1. CONDUCTOMETRIC TITRATION OF STRONG ACID AND …

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LIST OF EXPERIMENTS V+ TEAM

1. CONDUCTOMETRIC TITRATION OF STRONG ACID AND STRONG

BASE HCl vs NaOH

AIM

To determine the amount of strong acid (HCl) present in the given solution by

conductometric titration using standard NaOH of 0.1N.

PRINCIPLE

Solution of electrolytes conducts electricity due to the presence of ions. Since specific

conductance of a solution is proportional to the concentration of ions in it, conductance of the

solution is measured during titration.

When NaOH is added slowly from the burette to the solution HCl (strong

acid) gets neutralized first. Since the fast moving H+ ions are replaced by slow moving Na+

ions, decrease in conductance takes place until the end point is reached.

HCl + NaOH NaCl + H2O

When the end point is reached, addition of NaOH will cause sudden increase in the

conductance. This is due to the presence of fast moving OH ions.

MATERIALS REQUIRED

1) Conductivity bride 2) conductivity cell 3) Beaker 4) Standard NaOH 5) HCl

6) Burette, pipette, glass rod etc.,

PROCEDURE:

The burette is filled with NaOH solution up to the zero level. 50 ml of the given make

up HCl solution is pipetted out into a clean 100 ml beaker. The conductivity cell is placed in it.

The two terminals of the cell are connected with a conductivity bridge.

Now 1 ml of NaOH from the burette is added to the solution taken in the beaker,

stirred, and then conductivity is measured. This is continued up to the end point. (The

conductivity is going on decreasing up to the end point) After the end point, again NaOH is

gradually added and few more readings are noted.

Expt.No: Date:



.

Table

HCl vs NaOH



Volume of NaOH added (ml) Conductance (mho)



Calculation

Step: I Calculation of Normality of HCl

Volume of HCl V1 = 50 ml

Strength of the HCl N1 = ……N.

Volume of the NaOH V2 = ----------ml (From Graph)

Strength of the NaOH N2 = -----------N

According to the law of volumetric analysis

V1N1 = V2N2

V2N2

N = -----------

V1

………....ml x……..…...N

= -------------------------------------

50

Strength of HCl = ……………..…N

CO

ND

UC

TA

NC

E

VOLUME OF NaOH



Step : II

Calculation of amount of HCl:

The amount of HCl present in

one Litre of the given solution = ---------N x Eq.wt.of HCl (36.45)

= ……………..gms

RESULT

The amount of HCl present in 1 litre of the given solution =. …………gms.



2. CONDUCTOMETRIC TITRATION OF MIXTURE OF ACIDS (HCl

& CH3COOH) AGAINST NaOH

AIM:

To determine the amount of a strong acid and a weak acid (HCl and CH3COOH)

present in one litre of the given mixture of acid solution by conductometric titration using

standard NaOH of ………N.

PRINCIPLE:

Solution of electrolytes conducts electricity due to the presence of ions. Since specific

conductance of a solution is proportional to the concentration of ions in it, conductance of

the solution is measured during titration.

When the sodium hydroxide is added slowly from the burette to the solution, HCL

(strong acid) gets neutralised first. Since the fast moving H+ ions are replaced by slow

moving Na+ ions decrease in conductance take place until the end point is reached

HCl + NaOH NaCl + H2O (Ist neutralisation)

After the complete neutralisation of all HCl, the neutralisation of CH3COOH starts,

CH3COOH + NaOH CH3COONa + H2O ( 2nd neutralisation)

Since CH3COONa is stronger than CH3COOH conductivity slowly increases until all

CH3COOH is completely neutralised. When the end point is reached, addition of NaOH

will cause sudden increase in the conductance. This is due to the presence of fast moving

OH – ions.

MATERIALS REQUIRED:

1) Conductivity bridge 2) Conductivity cell 3) Beaker 4) Standard M/10 NaOH 5)

Approximately N/10 HCl & CH3COOH 6) Burette, pipette, glass rod etc.,

PROCEDURE:

The burette is filled with NaOH solution upto the zero level. 20 ml of the given mixture

of acids (HCl + CH3COOH) is pipetted out into a clean 100 ml beaker. The conductivity cell is

Expt.No: Date:



placed in it and then diluted to 50 ml by adding conductivity water, so that the electrodes are

well immersed in the solution. The two terminals of the cell are connected with a conductivity

bridge.

Now 1ml of NaOH from the burette is added to the solution, taken in the beaker stirred

for some time and then conductivity is measured. (The conductivity is going on decreasing

upto the end point) This process is repeated until atleast five readings are taken after the end

point (a) has been reached.

