CHEMISTRY MANUAL SEM I & II

7/25/2019 CHEMISTRY MANUAL SEM I & II

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CSEITQUESTIONS.BLOGSPOT.IN | CSEITQUESTIONS.BLOGSPOT.IN

KLNCE/Dept. of Chemistry Page 1

SEMESTER I


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LIST OF EXPERIMENTS

Expt. No. Name of the Experiments Page No.

1.Determination of DO Content of water sample by

Winklers method5

2.Determination of chloride content of water sample by

argentometric method11

3. Determination of strength of given HCl using pH meter 17

4.Determination of strength of acids in a mixture using

conductivity meter25

5.Estimation of iron content of the water sample using

spectrophotometer31

6.Determination of molecular weight of polyvinyl alcohol

using Ostwald viscometer37

7. Conductometric titration of strong acid Vs strong base 43


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TABLE OF CONTENTS

Ex. No. Date Name of the ExperimentsMarks

Obtained

Signature of the

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) Volume of sodiumthiosulphate (ml) IndicatorInitial Final

CALCULATION

Step1 : Standardization of sodium thiosulphate

Volume of potassium dichromate (V1) = 20 ml

Strength of potassium dichromate (N1) = -------------N

Volume of sodium thiosulphate (V2) = .ml

Strength of sodium thiosulphate (N2) = ?

According to the law of volumetric analysis,

V1N1 = V2N2

N2 = V 1N1 = 20 x

V2

Strength of sodium thiosulphate (N2) =..N


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Expt. No Date

DETERMINATION OF DO CONTENT OF WATER SAMPLE BY

WINKLERS METHOD

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 of strength

___________ N are supplied.

PRINCIPLE

Oxygen dissolves in water to an extent of 7-9 mg/lit at a temperature range of 25-35 C. 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 (White ppt)

Mn(OH)2 + O2 MnO(OH)2(Yellow brown ppt)

(DO in water) (Basic manganic oxide)

The precipitate dissolves in concentrated sulfuric acid liberating oxygen which in turn

oxidizes potassium iodide and liberates iodine. The liberated iodine is titrated against Na2S2O3using

starch indicator.

MnO(OH)2 + 2H2SO4 Mn(SO4)2 + 2H2O + [O]

2KI + H2SO4+ [O] K2SO4 + H2O + I2

2Na2S2O3 + I2 Na2S4O6 + 2NaI


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Titration II: Estimation of dissolved oxygen

Std. sodium thiosulphate Vs water sample

S. No.Volume of

water sample (ml)

Burette reading (ml)Volume of sodium

thiosulphate (ml) IndicatorInitial Final

CALCULATION

Step2 : Estimation of dissolved oxygen

Volume of sodium thiosulphate (V1) = ml

Strength of sodim thiosulphate (N1) = N

Volume of water sample (V2) = 100 ml

Strength of water sample (N2) = ?

V1N1 = V2N2

N2 = V 1N1 =

V2 100 x

= ___________ N

Amount of dissolved oxygen in

one litre of given water sample = Normality Eq. wt. of O2 1000 mg

= .N 8 1000

= mg.


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PROCEDURE

TITRATION I: STANDARDISATION OF SODIUM THIOSULPHATE

The burette is washed with water and rinsed with sodium thiosulphate solution. Then the

burette is filled with the given sodium thiosulphate solution. 20 ml of std. potassium dichromate

solution is pipetted out into a clean conical flask. To this, 5 ml conc. HCl and 5 ml 10% potassium

iodide are added. The liberated iodine is titrated against sodium thiosulphate solution. When the

solution becomes straw yellow colour, starch indicator is added and then titration is continued. The

end point of the titration is the disappearance blue colour and appearance of green colour. The titration

is repeated to get concordant values.

TITRATION II: ESTIMATION OF DISSOLVED OXYGEN

100 ml of the water sample is taken in the iodine flask, 2 ml of manganous sulphate and

2 ml of alkaline potassium iodide are added. The flask is stoppered and shaken several times for

complete mixing of reagents. The flask is left aside for some time. When half of the precipitate

settles down, the stopper is removed and 2 ml of concentrated sulphuric acid is added. The stopper is

replaced and the flask is inverted several times for complete dissolution of the precipitate. It is the

titrated against standardized sodium thiosulphate solution. Starch indicator is added when the

solution becomes light yellow. The titration is continued until the blue colour disappears. From the

titre value, the strength of dissolved oxygen is calculated and hence the amount of dissolved oxygen

in the water sample is found out.

RESULT

Amount of dissolved oxygen present in given water sample =..mg/litre


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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 (White ppt)

Mn (OH)2 + O2 MnO(OH)2(Yellow brown ppt)

(DO in water) (Basic manganic oxide)

MnO (OH)2 + 2H2SO4 Mn(SO4)2 + 2H2O + [O]

2KI + H2SO4+ [O] K2SO4 + H2O + I2

2Na2S2O3 + I2 Na2S4O6 + 2NaI

PROCEDURE

TITRATION I: STANDARDISATION OF SODIUM THIOSULPHATE

Burette: Sodium thiosulphate Indicator : Starch

Pipette: 20 ml std. potassium dichromate solution End point : Appearance of light green

Additional solution: 5ml conc.HCl + 5 ml 10% KI

Strength of sodium thiosulphate = Volume of K2Cr2O7x strength of K2Cr2O7Volume of Na2S2O3

