The pdf provides theory, procedure and calculations of physical and analytical chemistry experiments (Conductometry, Potentiometry and Chemical Kinetics) of semester IV as per revised syllabus of Mumbai University, effective from academic year 2017-18.

1. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
POTENTIOMETRY
Aim : To determine standard emf & the standard free energy change of Danial cell
Requirements :
Chemicals:
0.1M ZnSO4 soln., 0.1M CuSO4 soln., KCl or KNO3 Salt bridge, Distilled water.
Apparatus: Potentiometer, Std. cell , Cu electrode, Zn electrode, 50cm3 beakers(2),
5cm3 & 25cm3 pipette, 50cm3 volumetric flask (4).
Theory:
1. Cu+2 + Zn Cu + Zn+2
The above reaction can be studied by setting up the following chemical cell.
(-)Zn | ZnSO4 solution || CuSO4 solution | Cu (+)
2. The oxidation reaction at the left hand half cell is
Zn Zn+2 + 2e- ……. (1)
The reduction reaction at the right hand half cell is
Cu+2 + 2e- Cu …….. (2)
Adding 1 & 2 we get the cell reaction as
Cu+2 + Zn Cu + Zn+2
Using Nernst equation , the e.m.f of the cell is given by
2
2
10
0
log
2
0592
.
0






Cu
Zn
Cu
Zn
cell
cell
a
a
a
a
E
E
Where 0
cell
E is standard e.m.f of the cell.
By convention Cu
a = Zn
a = 1
2
2
10
0
log
2
0592
.
0





Cu
Zn
cell
cell
a
a
E
E
2
2
10
0
log
2
0592
.
0





Cu
Zn
cell
cell
a
a
E
E
The standard free energy change (G0 ) for this reaction is related to the e.m.f of the cell
( 0
cell
E ) by equation
G0 = - nF 0
cell
E also G0 = - RT lnK
The value of the 0
cell
E can be calculated by measuring e.m.f of the cell & by knowing the
activities of the ion in the solution. Then knowing G0 , the equilibrium constant can be
calculated.

2. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
Procedure:
1. Standardise the potentiometer using standard cell.
2.
a)
b)
c)
d)
Prepare the following solutions from 0.1M ZnSO4 & 0.1M CuSO4 solution in
50 cm3 std. flask.
25cm3 of 0.1M ZnSO4 soln. + 25cm3 distilled water  0.05M ZnSO4 soln .
5cm3 of 0.1M ZnSO4 soln. + 45cm3 distilled water  0.01M ZnSO4 soln .
25cm3 of 0.1M CuSO4 soln. + 25cm3 distilled water  0.05M CuSO4 soln .
5cm3 of 0.1M CuSO4 soln. + 45cm3 distilled water  0.01M CuSO4 soln .
3. Take a 50 cm3 beaker , rinse it with 0.01 M ZnSO4 & take approx. 30 to 40 cm3
of 0.01M solution of ZnSO4 in that beaker. Dip the zinc electrode in it & connect
it to the negative terminal of the potentiometer.
4. Take another 50 cm3 beaker , rinse it with 0.01 M CuSO4 & take approx. 30 to
40 cm3 of 0.01M solution of CuSO4 in that beaker. Dip the copper electrode in it
& connect it to the positive terminal of the potentiometer.
5. Connect the two beakers with a KNO3 salt bridge & measure the e.m.f of the cell.
6. Remove both the beakers & take 0.05M CuSO4 & 0.05M ZnSO4 solution. Wash
the Cu electrode with distilled water & dip inside 0.05M CuSO4 solution.
Similarly wash silver electrode & put inside 0.05M ZnSO4 solution & measure
the e.m.f of the cell. Determine the e.m.f of the cell as before.
7. Repeat the same procedure & determine e.m.f of the cell by using 0.1M CuSO4 &
0.1M ZnSO4 solution
Observations:
Cell : (-)Zn | ZnSO4 solution || CuSO4 solution| Cu(+)
Sr.
No.
ZnSO4
(m1)
CuSO4
(m2)
Mean
Activity
Coefficient
of ZuSO4
(1)
Mean
Activity
Coefficient
of CuSO4
(2)
Activity
of
Zn+2
( m11 )
Activity
of
Cu+2
( m22 )
e.m.f
of the
cell
Volts
0
cell
E
Volts
1 0.1M 0.1M 0.148 0.16
2 0.05M 0.05M 0.212 0.216
3 0.01M 0.01M 0.307 0.40
Mean = _____V

3. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
Calculations:
1) 0
cell
E (Standard e.m.f of the cell) :
2
2
10
0
log
2
0592
.
0





