ENGINEERING CHEMISTRY LAB MANUAL I B - T K...

ENGINEERING CHEMISTRY LAB MANUAL I B.Tech

Prepared by T K MOHAN Page 1

Expt. No. : 01

Date :

ESTIMATION OF HARDNESS OF WATER BY EDTA METHOD

AIM: To determine the hardness of given water sample by EDTA method.

APPARATUS: Burette, Pipette, Conical Flask, Beakers, Wash Bottle and Burette Stand.

CHEMICALS: EDTA solution, Standard Hard Water, Eriochrome Black-T, Ammonical Buffer

Solution.

PRINCIPLE : Hard water contains calcium and magnesium ions which form wine red colour

complex with the indicator, Eriochrome Black-T.

Ca2+

+ Eriochrome black T [Ca – Indicator Complex]

Indicator Wine Red Colour

EDTA forms a colourless complex with free metal ions.

Metal ion (Ca2+

or Mg2+

) + EDTA Metal ion - EDTA Complex

Stable complex

When free metal ions are not available EDTA extracts the metal ions from the metal ion

indicator complex there by releasing the free indicator,

EDTA + [Metal ion – indicator complex] [Metal ion – EDTA] + Indicator

Wine Red Colour Blue Colour

The reactions take place at a pH = 10in order to maintain it use a buffer made of ammonium

chloride and ammonium solution.

PROCEDURE:

Part A : - Standardization of EDTA Solution

i. Rinse and fill the burette with EDTA solution.

ii. Pipette out 20ml of Standard hard water solution into a clean conical flask.

iii. Add 5ml of ammonical buffer solution and 2-3 drop of EBT indicator, the solution

becomes wine red.

iv. Titrate against EDTA till wine red colour changes to blue.

v. Note the volume of EDTA used (V1ml).

vi. Repeat the titrations till the concordant reading are obtained.

Part B : - Estimation of Total Hardness

i. Pipette out 20ml of water sample into a clean conical flask.

ii. Add 5ml of ammonical buffer solution and 2-3 drop of EBT indicator, the solution

becomes wine red.

iii. Titrate against EDTA till wine red colour changes to blue.

iv. Note the volume of EDTA used (V2ml).

v. Repeat the titrations till the concordant reading are obtained.

Part C : - Estimation of Permanent Hardness

i. Pipette out 100ml of hard water sample into a beaker and boil the water till the volume

reduces to 50ml.

ii. Cool the solution and filter.

iii. Make up the solution up to the mark of 100ml.

iv. Pipette out 20ml of Standard boiled hard water into a clean conical flask.



v. Add 5ml of ammonical buffer solution and 2-3 drop of EBT indicator, the solution

becomes wine red.

vi. Titrate against EDTA till wine red colour changes to blue.

vii. Note the volume of EDTA used (V3ml).

viii. Repeat the titrations till the concordant reading are obtained.

RESULT:

i. Molarity of EDTA solution is____________________M.

ii. Molarity of standard hard water __________________M.

iii. The amount of total hardness in terms of CaCO3 equivalent weight is ________ ppm.

iv. The amount of permanent hardness in terms of CaCO3 equivalent weight is

__________ppm.

v. The amount of temporary hardness in terms of CaCO3 equivalent weight is

___________ppm.

OBSERVATION AND CALCULATIONS:

Part A : - Standardization of EDTA Solution

S.No. Volume of Standard Hard

Water

Burette Readings Volume of EDTA used

V1 ml Initial (ml) Final (ml)

M1V1 = M2V2

M1= Molarity of Standard Hard Water =_____________M

V1=Volume of Standard Hard Water =_____________V

M2=Molarity of EDTA =_____?_______M

V2= Volume of EDTA =_____________V

M2 = 2

11

V

VM

Part B : - Estimation of Total Hardness

S.No. Volume of Water Sample Burette Readings Volume of EDTA used


M2V2 = M3V3

M2= Molarity of EDTA = _____________M

V2= Volume of EDTA = _____________V

M3= Molarity of Standard Hard Water = _____?_______M

V3= Volume of Standard Hard Water = _____________V

M3 = 3

22

V

VM

= ___________M



Formula:

