in a clean conical flask.
2) To this add 1 ml. of Ammonia Buffer.
3) Add a pinch / 1 (or) 2 drops of EBT indicator.
4) The solution turns to Wine red colour.
5) Titrate with Std. EDTA till the colour changes from Wine red to sky Blue.
6) Note down the volume of EDTA rundown (v3 ml).
16
Step-2: Determination Total Hardness of water sample
Burette : Std. EDTA Solution
Concial Flask : 20 ml water sample + 1 ml. of Ammonia Buffer
Indicator : Erichrome Black-T Indicator
End Point : Wine red to Sky Blue.
S.No. Volume of Water
sample (ml)
Burette readings (ml) Volume of EDTA Rundown
(v3 ml) Initial Final
1
2
3
Calculation of Total Hardness:
= V3 x N2 × 50 ×1000
________________
Volume of Sample
= _____________ mg / litre.
Total Hardness is = _______________ mg / litre or ppm
17
Result:
Source of water sample Hardness (mg/l or ppm) Marks awarded Signature of the faculty
Significance: 1. The permissible limit of total hardness is 200 mg / litre as per W.H.O.
2. Absolutely softwater is tasteless and corrosion in nature.
3. Hardwater causes excessive consumption of soap.
4. Scales are formed in the boilers and reduce the heat efficiency of the boilers.
5. Important in determining the suitability of water for domestic and industrial use.
6. Determination of hardness serves as a basis for routine control of softening process
18
Observations & Calculations:
Step-1: Standardisation of Sodium thiosulphate soludion: Burette : Sodium thiosulphate solution.
Conical flask : Standard K2Cr2O7 + 10% KI + H2SO4
Indicator : 2 ml starch solution
End Point : Blue to colourless
S.No. Volume of Potassium
dichromate (ml)
Burette readings (ml) Volume of Hypo Rundown
(V ml) Initial Final
1
2
3
Potassium dichromate Sodium thiosulphate
N1 = N2 =
V1 = V2 =
N2 = 2
11
V
VN = …………….. N
19
EXPERIMENT NO - 5
ESTIMATE THE AMOUNT OF DISSOLVED OXYGEN BY MODERN
WINKLER’S METHOD
Aim: To find the quantity of Dissolved Oxygen present in given water sample.
Apparatus: BOD Bottles (300ml), Burette, Pipette, Conical Flask, Burette Stand, Wash bottle.
Reagents: 0.02N K2Cr2O7, 48% Manganese Sulphate, Alkali-Iodide-Azide Reagent, Starch
Indicator, Sodium Thiosulphate, Con. Sulphuric Acid, Sodium Hydroxide, sample water,
Distilled water.
Principle: Step-1: As the thiosulphate is a secondary standard solution, it has to be standardized by titrating
against a primary standard dichromate solution iodometrically using starch indicator.
K2Cr2O7 + 6KI + 7H2SO4 Cr2(SO4)3 + 4K2SO4 + 3I2 + 7H2O
Step-2:
Oxygen present in the water sample oxidizes divalent manganese (Mn2+) to its higher oxidation
state (Mn4+) in the presence of Alkali-Iodide-Azide solution. Oxidized manganese liberates
iodine from potassium iodide in acidic medium. The amount of Iodine liberated is equivalent to
dissolved oxygen present in water sample. The liberated iodine is estimated by titrating with
0.025N hypo using starch as an indicator (wrinkler titration).
OHMnOOOHMn 222212 2
(brown ppt)
22
2
2 242 IOHMnHIMnO
6423222 22 OSNaNaIOSNaI
(Sodium tetrathionate)
Formulas:
The strength of sodium thiosulphate N2 = 2
11
V
VN
The amount of dissolved oxygen in the given water sample
= Titre value x conc. of hypo x 8 x 1000 mg/lit.
Volume of sample
Procedure:
Step-1:
Standardisation of Sodium thiosulphate (hypo) solution:
1) Fill the burette with sodium thiosulphate solution.
2) Pipette 10ml of standard potassium dichromate solution into Conical Flask. Add 5 ml of
5 % KI solution and 5 ml of 5N H2SO4 solution.
3) Cover the conical flask with watch glass and put into dark place at 10 min.
4) Titrate the sample solution with thiosulfate until the solution becomes pale yellow.
Introduce 5 drops of starch indicator, and titrate with constant stirring to the
disappearance of the blue colour. Note down the burette reading.
5) Calculate the normality of Sodium thiosulphate.
20
Step-2: Determination of Dissolved oxygen in water sample: Burette : Std. Hypo solution (0.025N)
Concial Flask : 20 ml water sample consists of liberated Iodine
Indicator : 2 ml starch solution
End Point : Blue to colorless.
S.No. Volume of Water sample (ml) Burette readings (ml) Volume of Hypo Rundown
(V3 ml) Initial Final
1
2
3
Dissolved oxygen in the given water sample
= V3 x N2 x 8 x 1000
20
= _______________
= ________ mg /litre
21
Step-2:
Determination of dissolved oxygen present in water sample:
1) Collect 250 ml water sample in a 300ml capacity of BOD bottle.
2) Add 2 ml of manganese sulphate and 2 ml of Alkaline-Iodine-Azide solution.
3) Stopper the BOD bottle immediately.
4) Appearance of brown ppt. indicates the presence of DO.
5) Mix well by inverting the bottle 2 to 3 times and allow the brown ppt. to settle down.
6) Add 2 ml. of conc. H2SO4 solution to dissolve the precipitate.
7) Take 20 ml. of this solution into a clean conical flask.
8) Titrate the liberated iodine with standard hypo solution present in the burette.
9) Add 2 ml of starch solution when the colour of the solution becomes pale yellow. The
solution turns to blue colour.
10) Continue the titration till the blue colour is disappeared.
