Chemistry Manual Sem I & Ii
Chemistry Manual Sem I & Ii
Chemistry Manual Sem I & Ii
IN
SEMESTER - I
KLNCE/Dept. of Chemistry
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LIST OF EXPERIMENTS
Expt. No.
Page No.
1.
2.
11
3.
17
4.
25
5.
31
6.
37
7.
43
KLNCE/Dept. of Chemistry
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TABLE OF CONTENTS
Ex. No.
Date
KLNCE/Dept. of Chemistry
Marks
Signature of the
Obtained Staff with Date
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Titration I: Standardization of sodium thiosulphate
Standard potassium dichromate Vs sodium thiosulphate
S. No.
Volume of standard
potassium dichromate
(ml)
Burette reading
(ml)
Initial
Volume of sodium
thiosulphate (ml)
Final
Indicator
CALCULATION
Step 1 : Standardization of sodium thiosulphate
Volume of potassium dichromate
(V1)
= 20 ml
(N1)
= -------------N
(V2)
= .ml
(N2)
= ?
KLNCE/Dept. of Chemistry
V1 N1
= V2 N2
N2
= V 1 N1
V2
(N2)
=..N
20 x
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Expt. No
Date
PRINCIPLE
Oxygen dissolves in water to an extent of 7-9 mg/lit at a temperature range of 25-35C. The
amount of dissolved oxygen in water is estimated using Winklers reagent (manganous sulphate,
alkaline potassium iodide, concentrated sulphuric acid). Water sample is collected carefully avoiding
aeration/deaeration in ground stoppered flask. Initially manganous sulphate and alkali iodide
reagents are added and the reactions occur as follows:
Mn2+
2OH-
Mn(OH)2
O2
(DO in water)
The precipitate dissolves in concentrated sulfuric acid liberating oxygen which in turn
oxidizes potassium iodide and liberates iodine. The liberated iodine is titrated against Na2S2O3 using
starch indicator.
MnO(OH)2 + 2H2SO4
Mn(SO4)2
2H2O + [O]
2KI
+ H2SO4 + [O]
K2SO4
H2O
2Na2S2O3
+ I2
Na2S4O6
2NaI
KLNCE/Dept. of Chemistry
+ I2
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Titration II: Estimation of dissolved oxygen
Std. sodium thiosulphate Vs water sample
S. No.
Volume of
water sample (ml)
Volume of sodium
thiosulphate (ml)
Final
Indicator
CALCULATION
Step 2 : Estimation of dissolved oxygen
Volume of sodium thiosulphate
(V1)
= ml
(N1)
= N
(V2)
= 100 ml
(N2)
= ?
V1 N1
= V2 N2
N2
= V 1 N1
V2
=
100 x
= ___________ N
KLNCE/Dept. of Chemistry
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PROCEDURE
RESULT
Amount of dissolved oxygen present in given water sample =..mg/litre
KLNCE/Dept. of Chemistry
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Short Procedure
AIM
To estimate the amount of dissolved oxygen in the given water sample by Winklers method.
Sodium thiosulphate (link solution) and a std. solution of potassium dichromate are supplied.
PRINCIPLE
Mn2+
2OH-
Mn (OH)2
O2
(DO in water)
MnO (OH)2
+ 2H2SO4
Mn(SO4)2
+ 2H2O + [O]
2KI
+ H2SO4 + [O]
K2SO4
+ H2O
2Na2S2O3
+ I2
Na2S4O6
+ 2NaI
+ I2
PROCEDURE
TITRATION I: STANDARDISATION OF SODIUM THIOSULPHATE
Burette: Sodium thiosulphate
Indicator : Starch
Indicator : Starch
KLNCE/Dept. of Chemistry
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Viva Voce Questions and Answers
1. Name the method which is used to determine DO.
Winklers method
2. How much amount of dissolved oxygen normally present in water at 25C?
7 - 9 ppm
3. What is the link solution used in the determination?
Sodium thiosulphate
4. What is meant by winklers reagent?
Winklers reagent - Manganous sulphate, alkaline potassium iodide, concentrated sulphuric
acid.
5. Name the indicator used in this estimation.
Starch
6. Define iodometry and iodimetry.
Iodometry: Iodine is liberated in the titration
Iodimetry: Iodine is used in the titration
7. What is the end point in the dissolved oxygen determination?
The end point is the disappearance of blue colour.
8. What is the equivalent weight of oxygen?
Eight
9. Name the solutions which are used to determine the dissolved oxygen in a water sample.
(i) MnSO4
(ii) NaOH
(iii) KI
10. Write the formula that is used to calculate the amount of dissolved oxygen.
Amount of DO in one litre of tap water = 8 x N x 1000 mg/L or ppm
KLNCE/Dept. of Chemistry
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S. No.
Volume of
sodium chloride (ml)
Volume of
silver nitrate
(ml)
Final
Indicator
CALCULATION
Step 1 : Standardisation of silver nitrate
Volume of sodium chloride
(V1)
= 20 ml
(N1)
= .N
(V2)
= ..ml
(N2)
= ?
KLNCE/Dept. of Chemistry
20 x
= .N
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Expt. No..
Date.
NaCl
preferentially with chloride ion forming a white precipitate of silver chloride. But at the end point in
the absence of chloride ion, silver nitrate reacts with potassium chromate giving a red precipitate of
silver chromate.
2AgNO3
K2CrO4
(Yellow)
Ag2CrO4
(Red tinge)
+ 2KNO3
When end point is reached, a light red tinged precipitate is obtained due to the formation of silver
chromate.
PROCEDURE
TITRATION I: STANDARDISATION OF SILVER NITRATE
Burette: AgNO3
Indicator : 2 drops of 5% potassium chromate
Pipette: 20 ml of Std. NaCl
KLNCE/Dept. of Chemistry
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Titration II: Estimation of chloride ion
Std. silver nitrate Vs water sample
S. No.
Volume of given
water sample (ml)
Final
Volume of
silver nitrate (ml)
Indicator
CALCULATION
Step - 2 : Estimation of chloride ion content
Volume of water sample
(V1)
= 20 ml
(N1)
= ?
(V2)
= ..ml
(N2)
= N
= V2 N2
V1
x
20
= N
KLNCE/Dept. of Chemistry
35.5
10
= ____________ gm.
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The burette is washed with distilled water and rinsed with silver nitrate solution. Then it is
filled with same silver nitrate solution up to the zero mark. A 20 ml pipette is washed with water and
rinsed with small amount of sodium chloride solution and then 20 ml of same solution is pipetted out
into a clean conical flask. 2 drops of 5% potassium chromate solution is added as indicator. The
solution becomes yellow. It is titrated against silver nitrate taken in the burette. Near the end point
coagulation of the precipitate takes place. Silver nitrate is added in drops with constant shaking. The
end point is indicated by the appearance of pale red tinge on the precipitate. The titrations are
repeated to get concordant values and the normality of silver nitrate solution is calculated.
: AgNO3
Pipette
The burette is filled with same silver nitrate solution up to zero mark. The given water
sample is made up to 100 ml in standard measuring flask with distilled water. A 20 ml pipette is
washed with water and rinsed with made up water sample. 20 ml of this water sample is pipetted out
and 2 drops of 5% potassium chromate is added and titrated against AgNO3 solution. The end point
is the formation of red tinge on the precipitate. The titrations are repeated to get concordant values.
RESULT
The amount of chloride ion present in 100 ml of the given water sample = gm.
KLNCE/Dept. of Chemistry
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Short Procedure
AIM
To estimate the amount of chloride ion present in the given water sample by Mohrs method.
A std. solution of NaCl and approximately 0.01N AgNO3 solutions are provided.