Titre Value X Concentration of EDTA X 50 X 1000

Total Hardness =

Volume of the sample

=______________ppm

Part C : - Estimation of Permanent Hardness

S.No. Volume of Hard Water Sample Burette Readings Volume of EDTA used


Titre Value X Concentration of EDTA X 50 X 1000

Permanent Hardness =

Volume of the sample

Formula:

Total Hardness = Temporary Hardness + Permanent Hardness

Temporary Hardness = Total Hardness – Permanent Hardness

= _________________ppm.



Expt. No. : 02

Date :

DETERMINATION OF COPPER BY EDTA METHOD

AIM: To estimate the amount of copper present in the given solution by complexometric

method.

APPARATUS: Burette, Pipette, Beaker, Conical Flask, Burette Stand.

CHEMICALS: MgSO4, EDTA, fast sulphone black – F, ammonical buffer, Eriochrome black-T

PRINCIPLE: Copper forms complexes with EDTA in ammonia solution, which is colourless

and stable. Fast sulphone black – F is the indicator which is bright green in its colour forms

complex with copper ions to produce an unstable complex which is pale blue in its colour.

The copper solution is first treated with buffer and indicator to produce unstable pale blue

complex which is titrated against EDTA solution until the colour change from pale blue to dark

green takes place. Repeat the titration to get concordant values.

Cu2+

+ 2In [Cu – In]

Bright green pale blue

[Cu – In] + EDTA [Cu – EDTA] + In

Pale blue colourless Bright green

PROCEDURE:

Part A:

Standardisation of EDTA Solution:

Rinse the burette with EDTA solution and take EDTA solution upto the mark. Pipette out

20ml of MgSO4 solution into a conical flask, add 5ml of buffer solution and 2-3 drops of EBT

indicator. Titrate the wine red coloured complex with EDTA till a blue colour end point is

obtained. Repeat the titration to get concordant values. Let the titre value be x ml.

Part B:

Estimation of Copper Solution:

Make up the given solution upto the mark with distilled water and shake the flask well for

uniform concentration. Pipette out 20ml of copper solution into clean 250ml conical flask, add

20ml of distilled water, 5ml of ammonia and 5 drops of the fast sulphone black – F indicator.

Titrate the resulting pale blue coloured complex with EDTA till dark green colour endpoint is

obtained. Repeat the titration to get concordant values. Let the titre value be y ml.

RESULT:

i. Amount of copper present in the given solution is_________ g.


Molarity of MgSO4 (M1) = ________________M

Part A:

Standardisation of EDTA Solution:

S.No. Volume Of MgSO4

V1ml

Burette readings Volume Of EDTA

V2ml Initial Final



Molarity of EDTA (M2):

V1M1 V2M2

=

n1 n2

n1 =n2 = 1

V1 = Volume of MgSO4 =______________V

M1 = Molarity of MgSO4 =______________M

V2 = Volume of EDTA solution =______________V

20 x M1

M2 = = _________________M

Titre value (V2)

Part B:

Estimation of Copper Solution:

S.No. Volume Of Copper

V3ml

Burette readings Volume Of EDTA

V2ml Initial Final

Molarity of Copper Solution:

V2M2 V3M3

=

n2 n3

V2 = Volume of EDTA =______________V

M2 = Molarity of EDTA =______________M

V3 = Volume of Copper solution =______________V

V2M2 n3

M3 = x

V3 n2

n2 =n3 = 1

Titre value x M2

M3 = = ____________________M

20

Amount of copper present in the given solution

1000

solution of volumecopper ofweight Molecular M3 g

= ________________g



Expt. No. : 03

Date :

ESTIMATION OF DISSOLVED OXYGEN IN WATER SAMPLE

AIM: To estimate the amount of dissolved oxygen (DO) in the given water sample.

APPARATUS: Burette, Pipette, Conical Flask, Beakers, Wash bottle, Burette stand, BOD

bottle.