11) Note the volume of hypo used (V ml)
12) Repeat the titration till the concordant readings are obtained.
13) Calculate the amount of Dissolved Oxygen in the given water sample by using the
formula
Result:
Source of water sample Dissolved oxygen (mg/l) Marks awarded Signature of the faculty
Environmental Significance: 1. The level of dissolved oxygen in fresh water bodies is 8-15 mg/L at 250c
2. If the concentration of dissolved oxygen of water is below 6 ppm, the growth of fish gets
inhibited.
3. The dissolved oxygen is used by microorganisms to oxidise organic matter
4. Oxygen depletion helps in release of phosphates from bottom sediments and causes
eutrophication.
22
23
CYCLE II
24
Step-1: Observations & Calculations:
Burette : NaOH Solution
Conical Flask : 20 ml Oxalic Acid
Indicator : Phenolphthalein
End Point : Colour less to Pale pink.
S.No. Volume of Oxalic
acid (v1 ml)
Burette readings (ml) Volume of NaOH Rundown
(v2 ml) Initial Final
1
2
3
4
Calculations:
Oxalic Acid:
M1 = Molarity of Oxalic Acid = 0.2M
V1 = Volume of Oxalic Acid = 20 ml
n1 = Moles of Oxalic Acid = 1
Sodium Hydroxide:
M2 = Molarity of NaOH =?
V2 = Volume of NaOH =
n2 = Moles of NaOH = 2
Molarity of NaOH = ----------------- M
25
EXPERIMENT-1
pH METRIC TITRATION OF ACID BY BASE
Aim: To determine the Amount of unknown acid solution with standard base solution by pH
metric method.
Apparatus: pH Meter, Glass membrane electrode, 100ml Beaker, Burette, Volumetric Flask,
Glass Rod.
Chemicals: Stock acid solution, 0.2 M oxalic acid and Stock base solution.
Principle: When a glass surface is in contact with a solution it acquires a potential which
depends on H+ ion concentration of solution. This observation which has been made by Haber is
now used as basis of method of determining the pH of a solution where other electrode cannot be
used. The glass electrode has attained much attention in recent years because it can be used
almost in all solutions except those which are strongly acidic or strongly alkaline. It has been
observed that potential difference exists at the interface between glass and solution containing H+
ions. The magnitude of the difference of potential for a given variety of glass varies with its ions
concentration at 250C given by:
E = E0 + 0.0591 log [H+]; E0 = A constant for the given glass electrode.
H2C2O4 + 2 NaOH Na2C2O4 + 2H2O
HCl + NaOH NaCl + H2O
Formula:
Procedure:
Step 1: Standardization of sodium hydroxide by using oxalic acid
1. Rinse and fill the burette with the given NaOH solution
2. Pipette out 20 ml of 0.2 M oxalic acid solution into a clean conical flask
3. Add 1 or 2 drops of phenolphthalein indicator to oxalic acid solution.
4. Titrate the solution against sodium hydroxide solution drop wise with shaking till the
solution changes to pale pin
5. Note the volume of NaOH used. It is the end point
6. Repeat the titration until the concordant readings are obtained
7. Calculate the molarity of NaOH by using the formula mentioned above
Step 2: Determination of molarity of unknown HCl by using standard NaOH through pH
metric titration
1. Rinse and fill the burette with standard NaOH solution
2. Now you collect unknown acid in 100 ml volumetric flak and makeup with distilled water
26
Step-2: Observations and Calculations: pH Metric titration in between HCl and NaOH
VOLUME OF NaOH
ADDED
pH
Calculation of unknown molarity of HCl solution:
Sodium Hydroxide:
M2 = Molarity of NaOH =
V2 = Volume of NaOH =
n2 = Moles of NaOH = 1
HCl:
M3= Molarity of HCl =?
V3= Volume of HCl = 5ml
n3 = Moles of HCl = 1
Molarity of HCl = ----------------- M
Amount of HCl = ---------------- grs/l
27
3. Take 5ml of given unknown HCl solution into 100ml beaker and add 45ml of distilled
water. The contents are shaken thoroughly.
4. The glass membrane electrode is dipped into the beaker containing the solution.
5. Initially at “0” Burette reading of NaOH solution, pH of the unknown HCl solution can be
measured.
6. Then 0.5 ml of base is added from the burette to the acid solution and on stirring
thoroughly the pH of the resultant solution can be noted.
7. The pH is noted every time by the addition of 0.5 ml base and finally you observe the pH
jump is between V1 and V2 ml. After pH jump you need to note about 10 readings.
8. Plot the graph with the volume of base on X - axis versus pH on Y-axis. Identify the
suitable jump which changes the medium from acidic pH to Basic pH.
9. Take the average in-between the jump values and draw a line which intercepts X axis. The
intersection point gives value of the equivalence point (End point) of acid and base
Report:
S.No Given Amount of
unknown Acid
Reported Amount of
unknown Acid
% Error Marks Signature of the
Faculty
1
28
Step-1: Observations & Calculations:
Burette : NaOH Solution
Conical Flask : 20 ml Oxalic Acid
Indicator : Phenolphthalein
End Point : Colour less to Pale pink.
S.No. Volume of Oxalic
acid (v1 ml)
Burette readings (ml) Volume of NaOH Rundown
(v2 ml) Initial Final
1
2
3
4
Calculations:
Oxalic Acid:
M1 = Molarity of Oxalic Acid = 0.05M
V1 = Volume of Oxalic Acid = 20 ml
n1 = Moles of Oxalic Acid = 1 mole
Sodium Hydroxide:
M2 = Molarity of NaOH =?
V2 = Volume of NaOH =
n2 = Moles of NaOH = 2 mole
Molarity of NaOH = ----------------- M
29
EXPERIMENT-2
CONDUCTOMETRIC TITRATION OF AN ACID MIXTURE BY BASE
Aim: To determine the amount of unknown acids with standard base solution by conductometric
titration.