PRINCIPLE
AgNO3
NaCl
2AgNO3
K2CrO4
(Yellow)
AgCl
+
(White precipitate)
Ag2CrO4
(Red tinge)
NaNO3
+ 2KNO3
PROCEDURE
TITRATION I: STANDARDISATION OF SILVER NITRATE
Burette: AgNO3
: AgNO3
Pipette
KLNCE/Dept. of Chemistry
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Viva Voce Questions and Answers
1. What is the name of the method used to estimate the chloride ion in a water sample?
Mohr-s method
2. Name the link solution that is used in the determination of chloride ion.
Silver nitrate solution.
3. What is meant by argentometric method?
Water contains chloride ions in the form of dissolved NaCl, KCl, CaCl2 and MgCl2. The total
chloride content can be estimated by titration with standard AgNO3 solution (Argentometric
method or Mohrs method).
3. What is the indicator used in the determination?
Potassium chromate
4. What is the colour of the indicator in a water sample?
Yellow
5. What is the end point of titration?
Formation of red tinge precipitate (Due to the formation of Silver chromate)
6. Write the formula used to calculate the amount of chloride ion present in the water sample.
Amount of Cl- in the water sample = Equivalent weight of chloride ion x Normality
7. What kind of chlorides present in water?
NaCl, MgCl2, CaCl2.
8. Why does AgNO3 react first with chloride ions in the water and not with chromate ions?
The solubility product of Ag2CrO4 is high compared to that of AgCl, therefore AgNO3 reacts
first with chloride ions in the water and all the chlorides in solution have been precipitated as
AgCl.
9. The formation of red tinge is due to what?
It is due to the formation of silver chromate Ag2CrO4.
10. What is the equivalent weight of chloride ion?
35.5
KLNCE/Dept. of Chemistry
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Titration between HCl & NaOH
Volume of HCl (V1) = ______ ml
Volume of NaOH added (ml)
pH
0
2
4
6
8
10
12
14
16
18
20
KLNCE/Dept. of Chemistry
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Expt. No
Date.
PRINCIPLE
The pH of the solution is related to the H+ ion concentration by the following formula,
pH = -log [H+]
Measurement of pH of the solution gives the concentration of H+ ions in the solution. When
NaOH is added slowly from the burette to the solution of HCl, H+ ions are neutralized by hydroxide
ions. As a result, pH of the solution increases.
HCl + NaOH NaCl + H2O
The increase in pH values takes place until all the H+ ions are completely neutralized (up to
the end point). After the end point, further addition of NaOH increases pH sharply as there is an
excess of OH- ions.
PROCEDURE
The burette is filled with standard sodium hydroxide solution. Exactly 20 ml of the given
hydrochloric acid solution is pipetted out into a clean beaker and one test tube of distilled water is
added to it. The glass electrode is dipped in it and connected with a pH meter.
Now 2 ml of NaOH solution is added from the burette to the HCl solution taken in the beaker
and pH of the solution is noted for each addition. This process is continued until at least 5 readings
are taken after the end point. The observed pH values are plotted against the volume of NaOH
added. From the graph the end point is noted. This procedure gives the approximate end point.
Accurate end point can be determined by following the same procedure. Near the end point, the pH
values are noted for every 0.2 ml addition of NaOH.
KLNCE/Dept. of Chemistry
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KLNCE/Dept. of Chemistry
pH
pH
V (ml)
pH/V
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KLNCE/Dept. of Chemistry
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CALCULATION
Step 1 : Calculation of strength of HCl
Volume of HCl
(V1)
= 20 ml
Strength of HCl
(N1)
= ......................?
Volume of NaOH
(V2)
Strength of NaOH
(N2)
= .. N
= V2N2
N1
= V2 N2
V1
= ----- x ----20
= .................. N
= ....................... gm.
KLNCE/Dept. of Chemistry
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RESULT
Strength of given Hydrochloric acid
= .N
KLNCE/Dept. of Chemistry
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Short Procedure
AIM
To determine the strength of given HCl by pH meter and also calculate the amount of HCl.
You are provided with a standard solution of NaOH of strength ..N.
PRINCIPLE
pH of a solution is related to the H+ ion concentration as, pH = -log [H+]
HCl + NaOH
NaCl + H2O
PROCEDURE
Burette solution
Pipette solution
: 20 ml of the given HCl + one test tube of distilled water
Electrode
: Glass electrode
The gradual addition of NaOH from the burette (OH- ions) increases the pH value gradually.
At the end point (sharp increase) complete neutralization takes place (i.e.) all the fast moving
H+ ions are replaced by fast moving OH- ions.
After the end point there is only free OH-ions are present. So constant increase of pH occurs.
Graph 2: Volume of NaOH Vs pH /V
pH /V
pH
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Viva Voce Questions and Answers
1. Define pH
pH is defined as the negative to the base 10 logarithm (log) of the H+ ion
concentration. Mathematically, it is represented as
pH = -log [H+]
2. What is the pH value of a neutral solution?
7
3. Write the pH range for (i) an acidic solution (ii) a basic solution
(i)
(ii)
KLNCE/Dept. of Chemistry
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Titration between HCl, CH3COOH (mixture of acids) & NaOH
Volume of HCl (V) = 40 ml
Volume of
ConducNaOH
tance
added (v)
(mho)
ml
(V+v)
V
C x(V+v)
V
(mho)
Volume
Conducof NaOH
tance
added
(mho)
(v) ml
0.5
8.5
1.5
9.5
10
2.5
10.5
11
3.5
11.5
12
4.5
12.5
13
5.5
13.5
14
6.5
14.5
15
7.5
15.5
(V+v)
V
C x(V+v)
V
(mho)
C x (V+v) /V mho
Graph
(B)
(A)
Volume of NaOH (ml)
KLNCE/Dept. of Chemistry
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Expt. No
Date.
PRINCIPLE
Solutions of electrolytes conduct electricity due to the presence of ions. The specific
conductance of a solution is proportional to the concentration of the ions in it. When a mixture of
HCl and CH3COOH is titrated against NaOH, hydrochloric acid, being a much stronger acid will get
neutralized first. The neutralization of acetic acid will commence only after HCl has been
completely neutralized.
H+ + Cl-
Na+ + OH-
fast
slow
A graph is drawn between volume of NaOH added and the conductance of solution. There
are two points of intersection; first one corresponds to HCl and the second one corresponds to
CH3COOH.
PROCEDURE
The burette is filled with standard sodium hydroxide solution. Exactly 40 ml of the mixture
of hydrochloric acid and acetic acid is pipetted out into a clean beaker. A glass rod is placed in the
beaker for stirring the solution. The conductivity cell is kept immersed in the solution. The
conductance of the solution is measured by connecting the terminal of conductivity cell with a
conductivity bridge. 0.5 ml of standard sodium hydroxide solution is added each time from the
burette, stirred well and the conductance of the solution is noted after each addition. Conductance
decreases gradually due to the neutralization of hydrochloric acid, it reaches the first end point and
after that the conductance increases till it reaches the second end point. Further addition of sodium
hydroxide steeply increases the conductance value.
KLNCE/Dept. of Chemistry
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CALCULAION
Step-1: Strength of HCl
Volume of HCl (acid mixture)
(V1)
= 40 ml
Strength of HCl
(N1)
= .............. ?
Volume of NaOH
(V2)
Strength of NaOH
(N2)
= ____ N
= V2 N2
V1
x
40
= .................. N
Amount of HCl
The amount of HCl present in
1 litre of the given solution
(N1) =
.............. ?
(V2) =
Strength of NaOH
( N2) = ____ N
x
40
= .................. N
KLNCE/Dept. of Chemistry
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The values of observed conductivity or [C (V+v)/V] are plotted against volume of sodium
hydroxide added. The point of intersection at the two places gives the first and second end point
which corresponds to HCl and CH3COOH respectively. From this, the amount of HCl and
CH3COOH is calculated.
RESULT
The strength of hydrochloric acid in the mixture of acids
= ....N
= gm.
= ...N
KLNCE/Dept. of Chemistry
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Short Procedure
AIM
To determine conductometrically the strength of given hydrochloric acid and acetic acid in
the given mixture, by titrating with standard sodium hydroxide solution and calculate the amount of
each acid present in 1 litre of the solution.