CHEMICALS: 48% Manganese sulphate solution, Alkali-Iodide-Azide solution,

Conc.Sulphuric acid, Starch indicator, 0.025N sodium thio sulphate solution, 40% KCl solution,

Sample water, Distilled water.

PRINCIPLE: Dissolved Oxygen can be determined by iodometric titration. Dissolved Oxygen present in water oxidises by KI liberating an equivalent amount of iodine by which it is titrated against a standard

solution using starch as an indicator. An oxygen carrier like manganese hydroxide must be used to bring

about the reaction between KI and dissolved oxygen because of molecular oxygen in water is not capable

of reacting with KI. Manganese is produced by the action of potassium hydroxide or sodium hydroxide and manganese sulphate.

2KOH + MnSO4 Mn(OH)2 + K2SO4

2Mn(OH)2 + O2 2MnO(OH)2

Basic Manganic oxide

Oxygen carrier

2MnO(OH)2 + H2SO4 MnSO4 + 2H2O + (O) 2KI + H2SO4 + (O) K2SO4 + H2O + I2

2Na2S2O3 + I2 2NaI + Na2S4O6

Sodium Tetrathionate

The nitrates present in the water interfere with the dissolved oxygen of water, since these ions liberate iodine from KI. Sodium azide is added to water which reacts with nitrate to decompose it as

follows:

2NaN3 +H2SO4 2N3H + Na2SO4

HNO2 + N3H N2O + N2 + H2O

FORMULA :- The amount of dissolved oxygen in the given water sample

litmgsamplewaterofVoluime

hypoofConcvalueTitre/

10008.

Where 8 is the equivalent weight of oxygen.

PROCEDURE :-

1. Collect water sample in a 300ml capacity of BOD bottle and add 1ml of KCl solution.

2. Add 2ml of manganese sulphate and 2ml of Alkali-Iodide-Azide solution.

3. Stopper the BOD bottle immediately.

4. Appearance of brown precipitate indicates the presence of DO.

5. Mix well by inverting the bottle 2 to 3 times and allow the brown precipitate to settle

down.

6. Add 2ml of Conc. of H2SO4 solution to dissolve the precipitate.

7. Take 20ml of this solution in to a clean conical flask.

8. Titrate the liberated Iodine with standard Hypo solution present in the burette.

9. Add 2ml of starch solution when the colour of solution becomes pale yellow, the solution

turns to blue colour.

10. Continue the titration till the blue colour is disappeared.



11. Note the volume of hypo used (V ml)

12. Repeat the titration till concordant readings are obtained.

13. Calculate the amount of dissolved oxygen in the given water sample by using the

formula.

RESULT : -The amount of dissolved oxygen in the given water sample =_________mg/lit.

OBSERVATION and CALCULATIONS: -

Burette = Standard Hypo solution (0.025N).

Conical flask = 20ml water sample consists of liberated Iodine.

Indicator = 2ml Starch Solution.

End Point = Blue to colour less

S.No. Volume of Sample (ml) Burette Readings Volume of Hypo used

(Vml) Initial Final

Dissolved oxygen in the given water sample litmghypoofConcV

/20

1008.

=___________________ mg/lit



Expt. No. : 04

Date :

ESTIMATION OF COPPER BY IODOMETRY

AIM: To estimate the amount of Copper present in the whole of given solution by using

approximately N/20 solution of Sodium thiosulphate and few crystals of Potassium dichromate.

APPARATUS: Burette, Pipette, Conical Flask, Burette Stand, Standard Flask, Weighing bottle.

CHEMICALS: Copper solution, Ammonia, CH3COOH, Starch indicator, KI, HCl, K2Cr2O7.

PRINCIPLE : K2Cr2O7 is an oxidising agent. It liberates nascent oxygen atoms in the presence

of dil.H2SO4.

)(342CrCl2KCl8HClOCrK 23722 OOH

This nascent oxygen oxidizes ferrous salt to ferric salt in cold and in the presence of acid.