TITRATION II: ESTIMATION OF DISSOLVED OXYGEN

Burette: Sodium thiosulphate Indicator : Starch

Pipette: 100 ml water sample End point : Disappearance of blue colour

Additional solution: 2 ml manganese sulphate + 2 ml alkali iodide + 2 ml conc. H2SO4

Equivalent weight of oxygen = 8

Strength of water sample = Volume of Na2S2O3x Strength of Na2S2O3Volume of water sample

Amount of dissolved oxygen in water sample = Normality of water sample x Eq. wt. of O2

RESULT

Amount of dissolved oxygen present in given water sample =..mg/litre


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VivaVoce 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 (iv) conc. H2SO4

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


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Titration I: Standardization of silver nitrate

Standard sodium chloride Vs silver nitrate

S. No.

Volume of

sodium chloride (ml)

Burette reading (ml) Volume of

silver nitrate

(ml)

IndicatorInitial Final

CALCULATION

Step1 : Standardisation of silver nitrate

Volume of sodium chloride (V1) = 20 ml

Strength of sodium chloride (N1) = .N

Volume of silver nitrate (V2) = ..ml

Strength of silver nitrate (N2) = ?


V1N1 = V2N2

N2 = V1N1 = 20 x

V2

Strength of silver nitrate = .N


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Expt. No.. Date.

DETERMINATION OF CHLORIDE CONTENT OF WATER SAMPLE BY

ARGENTOMETRIC METHOD

AIM

To estimate the amount of chloride ion present in the given water sampleby Mohrs method.

A std. solution of ------------- N NaCl and approximately 0.01N AgNO3solutions are provided.

PRINCIPLE

Generally, 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).

AgNO3 + NaCl AgCl + NaNO3

(White precipitate)

Potassium chromate is used as indicator. In the presence of chromate ion, silver nitrate reacts

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 Ag2CrO4 + 2KNO3(Yellow) (Red tinge)

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 End point : Appearance of pale red tinge


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Titration II: Estimation of chloride ion

Std. silver nitrate Vs water sample

S. No.

Volume of given

water sample (ml)Burette reading (ml) Volume of

silver nitrate (ml) IndicatorInitial Final

CALCULATION

Step - 2 : Estimation of chloride ion content


Strength of chloride ion

in the water sample (N1) = ?

Volume of silver nitrate (V2) = ..ml

Strength of silver nitrate (N2) = N


V1N1 = V2N2

N1 = V2N2 = x

V1 20

Strength of chloride ion in the given water sample = N

Calculation of amount of chloride ion

Amount of chloride ion present in = Equivalent wt. of chloride ion Strength of chloride ion

100 ml of the given water sample 10

= 35.5 = ____________ gm.10


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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.

TITRATION II : ESTIMATION OF CHLORIDE ION

Burette : AgNO3 Indicator : 2 drops of 5% potassium chromate

Pipette : 20 ml of given water sample End point : Appearance of pale red tinge

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 AgNO 3solution. The end point

is the formation of red tinge on the precipitate. The titrations are repeated to get concordant values.

Equivalent weight of chlorine is 35.5

Equivalent weight of NaCl is 58.5

RESULT

The amount of chloride ion present in 100 ml of the given water sample = gm.


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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 AgNO3solutions are provided.

PRINCIPLE

AgNO3 + NaCl AgCl + NaNO3(White precipitate)

2AgNO3 + K2CrO4 Ag2CrO4 + 2KNO3(Yellow) (Red tinge)

PROCEDURE

TITRATION I: STANDARDISATION OF SILVER NITRATE

Burette: AgNO3 Indicator : 2 drops of 5% potassium chromate

Pipette: 20 ml of Std. NaCl End point : Appearance of pale red tinge

Strength of silver nitrate = Volume of sodium chloride x Strength of sodium chloride

Volume of silver nitrate

TITRATION II: ESTIMATION OF CHLORIDE ION

Burette : AgNO3 Indicator : 2 drops of 5% potassium chromatePipette : 20 ml of given water sample End point : Appearance of pale red tinge

Strength of water sample = Volume of silver nitrate x Strength of silver nitrate

Volume of water sample

Amount of chloride ion = Strength of chloride ion x Eq. Wt. of chloride ion

Equivalent weight of chlorine is 35.5

Equivalent weight of NaCl is 58.5

RESULTThe amount of chloride ion present in the given water sample = gm / litre


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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, CaCl2and MgCl2. The total

chloride content can be estimated by titration with standard AgNO3solution (Argentometric

method or Mohrs method).

3. What is the indicator used in the determination?

Potassium chromate4. 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. Whatkind of chlorides present in water?

NaCl, MgCl2, CaCl2.

8. Why does AgNO3react first with chloride ions in the water and not with chromate ions?

The solubility product of Ag2CrO4is high compared to that of AgCl, therefore AgNO3reacts

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


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Titration between HCl & NaOH

Volume of HCl (V1) = ______ ml

Volume of NaOH added (ml) pH0

2

4

6

8

10

12

14

16

18

20

Graph I : Volume of NaOH Vs pH Graph II : Volume of NaOH Vs pH/

V


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Expt. No Date.

DETERMINATION OF STRENGTH OF GIVEN HCl USING PH METER

AIMTo determine the strength of given HCl by using pH meter and also calculate the amount of

HCl. You are provided with a standard solution of NaOH of strength ..N.

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.


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Titration between HCl & NaOH

Volume of NaOH (ml) pH pH V (ml) pH/V


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The end point is obtained by plotting pH Vs volume of sodium hydroxide added or pH / V Vs

volume of sodium hydroxide added. From the end point the strength of hydrochloric acid is

calculated.