Apparatus: Conductivity meter (with cell), burette (10ml), volumetric flask (100 ml), beakers
(100 ml), stirrer / glass rod.
Chemicals: HCl solution, CH3COOH solution, 0.05 M oxalic acid and NaOH solution.
Principle: Conductometric titrations works on the principle of Ohm's law. As current is
inversely proportional to Resistance (R) and the reciprocal of resistance is termed as
Conductance, and its unit is Siemen (mho) cm-1. The electrical conductivity of a solution
depends on the number of ions and their mobility. In Conductometric titrations, the titrant is
added from the burette, and the conductivity readings are plotted against the volume of the
titrant. Upon adding a strong base to the strong acid and weak acid mixture, the conductance falls
until the strong acid is neutralized then raised slightly until weak acid neutralized then raised
rapidly. Such a titration curve consists of 3 lines which intersect at two particular points, known
as the End points or Equivalence points. First equivalence point corresponds to strong acid and
strong base titration. Second equivalence point corresponds to weak acid and strong base
titration. The method can be used for titrating coloured solutions or homogeneous, which cannot
be used with normal indicators.
Theory: In the titration addition of a strong base (NaOH) to mixture of a strong acid (HCl) and
weak acid (CH3COOH). Before NaOH is added, the conductance is high due to the presence of
highly mobile hydrogen ions. When the base is added, the conductance falls due to the
replacement of hydrogen ions by the added cation as H+ ions react with OH- ions to form
undissociated water. This decrease in the conductance continues till the first equivalence point.
After the first equivalence point, the conductance increases slightly due to neutralization of
CH3COOH and formation of CH3COONa and finally increased rapidly due to complete
neutralization of CH3COOH and excess addition of NaOH.
H2C2O4 + 2 NaOH Na2C2O4 + 2H2O
HCl + NaOH NaCl + H2O
CH3COOH + NaOH CH3COONa + H2O
30
Observations and Calculations: Conductometric titration in between mixture of HCl&
CH3COOH and NaOH
Volume of NaOH added Conductance Corrected conductance
C1= C[(v+V)/V]
31
Formula:
; ;
Procedure:
Step 1: Standardization of sodium hydroxide by using oxalic acid
1. Rinse and fill the burette with the given NaOH solution
2. Pipette out 20 ml of 0.05 M oxalic acid solution into a clean conical flask
3. Add 1 or 2 drops of phenolphthalein indicator to oxalic acid solution.
4. Titrate the solution against sodium hydroxide solution drop wise with shaking till the
solution changes to pale pin
5. Note the volume of NaOH used. It is the end point.
6. Repeat the titration until the concordant readings are obtained
7. Calculate the molarity of NaOH by using the formula mentioned above
Step 2: Determination of molarity of unknown HCl and CH3COOH by using standard
NaOH through conductometric titration
1. In 250 ml beaker take 25ml of given unknown HCl solution and 25ml of given unknown
CH3COOH solution add 50ml of distilled water. The contents are shaken thoroughly.
2. Now, the conductivity cell is immersed in the beaker and the initial conductance of the
solution is taken by stirring the solution and keeping it constant.
3. Then, 0.5 ml portions of base is added from the burette and stirred well. The conductance
of the solution for each addition is to be noted.
4. The conductivity is corrected by multiplying with the factor [(v+V)/V], where 'v' is the
volume of base added and 'V' is the volume of solution initially taken in the beake
5. Plot the graph with respect to the volume of base consumed versus corrected conductance.
From the intersection points on the graph which gives value represents the equivalence
points of acid and base.
32
Calculation of Unknown molarity of HCl solution:
Sodium Hydroxide:
M2 = Molarity of NaOH =
V2 = Volume of NaOH = n2 = Moles of NaOH = 1 mole
HCl:
M3= Molarity of HCl =?
V3= Volume of HCl = 25 ml
n3 = Moles of HCl = 1 mole
Molarity of HCl = ----------------- M
Amount of HCl = ---------------- gm/l
Calculation of Unknown molarity of CH3COOH solution
Sodium Hydroxide:
M2 = Molarity of NaOH =
V2 = Volume of NaOH = n2 = Moles of NaOH = 1 mole
CH3COOH:
M3= Molarity of CH3COOH =?
V3= Volume of CH3COOH = 25 ml
n3 = Moles of CH3COOH = 1 mole
Molarity of CH3COOH = ----------------- M
Amount of CH3COOH = ---------------- gm/l
33
Report:
S.No Given Amount of HCl &
CH3COOH
Reported Amount of HCl
& CH3COOH
% Error Marks Signature of
the Faculty
1
2
34
KNOWN:
Pilot Titration Accurate Titration
S.NO. Volume of K2Cr2O7 Potential in volts S.NO. Volume of K2Cr2O7 Potential in volts
35
EXPERIMENT NO-3
POTENTIOMETRIC TITRATION OF IRON (II) WITH POTASSIUM
DICHROMATE
Aim: To determine the concentration of Ferrous ammonium sulphate using standard potassium
dichromate following a potentiometric method.
Apparatus: Potentiometry, Standard calomel electrode,
Solutions required: 0.1N K2Cr2O7, 0.1N Ferrous ammonium sulphate and 1:1 H2SO4
General Principle:
Potentiometric Titrations: Potentiometric titrations involve the measurement of the potential of
a suitable indicator electrode with respect to a reference electrode as a function of titrant volume.
Potentiometric titrations provide more reliable data than data from titrations that use chemical
indicators and are particularly useful with colored or turbid solutions and for detecting the
presence of unsuspected species.