]2][2[3 22 IKOHOOHKI

]2[3 6422322 NaIOSNaIOSNa

K2Cr2O7 3[O] 6 642 OSNa

Any Cupric salt in neutral medium when treated with potassium iodide form a white precipitate

of Cuprous iodide and iodine is set free quantitatively. The liberated iodine is titrated against

Sodium thiosulphate using starch as an indicator.

OHSOKCuIKICuSO 24224 5222

22222 IICuCuI

NaIOSNaIOSNa 22 6422322

Sodium tetrathionate

From the above equation it is clear that 2Cu2+

= I2 = 2S2O32-

= 2e-

The equivalent weight of Copper = 63.5

PROCEDURE:

A) STANDARDISATION OF HYPO:

Burette is cleaned, rinsed and filled with Sodium thiosulphate. Initial reading of the burette is

noted. Pipette is cleaned and it is filled with K2Cr2O7 solution. 20 ml of K2Cr2O7 solution is

pipette out into clean conical flask and 5 ml of conc.HCl and 20 ml (or) 10 ml of 10% KI

solution. The liberated iodine is titrated by adding of hypo from burette until straw yellow

colour is obtained. Then 2 drops of freshly prepared starch indicator is added. The colour

changes to blue; the titration is continued until blue colour changes to light green. Repeat the

titration to get concurrent values. The normality of hypo is calculated.

B) ESTIMATION OF COPPER SOLUTION:

The given solution of copper is made up to the mark with distilled water and shaken well for

uniform concentration. The burette is filled with hypo and initial reading is noted. Pipette is

cleaned and filled with copper solution and it is pipette out into a clean conical flask and add

few drops of ammonia to neutralize the mineral acid present in the copper solution. The

precipitate is dissolved by adding drop by drop of acetic acid to get clear solution. To the

above solution 20 ml (or) 10 ml of 10% of KI is added. It is titrated by adding hypo from the

burette till straw yellow colour is obtained. Then 2 drops of freshly prepared starch solution

is added and titration is continued until white precipitate is obtained. The final reading is

noted, difference between the final and initial volume gives the volume of the hypo required



to estimate 20 ml of copper solution. Then calculate the normality and amount of copper is

present in the given solution.

RESULT : -

Amount of copper present in the given solution=______________g


Normality of K2Cr2O7 N1 = ________________N

STANDARDISATION OF HYPO:

S.No. Volume of K2Cr2O7

(V1ml)

Burette Readings Volume of Hypo (V2ml)

Initial Final

Volume of potassium dichromate V1 = _____________V

Normality of K2Cr2O7 N1 = _____________N

Volume of Hypo V2 = _____________V

Normality of Hypo N2 = _____?_______N

2

112

V

VNN

=_______________N

ESTIMATION OF COPPER SOLUTION:

S.No. Volume of Copper

(V3ml)

Burette Readings Volume of Hypo (V2ml)

Initial Final

Volume of Hypo V2 = _______________V

Normality of Hypo N2 = _______________N

Volume of Copper solution V3 = _______________V

Normality of Copper solution N3 =______?_________N

3

223

V

VNN

=___________________N

Amount of copper present in the given solution1000

..3 solutionofVolumecopperofWtEqNg

=___________________g



Expt. No. : 05

Date :

ESTIMATION OF IRON (II) USING DIPHENYLAMINE INDICATOR

(DICHROMETRY – INTERNAL INDICATOR METHOD)

AIM: To estimate the amount of ferrous ion present in the whole of the given solution using a

standard solution of potassium dichromate.

APPARATUS: Burette, Pipette, Beaker, Conical Flask, Burette Stand, Standard Flask, etc.

CHEMICALS REQUIRED: k2Cr2O7, H2SO4, Phosphoric acid, Iron Solution, Internal Indicator

(Diphenylamine).

PRINCIPLE: K2Cr2O7 is an oxidizing agent; it liberates nascent oxygen atoms in the presence

of dil.H2SO4 .