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CALCULATION

Step1 : Calculation of strength of HCl

Volume of HCl (V1) = 20 ml

Strength of HCl (N1) = ......................?

Volume of NaOH (V2) = ........................ (titre value from graph)

Strength of NaOH (N2) = ..N


V1N1 = V2N2

N1 = V2N2 = ----- x -----

V1 20

Strength of given HCl = .................. N

Step2 : Calculation of amount of HCl

The amount of HCl present in

1 litre of the given solution = Normality of HCl x Equivalent wt. of HCl (36.5)

= ....................... gm.


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RESULT

Strength of given Hydrochloric acid = .N

Amount of HCl present in 1 litre of the solution = gm.


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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 : Sodium hydroxide 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 1: Volume of NaOH Vs pH Graph 2: Volume of NaOH Vs pH /V

pH pH /V

Volume of NaOH (ml) Volume of NaOH (ml)

Strength of HCl = Volume of NaOH x Strength of NaOH

Volume of HCl

Amount of HCl = Strength of HCl x Equivalent weight of HCl (36.5)

RESULT

Strength of given hydrochloric acid =.NAmount of HCl present in 1 litre of the solution = gm.


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1. Define pH

pH is defined as the negative to the base 10 logarithm (log) of the H+ion

concentration. Mathematically, it is represented aspH = -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) Acidic solution : < 7

(ii) Basic solution : > 7

4. Write the formula for pH.

pH = -log [H+] = log 1 / [H

+]

5. What is the unit for pH?

No unit. Mere number.

6. How is the pH of the solution found out?

Using pH meter, indicators or pH paper.

7. Name the electrodes used in pH metry.

Glass electrode and calomel electrode.

8.

Explain why the pH value increases steeply at the end point.

At the end point, all H+ions are neutralized and the basic hydroxide ion will be

present. Hence pH is increased.

9. What is the equivalent weight of HCl?

36.5

10.Write the formula used to calculate the amount of HCl.

Amount of HCl = Equivalent weight of HCl (36.5) x Normality of HCl

11.pH + pOH = ?

14


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Titration between HCl, CH3COOH (mixture of acids) & NaOH

Volume of HCl (V) = 40 ml

Volume

of NaOHadded

(v) ml

Conduc-

tance(mho)

(V+v)V

C x(V+v)

V(mho)

8

8.5

9

9.5

10

10.511

11.5

12

12.5

13

13.5

14

14.5

15

15.5

Graph

Volume of

NaOHadded (v)

ml

Conduc-

tance(mho)

(V+v)V

C x(V+v)

V(mho)

0

0.5

1

1.5

2

2.53

3.5

4

4.5

5

5.5

6

6.5

7

7.5

Volume of NaOH (ml)

Cx(V+v)/Vm

ho

(A)

(B)


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Expt. No Date.

DETERMINATION OF STRENGTH OF ACIDS IN A MIXTURE USING

CONDUCTIVITY METER

AIM

To determine conductometrically the strength of given hydrochloric acid and acetic acid

present in the given mixture, by titrating with standard..N sodium hydroxide solution and

calculate the amount of each acid present in 1 litre of the solution.

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 ofHCl 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 Na

+ + Cl

- + H2O (l)

CH3COOH(aq) + Na+

(aq) + OH-aq

slow CH3COO

-(aq)+ Na

+(aq) + H2O(l)

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 aconductivity 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

decreasesgradually due to the neutralization of hydrochloric acid, it reaches the first end point and

after that the conductance increasestill it reaches the second end point. Further addition of sodium

hydroxide steeply increases the conductance value.


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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) = V2= ........... (A) ml (Ist titre value from graph)

Strength of NaOH (N2) = ____ N


V1N1 = V2N2

N1 = V2N2 = x

V1 40

Strength of HCl = .................. N

Amount of HCl

The amount of HCl present in1 litre of the given solution = Strength of HCl x Eq.wt.of HCl (36.5)

= ................. N x 36.5

= ................... gms

Step2 : Strength of CH3COOH

Volume of CH3COOH (acid mixture) ( V1) = 40 ml

Strength of CH3COOH (N1) = .............. ?Volume of the NaOH (V2) = (B-A) ml (II titre value from graph)

Strength of NaOH ( N2) = ____ N


V1N1= V2N2

N1 = V2N2 = x

V1 40

Strength of CH3COOH = .................. N

Calculation of amount of CH3COOH

The amount of CH3COOH present in = Normality of CH3COOH x Eq. wt. CH3COOH1 litre of the given solution

= .......................... x 60

= ............................ gm.


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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

The amount of HCl in 1 litre of the mixture of acids = gm.

The strength of Acetic Acid in the mixture of acids = ...N

The amount of CH3COOH in 1 litre of the mixture of acids = gm.


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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 Na

++ Cl

- + H2O (l)

CH3COOH (aq) + Na+

(aq) +OH-aq

slow CH3COO

-(aq)+ Na

+(aq) + H2O(l)

PROCEDUREBurette solution: standard sodium hydroxide solution

Pipette solution: 40 ml of the mixture of hydrochloric acid

and acetic acid

Cell: The conductivity cell

Initially the conductance is high due to the presence of fast moving H+

ions from HCl in the

mixture of acids.

NaOH is added slowly from the burette, 0.5 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 HCl is completely neutralized (end point A),the conductance increases gradually with

the gradual (0.5ml) addition of NaOH. This is due to the neutralization of CH 3COOH by

NaOH.

After the neutralization of CH3COOH (end point B) the conductance increases sharply due

to presence of free OH-ions.