A typical cell for potentiometric analysis consists of a reference electrode, an indicator electrode
and a salt bridge. This cell can be represented as
A reference electrode, Eref, is a half-cell having a known potential that remains constant at
constant temperature and independent of the composition of the analyte solution. The reference
electrode is always treated as the left-hand electrode in potentiometric measurements. Calomel
electrodes and silver/silver chloride electrodes are types of reference electrodes. An indicator
electrode has a potential that varies with variations in the concentration of an analyte. Most
indicator electrodes used in potentiometry are selective in their responses. Metallic indicator
electrode and membrane electrodes are types of indicator electrodes. The third component of a
potentiometric cell is a salt bridge that prevents the components of the analyte solution from
mixing with that reference electrode. A potential develops across the liquid junctions at each end
of the salt bridge. The junction’s potential across the salt bridge, Ej, is small enough to be
neglected. The potential of the cell is given by the equation;
Ecell= Eind – Eref + Ej
36
UNKNOWN:
Pilot Titration Accurate Titration
S.NO. Volume of K2Cr2O7 Potential in volts S.NO. Volume of K2Cr2O7 Potential in volts
Calculation:
Known: 20 ml of Fe(II) solution consume x ml of Cr(VI)
Unknown: x ml of Cr(VI) solution consume 20 ml of Fe(II) solution
y ml of Cr(VI) solution consume ? ml of Fe(II) solution
? ml = y×20/x
20 ml of Fe(II) solution contains = ? ml of Iron
100 ml of Fe(II) solution contains = ?? ml of Iron
?? = ?×100/20
37
Procedure:
1. 20.00 ml of ferrous ammonium sulphate is transferred to a 100 ml beaker by means of
burette.
2. To this 15.00 ml of distilled water and 5.00 ml of 1:1 H2SO4 are added and the contents
are shaken thoroughly.
3. The platinum electrode placed in the beaker the being connected to the positive end of the
potentiometer.
4. After measuring the emf of the solution in the beaker, the standard K2Cr2O7 add from the
burette in one ml portion while stirring the content in the beaker with glass rod and
corresponding e.m.f are note down.
5. At a certain point a sudden raise in potential is observed this shows that the end point is
in between these additions where the jump is observed thus titrated readings are called
point readings by which are can get the end point approximately.
6. Again taking the same content in the beaker and titrated with standard K2Cr2O7 in the
same manner by the addition of 1 ml dichromate solution up to the jump reading. But in
between point readings add only 0.1 ml dichromate and note down potential readings.
7. After the end point 4-5 ml of dichromate is added in ml portion and the corresponding
readings are note down.
8. The plot is made between the e.m.f values as ordinate (Y-axis) and volume of dichromate
added as the abscissa (X-axis). The volume of dichromate corresponding to the point of
inflection gives the equilibrium point for the titration.
9. The same procedure is repeat for the unknown solution of Fe(II). From their equilibrium
points the unknown volume is calculate from this its concentration can also calculate.
Report:
S.NO. Volume of Fe(II) % Error Faculty
Signature Given Reported
38
Observation and Calculation:
Step-1: Standardization of KOH Solution.
Burette : KOH solution
Conical flask : 20 ml. of Oxalic Acid
Indicator : Phenolphthalein
End point : Colourless to pink.
S. No. Vol. of Oxalic Acid
Burette readings Vol. of KOH
rundown Initial Final
Normality of KOH N2 = 2
11
V
VN
N1 = Normality of oxalic acid
V1 = Volume of the oxalic acid
V2 = Volume of the KOH
39
EXPERIMENT NO-4
DETERMINATION OF ACID NUMBER OF LUBRICATING OIL
Aim: To determine the Acid number of lubricating oil.
Apparatus: 50 ml burette, 20ml pipette, 250 ml conical flask, 100ml beaker, 250 ml beaker, 50
ml beaker and 50 ml measuring jar.
Chemicals: KOH solution, 0.02N oxalic acid, Oil sample, Phenolphthalein indicator, Ethyl
alcohol.
Principle:
The Acid number of lubricating oil is defined as the number of milligrams of potassium
hydroxide required to neutralize the free acid present in 1 g of the oil sample. In good lubricating
oils, the acid number should be minimum (<0.1). Increase in acid value should be taken as an
indicator of oxidation of the oil which may lead to gum and sludge formation besides corrosion.
Since free fatty acids present in the oil react with base, their quantity can be estimated by
titrating the known weight of the oil sample dissolved in a suitable solvent with a standard
alcoholic solution of KOH to a definite end point
RCOOH + KOH RCOOK + H2O
Procedure:
Step 1: Standardization of KOH
20 ml standard oxalic acid solution is pipette out into a 250 ml of conical flask and few
drops of Phenolphthalein indicator is added.
The above solution is titrated with standard KOH solution taken in the burette until the
solution changes from colorless to light pink colour
The same procedure is repeated until any two readings coincide
The concentration of KOH is calculated.
40
Step-2: Determination of Acid Number:
Burette : Std. KOH solution
Conical flask : 1gm. of lubricating oil + 5 ml of alcohol.
Indicator : Phenolphthalein
End point : Colourless to pink.
S. No. Vol. of
lubricating oil
Burette readings Vol. of KOH
Initial Final
Acid Number of given oil sample is
(mg of KOH required to neutralize the acid present in 1 gm of oil) =
1000
100..2 KOHofVolKOHofwtEqN
N2 = Normality of KOH
Eq. wt of KOH = 56.01
Vol. of KOH = titer value in the above titration
=------------------------
41
Step 2: Determination of Acid Number of given oil sample
1 gram (1.1 ml) of oil sample is taken in a 250 ml conical flask and dissolved in 5 ml of
Ethyl alcohol.
One or two drops of Phenolphthalein indicator is added and the solution is titrated with
KOH taken in the burette until the solution changes from colorless to light pink
The same procedure is repeated until any two readings coincide
The Acid Number of oil sample is calculated.