After the end point, again NaOH is gradually added, which causes increase in

conductance. The increase in conductance is observed until the end point (b) is reached.

After the second end point, sudden increase in conductance is observed on further

addition of NaOH. The reading (conductivity) is continuously measured for each addition of

NaOH and are tabulated. Now the graph is plotted between the volumes of NaOH versus

conductivity. From the graph the first end point (a) and the second end point (b) is noted. From

the end points the amount of HCl and CH3COOH present in 1 litre of the mixture of solution is

calculated.



Table

Volume of mixture (HCl + CH3COOH) Vs NaOH

Volume of NaOH

added (ml)

Conductance (mho)



Calculation

StepI: Calculation of normality of HCl

Volume of the mixture (HCl) V1 = 20ml

Strength of the mixture (HCl) N1 = ………N

Volume of the NaOH V2 = …….. ml (Ist titre value)

Strength of the NaOH N2 = ……… N


V1N1 = V2N2

V2N2

N1 = _____

V1



…..ml x ….N

= _________

20

Strength of HCl = ……..N

Step II: Calculation of amount of HCl :

The amount of HCl Present in

1000 ml of the given solution = Normality of HCl x Eq, wt Hcl

= ……..gms

Step III : Calculation of normality of CH3COOH

Volume of the mixture (CH3COOH) V1 = 20 ml

Strength of the mixture (CH3COOH) N1 = …….N

Volume of the NaOH V2 = ………ml (2nd titre value)

Normality of NaOH V2 = 0.1 N


V1N1 = V2N2

V2N2

N1 = ____

V1

…….ml x…N

= __________

20

Normality of CH3COOH = ……..N

Step IV : Calculation of amount of CH3COOH

The amount of CH3COOH Present in = Normality of CH3COOH x eq.wt.CH3COOH

1000 ml of the given solution

= ……gms

RESULT: i) The amount of HCL present in 1 litre of the given solution ………gms

ii) The amount of CH3COOH present in 1 litre of the given solutions

……….….gms

3. CONDUCTOMETRIC PRECIPITATION TITRATION USING BaCl2

AGAINST Na2SO4

Expt.No: Date:



AIM :

To determine the amount of BaCl2 present in one litre of the given solution by

conductometric titration using standard Na2SO4 of ……….N

PRINCIPLE:

Solution of electrolytes conducts electricity due to the presence of ions since specific

conductance of a solution is proportional to the concentration of ions in it, conductance of the

solution is measured during titration.

In the precipitation titration, the ions are converted to insoluble precipitate, which will

not contribute in the conductance.

When Na2SO4 is added slowly from the burette to the solution of BaCl2, BaSO4 gets

precipitated while the chloride ions remain unchanged.

[Ba2+ + 2Cl]+[2Na+ + SO24] BaSO4 + 2Na+ + 2Cl-

The Ba2+ ions in the solution are replaced by free Na+ ions. Since the mobility of Na+

ions is less that of Ba2+ ions the conductance of the solution decreases.

After the end point, when all the Ba2+ ions are replaced, further addition of Na2SO4

increases the conductance. This is due to the increase of Na+ and SO24 ions in the solution.

MATERIALS REQUIRED

1) Conductivity bride 2) conductivity cell 3) Beaker 4) Standard N/10Na2SO4 5)

Approximately N/10BaCl2 solution 6) Burette, pipette, glass rod etc.,

PROCEDURE:

The burette is filled with Na2SO4 solution upto the zero level. 20 ml of the given BaCl2

solution is pipetted out into a clean 100 ml beaker. The conductivity cell is placed in it and then

diluted to 50 ml by adding conductivity water. The two terminals of the cell are connected with

a conductivity bridge.

Now 1 ml of Na2SO4 from the burette is added to the solution taken in the beaker,

stirred, and then conductivity is measured. This is continued upto the end point. (The

conductivity is going on decreasing upto the end point) After the end point, again Na2SO4 is

gradually added and few more reading are noted.

Thus the conductivity is continuously measured for each addition of Na2SO4 and are

tabulated. Now the graph is plotted between the volume of Na2SO4 and conductivity. From the

graph end point is noted and hence amount of BaCl2 present in 1 litre is calculated.



Table



BaCl2 Vs Na2SO4

Volume of

Na2SO4 added

(ml)

Conductance

(mho)

Calculation

Step: I Calculation of normality of BaCl2



Volume of BaCl2 mixture (HCL) V1 = 20 ml

Strength of the BaCl2 solution N1 = ……N.