Volume of HCl


Strength of CH3COOH = Volume of NaOH x Strength of NaOH

Volume of CH3COOHAmount of CH3COOH = Strength of CH3COOH x Eq. wt. of CH3COOH (60)

RESULTThe strength of hydrochloric acid in the mixture of acids = ....N

The amount of HCl in 1 litre of the mixture of acids = gm.

The strength of Acetic Acid in the mixture of acids = ...N

The amount of CH3COOH in 1 litre of the mixture of acids = gm.

ho


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Vivavoce 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)


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Graph: Calibration curve (Absorbance Vs Concentration)

Absorbance

Concentration in ppm

TABLEI

Preparation of various concentration of Fe3+

solution

Volume of iron

solution (ml)

Volume of

HNO3(ml)

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, A = absorbance

I0 = Intensity of incident light

It = Intensity of transmitted light

c = Concentration of solution

l = Thickness of the cell

= Molar absorption co-efficient

Fe3+

+ 6SCN- [Fe (SCN) 6]

3-

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.


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TABLEII

Measurement of absorbance

Blank (distilled water): Zero absorbance; max= 480 nm

S. No. Concentration Absorbance

1 1ppm 0.01 N

2 2ppm 0.02 N

3 3ppm 0.03 N

4 4ppm 0.04 N

5 5ppm 0.05 N

6 6ppm 0.06 N

7 Unknown -----

Calculation of amount of iron content

Amount of iron present in the given solution = Normality Eq. wt. of iron 1000 ppm

= . 55.85 1000 ppm

= . ppm


33/104



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


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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 formof 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 = Normality Eq. wt. of iron 1000 ppm

RESULT

The amount of ferric iron present in the given sample of water = ppm


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1. What is spectroscopy?

Spectroscopy is the measurement and interpretation of electromagnetic radiationabsorbed or emitted when the molecules or atoms or ions of a sample move from one allowed

energy state to another.

2. State Beer Lamberts law.

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.

Log I0/ It= cl = A

I0 = Intensity of incident light

It = Intensity of transmitted light

Log I0/ It = cl = A

c = Concentration in moles/litre

l = Thickness of the cell

A = Absorbance or optical density

= Molar absorption co-efficient

3. Write any two applications of Beer-Lamberts law

1. The concentration of colored solutions can be estimated.

2. The concentration of various industrial products can be determined.

3. The purity of chemical species can be determined.

4. Write any two reasons for the deviations from Beer-Lamberts Law.

Deviations from the law will occur when-

1.

Monochromatic light is not used.2. The colored solute ionized, dissociated or associated in the solution

3. Impurities are present in the solution

4. The solution undergoes polymerization.


36/104



TABLEI

Preparation of various concentrations of polymer solution

Stock solution N1= 1% Total volume V2= 30 ml

S. No.Volume of 1% polymer solution

(stock solution) (V1 ml)Volume of water (ml)

Concentration in % (N2)

N2= V1N1/ V 2

1.

2.

3.

4.

5.

TABLEII

Viscosity data for a polymer / solvent

Flow time of the pure solvent (t0) = sec

S. No.Concentration in %

(N2)

Flow time (t)

sec(r) = t/t0 sp= r1 sp/C = red

1.

2.

3.

4.

5.

Graph

Concentration (%)

i

sp/C


37/104



Expt.No Date

DETERMINATION OF MOLECULAR WEIGHT OF POLYVINYL ALCOHOL USING

OSTWALD VISCOMETER

AIM

To determine the molecular weight and degree of polymerization of the given polyvinyl

alcohol polymer solution (1%) using Ostwalds viscometer.

PRINCIPLE

Molecular weight of a polymer is nothing but the average molecular weight. This can be

determined by measuring the intrinsic viscosity (i) of a dilute polymer solution. This intrinsic

viscosity is related to the molecular weight of the polymer by the following relationship.

i = KMa(Mark - Hownik equation)

Where, i = Intrinsic viscosity

K & a = Constants for a given polymersolvent combination at a giventemperature

M = Average molecular weight

Degree of polymerization (Dp) provides another way of expressing the molecular weight as

follows.

M = Dp x m

Where, M = molecular weight of the polymer

Dp = degree of polymerization

m = molecular weight of the monomer or the repeat unit

Important viscosity definitions:

Relative viscosity (r) = t/t0

Specific viscosity (sp) = r1

Reduced viscosity (red) = sp/C

Intrinsic viscosity (i) = lim sp/Cc0

Where, t = Flow time for the polymer solution

t0 = Flow time for the solvent

C = Concentration of the polymer


38/104



CALCULATION

MarkHownik equation is given by,

i = KMa

log i = log K + a log M

log i - log Klog M =

a

log i - log KM = A. log

a

where, M = Molecular weight of the polymer.

Values of constants K anda for water polyvinyl alcohol system

K = 4.53 x 10-4

and

a = 0.64


39/104



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

STEP II: Flow time of solvent

20 ml of the solvent (water) is taken into the viscometer and is sucked through the capillary

tube up to the upper mark, without any air bubbles. Now note the flow time of the solvent to

flow from the upper mark (M1) to lower mark (M2) in the viscometer.

STEP III: Flow time of polymer solutions

The water is drained out completely and 20 ml solutions of polymer of different

concentrations are taken in the viscometer one after another by following the above procedure.

The flow time for various concentrations is determined.