Result:
Name of
lubricating
oil sample
Weight of oil
sample
Acid number Marks awarded Signature of the faculty
42
Step 1: Standardization of KMnO4
S. No. Vol. of H2C2O4 Burette Readings
Vol. of KMnO4 Initial Final
Concentration of KMnO4 is calculated by using the formula
NKMnO4 =
4
422422
KMnOV
VNOCHOCH
43
EXPERIMENT – 5
ESTIMATION OF PERCENTAGE OF PURITY OF PYROLUSITE ORE
Aim: To determine the amount of manganese dioxide present in a given pyrolusite ore sample.
Principle:
Pyrolusite is one of the manganese ores. Manganese is present in the ore as its oxidised form,
manganese dioxide. It is widely used in industries for making of manganese alloys. Pyrolusite
powder is boiled with excess of known oxalic acid in the presence of conc. sulphuric acid to
reduce the Mn4+ in MnO2 to Mn2+. The excess of oxalic acid is back titrated with standard
potassium permanganate solution.
MnO2 + H2C2O4 + H2SO4 MnSO4 + 2H2O + 2CO2
5H2C2O4 + 2KMnO4 + 3H2SO4 K2SO4 + 2MnSO4 + 8H2O + 10CO2
Required Solutions:
0.25 N stock oxalic acid: Nearly 3.2g (=W) of oxalic acid is weighed and dissolved in 250
ml (=V) of water in a volumetric flask.
Vacid oxalic ofEq.wt
1000acid oxalicStock
WN
0.05 N oxalic acid solution: 20 ml (=Vx) of stock oxalic acid solution is taken in a 100 ml of
volumetric flask and the solution is made upto the mark. The concentration of the solution is
calculated by the formula.
)100(V acid Ooxalic
acid OoxalicStock
acid Ooxalicml
VNN x
0.05 N KMnO4 solution: Nearly 0.4gms of KMnO4 is weighed and dissolved in a small
portion of water, it is boiled for 5mins and filtered by using glass wool. The solution is made
upto 250ml. As KMnO4 cannot be obtained in pure form, it is standardized with standard
oxalic acid solution.
Conc. H2SO4 is required
44
Step 2: Preparation & Standardization of pyrolusite sample Solution
S. No. Vol. of
Sample
Burette Readings Vol. of KMnO4
Initial Final
Concentration of unreacted oxalic acid
acid Oxalic
44acid Oxalic
V
VNN
KMnOKMnO
Amt of oxalic acid unreacted (gr) = W2 = 1000
Vacid oxalic of Eq.wt. 2acid Oxalic N
Amt of oxalic acid taken (gr) = W3 = 1000
V acid oxalic of Eq.wt. 1acid OxalicStock N
Amt of oxalic acid reacted = W4 = W3-W2
Amt of MnO2 present in pyrolusite = 42
acid Oxalic of Wt.M.
MnO of Wt.Mol.W
% of MnO2 is pyrolusite = 100 sample of Wt.
MnO of Wt. 2
45
Procedure:
Step 1: Standardization of KMnO4:
20ml of the 0.05N oxalic acid solution is taken in a conical flask and add 5 ml of conc.
H2SO4 and dilute the mixture to 100ml.
The mixture is heated to 70-80oC
The solution is titrated with KMnO4 solution to be standardized in hot condition until the
solution becomes light pink.
The same procedure is repeated till two successive readings are coinciding.
Step 2: Preparation & Standardization of pyrolusite sample Solution
Nearly 0.6 g (=W1) of powdered pyrolusite sample is weighed accurately and transferred into
a 250 ml conical flask.
Exactly 50 ml (=V1) of stock oxalic acid is added with the help of burette.
10 ml of concentrated H2SO4 is also added.
Now the mixture is boiled gently to reduce MnO2 to MnSO4 until no black particles of
pyrolusite ore remain undissolved.
The mixture is cooled and diluted to 250 ml (=V2) in a volumetric flask.
20 ml of this solution is taken in a conical flask and diluted to 100 ml
5 ml of concentrated H2SO4 is added and heated to 70O-800C.
Now it is titrated with standard KMnO4 solution until light pink colour is formed.
The same procedure is repeated till two successive readings coincide.
The concentration of unreacted oxalic acid is calculated.
The percentage purity of pyrolusite is calculated.
Report:
Substance Percentage Reported Percentage Given Error
46
47
CYCLE III
48
Step-1: formation of o,p methyl phenol:
Step-2; formation of novalac:
Step-3: formation of bakelite:
49
EXPERIMENT – 1
PREPARATION AND CALCULATION OF THE YIELD OF PHENOL-
FORMALDEHYDE RESIN
Aim: To prepare the phenol – formaldehyde resin
Principle:
Phenol formaldehyde resins (PF) include synthetic thermosetting resins such as bakelite
obtained by the reaction of phenols with formaldehyde. Phenol-formaldehyde resins are formed
by a step-growth polymerization reaction that can be either acid- or base-catalysed.
Phenol is reactive towards formaldehyde at the ortho and para sites (sites 2, 4 and 6)
allowing up to 3 units of formaldehyde to attach to the ring. The initial reaction in all cases
involves the formation of ortho & para hydroxymethyl phenols. The ortho hydroxymethyl phenol
undergoes linear polymerization to form novalac which is fusible and soluble in most of the
organic solvents. Novolacs are phenol-formaldehyde resins made where the molar ratio of
formaldehyde to phenol of less than one. Hexamethylene tetramine or "hexamine" is a hardener
that is added to crosslink novolac. At ≥180 °C, the hexamine forms crosslink’s to form
methylene and dimethylene amino bridges. Base-catalysed phenol-formaldehyde resins are made
with a formaldehyde to phenol ratio of greater than one (usually around 1.5). These resins are
called resols.