Cu
Zn
cell
cell
a
a
E
E where , 2

Zn
a = m11 & 2

Cu
a = m22
2
2
1
1
10
0
log
2
0592
.
0


m
m
E
E cell
cell



2) G0 (Standard free energy change) :
G0 = - nF 0
)
(Mean
cell
E
G0 = - 2 x 96,500 x 0
)
(Mean
cell
E
G0 = _________ Joules
Result :
1) Mean 0
)
(Mean
cell
E = _________ Volts
2) Standard free energy change G0 = ________ Joules

4. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
POTENTIOMETRY
Aim : To determine the strength of a given strong acid (HCl) by potentiometric
titration using quinhydrone electrode.
Requirements :
Chemicals: Stock HCl solution, Quinhydrone powder, KCl Salt bridge, 0.1N NaOH,
Saturated KCl solution, Distilled water.
Apparatus: Potentiometer, Std. cell , Saturated calomel electrode, Pt electrode,
10cm3 pipette, 25cm3 burette, 100cm3 volumetric flask (1),
100cm3 beakers(2).
Theory:
1. In this method , the cell is set up which consists of two half cells, one is called
as indicator electrode and other as reference electrode.
2. The concentration of H+ ions cannot be measured directly with normal
potentiometer. Hence, to convert this system to reversible redox system
quinhydrone powder is added to it. Thus the indicator electrode is quinhydrone
electrode, which is reversible to H+ ion concentration of the acid solution.
3. Saturated calomel electrode is used as reference electrode.
4. The cell so set up is represented as
(-)SCE || Quinhydrone, H+(from acid) | Pt(+)
The cell emf depends on H+ ion concentration, hence , pH of the solution
Ecell = E0Q – 0.0592pH – Ecalomel
0.0592
E
-
0.457
0.0592
E
-
0.242
-
0.699
E
-
E
pH cell
cell
SCE
0
Q





0592
.
0
cell
E
From the value of pH , concentration of H+ can be calculated.
5. The potential of calomel electrode being constant, emf of the cell depends upon
Quinhydrone electrode which in turn depends upon the concentration of H+
ions.
6. As the titration proceeds, the H+ ion concentration decreases, decreasing the
emf of the cell. In the beginning the decrease is slow but towards the end point it
is comparatively rapid and sharp.
7. After the end point on further addition of alkali, again a small change in emf is
found and now the values of emf becomes negative.
Procedure:
1. Set up the potentiometer and standardize it with standard cell.
2. Dilute the given solution of acid to 100ml with distilled water in a standard flask
and pipette out 10ml of diluted solution in a 100ml beaker.
3. Add a pinch of quinhydrone powder. Stir the solution with the help of glass rod.
Insert platinium electrode in it. Connect it to the positive terminal of
potentiometer.

5. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
4. In another beaker take saturated KCl solution and insert calomel electrode in it.
Connect it to the negative terminal of potentiometer.
5. Put a KCl salt bridge connecting the two beakers. Now measure the emf of the
cell set up.
6. Fill the burette with 0.1N NaOH solution. Add 0.5ml of NaOH, stir the solution
well. Record the emf of the cell.
7. Take readings after addition of 0.5 ml of 0.1N NaOH solution each time.
8. Repeat the procedure in step no. 7. After the cell attains negative emf take 6-7
readings more.
Observations :
1. Cell : (-)SCE || Quinhydrone, H+(from acid) | Pt(+)
2. Volume of diluted acid taken for titration = 10 cm3
3. Burette solution = 0.1N NaOH
4. Temperature T = ( 273 + Room temp ) = ______ K
Volume of
NaOH added
(cm3)
e.m.f.
( Volt )
∆ E
Volt
∆ V
(cm3)
∆ E /∆ V
( Volt /cm3 )
Mean
Volume
(cm3)
pH
0.0
0.5
.
.
Calculations and Graph:
1. Calculate pH using equation
2. Plot the graph of ∆ E /∆ V against mean volume. Determine equivalence
point.
3. Plot the graph of pH against Volume. Determine equivalence point.
0.0592
E
-
0.457
pH cell

E/V
(V/cm3)
Mean Volume of 0.1N NaOH Volume of NaOH
pH

6. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
Note : Use mean equivalence point (Vx) for calculating amount of acid in the
given solution.
Calculations :
Volume of NaOH solution required for equivalence point (mean equivalence
point from graph ‘Vx’ ) = ____ cm3
10
Vx
0.1
N
HCl
of
Normality NaOH 