PRINCIPLE
H+ + Cl-
Na+ + OH-
fast
slow
PROCEDURE
C x (V+v) /V mho
= ....N
= gm.
= ...N
= gm.
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Viva voce Questions and Answers
1. Give an example for strong acid and weak acid.
Strong acid: HCl
Weak acid: CH3COOH
2. Give an example for a strong base.
NaOH
3. Give reasons for decrease in conductance when a base is added to a mixture of acids initially.
At the beginning of the titration, the conductance of mixture of acids is due to H+
ions. When a base is added, the fast moving H+ ions are replaced by slow moving Na+ ions.
Hence, the conductance decreases.
4. Why the conductance increases linearly after the first end point?
The first end point corresponds to the neutralization of strong acid (HCl) by strong
base. After the first end point, the added base neutralizes the weak acid (CH3COOH). This
acid is feebly ionized and less amount of H+ ions are available to OH- ions for neutralization.
Hence, the conductance increases linearly.
5. Why the conductance increases sharply after the second end point when a base is added?
After the neutralization of mixture of acids, addition of base gives free OH- ions
which has high conductance and hence the conductance increases sharply..
6. Write the equivalent weights of HCl and CH3COOH
HCl
- 36.5
CH3COOH - 60
7. Write the formula used to calculate the amount of HCl and CH3COOH.
The amount of HCl = Normality x Equivalent weight of HCl (36.5)
The amount of CH3COOH = Normality x Equivalent weight of CH3COOH (60)
KLNCE/Dept. of Chemistry
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Graph: Calibration curve (Absorbance Vs Concentration)
Absorbance
Concentration in ppm
TABLE I
Preparation of various concentration of Fe3+ solution
Volume of iron
solution (ml)
Volume of
HNO3 (ml)
KLNCE/Dept. of Chemistry
Volume of
NH4SCN (ml)
Volume of distilled
H2O (to make equal
volume)
Concentration of
iron in ppm
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Expt. No
Date.
ESTIMATION OF IRON CONTENT OF THE WATER SAMPLE USING
SPECTROPHOTOMETER
AIM
To estimate the amount of ferric iron present in the given sample by thiocyanate method
using spectrophotometer.
PRINCIPLE
The estimation is based on Beer- Lamberts law, which states that When a beam of
monochromatic radiation is passed through a solution of an absorbing substance, the rate of decrease
of intensity of radiation with thickness of the absorbing solution is proportional to the intensity of
incident radiation as well as to the concentration of the solution. The law can be written in the form
of mathematical equation as,
Log I0 / It = cl = A
Where,
absorbance
I0
It
Concentration of solution
Fe3+ + 6SCN-
Fe3+ ions do not give any colour in solution. However a red colour can be produced, when it
reacts with a thiocyanate solution due to the formation of complex and complementary for this
colour will be in the blue region ( = 480 nm). So in the spectrophotometer, this radiation is allowed
to pass through the solution. The spectrophotometer will measure the incident radiation and the
transmitted radiation. To measure (I0), a blank solution (without Fe3+) is taken and the transmitted
light is measured, its absorbance is measured for various concentrations and a calibration graph can
be drawn with absorbance Vs concentration.
KLNCE/Dept. of Chemistry
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TABLE II
Measurement of absorbance
Blank (distilled water): Zero absorbance; max = 480 nm
S. No.
Concentration
Absorbance
1ppm
0.01 N
2ppm
0.02 N
3ppm
0.03 N
4ppm
0.04 N
5ppm
0.05 N
6ppm
0.06 N
Unknown
-----
KLNCE/Dept. of Chemistry
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PROCEDURE
The spectrophotometer is switched on and warmed up for 10 minutes. The monochromator is
adjusted for max = 480 nm. The blank solution is kept in the cell and transmittance (I0) is measured.
Usually the instrument is calibrated for transmittance 100, for which absorbance is zero. Similarly
various known concentration of ferric iron solutions, after adding nitric acid and thiocyanate
solution, are kept in the instrument one by one and absorbance is measured in each case.
Now, the unknown solution (water sample) is treated with thiocyanate and nitric acid, kept in
the spectrophotometer and absorbance is measured. A calibration graph is drawn between
concentration and absorbance. From this, the unknown concentration is found out.
RESULT
The amount of ferric iron present in the given sample of water = ppm
KLNCE/Dept. of Chemistry
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Short Procedure
AIM
To estimate the amount of ferric iron present in the given sample by thiocyanate method
using spectrophotometer.
PRINCIPLE
The estimation is based on Beer- Lambert law, which states that When a beam of
monochromatic radiation is passed through a solution of an absorbing substance, the rate of decrease
of intensity of radiation with thickness of the absorbing solution is proportional to the intensity of
incident radiation as well as to the concentration of the solution. The law can be written in the form
of mathematical equation as,
Log I0 / It = cl = A
PROCEDURE
The monochromator is adjusted for = 480 nm.
The blank solution is kept in the cell and transmittance (I0) is measured.
Similarly for various known concentration of ferric iron solution, after adding nitric
acid and thiocyanate solution, the absorbance is measured.
Similarly, the absorbance for unknown solution is measured.
A calibration graph is drawn between concentration and absorbance. From this, the
unknown concentration is found out.
CALCULATION
Amount of iron present in the given solution
RESULT
The amount of ferric iron present in the given sample of water = ppm
KLNCE/Dept. of Chemistry
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It
Log I0 / It
= cl =
Concentration in moles/litre
KLNCE/Dept. of Chemistry
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TABLE I
Preparation of various concentrations of polymer solution
Stock solution N1 = 1%
S. No.
Total volume V2 = 30 ml
Concentration in % (N2)
N2 = V1N1 / V 2
1.
2.
3.
4.
5.
TABLE II
Viscosity data for a polymer / solvent
Flow time of the pure solvent (t0) = sec
S. No.
Concentration in %
(N2)
(r) = t/t0
sp = r 1
sp/C = red
1.
2.
3.
4.
5.
Graph
sp/C
i
Concentration (%)
KLNCE/Dept. of Chemistry
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Date
Expt.No
Where, i
Intrinsic viscosity
K&a
Degree of polymerization (Dp) provides another way of expressing the molecular weight as
follows.
Where,
Dp x m
Dp
degree of polymerization
t/t0
r 1
sp/C
lim sp/C
c0
Flow time for the polymer solution
t0
KLNCE/Dept. of Chemistry
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CALCULATION
Mark Hownik equation is given by,
i
KMa
log i
log K + a log M
log i - log K
log M
=
a
log i - log K
= A. log
a
KLNCE/Dept. of Chemistry
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PROCEDURE
STEP I: Preparation of polymer solution of different concentration
Polymer solutions of different concentrations, say 0.1%, 0.2%, 0.3%, 0.4% and 0.5% are
prepared from the given polymer stock solution as shown in the table-I
viscosity (i). From the value, the molecular weight of the polymer is calculated using
Mark Hownik equation.
RESULT
Molecular weight of the given polymer
=..
Degree of polymerization
=..
KLNCE/Dept. of Chemistry
Page 39
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Short Procedure
AIM
To determine the molecular weight and degree of polymerization of the given 1% polymer
solution (polyvinyl alcohol) using Ostwalds viscometer.
PROCEDURE
First water is taken in the viscometer upto the mark and flow time is noted in seconds.
Then Polymer solution (Poly vinyl alcohol) of different concentration are taken in the
viscometer and flow time is recorded for each concentration.
CALCULATION
By using, Mark Hownik equation, the molecular weight of the polymer is calculated.
i
=
log i =
KMa
log K + a log M
log i - log K
log M =
a
Where, M
i
Where, i
K&a
sp / C
i
Concentration (%)
RESULT
Molecular weight of the given polymer
Degree of polymerization
KLNCE/Dept. of Chemistry
=..