When the molar ratio of formaldehyde: phenol reaches one, every phenol is linked
together via methylene bridges, generating one single molecule, and the system is entirely
crosslinked. In bakelite, this ratio is greater than one, and so it is very hard
Procedure:
5.0 ml of phenol is weighed and transferred into a clean and dried 250 ml beaker.
7 ml of formaldehyde solution is added carefully. (Caution : Avoid the inhaling the
vapours and spilling of it on body)
5.0 ml of glacial acetic acid and one or two spatula of hexamine (hexamethylene
tetramine) are added
The contents of the beaker are heated gently in a water bath.
The beakers is removed from the water bath and add conc. hydrochloric acid slowly
drop-wise with constant stirring.
After the addition of hydrochloric acid a white substance precipitates first. Finally, a pink
colored plastic clump is formed at the bottom of the beaker while stirring.
The plastic clump is now washed with distilled water for several times and then dried in a
oven
Now the sample is weighed and its weight is reported
50
51
Precautions:
Formaldehyde solution 37 % is very toxic by inhalation, ingestion and through skin
absorption. Readily absorbed through skin. Probable human carcinogen. Mutagen. May
cause damage to kidneys, allergic reactions, sensitization and heritable genetic damage.
Phenol is acute poisoning by ingestion, inhalation or skin contact may lead to death.
Phenol is readily absorbed through the skin. Highly toxic by inhalation.
Safety glasses and protective gloves required. The experiment should be performed under
a portable fume cupboard giving all-round visibility
Report:
Amount of Resin
formed
Marks awarded Sign. of the faculty
52
Observations & Calculations:
Water
Time of flow Average time
Benzene /
Aniline/
Toulene
Time of
flow
Average
time
Densities of different organic liquids:
Name of the Organic
Liquid
Density in gr./cc
BENZENE 0.8737
TOULENE 0.8625
ANILINE 1.02
Calculations:
l
w
l
w
l
w
t
t.
?...................l
53
EXPERIMENT – 2
ESTIMATION OF VISCOSITY OF AN ORGANIC SOLVENT BY USING
OSTWALD VISCOMETER
Aim: To determine the coefficient of viscosities of various liquids like benzene, Aniline and
acetic acid.
Principle: Resistance to flow of a liquid is known as viscosity. The retarding force is
proportional to the area of contact and the velocity gradient. The proportionality constant ‘η’ is
the coefficient of viscosity, and is characteristic of a liquid. When a liquid flows through a
capillary tube, η = πρr4t/8Vl, where ‘ρ’ is the density of liquid, ‘r’ is the radius of the tube, ‘t’ is
the time of flow, ‘l’ is the length of the capillary tube and V is the volume of liquid. This is
called absolute viscosity determination. But if the same volumes of two liquids are allowed to
flow through the same tube, then we have the relation.
l
w
l
w
l
w
t
t.
ηw =Viscosity of water; ηl = Viscosity of given organic liquid;
ρw= Density of water; ρl= Density of given organic liquid
tw= Time of flow of water; tl= Time of flow of given organic liquid
If the viscosity of one liquid is known, then the other can be calculated. An Ostwald’s viscometer
is used to determine relative viscosity of liquids
Procedure:
1. An Ostwald viscometer is cleaned by rinsing three times with small volumes of acetone and
dried.
2. The lower bulb is filled with distilled water and clamped vertically.
3. A rubber bulb is attached to the tip of the narrow limb and the liquid drawn up to a level
much above the upper mark.
4. The rubber bulb is removed and the liquid allowed to flow down freely.
5. A stop watch is started just as the liquid meniscus passes the upper mark and stopped when it
passes the lower mark.
6. The time of flow ‘tw’ is noted. Density of water is 0.9971 gr./cc and ‘ηw’ for water is 8.94
milli poise at 250C.
7. The viscometer is again cleaned and dried as before. It is then filled with the experimental
liquid (Benzene, Toluene, Aniline and Glacial acetic acid). 8. Time of flow ‘tl’ is determined as in the case of water. The densities of the liquids may either
be determined or taken from literature. 9. The coefficient of viscosity ‘ηl’ of the liquid can be calculated.
54
55
Result:
Name of the
Given Organic
solvent
Coefficient of viscosity (m.p) Marks awarded Signature of the faculty
56
Observation:
Sl.No. Source of Water
Sample
Turbidity (in NTU)
1
2
3
4
5
57
EXPERIMENT-3
NEPHELOMETRIC DETERMINATION OF TURBIDITY PRESENT IN THE
GIVEN WATER SAMPLE
Aim: To determine the turbidity of the various water samples.
Apparatus: Nephelometer, Test tubes
Chemicals: Standard turbidity suspensions 400 NTU, 100 NTU, 40 NTU, 20 NTU, and 10
NTU, Distilled water, sample water.
Principle: Turbidity is an equal property of sample that cause light to be absorbed and scattered
rather than thransmitted through the sample. It depends on the amount and particle size of the
suspended matter present in water. The method of determination is based on a comparison of the
internsity of the light scattered by the sample under defined conditions with the intensity of light
scattered by a standard reference suspension. The higher the intensity of scattered light, the
higher the turbidity. It is expressed as NTU. This can be measured by Nephelometer.
Procedure:
1) Connect the three pin plug to a main outlet and switch the instrument to ON position. Let
the instrument be warm up for about 2 to 3 minutes.
2) Set the Mode Selector to the 100 NTU. Take 4 to 5 ml distilled water in a test tube and
insert into cuvette holder. Adjust the Set Zero control to obtain zero reading on DPM
display.