HCl
V
Vx
= _______ N.
10 cm3 of the diluted acid solution required Vx cm3 of 0.1N NaOH solution.
 100 cm3 of the diluted solution will require 10 Vx cm3 of 0.1N NaOH soln.
1000 cm3 of 1N NaOH = 36.5 g of HCl
HCl
of
g
_______
0.1
3
NaOH
N
0.1
of
cm
Vx
10 3
_
1000
10
5
.
6





Vx
100
% 









value
)
al
(Theoretic
Accepted
value
)
al
(Theoretic
Accepted
value -
al
Experiment
Error
Results :
1. Equivalence point from the graph of ∆E/∆V against mean volume _____ cm3
2. Equivalence point from the graph of pH against mean volume _____ cm3
3. Amount of HCl present in the given solution ______ g
4. Percentage Error : ________%

7. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
POTENTIOMETRY
Aim : To determine the amount of Fe (II) and formal redox potential in the given
solution by potentiometric titration against a standard solution of potassium
dichromate.
Requirements :
Chemicals: 0.1N K2Cr2O7 solution, a solution of Fe(II) ions, 2N H2SO4,
KCl Salt bridge, saturated KCl solution.
Apparatus: Potentiometer, Std. cell , Saturated calomel electrode, Pt electrode,
10cm3 pipette, 25 cm3 burette, 100 cm3 standard volumetric flask,
100 cm3 beakers (2), glass rod etc.
Theory:
1. In this method a cell is set up which consists of calomel as reference & a redox
system as the indicator electrode.
2. In this case the redox system contains excess of Fe+2 ions which are oxidized to
Fe+3 by K2Cr2O7 during the course of titration. This changes the activities of Fe+2
in the solution, thus changing the electrode potential of Fe+2/Fe+3 system
accordingly.
3. Since the electrode potential of calomel electrode is constant, the emf of the cell
depends upon the electrode potential of Fe+2/Fe+3 system.
4. The cell to be constructed in this experiment is
(-)Hg | Hg2Cl2 , KCl (Saturated ) || Fe+3 , Fe+2 | Pt(+)
5. The e.m.f of this cell , neglecting activity co-efficient terms is
Ecell = E Fe+3/Fe+2 - Ecal
But E Fe+3/Fe+2 = E0 Fe+3/Fe+2 -
3
2
ln
1 

Fe
Fe
a
a
F
RT
 Ecell = E0 Fe+3/Fe+2 -
3
2
ln
1 

Fe
Fe
a
a
F
RT
- Ecal(red) ……… (1)

8. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
6. When K2Cr2O7 is added, aFe+2 decreases & aFe+3 increases as a result the e.m.f
of cell increases.
7. At the end point, an inflection is observed due to sudden increase in e.m.f of the
cell. This is again due to rapid and sharp decreases in aFe+2 .
8. At half equivalence point aFe+3 = aFe+2
1
3
2




Fe
Fe
a
a
0
ln
3
2




Fe
Fe
a
a
 equation (1) becomes
 E(obs)1/2 = E0 Fe+3/Fe+2 - Ecal(red)
Formal redox potential E0 Fe+3/Fe+2 = Ecal(red) + E(obs)1/2
E0 Fe+3/Fe+2 = 0.242 + E(obs)1/2
Procedure:
1. Standardise the potentiometer using standard cell.
2. Dilute the given Fe(II) solution to 100cm3 with distilled water & take 10 cm3 of
this solution in a 100 cm3 beaker for titration. Add 2 test tube of 2N H2SO4 to it.
Insert a clean, polished platinum electrode and connect it to the positive
terminal of the electrode.
3. In another beaker take saturated KCl solution and place a calomel electrode in
it. Connect this to negative terminal of the potentiometer.
4. Connect the two beakers with a KCl salt bridge.
5. Measure the e.m.f of the cell.
6. Fill up the burette with 0.1N K2Cr2O7 solution & add 0.5 cm3 of K2Cr2O7
solution to the beaker containing Fe(II) solution. Stir the solution gently and
allow it to stand for about 10 seconds for stabilization. Then measure the e.m.f.
7. Repeat the procedure in step No. 6. Towards the equivalence point there will be
a sudden sharp increase in the e.m.f. Repeat the procedure till the changes in
e.m.f i.e. the increase in e.m.f between two successive reading is very small or
the e.m.f values are nearly constant.

9. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
Observations:
Cell : (-)Hg | Hg2Cl2 , KCl (Saturated ) || Fe+3 , Fe+2 | Pt(+)
Volume of Fe(II) solution taken for titration = 10 cm3
Normality of K2Cr2O7 solution taken in the burette = 0.1 N
Volume of
titrant added in
cm3
(V)
Mean
volume
( cm3 )
e.m.f in V
( E )
E
( Volt )
V
( cm3 )
E/V
( Volt/ cm3 )
0.0
0.5
1.0
1.5
.
.
.
Graphs : Plot
1) Plot graph of E/V against mean Volume. Determine the equivalence point of
the titration(Vx). Determine the normality & amount of given Fe(II) solution.
2) Plot Graph of E against V, using (Vx) , calculate the formal redox potential.
Volume of K2Cr2O7
E
(Volts)
Volume of K2Cr2O7
E/V
(V/cm3)

10. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
Calculations :
1) Normality of Fe(II) solution :
From graph of E/V against V , the volume of 0.1N K2Cr2O7 solution
required for equivalence point = Vx cm3
Fe(II) = K2Cr2O7
N1V1 = N2V2
 N1 = N2V2
V1
 N1 = 0.1 x Vx = A N = Normality of diluted Fe(II) solution.
10
2) Amount of Fe(II) in the given solution :
10 cm3 of Fe(II) solution required Vx cm3 of 0.1N K2Cr2O7 solution
 100 cm3 of diluted Fe(II) solution = 10Vx cm3 of 0.1N K2Cr2O7 solution
1000 cm3 of 1N K2Cr2O7 = 55.85g of Fe(II)
 10Vx cm3 of 0.1N K2Cr2O7 = 55.85 x 10Vx x 0.1
1000
= ___________ g of Fe(II)
3) Percentage Error : Given weight of Iron (II) = _________ g
100
% 









value
)
al
(Theoretic
Accepted
value
)
al
(Theoretic
Accepted
value -
al
Experiment
Error
Results :
1. Volume of 0.1N K2Cr2O7 required for titration : Vx = _________ cm3.
2. Normality of the diluted Fe(II) solution : A N = ________ N.
3. Amount of Fe(II) in the given solution : ___________g.
4. Percentage Error : _______ %.

11. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
CONDUCTOMETRY
Aim : To determine the amount of strong acid in the given solution by titrating
it against a strong base conductometrically.
Requirements :
Chemicals : 0.25N NaOH, strong acid in a 100 cm3 standard volumetric flask,
conductivity water.
Apparatus: 25cm3 pipette, 25cm3 burette, 100cm3 beaker , glass rod,
Conductometer, Conductivity cell.
Theory :
H+ & OH- ions have very high molar conductivity . When strong acid
(HCl) is titrated against strong bade (NaOH) , the initial conductance is high
due to the complete dissociation of strong acid (HCl) . As NaOH is added, H+
ions combine with OH- ions to form undissociated water molecules and H+ ions
get replaced by Na+ ions having much lower conductance value. Hence, the
conductance of the solution goes on decreasing and reaches a minimum (at
equivalence point).
After the equivalence point, NaOH added remains unreacted. Being a
strong base, it ionizes completely producing OH- ions of high conductivity.
Hence, conductance increases again.
On plotting a graph of conductance against the volume of base added, a
‘V’ shaped curve is obtained. The point of intersection of the two segments in
the graph gives the end point of the titration.
Procedure :
1. Dilute the given acid to 100cm3 with conductivity water.
2. Rinse and fill the burette with 0.25N NaOH.
3. Pipette out 25 cm3 of the diluted solution in a 100 cm3 beaker.
4. Place a conductivity cell in the beaker after rinsing with distilled water. Put a glass
rod in the beaker to stir the solution.
5. Measure the conductance of the solution.
6. Add 0.5 cm3 of NaOH solution from the burette. Stir the solution & note down the
conductance value.
7. Repeat the step No. 6 each time adding 0.5 cm3 of NaOH solution until the
conductance values show a sharp increase. Take 6-8 more reading after the sharp
increase.

12. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
Observations :
1. Volume of the diluted acid mixture taken in the beaker = 25 cm3
2. Normality of NaOH solution used = 0.25N
Obs.
No.
Volume of NaOH
(cm3)
Conductance
(µS)
1 0.0
2 0.5
3 1.0
4 1.5
. .
. .
Graph :
Plot a graph of conductance against volume of NaOH added.
Calculations :
Volume of NaOH solution required for equivalence point (Vx) = ____ cm3 from
graph
25
Vx
0.25
N
HCl
of
Normality NaOH 




HCl
V
Vx
= _______ N.
25 cm3 of the diluted acid solution required Vx cm3 of 0.25N NaOH solution.
 100 cm3 of the diluted solution will require 4 Vx cm3 of 0.1N NaOH solution
1000 cm3 of 1N NaOH = 36.5 g of HCl
HCl
of
g
_______
_
1000
0.25
4
5
.
6
3
NaOH
0.25N
of
cm
Vx
4 3