(Wash and rinse the viscometer with water before taking the polymer solution of different

concentrations)

From the flow time, relative viscosity (r) can be calculated. A Graph is plotted between

sp/C and concentration (%). A straight line is obtained with an intercept called intrinsic

viscosity (i). From the value, the molecular weight of the polymer is calculated using

MarkHownik equation.

RESULT

Molecular weight of the given polymer =..

Degree of polymerization =..


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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.

CALCULATIONBy using, MarkHownik equation, the molecular weight of the polymer is calculated.

i = KMa

log i = log K + a log M

log i - log Klog M = M = A.log log i - log K

a a

Where, M = Molecular weight of the polymer.

i = KMa(Mark - Hownik equation)

Where, i = Intrinsic viscosityK & a = Constants for a given polymersolvent combination at a given

temperature


Degree of polymerization (Dp) = Molecular weight of the polymerMolecular weight of the monomer (vinyl alcohol)

Graph

sp / C

i

Concentration (%)

RESULTMolecular weight of the given polymer =..

Degree of polymerization =..


41/104




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,

i = Intrinsic viscosity

K & a = Constants for a given polymersolvent combination at a giventemperature


4. What is the polymer used in this experiment?

CH2 CH

OH n

Polyvinyl alcohol

5. What are the values of Kand afor water-polyvinyl alcohol system?

K = 4.53 x 10-4

and a = 0.64

6. What is the molecular weight of vinyl alcohol?

44

7. What is degree of polymerization? How is it calculated?

The number of monomer or repeat unit present in ae polymer is called

degree of polymerization, n or Dp. It is calculated as, Dp =M / m

Where, M = Average molecular weight of the polymer

m = molecular weight of the monomer

8. Write the relationship between different viscosities and flow time.

Relative viscosity (r) = t/t0

Specific viscosity (sp) = r1 = (t/t0) - 1

Reduced viscosity (red) = sp/C = ((t/t0)1)/C

Intrinsic viscosity (i) = lim sp/Cc0

Where, t = Flow time for the polymer solution.

t0 = Flow time for the solvent.

C = Concentration of the polymer.


42/104



Titration between standard HCl & NaOH

Volume of HCl (V) = 40 ml

Volume of NaOH

added (v) mlConductance (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


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Expt. No.. Date

CONDUCTOMETRIC TITRATION OF STRONG ACID VS STRONG BASE

AIM

To determine the strength of given hydrochloric acid by conductometric titration against

N standard sodium hydroxide solution and calculate the amount of hydrochloric acid present

in 1 litre of the solution.

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.


44/104



Graph

CALCULATION

Strength of HCl

Volume of HCl (V1) = 40 ml

Strength of HCl ( N1) = ................ ?

Volume of NaOH (V2) = ........... ...... ml (titre value)

Strength of NaOH (N2) = _________ N

According to the law of Volumetric analysis,

V1N1 = V2N2

N1 = V2x N2 = x

V1 40

Strength of HCl = .................N

Amount of HCl

Amount of HCl present in

1 litre of the given solution = Normality of HCl x Equivalent weight of HCl

= .................. x 36.5

= --------------- gm.


45/104



RESULT

Strength of given hydrochloric acid = .N

Amount of HCl present in 1 litre of the solution = .gm.


46/104



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

H++ Cl

- + NaOH Na

+ + Cl

- + H2O

PROCEDURE

Burette solution: Std. NaOH solution

Pipette solution: 40 ml of given HCl

Cell : Conductivity cell

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).


Volume of HCl


RESULT

Strength of given hydrochloric acid = .N

Amount of HCl present in 1 litre of the solution = .gm.

Volume of NaOH (ml)

Cx

(V+

v)/

Vm

ho


47/104



Vivavoce Questions and Answers

1. What is conductance? Write its SI unit.

The current conducting capacity of a material is called conductance.

Its SI unit is Siemens, S.

2. What are specific conductance and equivalent conductance?

The conductance of one metre cube of an electrolyte solution is called specific

conductance and it is denoted by (kappa).

The conductance of an electrolyte solution containing one gram equivalent of the

electrolyte is known as equivalent conductance. It is denoted by (lambda).

3. How is specific conductance related with equivalent conductance?

= / V

where V is the volume of the solution containing one gram equivalent of the electrolyte.

4. Give examples for strong acid and weak acid.

Strong acid: HNO3, HCl, H2SO4

Weak acid : Acetic acid (CH3COOH), oxalic acid (C2H2O4), formic acid (HCOOH)

5. Explain why hydrochloric acid is a strong acid.

Hydrochloric acid is completely dissociated in all concentrations. Hence it is regarded

as strong acid.

HCl 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.


48/104



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 volumetricmethods 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.


49/104



SEMESTER II


50/104



LIST OF EXPERIMENTS

Expt. No. Name of the Experiments Page No.

1. Determination of alkalinity in water sample 53

2.Determination of total, permanent & temporary

permanent hardness of water by EDTA method61

3.Estimation of copper content of the given solution by

EDTA method71

4.Estimation of iron content of the given solution using

potentiometer77

5. Estimation of sodium present in water usingflame photometer

85

6. Corrosion experimentweight loss method 89

7. Conductometric precipitation titration usingBaCl2and Na2SO4

95

8. Determination of CaO in cement 101


51/104



TABLE OF CONTENTS

Ex. No. Date Name of the ExperimentsMarks

Obtained

Signature of the

Staff with Date


52/104



TABLEI

Titre values and different alkalinities

Result of titration OH- CO3

- HCO3

-

If P = 0 0 0 [M]

If P = M [P] or [M] 0 0

If P = 1/2M 0 2[P] or [M] 0

If P > 1/2M 2[P] - [M] 2[M-P] 0

If P < 1/2M 0 2[P] [M]2[P]

Titration I: Standardization of hydrochloric acid

HCl Vs Std. NaOH

S. No.Volume of sodium

hydroxide (ml)


hydrochloric acid (ml)Indicator

Initial Final

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 = 20 x

V2

Strength of HCl (N2) = _______________ N


53/104



Expt. No Date

DETERMINATION OF ALKALINITY IN WATER SAMPLE

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

The presence of hydroxide, carbonate and bicarbonate ions mainly makes the water alkaline.