=..
Page 40
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Viva Voce Questions and Answers
1. Name the method used to determine the molecular weight of a polymer
Viscometry
2. What is the name of the viscometer used in this experiment?
Ostwalds Viscometer
3. Write Mark-Hownik equation.
i
KMa
Where,
Intrinsic viscosity
Constants for a given polymer solvent combination at a given
temperature
Average molecular weight
K&a =
M
CH
OH
t/t0
r 1 = (t/t0) - 1
lim sp/C
c0
Flow time for the polymer solution.
Flow time for the solvent.
Concentration of the polymer.
Where,
t
t0
C
KLNCE/Dept. of Chemistry
=
=
=
=
Page 41
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Titration between standard HCl & NaOH
Volume of HCl (V) = 40 ml
Volume of NaOH
added (v) ml
Conductance (mho)
C x [(V+v)/V]
(mho)
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
KLNCE/Dept. of Chemistry
Page 42
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Expt. No..
Date
PRINCIPLE
When hydrochloric acid (H+ + Cl-) is titrated against sodium hydroxide, there is decrease in
the conductance. This is because of the replacement of fast moving H+ ions by the slow moving Na+
ions.
H+ + Cl-
+ NaOH
Na+ + Cl-
+ H2O
The decrease in the conductance continues up to the neutralization point. After all the H+ ions
are replaced, the added sodium hydroxide introduces excess of fast moving OH - ions. Hence,
conductance slowly increases after the neutralization point. A graph is drawn between the volume of
NaOH added (v) and C (V+v/V), where C is the conductance of the solution, V is the volume of HCl
taken and v is the volume of sodium hydroxide added. The end point is the exact point of
intersection of two straight lines.
PROCEDURE
The burette is filled with std. NaOH solution. Exactly 40 ml of given HCl is pipetted out into
a clean beaker. A glass rod is placed in the beaker. The conductivity cell is dipped into it. The
conductance of the solution is measured by connecting the terminals of conductivity cell with a
conductivity bridge. 2 ml of standard NaOH solution is added from the burette each time and
conductance of the solution is noted after each addition. This process is repeated until at least 5
readings are taken beyond the end point. The values of observed conductivity or C [(V+v)/V] are
plotted against the volume of NaOH added (v). The point of intersection in the graph gives the end
point. From the end point, the strength of acid is calculated and hence the amount of hydrochloric
acid in 1 litre is also calculated.
KLNCE/Dept. of Chemistry
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Graph
CALCULATION
Strength of HCl
Volume of HCl
(V1)
= 40 ml
Strength of HCl
( N1)
= ................ ?
Volume of NaOH
(V2)
Strength of NaOH
(N2)
= _________ N
Strength of HCl
Amount of HCl
Amount of HCl present in
1 litre of the given solution
= V2 x N2
V1
x
40
= .................N
C x (V + v ) / V mho
N1
KLNCE/Dept. of Chemistry
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RESULT
Strength of given hydrochloric acid
= .N
= .gm.
KLNCE/Dept. of Chemistry
Page 45
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Short Procedure
AIM
To determine the strength of given hydrochloric acid by conductometric titration against
standard sodium hydroxide solution and calculate the amount of hydrochloric acid present in 1 litre
of the solution.
PRINCIPLE
+ NaOH
Na+ + Cl-
+ H2O
C x (V + v ) / V mho
H+ + ClPROCEDURE
Initially the conductance is high due to the presence of fast moving H+ions from HCl.
NaOH is added slowly from the burette, 2 ml at a time. The conductance is noted for each
addition of NaOH. The conductance decreases for each addition, due to the replacement of
fast moving H+ ions from HCl by slow moving Na+ ions.
When the end point is reached, addition of NaOH increases the conductance.
End point is found out by plotting a graph between Volume of NaOH (X axis) and
Cx(V+v/V) (Y axis).
Strength of HCl
Amount of HCl
RESULT
Strength of given hydrochloric acid
= .N
= .gm.
KLNCE/Dept. of Chemistry
Page 46
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H+ + Cl-
6. Explain why the strength of burette solution is always greater than the pipette solution in
conductometric titration.
We know that the dilution of the solution decreases the conductance value markedly.
Hence, burette is filled with concentrated solution to avoid dilution effect.
7. Give reasons for the decrease of conductance of an acid when a base is added initially to it.
At the beginning of the titration, the conductance of an acid is due to H+ ions. When
the base is added, the fast moving H+ ions are replaced by slow moving Na+ ions. Hence, the
conductance decreases.
8. Why the conductance increases when NaOH is added after the end point?
After the end point, further addition of NaOH results in the presence of free OH- ions,
which increases the conductance.
KLNCE/Dept. of Chemistry
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9. What is the equivalent weight of (i) HCl (ii) NaOH
(i)
HCl : 36.5
(ii) NaOH : 40
10. What is the principle used to fix the end point of the titration?
When the conductance values are plotted against the volume of the burette solution,
we obtain two straight lines. The intersection of these two lines will be the end point of the
titration.
11. Write the formula that is used to calculate the amount of hydrochloric acid.
Weight of HCl per litre = Normality x Equivalent weight of HCl
= N x 36.5 g
12. What are the advantages of conductometric titrations?
1. Coloured solutions, which cannot be titrated by ordinary volumetric
methods with the help of indicators, can be successfully titrated
conductometrically.
2. The method can be employed in the case of very dilute solutions and also
for weak acids and bases.
3. No special care is necessary near the end point as it is determined
graphically.
13. What is cell constant?
Cell constant = l / a, where l is the length of the conductor and a is the area
of the electrode.
KLNCE/Dept. of Chemistry
Page 48
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SEMESTER - II
KLNCE/Dept. of Chemistry
Page 49
CSEITQUESTIONS.BLOGSPOT.IN | CSEITQUESTIONS.BLOGSPOT.IN
LIST OF EXPERIMENTS
Expt. No.
Page No.
1.
53
2.
61
3.
71
4.
77
5.
6.
7.
8.
KLNCE/Dept. of Chemistry
85
95
89
101
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TABLE OF CONTENTS
Ex. No.
Date
KLNCE/Dept. of Chemistry
Marks
Signature of the
Obtained Staff with Date
Page 51
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TABLE I
Titre values and different alkalinities
OH0
[P] or [M]
0
2[P] - [M]
0
Result of titration
If P = 0
If P = M
If P = 1/2M
If P > 1/2M
If P < 1/2M
CO320
0
2[P] or [M]
2[M-P]
2[P]
HCO3[M]
0
0
0
[M] 2[P]
S. No.
Volume of sodium
hydroxide (ml)
Final
Volume of
hydrochloric acid (ml)
Indicator
CALCULATION
Strength of HCl
Volume of NaOH
(V1)
= 20 ml
Normality of NaOH
(N1)
= ________
Volume of HCl
(V2)
= ________ ml
Normality of HCl
(N2)
= ?
V1N1
= V2N2
N2
= V1N1
V2
20 x
= _______________ N
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Expt. No
Date
OH
CO3
+ HCO3
+ H2O
The alkalinity in water may be determined by titrating the water against standard acid using
phenolphthalein and methyl orange indicators in which the following reactions take place.
a)
OH
b)
2
CO 3
+ H
+ H
H2O
HCO3
Phenolphthalein
end point (P)
The remaining half of the carbonate alkalinity and bicarbonate alkalinity are determined by
methyl orange indicator.
c) HCO3 + H+
H2O + CO2
The volume of acid consumed upto the colour change of phenolphthalein (P) shows the
completion of reactions a and b whereas that of methyl orange (M) shows the completion of
reaction c. The values (P) and (M) show the presence of alkaline constituents and their amount in
terms of volume of hydrochloric acid.
Alkalinity is expressed in terms of CaCO3 equivalent.
KLNCE/Dept. of Chemistry
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S. No.