Volume of the Na2SO4 V2 = --------ml (titre value)

Strength of the Na2SO4 N2 = --------N


V1N1 = V2N2

V2N2

N1 = _____

V1

….ml x….N

= ___________

20

Strength of BaCl2 = ……N

Step : II Calculation of amount of BaCl2:

The amount of BaCl2 present in

Litre of the given solution = ------- x Eq.wt.of BaCl2 (122.14)

= ………..gms

RESULT:

The amount of BaCl2 present in 1 litre of the given solution = …..……gms.



4. ESTIMATION OF FERROUS IRON BY

POTENTIOMETRIC TITRATION

AIM:

To estimate the amount of ferrous iron present in whole of the given solution by

potentiometric method. A standard solution of potassium dichromate of strength ……..N is

provided.

PRINCIPLE:

Potentiometric titrations depend on measurement of emf between reference electrode

and an indicator electrode. When a solution of ferrous iron is titrated with a solution of

potassium dichromate, the following redox reaction takes place.

6Fe24 + Cr2O7 2 + 14H 6Fe3+ + 7H2O

During this titration Fe2+, whose concentration increases, at the end point, there will be

a sharp change due to sudden removal of all Fe2+ ions.

Connecting this redox electrode with a calomel electrode as shown below sets up the

cell:

Hg,HgCl2(s) KCl // Fe2+, Fe3+, Pt

A graph between emf measured against the volume of K2Cr2O7 added is drawn and the

end point is noted from the graph.

MATERIALS REQUIRED: (i)Potentiometer (ii) pt electrode (iii) saturated calomel electrode (iv) standard K2Cr2O7

solution (v) given ferrous iron solution.

PROCEDURE:

Expt.No: Date :



The given ferrous iron solution is made upto 100 ml in a standard flask. 20 ml of this

made up solution is pipetted out into a clean 100 ml beaker. About 10 ml of dil H2SO4 and 20

ml of distilled water are added in it. A platinum electrode is dipped into the solution. This

electrode is then coupled with a saturated calomel electrode and the cell is introduced into

potentiometric circuit. The std K2Cr2O7 solution is taken in the burette and added.

Titration – I First adding std. K2Cr2O7 solution in portions of 1 ml and the emf carries out a

Preliminary titration. of the cell is measured after each addition. The addition of K2Cr2O7 is

continued even after the end point and the range at which end point lies is found out by plotting

volume of K2Cr2O7 added emf (graph I).

Titration –II

Another titration is carried out by adding std. K2Cr2O7 solution in portions of 0.1 ml

near the end point and the emf of the cell is measured after each addition. The addition of

K2Cr2O7 is continued even after the end point for further 1 ml. The accurate end point is

determined by plotting E / V Vs Volume of K2Cr2O7 added (graph ∏). From the end point,

the strength of ferrous iron solution and hence its amount can be calculated.

Observations and calculations:

Figure 1 – Potentiometric titration

Step: I

Table – I

Volume of ferrous iron solution = 20 ml

Volume of K2Cr2o7

(ml) Emf (Volts)



Graph – 1 - Graph-2

Step : II

Table – II

Volume of ferrous iron solution = 20 ml

Volume of K2Cr2O7

(ml)

Emf

(Volts)

E

(Volts)

Vml E / V volts / ml



Step – III

Calculation of normality of ferrous iron solution:

Volume of ferrous iron solution V1 = 20 ml

Normality of ferrous iron solutionN1 = --------N

Volume of K2Cr2O7 V2 = …………ml(from graph 2)

Normality of K2Cr2O7 N2 = ………..N

According to the law of volumetric analysis V1N1 = V2N2

V2 x N2

N1 = _______

20

Normality of ferrous iron solution N1 = ………….. N

RESULT:

The amount of ferrous iron present in 100 ml of the given solution is = ………gms.



5. DETERMINATION OF STRENGTH OF HCL AGAINST NaOH BY

pH METRY

AIM:

To determine the strength of given HCL by pH metry, a standard solution of

NaOH……..N is provide.

PRINCIPLE:

Since 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+ ion is the solution.

When NaOH is added slowly of m the burette to the solution of HCL, the fast moving H+

Ions are progressively replaced by slow moving Na+ ions. As a result pH of the solution

Increases.

H+ CL- + Na+ OH- → NaCl+H2O

The increase in pH takes place until the all the H+ ions are completely neutralized

(upto the end point) after the end point, further of NaOH increases pH sharply as there

is an excess of fast moving OH- ions.

MATERIALS REQUIRED:

Expt.No: Date:



1pH meter 2) Glass electrode 3) Beaker 4) Standard n/10 NaOH

5) Approximately N/10 HCI 6) Burette, Pipette, Glass rod etc.,

PROCEDURE:

The burette is filled with std. NaOH Solution. Exactly 20m of the given HCL

Solution is pipetted out into a clean beaker. It is then diluted to 20 ml by adding distilled

Water. The glass electrode is dipped in it and connected with a pH meter.