Therefore the alkalinity of a water sample is due to the presence of either one or more of the above

constituents. But hydroxide and bicarbonate ions cannot co-exist due to the following reaction.

OH + HCO3 CO3 + H2O

2

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.

OH + H H2O+

CO3 +

2

H

+

HCO3

a)

b)

Phenolphthaleinend point (P)

The remaining half of the carbonate alkalinity and bicarbonate alkalinity are determined by

methyl orange indicator.

c) HCO3

+ H+ H2O + CO2 Methyl orange end point (M)

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 CaCO3equivalent.


54/104



Titration I: Estimation of alkalinity

Standard HCl Vs water sample

S. No. Volume ofwater sample (ml)

Volume of HCl Concordant Value

Phenolphthaleinend point (P)

Methyl orangeend point (M)

[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]

= 2 x---------- - ---------ml

ii) Volume of HCl required for CO32-

alkalinity = 2[M]2[P]

= 2 x---------- -2 x ---------ml

iii) HCO3-

is not present.

1. Calculation of OH- alkalinity:

Volume of HCl (V1) = ------- ml

Strength of HCl (N1) = ------- N


Strength of water sample (N2) = ?

Due to OH-alkalinity

N2 = V1x N1 = x

20 20

Strength of hydroxide alkalinity = `------------- N


55/104



PROCEDURE

TITRATION I: STANDARDISATION OF HYDROCHLORIC ACID

Burette : Hydrochloric acid Indicator: Phenolphthalein

Pipette: 20 ml of standard sodium hydroxide End point: Just disappearance of pink colour

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

Pipette : 20 ml of given water sample End point : 1. Just disappearance of pink colour2. Golden yellow to pale pink colour

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.


56/104



Amount of OH-content in1 lit of water sample,

in terms of CaCO3equivalent = strength of hydroxide alkalinity x

equivalent of CaCO3

OH-alkalinity interms of CaCO3equivalent = ------------------ x 50 x 10

3ppm

Alkalinity due to OH- ion = ---------------- ppm

2. Calculation of CO32-

alkalinity:

Volume of HCl (V1) = ------- ml

Strength of HCl (N1) = ..N


Strength of water sample

Due to carbonate alkalinity (N2) = ?

N2 = V1x N1 = x

20 20

Strength of carbonate alkalinity = N

Amount of CO32-

content in1 lit of water sample,

in terms of CaCO3equivalent = strength of carbonate alkalinity x

equivalent of CaCO3

Carbonate alkalinity in terms of

CaCO3equivalent = ------------------ x 50 x 103ppm

Alkalinity due to carbonate = ---------------- ppm


57/104



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 CaCO3equivalent.

RESULT

The given water sample contains the following alkalinity

Hydroxide alkalinity = .ppm

Carbonate alkalinity = .ppm

Bicarbonate alkalinity = .ppm


58/104



Short Procedure

AIM

To determine the alkalinity of given water sample. A standard solution of N sodiumhydroxide 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 End point: Just disappearance of pink colour


Volume of HCl

TITRATION II: ESTIMATION OF ALKALINITY

Burette : Hydrochloric acid Indicator : 1. Phenolphthalein 2. Methyl orange

Pipette : 20 ml of given water sample End point : 1. Just disappearance of pink colour

2. Golden yellow to pale pink colour

if the data satisfies the condition P > M, both hydroxide & carbonate alkalinity are present.

i) OH-alkalinity = 2[P][M] (Phenolphthalein end point)

Strength of OH-alkalinity = Volume of HCl x Strength of HCl

Volume of water sample

Amount of OH-alkalinity = Strength of water sample x Eq. wt. of CaCO3(50)

ii) CO32-

alkalinity = 2[M]2[P] (Methyl Orange end point)

Strength of CO32-

alkalinity = Volume of HCl x Strength of HClVolume of water sample

Amount of CO32-

alkalinity = Strength of water sample x Eq.Wt. of CaCO3(50)

RESULTThe given water sample contains the following alkalinity

Hydroxide alkalinity = .ppmCarbonate alkalinity = .ppm

Bicarbonate alkalinity = .ppm


59/104




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 CO3

2-(ii) HCO3

-and CO3

2-

4. 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.

OH + HCO3 CO3 + H2O

2

7. What type of alkalinity is present if P = 0?

Bicarbonate alkalinity

8. What is the burette solution used?

HCl

9. Explain why two indicators are used in the same titration?

Phenolphthalein indicator is not suitable for bicarbonate alkalinity and hence

methyl orange indicator is also used to find out bicarbonate alkalinity in the same titration.

10. How is the amount of alkalinity expressed?

The alkalinity is expressed in terms of CaCO3equivalent and the unit is ppm or mg/L.

11. Which is strong? OH-or HCO3

-ions.