Volume of
water sample (ml)
Volume of HCl
Phenolphthalein
Methyl orange
end point (P)
end point (M)
Concordant Value
[P]
[M]
Calculation:
From the values of [P] & [M] calculate the types of alkalinity present in the water sample using the
table 1.
Types of alkalinity
For example, if the data satisfies the condition P > M, both hydroxide & carbonate alkalinity are
present.
i)
Volume of HCl required for OH- alkalinity
=
2[P] [M]
ii)
iii)
1.
2 x---------- - ---------ml
2[M] 2[P]
2 x---------- -2 x ---------ml
(V1)
(N1)
(V2)
(N2)
=
=
=
=
------- ml
------- N
20 ml
?
N2
V1 x N1
20
`------------- N
3-
KLNCE/Dept. of Chemistry
x
20
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PROCEDURE
TITRATION I: STANDARDISATION OF HYDROCHLORIC ACID
Burette : Hydrochloric acid
Indicator: Phenolphthalein
The burette is washed with water and rinsed with distilled water and hydrochloric acid. Then
it is filled with hydrochloric acid up to zero level mark and the initial burette reading is noted.
A 20 ml pipette is washed with water, rinsed with distilled water followed by the given
sodium hydroxide solution. Then 20 ml of sodium hydroxide solution is pipetted out into a clean
conical flask. A drop of phenolphthalein indicator is added. The solution becomes pink colour. It is
then titrated against hydrochloric acid taken in the burette. The end point is the just disappearance of
pink colour. The final burette reading is noted. Titrations are repeated for concordant values. The
readings are tabulated. From the titre value the strength of hydrochloric acid is calculated.
TITRATION II: ESTIMATION OF ALKALINITY
Burette : Hydrochloric acid
Indicator : 1. Phenolphthalein
2. Methyl orange
The burette is filled with hydrochloric acid up to zero level mark and the initial
reading is noted. A 20 ml pipette is washed with water and rinsed with the given water sample, then
exactly 20 ml of water sample is pipetted out into a clean conical flask. 2 drops of phenolphthalein
indicator are added. Pink colour is observed. The solution is titrated against the standard acid until
just disappearance of pink colour. The end point is noted (P). Then one drop of methyl orange
indicator is added to the same solution and the titration is continued against hydrochloric acid till the
appearance of pale pink colour. The final burette reading is noted (M). Titrations are repeated for
concordant values and the readings are tabulated.
KLNCE/Dept. of Chemistry
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Amount of OH- content in1 lit of water sample,
in terms of CaCO3 equivalent
= ---------------- ppm
= ------- ml
Strength of HCl
(N1)
= .. N
(V2)
= 20 ml
(N2)
= ?
N2
V1 x N1
20
Strength of carbonate alkalinity
x
20
= N
= ---------------- ppm
KLNCE/Dept. of Chemistry
Page 56
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From the values of P & M, the type of alkalinity present in the water sample is found out and
then the amount of each alkalinity is calculated in terms of CaCO3 equivalent.
RESULT
The given water sample contains the following alkalinity
Hydroxide alkalinity
= .ppm
Carbonate alkalinity
= .ppm
Bicarbonate alkalinity
= .ppm
KLNCE/Dept. of Chemistry
Page 57
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Short Procedure
AIM
To determine the alkalinity of given water sample. A standard solution of N sodium
hydroxide and an unknown strength of hydrochloric acid are supplied.
PRINCIPLE
Alkalinity is the ability to neutralize acid. Water may be alkaline due to the presence
of hydroxide , carbonate or bicarbonate ions. The type of alkalinity and amount can be found out by
titration with standard HCl using phenolphthalein and methyl orange indicators.
PROCEDURE: TITRATION I: STANDARDISATION OF HYDROCHLORIC ACID
Burette : Hydrochloric acid
Indicator: Phenolphthalein
Pipette: 20 ml of standard sodium hydroxide
Strength of HCl
2. Methyl orange
RESULT
The given water sample contains the following alkalinity
Hydroxide alkalinity
= .ppm
Carbonate alkalinity
= .ppm
Bicarbonate alkalinity
= .ppm
KLNCE/Dept. of Chemistry
Page 58
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Viva Voce Questions and Answers
1. What is meant by alkalinity of water?
Capacity to neutralize acid
2. What are the ions causing alkalinity?
Hydroxide, carbonate and bicarbonate ions.
3. What are the possible combinations of ions for alkalinity?
(i) OH- and CO32- (ii) HCO3- and CO324. What is the role of phenolphthalein indicator?
It is used to find the alkalinity caused by hydroxide ions and half of the carbonate ions.
5. What is the role of methyl orange indicator?
To find out the alkalinity caused by bicarbonate and other half of the carbonate ions.
6. Hydroxides and bicarbonates cannot exist together. Why?
Hydroxides and bicarbonates cannot exist together due to the following reaction.
2
OH
+ HCO3
CO3
+ H2O
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Titration I: Standardization of EDTA
Standard hard water Vs EDTA
S. No.
Volume of standard
hard water (ml)
Volume of
EDTA (ml)
Final
Indicator
CALCULATION
Strength of EDTA
Volume of standard hard water
= 20 ml
= ml
= 20 mgs of CaCO3
= 20
mg of CaCO3 equivalent
V1
KLNCE/Dept. of Chemistry
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Expt. No
Date.......
producing
Ca2+
ions
and
Mg2+
form
complexes
with
both
EDTA
(Ethylenediaminetetraacetic acid) and EBT (Eriochrome black T). But EDTA metal ion complex
is more stable than EBT metal ion complex.
Ca2+ or Mg2+ + EB
(In water)
pH 9-10
pH 9-10
KLNCE/Dept. of Chemistry
Page 61
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Titration II: Estimation of total hardness
Given water sample Vs EDTA
Volume of given
S. No.
water sample
(ml)
Final
Volume of
EDTA
(ml)
Indicator
= 20 ml
= . ml
V2 ml of EDTA
20 x V21000
mg of CaCO3
V1 x 20
= 1000 V2 / V1 mg of CaCO3
=
=..ppm
KLNCE/Dept. of Chemistry
= ..ppm
Page 62
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To this 5 ml of ammonia buffer solution (approximately half test tube) and 2 drops of EBT
indicator are added. The solution becomes wine red colour. This solution is titrated against EDTA
solution taken in the burette. End point is change of colour from wine red to purple blue. Final
burette reading is noted. The titration is repeated for concordant values. The readings are tabulated.
Let the volume of EDTA be V1 ml.
TITRATION II: ESTIMATION OF TOTAL HARDNESS
Burette : EDTA
Indicator : EBT
Pipette
KLNCE/Dept. of Chemistry
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Titration III: Estimation of permanent hardness
Boiled water sample Vs EDTA
S. No.
Volume of given
boiled water
sample (ml)
Volume of
EDTA (ml)
Final
Indicator
= 20 ml
20 x V3
mg of CaCO3
V1
1000 ml of boiled hard water sample contains
= 20 x V3 x 1000
V1 x 20
= 1000 V3 / V1 mg of CaCO3
=
= ppm
Permanent hardness
=
=
= -------------ppm
KLNCE/Dept. of Chemistry
Page 64
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20 ml of boiled, cooled and filtered hard water sample (temporary hardness removed water
sample) is pipetted out into a clean conical flask. 5 ml of buffer solution and 2 drops of EBT are
added to it. It is titrated against EDTA. The end point is change of colour from wine red to purple
blue. Let the volume of EDTA be V3 ml. From the titre value, the permanent hardness can be
calculated.
The temporary hardness of the water sample can be found out by subtracting the value of
permanent hardness from the value of total hardness.
RESULT
Total hardness of given water sample
ppm
. ppm
. ppm
KLNCE/Dept. of Chemistry
Page 65
CSEITQUESTIONS.BLOGSPOT.IN | CSEITQUESTIONS.BLOGSPOT.IN
Short Procedure
AIM
To determine the total hardness and also the permanent hardness of the given water sample.