Vx
Volume of NaOH
Conductance
Vx

13. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
Percentage Error : Given weight of HCl = _________ g
100
% 









value
)
al
(Theoretic
Accepted
value
)
al
(Theoretic
Accepted
value -
al
Experiment
Error
Results :
1. Volume of 0.25N NaOH required to neutralize strong acid Vx = ________ cm3
2. Normality of HCl = ___________N
3. Amount of HCl present in the given solution = ________ g.
4. Percentage Error = ___________ %

14. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
CHEMICAL KINETICS
Aim : To compare the strength of HCl and H2SO4 by studying kinetics of acid
catalysed hydrolysis of Methyl acetate.
Requirements :
Chemicals : 0.5NHCl, 0.5N H2SO4 ,0.1N NaOH, Methyl acetate, Phenolphthalein
indicator, ice.
Apparatus: 5cm3 pipette, 50 cm3 burette, 100 cm3 measuring cylinder,
water bath, Thermostat, reagent bottle, 250 cm3 conical flask .
Theory:
1. The hydrolysis of methyl acetate being a very slow reaction is catalysed by acid.
2. The reaction is represented as :
CH3COOCH3 + H2O 


H
CH3COOH + CH3OH
3. In this reaction water is present in such a large excess that its concentration
does not change appreciably over the course of the reaction. Hence the reaction
rate is solely dependant on the concentration of methyl acetate. The reaction
though bimolecular is of first order. Such reactions are called pseudo-
unimolecular reaction.
4. The rate of hydrolysis of ester is found to be directly proportional to the
concentration of H+ ions used as catalyst. Hence the specific reaction rate of acid
catalysed hydrolysis of an ester can be used to compare the concentration of H+
ion in equimolar solutions of two acids.
5. If equal volume of two acids HCl and H2SO4 of same normalities are used for
catalyzing the hydrolysis of an ester and if K1 and K2arethe specific reaction
rates of with HCl and H2SO4 respectively then the ratio of their strength (S) is
given by
2
1
K
K
S 
Procedure : Set I ( At Room temperature )
1. In one dry & clean glass stoppered bottle pipette 5 cm3 of Methyl acetate
and in another bottle take 100cm3 0.5N HCl. Keep them in a water
bath to attain the room temperature.

15. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
2. Fill the burette with 0.1N NaOH solution.
3. At a known time add 100cm3 of HCl to the bottle containing Methyl
acetate. Shake this reaction mixture gently.
4. Immediately pipette out 5cm3 of the reaction mixture & transfer it to a
conical flask containing a few ice pieces and a drop of phenolphthalein
indicator.
5. Titrate this with 0.1N NaOH from the burette. The end point is from
colourless to faint pink colour. This is the reading at zero minute.
6. Repeat this procedure by titrating 5cm3 of the same reaction mixture at
the interval of 5, 10, 15, 20, 25 & 30 minutes from the start ( zero time ).
Set II : ( At Room temperature )
Repeat the procedure of Set I using 0.5N H2SO4
Set II : 5cm3 of methyl acetate + 100cm3 of 0.5N H2SO4
Observations :
Given : T = ____ cm3
Set I : 5cm3 of methyl acetate + 100cm3 of 0.5N HCl
Temperature , T1 = _______ 0C = ( _____ + 273 ) = ________ K
Time
‘t’
(min)
Titration
reading
Tt ( cm3 )
a =
T -
T0
(cm3 )
log10 a a - x =
T - Tt
(cm3 )
log10 (a-x)
x
a
a
t
k
)
(
log
303
.
2
10


min-1
0 ____ = T0 - - -
5 - -
10 - -
15 - -
20 - -
25 - -
30 - -
Mean K1=_______
Similarly prepare the observation table for Set II and calculate K2 for
reaction with 0.5N H2SO4

16. [SEM IV : PHYSICAL & ANALYTICAL CHEMISTRY PRACTICAL MANUAL] Dr. Aqeela Sattar, Royal College
Graphs : Plot a graph of log(a-x) against time ‘t’ for both the sets.
time ‘t’ (min )
Calculations:
For both Sets, plot graph of log(a –x) vs t, and determine ‘k’
Rate constant K = - 2.303 x slope
Calculate relative strength using relation ,
2
1
K
K
S 
Results :
1) Values of specific reaction rate:
Set I Set II
By Calculation K1 = _________ min-1 K2 = _________ min-1
By Graph K1 = _________ min-1 K2 = _________ min-1
2) Relative strength of HCl and H2SO4
S= ________ by Calculation S= ________ by Graph
log(a-x)