OH-ions

12. What type of indicators is used in this experiment?Acid - base neutralization indicators.

13. What are the ions responsible for alkalinity if P = M?

Only OH-ions

14. When P > 1/2M, what are the ions responsible for alkalinity?

OH-and CO3

2-ions


60/104



Titration I: Standardization of EDTA

Standard hard water Vs EDTA

S. No.

Volume of standard

hard water (ml)

Burette reading (ml)

Volume ofEDTA (ml)


CALCULATION

Strength of EDTA

Volume of standard hard water = 20 ml

Volume of EDTA solution consumed, V1 = ml

1ml of standard hard water contains 1mg of calcium carbonate

20 ml of standard hard water contains 20 mgs of CaCO3

20 ml of standard hard water consumes V1ml of EDTA

i.e., V1ml of EDTA solution = 20 mgs of CaCO3

1ml of EDTA solution = 20 mg of CaCO3equivalent

V1


61/104



Expt. No Date.......

DETERMINATION OF TOTAL, TEMPORARY & PERMANENT HARDNESS OF

WATER SAMPLE BY EDTA METHOD

AIM

To determine the total hardness, permanent hardness and temporary hardness of the given

water sample by EDTA method. Standard hard water and a link solution of EDTA are supplied.

PRINCIPLE

Hardness producing ions Ca2+

and Mg2+

form complexes with both EDTA

(Ethylenediaminetetraacetic acid) and EBT (Eriochrome blackT). But EDTAmetal ion complex

is more stable than EBTmetal ion complex.

Ca2+

or Mg2+

+ EB CaEBT or MgEBT + 2H+

(In water) (Less stable, wine red)

CaEBT or MgEBT + EDTA2-

[CaEDTA]2-

or [MgEDTA]2-

+ EBT

(More stable, colourless) (Purple blue)

The end point of the titration is colour change from wine red to purple blue.

PROCEDURE

TITRATION I: STANDARDISATION OF EDTA

Burette : EDTA Indicator : EBT

Pipette : 20 ml std. hard water End point : Wine red to purple blue colour

Condition: 5 ml of ammonia buffer solution

A 50 ml burette is washed with water and rinsed with EDTA. Then it is filled with EDTA solution

up to zero mark and the initial burette reading is noted. A 20 ml pipette is washed with

water and rinsed with the given std. hard water (1 ml 1 mg of CaCO3), then exactly 20 ml of std.

hard water is pipetted out into a 250 ml clean conical flask.

H 9-10

H 9-10


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Titration II: Estimation of total hardness

Given water sample Vs EDTA

S. No.

Volume of given

water sample

(ml)


EDTA

(ml)


Calculation of the total hardness of the given water sample

Volume of hard water = 20 ml

Volume of EDTA consumed, (V2) = . ml

20 ml of the given hard water sample consumes V2ml of EDTA

1000 ml of the given hard water sample contains 20 x V21000 mg of CaCO3

V1 x 20

= 1000 V2/ V1mg of CaCO3

=

=..ppm

Total hardness of the given sample = ..ppm


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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


Pipette : 20 ml of given water sample End point : Wine red to purple blue

Condition : 5 ml of ammonia buffer solution

The burette is filled with the same EDTA solution up to zero mark and the initial burette

reading is noted. The pipette is washed with distilled water and rinsed with the given water sample.

Then 20 ml of water sample is pipetted out into a clean conical flask. To this 5 ml of buffer solution

and 2 drops of EBT indicator are added. The solution becomes wine red in 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 this volume of EDTA be V2 ml. From the titre value V2,the total hardness

of water sample is calculated.

TITRATION III: ESTIMATION OF PERMANENT HARDNESSBurette : EDTA Indicator : EBT

Pipette : 20 ml of boiled water sample End point : Wine red to purple blue


The burette is filled with EDTA up to zero mark and the initial burette reading is noted.


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Titration III: Estimation of permanent hardness

Boiled water sample Vs EDTA

S. No.

Volume of given

boiled water

sample (ml)

Burette reading (ml)Volume ofEDTA (ml)


Calculation of permanent hardness

Volume of the boiled hard water = 20 ml

Volume of the EDTA solution consumed , (V3) =..ml (titre value)

20 ml of the boiled hard water sample consumes V3ml of EDTA

i.e. 20 ml of boiled hard water sample contains = 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/ V1mg of CaCO3

=

Permanent hardness = ppm

Estimation of temporary hardness

Temporary hardness = Total hardness - Permanent hardness

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

= -------------ppm


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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

Permanent hardness of given water sample = . ppm

Temporary hardness of given water sample = . ppm


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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 CaEBT or MgEBT + 2H+

(In water) (Less stable, wine red)

CaEBT or MgEBT + EDTA2-

[CaEDTA]2-

or [MgEDTA]2-

+ EBT

(More stable, colourless) (Purple blue)

PROCEDURE



Pipette : 20 ml of std. hard water End point : Wine red to purple blue colour


Strength of EDTA = Volume of std. hard water x Strength of std. hard water

Volume of EDTA

TITRATION II: ESTIMATION OF TOTAL HARDNESS

Burette : EDTA Indicator : EBTPipette : 20 ml of given water sample End point : Wine red to purple blue


Total hardness = Volume of EDTA for 20 ml of given hard water x 1000Volume of EDTA for 20 ml of std. hard water

TITRATION III: ESTIMATION OF PERMANENT HARDNESS


Pipette : 20 ml of boiled water sample End point : Wine red to purple colour


Permanent hardness = Volume of EDTA for 20 ml of boiled hard water x 1000

Volume of EDTA for 20 ml of std. hard water

RESULT

Total hardness of given water sample = ppm

Permanent hardness of given water sample = .ppm

Temporary hardness of given water sample = . ppm


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1. What are the different sources of water?