3) Fill up another tet tube with 4 to 5 ml of 100 NTU standard suspension and insert in the
cuvette holder. Adjust the level control such that the DPM display reads a value of 100
NTU. Now the instrument is calibrated for the 100 NTU range, and any suspension in
this range can be measured.
4) Fill up another test tube with 4 to 5 ml of the unknown turbidity sample and insrt in the
cuvette holder. Note the reading on the DPM display. This reading directly corresponds
to the turbidity of the unknown sample in NTU.
5) Adjust the Mode selector to 40, 20 and 10 NTU range and repeat the procedure for any
unknown solution whose turbidity is expected to be within 40, 20 and 10 NTU
respectively.
Result:
Source of water sample Turbidity (NTU) Marks awarded Signature of the faculty
58
OBSERVATION TABLE: Determination of Flash point and Fire point:
Sl.No. Temperature in Degree Celcius Inference
59
EXPERIMENT – 4
DETERMINATION OF FLASH AND FIRE POINTS OF GIVEN OIL
SAMPLES
Aim: To determine the flash and fire point of a given fuel oil.
Apparatus required: Pensky marten’s apparatus, thermometer and beaker
Oli used: Kerosene / Diesel
Significance of experiment:
In industries oils are generally used for combustion, lubricant and cooling purposes. To
use oil for any purpose, we are required to know the flash point and fire point, to eliminate the
fire hazards and to set working temperature conditions. Flash point is also used as a means to
identify the presence of impurities in the lubricant oil.
Theory:
Flash point: it is the lowest temperature at which the oil gives off enough vapors to ignite for a
moment, when a test flame is brought near it.
Fire Point: It is the lowest temperature at which the oil burns continuously for atleast 5 seconds
when a test flame is brought near it.
DESCRIPTION OF THE APPARATUS: This is the closed cup apparatus and it consists the following given parts. The oil cup is
cylindrical vessel with lid. It is fitted on the outside with a flat circular flange. There are three
ports cut in the lid. The central one for admitting test flame and one on either side of it for
observing the flash over the lid. There is a shutter, which opens the port on the lid for
introducing test-flame over the oil surface. A stirrer is placed in the cup for stirring the oil to get
uniform temperature.
PROCEDURE: 1) Fill the oil cup to the mark, place the cover on the cup, then start heating.
2) Stirr the oil to main the uniform temperature.
3) Apply the test flame in the oil cup for every 5 seconds.
4) Repeat the same til the momentary flash occurs, then note the temperature as a flash
point.
5) Rise the temperature slowly and check for fire point for every degree rise of temperature.
Note the temperature as fire point, when oil glows for more than five seconds.
60
61
PRECAUTIONS: 1) Thermometer should not touch the bottom of the cup.
2) The energy regular should be adjusted so that the heating of oil is uniform slow and
steady.
3) After every trial allow the cup and thermometer to reach room temperature before
introducing the next sample.
4) The bulb of the thermometer should dip inside the oil sample.
RESULT:
Name of the
Given oil sample
Flash point(0C) Fire point (0C) Marks awarded Signature of the faculty
62
OBSERVATION TABLE: Determination of Flash point and Fire point:
Sl.No. Temperature in Degree Celcius Inference
63
CLEVELANDS APPARATUS
Aim: to determine the flash and fire point of a given fuel oil
Apparatus required: Clevelands apparatus, thermometer and beaker
Oli used: Kerosene / Diesel.
Significance of experiment:
In industries oils are generally used for combustion, lubricant and cooling purposes. To
use oil for any purpose, we are required to know the flash point and fire point, to eliminate the
fire hazards and to set working temperature conditions. Flash point is also used as a means to
identify the presence of impurities in the lubricant oil.
Theory:
Flash point: it is the lowest temperature at which the oil gives off enough vapors to ignite for a
moment, when a test flame is brought near it.
Fire Point: It is the lowest temperature at which he oil burns continuously for atleast 5 seconds
when a test flame is brought near it.
DESCRIPTION OF THE APPARATUS: The clevelands apparatus consists of an open cup to which a handle is fitted. There is a
mark for the maximum level of oil inside the cup. The cup is heated electrically by separate
arrangement of integrated device consist of a heater and power supply regulator. The test flame
is supplied over the surface of oil by simple mechanism. The thermometer dipped manually in
the oil to read the temperature.
PROCEDURE: 1) Fill the oil cup to the mark, place the cover on the cup, then start heating.
2) Stirr the oil to main the uniform temperature.
3) Apply the test flame in the oil cup for every 5 seconds.
4) Repeat the same til the momentary flash occurs, then note the temperature as a flash
point.
5) Rise the temperature slowly and check for fire point for every degree rise of temperature.
6) Note the temperature as fire point, when oil glows for more than five seconds.
7) After reaching the fire point, put off the heater.
PRECAUTIONS: 1) Thermometer should not touch the bottom of the cup.
2) The energy regular should be adjusted so that the heating of oil is uniform slow and
steady.
3) After every trial allow the cup and thermometer to reach room temperature before
introducing the next sample.
4) The bulb of the thermometer should dip inside the oil sample.
5) Breathing over the surface of the oil should be avoided.
RESULT:
Given oil sample Flash point(0C) Fire point (0C) Marks awarded Signature of the faculty
64
Step-1: Observations & Calculations:
Burette : NaOH Solution
Conical Flask : 20 ml Oxalic Acid
Indicator : Phenolphthalein
End Point : Colour less to Pale pink.
S.No. Volume of Oxalic
acid (v1 ml)
Burette readings (ml) Volume of NaOH Rundown
(v2 ml) Initial Final
1
2
3
4
Calculations:
Oxalic Acid:
M1 = Molarity of Oxalic Acid = 0.2M
V1 = Volume of Oxalic Acid = 20 ml
n1 = Moles of Oxalic Acid = 1
Sodium Hydroxide:
M2 = Molarity of NaOH =?