Standard hard water and a link solution of EDTA are supplied.
PRINCIPLE
Ca2+ or Mg2+ + EBT
(In water)
Ca EBT or Mg EBT + EDTA2-
PROCEDURE
TITRATION I: STANDARDISATION OF EDTA
Burette : EDTA
Indicator : EBT
Pipette
. ppm
. ppm
KLNCE/Dept. of Chemistry
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Viva Voce Questions and Answers
HOOC - H2C
N - CH2 - CH2 - N
HOOC - H2C
CH2 - COOH
8. Mention the link solution that is used in the determination of hardness of water
EDTA solution
9. What are the types of hardness?
(i) Carbonate hardness (or) Temporary hardness
(ii) Non- carbonate hardness (or) Permanent hardness
10. What is temporary hardness?
Hardness caused by bicarbonates of Ca & Mg is called temporary hardness.
11. What is permanent hardness?
Hardness caused by chlorides and sulphates of Ca & Mg is called permanent hardness.
12. What happens on boiling the hard water?
If temporary hardness is present, it is removed.
KLNCE/Dept. of Chemistry
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13. How hardness of water is expressed?
It is expressed in terms of CaCO3 equivalent.
14. Why CaCO3 is chosen as standard to express hardness?
Its solubility is less. Its Molecular weight is 100 and Equivalent weight is 50 which is easy
for calculation.
15. Define pH
pH of any solution is defined as
pH = -log10 aH+, where aH+ is the activity of the hydrogen ions. For dilute solutions,
activity can be replaced by concentration. i.e., pH = -log10 [H+]
16. What is the function of buffer solution?
To maintain the pH of a reaction medium.
17. What are buffer solutions?
A mixture of weak acid and its salt or weak base and its salt is known as buffer solution.
18. What are the units of hardness?
1. ppm Parts per million. It is the number of parts of hardness producing substances (in
terms of CaCO3) present per million parts of water.
2. mg/L: milligram per litre. It is the number of milligram of calcium carbonate equivalent
hardness present in 1litre of water.
3. Clarkes Degree (0Cl): It is the number of parts of CaCO3 equivalent hardness present in
70,000 parts of water. (i.e., in one gallon of water).
4. Degree of French (0Fr): It is the number of parts of CaCO3 equivalent hardness present in
105 parts of water.
The above 4 units are correlated as given below.
1 ppm = 1 ml / L = 0.07 0Cl = 0.1 0Fr
19. Why the pH of the solution is maintained between 6 to10?
Eriochrome Black - T is the metal ion indicator. This dyestuff tends to polymerize in strongly
acidic solution to a red brown product and hence the indicator is generally
used with
KLNCE/Dept. of Chemistry
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20. Name other indicators beside Eriochrome Black T which could be used for the
determination of hardness.
Murexide (ammonium salt of purpuric acid), solochrome dark blue or calcon, Zincon and
Xylenol orange.
21. Name the different methods of determining hardness.
1. O-Hehners method
2. Soap titration method
3. EDTA method
KLNCE/Dept. of Chemistry
Page 69
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Titration I: Standardization of EDTA
Standard CaCO3 Vs EDTA
S. No.
Volume of standard
calcium carbonate (ml)
Final
Volume of
EDTA (ml)
Indicator
CALCULATION
Strength of EDTA
Volume of std. CaCO3 solution V1
20 ml
-------- N
Volume of EDTA
V2
-------- ml
Strength of EDTA
N2
N2
V1 x N1
20 x
V2
Strength of EDTA
KLNCE/Dept. of Chemistry
N2
--------- N
Page 70
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Expt. No.
Date .
FSB-F
[Cu-FSB-F] +
EDTA2-
[Cu-FSB-F] + 2H+
(Purple colour)
[Cu- EDTA]2(Colourless)
+ FSB-F
(Green colour)
PROCEDURE
TITRATION I: STANDARDISATION OF EDTA
Burette : EDTA
Indicator : EBT
Pipette
: 20 ml of std. CaCO3
KLNCE/Dept. of Chemistry
Page 71
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Volume of given
brass solution (ml)
Initial
Volume of
EDTA (ml)
Final
Indicator
CALCULATION
Strength of brass solution
Volume of EDTA
V1
---------- ml
Strength of EDTA
N1
-------N
V2
20 ml
N2
N2
V1 x N1
= x
V2
Strength of copper in brass solution
20
= N
------------------ x 63.54
= gm.
10
KLNCE/Dept. of Chemistry
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5 ml of ammonia buffer solution and 2 drops of EBT indicator are added. The solution turns wine
red in colour and it is then titrated against EDTA taken in the burette. The change of wine red colour
to purple blue colour is the end point. The final reading is noted. The titration is repeated to get
concordant values. From the volume of EDTA consumed, strength of EDTA solution is calculated.
TITRATION II: ESTIMATION OF COPPER
Burette : EDTA
Indicator :
Pipette
End point:
RESULT
Strength of copper solution
= ..N
= ..gm.
KLNCE/Dept. of Chemistry
Page 73
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Short Procedure
AIM
To estimate the amount of copper present in 100 ml of the given brass solution. A standard
solution of CaCO3 of strength -------M and an approximately 0.01 M EDTA solution are provided.
PRINCIPLE
Cu2+
+
(Copper ions)
FSB-F
[Cu-FSB-F] + 2H+
(Purple colour)
[Cu-FSB-F] +
EDTA2-
[Cu- EDTA]2(Colourless)
+ FSB-F
(Green colour)
PROCEDURE
TITRATION I: STANDARDISATION OF EDTA
Burette : EDTA
Indicator : EBT
Pipette
: 20 ml of std. CaCO3
End point:
RESULT
Strength of copper solution
= ..N
= ..gm.
KLNCE/Dept. of Chemistry
Page 74
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Viva Voce Questions and Answers
1. What is an alloy?
Alloy is a homogenous mixture of a metal with another metal or a metal with a non
metal or two metals with a non metal.
2. How are alloys classified?
Alloys are classified into (i) ferrous alloys (ii) non ferrous alloys
3. What is amalgam?
An alloy of mercury with a metal is called by the special name amalgam.
4. Name two ferrous alloys.
Steel, nichrome
5. Give examples for non ferrous alloys.
Brass, bronze and german silver
6. Name few alloys of copper.
Brass, bronze, solder, gun metal
7. What are the constituents present in brass?
Cu & Zn
8. What is the basic difference between brass and bronze?
Brass is an alloy of copper with zinc, whereas bronze is an alloy of copper with tin.
9. Name the metal ion indicator that is used in the estimation of copper in brass.
Fast sulphon black F (FSB - F)
10. What is the end point of the titration?
Purple colour is changed to green colour.
11. Mention the link solution that is used in the brass analysis
EDTA
12. What is the equivalent weight of copper?
63.54
14. What is the percentage of copper in brass?
From 55 60 %
15. What type of indicator does fast sulphon black-F belong to?
Metal ion indicator
KLNCE/Dept. of Chemistry
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0
2
4
6
8
10
12
14
16
18
20
KLNCE/Dept. of Chemistry
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Expt. No
Date.
PRINCIPLE
Potentiometric titration depends on the measurement of emf between reference electrode and
an indicator electrode. When a solution of ferrous salt is titrated with a solution of K2Cr2O7 the
following redox reaction takes place.
The ferrous ions are oxidized by Cr6+ ions. The potential slowly increases and at the end
point, there will be a sudden increase in potential. The cell is set up by connecting this redox
electrode with a calomel electrode as shown below:
Hg / Hg2Cl2(s), KCl // Fe2+,Fe3+ / Pt
The emf of the cell is measured by means of a potentiometer. The end point is obtained by
plotting the measured emf against volume of the K2Cr2O7 solution added.