Ocean, River, Lake, Spring, Rain water.

2. Why rain water is considered to be the purest form of natural water?

Since the rain water is free from any dissolved salt, it is considered to be the purest form of

natural water.

3. Name the impurities present in natural water.

(i) Suspended impurities (ii) Dissolved impurities (iii) Colloidal impurities

4. Define hard water.

The water which does not readily lather with soap but forms precipitate is known as hard

water.

5. What is soft water?

The water which gives lather easily with soap is called soft water.

6. What is EDTA?

EDTA is, Ethylenediaminetetraacetic acid.

7. Give the structure of EDTA

N - CH2- CH2- N

CH2 - COOH

CH2- COOH

HOOC - H2C

HOOC - H2C

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.


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13. How hardness of water is expressed?

It is expressed in terms of CaCO3equivalent.

14. Why CaCO3is 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 = -log10aH+, 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. ppmParts 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 CaCO

3equivalent 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 CaCO3equivalent hardness present in

105parts of water.

The above 4 units are correlated as given below.

1 ppm = 1 ml / L = 0.070Cl = 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

solution having pH greater than 6.0.


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20. Name other indicators beside Eriochrome BlackT 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


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Titration I: Standardization of EDTA

Standard CaCO3 Vs EDTA

S. No.

Volume of standard

calcium carbonate (ml)


Volume ofEDTA (ml)


CALCULATION

Strength of EDTA

Volume of std. CaCO3 solutionV1 = 20 ml

Strength of std. CaCO3solutionN1 = -------- N

Volume of EDTA V2 = -------- ml

Strength of EDTA N2 = ?

N2 = V1x N1 = 20 x

V2

Strength of EDTA N2 = --------- N


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Expt. No. Date .

ESTIMATION OF COPPER CONTENT OF THE GIVEN SOLUTION BY

EDTA METHOD

AIM

To estimate the amount of copper present in 100 ml of the given brass solution. A standard

solution of CaCO3of strength -------N and an approximately 0.01 M EDTA solution are provided.

PRINCIPLE

Brass is an alloy of 55% copper and 33% zinc. The purpose of alloying copper and zinc is to

improve machinability and strength. Brass may also contain small amounts of lead, tin or

aluminium.

In this method, brass is dissolved in con. HNO3, so that copper present in the brass is brought

into solution in the form of cupric ions. Then the Cu2+

ion is determined by the usual

complexometric method using EDTA.

Cu2+

+ FSB-F

[Cu-FSB-F] + 2H+

(Copper ions) (Purple colour)

[Cu-FSB-F] + EDTA2-

[Cu-EDTA]

2- + FSB-F

(Colourless) (Green colour)

PROCEDURE



Pipette : 20 ml of std. CaCO3 End point : Wine red to purple blue colour


The burette is washed well with distilled water and rinsed with the small amount of given

EDTA solution. It is then filled with the same solution up to the zero mark without any air bubbles.

The pipette is washed with distilled water and rinsed with the small amount of standard CaCO3solution. 20 ml of this solution is pipetted out into a clean conical flask.


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Titration II: Estimation of copper

Standard EDTA Vs brass solution

S. No.Volume of given

brass solution (ml)


Volume of

EDTA (ml)Indicator

Initial Final

CALCULATION

Strength of brass solution

Volume of EDTA V1 = ---------- ml

Strength of EDTA N1 = -------N

Volume of brass solution V2 = 20 ml

Strength of brass solution N2 = ?N2 = V1x N1 = x

V2 20

Strength of copper in brass solution = N

Amount of copper in Brass

Amount of Cu in 100 ml of given

brass solution = strength of Cu x equivalent wt. of Cu x 100

1000

= ------------------ x 63.54 = gm.

10


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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 : Fast sulphon black - F

Pipette : 20 ml of given brass solution End point: Blue to green colour


The given brass solution is made up to 100 ml in a standard measuring flask. 20 ml of this

solution is pipetted out into a clean conical flask. 5 ml of ammonia buffer solution and 3-4 drops of

fast sulphon black - F indicator are added. The solution turns blue in colour. The solution is titrated

against standard EDTA solution taken in the burette. The change of blue colour into green colour is

the end point. The titration is repeated to get concordant values.

Equivalent weight of copper = 63.54

RESULT

Strength of copper solution = ..N

Amount of copper present in 100 ml of the given solution = ..gm.


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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 CaCO3of strength -------M and an approximately 0.01 M EDTA solution are provided.

PRINCIPLE

Cu2+

+ FSB-F

[Cu-FSB-F] + 2H+

(Copper ions) (Purple colour)

[Cu-FSB-F] + EDTA2-

[Cu-EDTA]

2- + FSB-F

(Colourless) (Green colour)

PROCEDURETITRATION I: STANDARDISATION OF EDTA


Pipette : 20 ml of std. CaCO3 End point : Wine red to purple blue colour


Strength of EDTA = Volume of CaCO3 x Strength of CaCO3

Volume of EDTA

TITRATION II: ESTIMATION OF COPPER

Burette : EDTA Indicator : Fast sulphon black - F

Pipette : 20 ml of given brass solution End point: Blue to green colour


Strength of copper = Volume of EDTA x Strength of EDTA

Volume of brass solution

Amount of copper in 100 ml = Strength of copper x Eq. wt. of copper (63.54)

10

RESULT

Strength of copper solution = ..N

Amount of copper present in 100 ml of the given solution = ..gm.


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