Now NaOH solution is gradually added form the burette to the HCL solution taken

In the beaker and pH of the solution is noted each addition. This process is continued

Until atleast 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.

Table : I

Volume of HCL Vs NaOH

Volume of NaOH

ml

pH



Table :

Volume of NaOH Vs pH / V



Volume of

NaOH ml pH

pH V (ml) pH/ /v

Calculation

Step: I Calculation of Strength of HCL

Volume of HCL V1 = 20 ml

Strength of the HCL N1 = ………N

Volume of the NaOH V2 = ……… (titre value)

Strength of the NaOH N2 = --------- N


V1N1 = V2N2

V2N2

N1 = _____

V1



…. X……N

= ____________

20

Strength of HCL present in 1 litre of the given solution = …………. N

Step :II Calculation of amount of HCL :

The amount of HCL Present in

1000ml of the given solution = …… N x Eq.wt.of HCL (36.45)

= ……….. gms

RESULT:

1. Strength of the given HCL solution = ………………N

2. Amount of HCl present in 1 litter of the solution = ………………gms

6. DETERMINATION OF WATER OF CRYSTALLIZATION OF A

CRYSTALLINE SALT (CuSO4)

AIM:

Expt.No: Date:



7. DETERMINATION OF THE IRON CONTENT BY

SPECTROPHOTOMETRY

AIM:

To determine the amount of iron content of an unknown solution by spectrophotometry.

PRINCIPLE :

Expt.No: Date:



Spectrophotometer is an instrument used to measure the intensity of the light absorbed

by a substance. The relationship between absorbance (A) and the concentration of the solution

(C) is given by Beer - Lambert’s law.

Io

log __ = A = ECX

I

Where

Io = Intensity of incident light

I = Intensity of transmitted light

E = Molar absorption co-efficient

X = Thickness of the cell

C = Concentration of the solution

From the equation it is seen that, the absorbance (A) is directly proportional to the

molar concentration and thickness of the cell.

In the determination of iron, ferrous iron solution is acidified with HNO3 to convert

Fe2+ into Fe3+. Fe3+ ions does not give any colour in solution. But, it gives red colour when it

reacts with potassium thiocyanate (KSCN) (or) Ammonium thiocyanate (NH4SCN) solution.

Fe3+ 6KSCN [Fe(SCN)6]

3- + 6K+

red colour complex

This complex has maximum absorption in the region = 480 nm. A calibration curve is drawn

by measuring the absorbance of known solution. Then the absorbance of unknown solution is

measured, using which the concentration can be determined from the calibration curve.

MATERIAL REQUIRED:

1) Spectrophotometer 2) Ferrous ammounium sulphate 3) HNO3 4) 25 ml standard

flask 5 Nos. 6) Graduate pipette 7) Potassium thiocyanate (or) Ammonium

thiocyanate.

PROCEDURE :



1.Preparation of stock Fe3+ iron solution

A stock solution of Fe3+ is prepared by dissolving 0.0838 gms of ferrous ammonium

sulphate with 1 ml of con. HNO3 in 1 litre of distilled water.

From the stock solution, various concentrations are prepared as shown in table I

1. Determination of iron content

The spectrophotometer is switched on and warmed up for 10 minutes. The monochromater is

adjusted for = 480 nm

The blank solution (distilled water ) is kept in the cell and the absorbance is measured

for which the absorbance is zero and transmittance is 100.

Now the absorbance of all the standard solutions are similarly measured. Also the

absorbance of unknown solution is measured.

The calibration graph is drawn between the concentration and absorbance from which

the concentration of unknown solution is measured.

Table I : Preparation of various concentration of Fe3+ solution

Volume of iron

solution

Volume of

HNO3

Volume of

KCN3

Concentration of Iron

ppm

2

4

6

8

Unknown

3

3

3

3

3

5

5

5

5

5

2 ppm

4 ppm

6 ppm

8 ppm

-

1 ml of stock solution = 1 ppm of iron



Calibration curve (absorbance Vs concentration)

Table II: Measurement of absorbance

Blank (distilled water) : zero absorbance

Concentration

ppm

Absorbance

2 ppm

4 ppm

6 ppm

8 ppm

Unknown

RESULT:

Amount of iron present in the solution = ……..ppm

CONCENTRATION

AB

SO

RB

AN

CE



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1. CONDUCTOMETRIC TITRATION OF STRONG ACID AND …