PROCEDURE
Exactly 20 ml of given ferrous sulphate solution is pipetted out into a clean 100 ml beaker
and one test tube of dil. H2SO4, is added to it. A bright platinum electrode and calomel electrode are
dipped into the solution. This cell is connected to the potentiometer and the potentials are measured
for every 2 ml addition of potassium dichromate from the burette. This procedure gives the
approximate end point. Accurate end point can be determined by adding 0.2 ml K2Cr2O7 solution
near the end point. The end point is obtained by plotting emf (E) Vs volume of K2Cr2O7 or
KLNCE/Dept. of Chemistry
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TABLE II
Volume of K2Cr2O7
(ml)
KLNCE/Dept. of Chemistry
Emf
(mv)
E
(mv)
V
(ml)
E / V
(mv/ ml)
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E / V Vs volume of K2Cr2O7. From the end point the strength and the weight of ferrous iron
present in the whole of the given solution are calculated.
KLNCE/Dept. of Chemistry
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Graph 2: volume of K2Cr2O7 Vs E/V
E/V
mv/ml
Emf
(mv)
CALCULATION
Strength of ferrous iron solution
Volume of ferrous iron solution
V1
= 20 ml
N1
= ?
Volume of K2Cr2O7,
V2
= ............. ml
Strength of K2Cr2O7,
N2
= . N
N1
= V2 x N2
= .. x ..
V1
Strength of ferrous iron solution
N1
20
= ..................... N
. N x 55.85
= ..gm.
KLNCE/Dept. of Chemistry
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RESULT
The amount of ferrous iron present in one litre of the given solution =.gm.
KLNCE/Dept. of Chemistry
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Short Procedure
AIM
To estimate the amount of ferrous iron in one litre of the given solution by potentiometric
titration. A standard solution of potassium dichromate of strengthN is provided.
PRINCIPLE
K2Cr2O7 + 6FeSO4 + 7H2SO4
PROCEDURE
Burette
Pipette
: 20 ml of ferrous sulphate
End point
Condition
Electrode
Amount of ferrous iron = Strength of ferrous iron x Equivalent weight ferrous iron (55.85)
Graph 1: Volume of K2Cr2O7 Vs. emf
E/V
mv/ml
Emf
(mv)
RESULT
The amount of ferrous iron present in one litre of the given solution =.gm.
KLNCE/Dept. of Chemistry
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Viva voce Questions and Answers
1. Define emf.
The amount of charge produced in any material or cell is called emf.
2. What is the unit for emf?
------- Volt
: Eg. H2 electrode
KLNCE/Dept. of Chemistry
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0.6
Absorbance
(or)
Emission
Concentration
Determination of Na+ ion in the sample water
max = 589 nm
Volume of solution
Concentration
(Standard)
(Unknown)
KLNCE/Dept. of Chemistry
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Expt. No
Date.
ESTIMATION OF SODIUM PRESENT IN WATER USING
FLAME PHOTOMETER
AIM
Determine the amount of sodium present in the given water sample. A standard solution of
sodium chloride containing 100 mg of sodium per litre (1000 ppm) is provided.
PRINCIPLE
When a solution containing a metallic compound is aspirated into a flame, a vapour
containing the metal atoms will be formed. Some of these metal atoms in gaseous state, may be
raised to an energy level, which is sufficiently high to permit the emission of radiation, which is
characteristic to the metal under test. Since metals like sodium, potassium and lithium have an easily
excited flame spectrum of sufficient intensity for detection by a photocell, they are usually analysed
by flame photometer.
PROCEDURE
Exactly 2.54 g of pure dry sodium chloride is dissolved in 1 litre of distilled water. This
solution contains 1000 mg of sodium per litre. This stock solution is diluted to prepare four standard
solutions of 10, 5, 2.5 and 1 ppm sodium ions respectively.
A calibration curve is plotted, using solutions of known concentrations. The test solutions
may be suitably diluted to get readings in the range 0.1 to 0.4. The absorbance emission readings are
taken in the flame photometer for different concentrations at 589 nm wavelength. The unknown
concentration of sodium in water sample is found out with the help of the calibration curve.
Note: The amount of potassium in a given water sample may also be estimated by similar procedure
and measuring the absorbance at 766 nm.
RESULT
The amount of sodium present in the given water sample
KLNCE/Dept. of Chemistry
= ppm
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Short Procedure
AIM
Determine the amount of sodium present in the given water sample. A standard solution of
sodium chloride containing 100 mg of sodium per litre (1000 ppm) is provided.
PRINCIPLE
When a solution containing a metallic compound is aspirated into a flame, a vapour
containing the metal atoms will be formed. Some of these metal atoms in gaseous state, may be
raised to an energy level, which is sufficiently high to permit the emission of radiation, which is
characteristic to the metal under test. Since metals like sodium, potassium and lithium have an easily
excited flame spectrum of sufficient intensity for detection by a photocell, they are usually analysed
by flame photometer.
PROCEDURE
A standard solution of sodium chloride is prepared.
It is diluted to four or five different concentrations.
The absorbance at 589 nm is measured in a flame photometer for various concentrations.
The absorbance is measured for unknown concentration.
A graph is plotted between absorbance and concentration.
From the graph concentration is found out.
CALCULATION
Amount of sodium = Concentration of NaCl solution x Eq. wt of sodium (23)
RESULT
The amount of sodium present in the given water sample
KLNCE/Dept. of Chemistry
= ppm
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KLNCE/Dept. of Chemistry
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S. No.
Final weight of
steel specimen
(W2)
Percentage of
HCl
Weight loss
(W1-W2) gm
CALCULATION
Rate of corrosion
KLNCE/Dept. of Chemistry
( )
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Expt. No
Date.
CORROSION EXPERIMENT- WEIGHT LOSS METHOD
AIM
To determine the corrosion rate of mild steel specimen in HCl different concentrations at
constant temperature.
PRINCIPLE
Generally metal surfaces are covered with impurities like rust and scales. These impurities if
present at the time of coating will produce porous and discontinuous coatings. In order to get a
uniform, smooth, and adherent coating, these substances must be removed by acid cleaning (or)
pickling for which dil.HCl is used as pickling solution.
PROCEDURE
The mild steel specimens (1-5) of known dimensions are taken, washed with distilled water
and air dried. Initial weights of the specimens (1-5) are noted as W1 gm.
Similarly HCl of different concentration (3%, 6%, 9%, 12%, 15%) are prepared (called
pickling solution). Now each steel specimen is dipped in 100 ml of the above five different
concentration of HCl for 30 minutes. After that they are taken out and washed with distilled water
thoroughly and air dried. Again the specimens (1-5) are weighted and the final weights are noted as
W2 gm.
KLNCE/Dept. of Chemistry
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KLNCE/Dept. of Chemistry
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From the weight loss, the rate of corrosion can be calculated using the following relation.
Rate of corrosion
( )
RESULT
Rate of corrosion
KLNCE/Dept. of Chemistry
= --------------- .
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Short Procedure
AIM
To determine the corrosion rate of mild steel specimen in HCl of different concentrations at
constant temperature.
PRINCIPLE
Corrosion is a process of gradual destruction of metals by the chemical or
electrochemical reaction with its environment. Experimentally corrosion can be measured by dipping
a mild steel plate in a known concentration of an acid for a particular time. The rate of corrosion can
be determined with different concentration and different time.
PROCEDURE
The mild steel specimens of known dimensions are taken, washed with distilled water and air
dried. Initial weights of the specimens are noted as W1 gm.
Similarly HCl of different concentration (3%, 6%, 9%, 12%, 15%) are prepared (called
pickling solution). Now the steel specimens are dipped in 100 ml of various HCl solutions for 30
minutes. After that they are taken out and washed with distilled water thoroughly and air dried.
Again the specimens are weighted and the final weights are noted as W2 gm.
Rate of corrosion
( )
RESULT
Rate of corrosion
KLNCE/Dept. of Chemistry
= --------------- .
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Viva voce question and answers.