V2 = Volume of NaOH =
n2 = Moles of NaOH = 2
Molarity of NaOH = ----------------- M
65
EXPERIMENT-5
POTENTIOMETRIC DETERMINATION OF STRONG ACID USING
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, 100ml
Beaker, Burette, Volumetric Flask, Glass Rod.
Chemicals: 0.1 M Hcl, 0.1 M NaOH , 0.1M Oxalic acid and Phenolpthalein indicator.
Principle: In the titration of 0.1M HCl with 0.1M 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/PtFormula:
Procedure:
Step 1: Standardization of sodium hydroxide by using oxalic acid
8. Rinse and fill the burette with the given NaOH solution
9. Pipette out 20 ml of 0.1 M oxalic acid solution into a clean conical flask
10. Add 1 or 2 drops of phenolphthalein indicator to oxalic acid solution.
11. Titrate the solution against sodium hydroxide solution drop wise with shaking till the
solution changes to pale pin
12. Note the volume of NaOH used. It is the end point
13. Repeat the titration until the concordant readings are obtained
14. Calculate the molarity of NaOH by using the formula mentioned above
66
Step-2: Observations and Calculations: pH Metric titration in between HCl and NaOH
VOLUME OF NaOH
ADDED
EMF (mv)
1ml
2ml
3ml
4ml
5ml
6ml
7ml
8ml
Calculation of unknown molarity of HCl solution:
Sodium Hydroxide:
M2 = Molarity of NaOH =
V2 = Volume of NaOH =
n2 = Moles of NaOH = 1
HCl:
M3= Molarity of HCl =?
V3= Volume of HCl = 20ml
n3 = Moles of HCl = 1
Molarity of HCl = ----------------- M
Amount of HCl = ---------------- grs/l
67
Step 2: Determination of Strength of unknown HCl by using standard NaOH through
Potentiometric titration
1. Pipette out 20ml of 0.1M HCl solution into beaker and saturate it with quinhydrone.
2. Dip the indicator electrode (platinum electrode) and connect the indicator electrode and
saturated calomel electrode (reference electrode) to the potentiometer.
3. The two half cells are connected by means of a salt bridge.
4. The potentiometer is standardized and used for measuring the emf directly.
5. Take 20 ml of HCl solution in the beaker and add 1 ml of 0.1M NaOH from the burette
every time.
6. Shake well after each addition and measure the cell emf.
7. 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.
8. 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.
Report:
S.No Given Amount of
unknown Acid
Reported Amount of
unknown Acid
% Error Marks Signature of the
Faculty
1
68
69
TOPIC BEYOND THE
SYLLABUS
70
Step1: Observations and calculations
Preparation of standard solution of Disodium salt of EDTA
Weight of weighing bottle + EDTA salt (W1) = ----------------- g
Weight of empty weighing bottle (W2) = ----------------- g
Weight of EDTA transferred (W1 – W2) = ---------------- g
Molarity of EDTA = Weight of EDTA taken X 4
Mol. Weight of EDTA(372)
71
EXPERIMENT NO 1
DETERMINATION OF CALCIUM OXIDE IN THE GIVEN SAMPLE OF
CEMENT SOLUTION
Aim: To determine the percentage of calcium oxide in the given cement solution using standard
Std EDTA.
Principle:
Cement contains compounds of Calcium, Aluminium, Magnesium, Iron and insoluble silica.
When dissolved in acid , silica remains undissolved. On treating with ammonia, aluminium and
iron can be precipitated as their hydroxides and separated. The provided cement solution
contains calcium and magnesium ions. The constituents of Portland cement are CaO (60-67%),
SiO2 ( 17- 25%), Al2O3 (3-8%), Fe2O3 (0.5-6%), MgO(0.1 -4%).SO3(1-3%), K2O & Na2O
(0.5-1.5%) , CaSO4 (3-5%)
To estimate the calcium content in the given solution, a known volume of cement
solution is titrated with standard EDTA solution in presence of Mg, Calcium ions present in the
cement solution can be titrated against EDTA using Patton & Reader’s indicator in the pH range
12-14. The indicator combines with Calcium ions to form a wine red coloured complex. Ca+2 +
Indicator Calcium-Indicator complex (Wine red)
Near the end point, when free calcium ions are exhausted in the solution, further addition
of EDTA, dissociates Calcium-Indicator complex, consumes the calcium ions and release free
indicator which is blue in colour. Therefore colour change is wine red to blue.
Procedure:-
Step1. Preparation of standard solution of Na2 EDTA
Weigh out the given EDTA crystals accurately into a 250 ml volumetric flask. Add quarter test
tube of ammonia. Dissolve in distilled water and dilute up to the mark, mix well.
Step 2. Determination of calcium oxide
Pipette out 25 Cm3 of the given Cement solution into a clean conical flask. Add 5ml of glycerol,
5ml of diethyl amine and and 10ml of 4N NaOH solution. Add 3-4 drops of Patton and Reeder’s
indicator. Titrate against standard EDTA solution till the colour change from wine red to clear
blue. Repeat the titration to get concordant values.
72
Step 2: Observations and calculations
S. No. Vol. of Sample
Burette readings Vol. of EDTA
Initial Final
Volume of EDTA consumed by 25 Cm3 of cement solution = -------------------
Weight of Cement in 25ml =0.1g
1000 Cm3 of 1 M EDTA = 56.08 g of CaO (Molecular mass of CaO =56.08)
-------- x -------- x 56.08
------- Cm3 of -------- M EDTA = --------------------- g of CaO
1000 x 1
= -----------------------
25 Cm3 of the Cement solution contains ------------- g of CaO
.............100
% of CaO in the given sample of cement solution = --------------------- = ---------------------
0.1
73
Result:
S.NO Sample % of CaO Signature of
the Faculty
74
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