( )
W1
W2
= Time of exposure
KLNCE/Dept. of Chemistry
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Titration between BaCl2 Vs Na2SO4
Volume of Na2SO4 added (ml)
Conductance (mho)
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
KLNCE/Dept. of Chemistry
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Expt. No
Date.
CONDUCTOMETRIC PRECIPITATION TITRATION USING
BaCl2 AND Na2SO4
AIM
To determine conductometrically the strength of given sodium sulphate solution by titrating
it with a standard solution of .N barium chloride.
PRINCIPLE
The reaction between sodium sulphate and barium chloride may be represented as,
2Na+ + SO42-
Ba2+ + 2Cl-
When a solution of sodium sulphate is added to barium chloride solution, the sodium ions are
replaced by barium ions, due to the formation of BaSO4 precipitate. As a result of precipitation, the
conductance decreases. After completion of precipitation, conductance value increases due to free
SO42- ions.
PROCEDURE
Exactly 40 ml (2 x 20 ml) of barium chloride solution is pipetted out into a clean beaker. The
conductivity cell is dipped into it. The terminals of the conductivity cell are connected to a
conductivity bridge and the conductance of the solution is measured. To this solution, sodium
sulphate solution (2 ml) from a burette is added at each time.
The conductance of the solution after each addition of sodium sulphate is measured. The
conductance goes on decreasing, reaches a minimum and then increases. The values of observed
conductance are plotted against the volume of Na2SO4 added. The point of intersection of the two
lines gives the end point. From the end point, the strength of sodium sulphate can be calculated.
KLNCE/Dept. of Chemistry
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CALCULATION
Strength of Na2SO4
Volume of barium chloride solution (V1)
= 40 ml
= .N
Volume of Na2SO4
(V2)
Strength of e Na2SO4
(N2)
= ?
N2
= 40 x N1
V2
= N
Conductance (mho)
Strength of Na2SO4
= 40 x ----------
KLNCE/Dept. of Chemistry
Page 96
Volume of Na2SO4 (ml)
CSEITQUESTIONS.BLOGSPOT.IN | CSEITQUESTIONS.BLOGSPOT.IN
RESULT
Strength of given sodium sulphate solution = ..N
KLNCE/Dept. of Chemistry
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Short Procedure
AIM
To determine conductometrically the strength of given sodium sulphate solution by titrating
it with a standard solution of .N barium chloride.
PRINCIPLE
The reaction between sodium sulphate and barium chloride may be represented as
2Na+ + SO42-
Ba2+ + 2Cl-
PROCEDURE
Burette
: Sodium sulphate
Pipette
Cell
: conductivity cell
For each addition of Na2SO4 to BaCl2 solution, the conductance goes on decreasing.
At the end point, the conductance reaches the minimum and on further addition of sodium
sulphate, the conductance increases.
By plotting a graph between conductance and volume of Na2SO4, the end point can be found
out.
KLNCE/Dept. of Chemistry
Conductance (mho)
Page 98
Volume of Na2SO
CSEITQUESTIONS.BLOGSPOT.IN | CSEITQUESTIONS.BLOGSPOT.IN
BaCl2 Vs Na2SO4
(ii)
AgNO3 Vs NaCl
KLNCE/Dept. of Chemistry
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Volume of cement
solution (ml)
Initial
(ml)
Final
(ml)
Volume of
Concordant
EDTA (ml)
value (ml)
Indicator
Patton and
Reeders
indicator
CALCULATION
Volume of EDTA
= ..ml
Strength of EDTA
= N
= 20 ml
= 56
= 28
Weight of CaO in 1L
=___________g.
KLNCE/Dept. of Chemistry
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Expt. No
Date.
DETERMINATION OF CaO IN CEMENT
AIM
Determine the amount of calcium oxide present in the given cement sample. You are
provided with . N EDTA solution.
PRINCIPLE
The general composition of Portland cement is
CaO = 60-66%, SiO3 = 17-25, Al2O3= 3-8%, Fe2O3 =2-6%, MgO = 0.1-5.5%, SO3 = 1-3%, Na2O =
0.5-1.5%.
Thus, calcium oxide is the prime constituent of cement. Excess CaO in cement reduces the
strength of cement. Lesser amount of CaO also reduces the strength of cement and result in the
quick.
PROCEDURE
A known weight of the cement sample is treated with con. HCl and the insoluble residue,
SiO2 is removed by filtration. Iron and aluminium in the filtrate are precipitated as their hydroxides
and filtered. The resulting cement solution containing calcium and magnesium is diluted to 250 ml
using distilled water in a standard flask.
Calcium of the oxide or calcium ion present in the cement solution is determined by titrating
a known volume of the cement solution with standard EDTA . The cement solution is treated with
diethyl amine to maintain the pH at 12.5. 4 N NaOH solution is added to precipitate Mg (II) present
in the cement solution as magnesium hydroxide, so that Mg (II) does not interfere in the
determination of CaO .
The solution is then titrated against standard EDTA solution using Patton and Reeders
indicator. The indicator can be used in the determination of calcium in the presence of magnesium as
the latter is the form of precipitate. The end point is the colour change from wine red to clear blue
without any reddish tinge. Glycerol is added to get a sharp end point.
KLNCE/Dept. of Chemistry
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RESULT: Weight of calcium oxide in the given cement solution = ___________ gm.
Short procedure
AIM
Determine the amount of calcium oxide in the given cement sample. You are provided with
standard EDTA solution.
PRINCIPLE
Cement is a mixture of oxides of Ca, Mg, Fe, Al, Si with calcium oxide as major
constituent. The calcium in cement can be estimated by EDTA method or by oxalate method. In the
EDTA method, calcium is determined by titration with EDTA at a pH of 12.5.
PROCEDURE
A Known weight of the cement sample is treated with con,HCl and filtered.
The filtrate is made upto 250 ml standard measuring flask.
20 ml this made up solution is pipetted out into a conical flask.
Magnesium in the cement solution is precipitated as Mg(OH)2 by adding 4N NaOH.
It is titrated with standard EDTA at a pH of 12.5, using Patton and Reeder indicator.
The end point is colour change from wine red to blue.
CALCULATION
Strength of cement solution (CaO) = Volume of EDTA x Strength of EDTA
Volume of cement solution
Amount of CaO
RESULT
Weight of calcium oxide in the given cement solution = ___________ gm / litre.
KLNCE/Dept. of Chemistry
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KLNCE/Dept. of Chemistry
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Viva voce Questions and Answers
1. What is cement?
Cement is a lime based building material used to bind together coarse aggregates.
2. What are the constituents of cement?
Oxides of Ca, Mg, Fe, Al, Si are the constituents.
3. How is cement solution prepared?
Accurately weighed amount of cement is warmed with moderately conc. HCl till cement
dissolves. Insoluble silica is filtered off and the filtrate is the cement.
4. What is the role of glycerol?
Glycerol is added to get the sharp end point and also to mask the magnesium ions.
5. Which is the indicator used in the determination of CaO in cement solution?
Patton and Reeders indicator is used. It gives sharp color change in the pH range 12-14.
6. What is the function of diethylamine?
It is added to maintain the pH around 12-14.
7. Why is this titration called rapid EDTA method?
The titration is called rapid EDTA method because, by this method, Ca2+ ions in cement
solution are estimated directly and quickly by using standard EDTA solution without removing the
other metal ions.
8. What are the molecular formula and molecular weight of Patton and Reeders indicator?
Molecular formula = C21 H 14 N2 O7 S
Molecular weight = 438
9. What is the role of NaOH in this titration?
Sodium hydroxide reacts with Mg2+ ions and precipitates it out in the form of Mg(OH)2 from
the cement solution.
10. What is the chemical name of Patton and Reeders indicator?
The chemical name of Patton and Reeders indicator is 2- hydroxy-1-(2-hydroxy-4-sulpholnaphthylazo)-3 naphthoic acid (HHSNNA).
KLNCE/Dept. of Chemistry
Page 104