K2Cr2O7 +4H2SO4 K2SO4 + Cr2 (SO4)3 + 4H2O + 3(O)

The nascent oxygen oxidizes ferrous salt to ferric salt in cold in presence of an acid.

3[2FeSO4 + H2SO4 + (O) Fe2 (SO4)3 + H2O]

K2Cr2O7 + 6FeSO4 + 7H2SO4 K2SO4 + 3Fe2 (SO4)3 + Cr2 (SO4)3 + 7H2O

1 K2Cr2O7 = 3(O) = 6Fe2+

.

Equivalent weight of iron is 55.85 and K2Cr2O7 is 49 i.e.1/6 of its molecular weight. Indicator

diphenyl amine employes appearance of blue violet colour in end point of titration.

PROCEDURE:

Estimation of Ferrous Ion:-

The given solution of ferrous ion is made upto the mark with distilled water and shaken

well for uniform concentration. Then burette is filled with K2Cr2O7 solution, then pipette out 20

ml of ion solution into a clean conical flask.

Add 5ml of acid mixture (dil. H2SO4 + H3PO4) and few drops of diphenyl amine

indicator. It is titrated against K2Cr2O7 which is taken in the burette until violet blue colour is

observed. Final reading is noted and experiment is repeated to get concordant value.

RESULT:

Amount of ferrous ion present in the given solution = __________________ g




Normality of K2Cr2O7, N1 =__________________N

Estimation of Ferrous Ion Vs K2Cr2O7:-

S.No. Volume of ferrous

solution (V2 ml)

Burette Readings Volume of K2Cr2O7

(V1ml) Initial Final

Volume of Potassium dichromate solution V1 = __________ml

Normality of Potassium dichromate solution N1 = __________N

Volume of ferrous ion V2 = __________ml

Normality of ferrous ion N2 = ___?______N

2

112

V

NVN =__________N

Amount of ferrous ion present in the given solution

1000

.. )(Nion ferrous ofNormality 2

2 solutionofvolumeFeofwtEqg

=___________ g



Expt. No. : 06

Date :

DETERMINATION OF ALKALINITY OF WATER SAMPLE

AIM: To determine the alkalinity of the given water sample.

APPARATUS: Burette, Pipette, Conical flask, Beakers, Glazed tile, Wash bottle, Burette stand.

CHEMICALS: Standard sulphuric acid (0.02 N), Phenolphthalein indicator, Methyl orange

indicator, Water sample, Distilled water.

PRINCIPLE: The alkalinity of a solution is a measure of its capacity to neutralize bases.

Alkalinity is due to3

2

3 ,, HCOCOOH ions. It is determined by titration against standard

sulphuric acid using phenolphthalein and methyl orange indicators.

Titration to pH 8.3 with phenolphthalein or the disappearance of pink will indicate

phenolphthalein alkalinity.

OHHOH2

3

2

3 HCOHCO

Titration to pH 4.5 with methyl orange or the appearance of pink colour will indicate total

alkalinity. (i.e., complete neutralization of OH-, CO3

2-and HCO3

- ions).

3

HCO H 2232 COOHCOH

FORMULA:

The amount of alkalinity interms of CaCO3 equivalents

ppmsampleofVolume

SOHofConcvalueTitre 100050. 42

PROCEDURE:

Part A: Phenolphthalein Alkalinity:

1. Rinse and fill the burette with 0.02N H2SO4 solution.

2. Pipette out 20 ml of water sample into a clean conical flask.

3. Add a drop of phenolphthalein indicator to the sample.

4. If pink colour develops titrate against 0.02N H2SO4 till the pink colour disappears

indicating pH 8.3.

5. Note the volume of H2SO4 used (V1 ml).

6. Repeat the titration to obtain concordant readings.

7. Calculate phenolphthalein alkalinity by using given below.

Phenolphthalein alkalinity ppmV

20

10005002.01

Part B: Total Alkalinity:

1. Rinse and fill the burette with 0.02 N H2SO4 solution.


3. Add 2 to 3 drops of methyl orange indicator to the sample.

4. If yellow colour develops, titrate against 0.02 N H2SO4 solution till the pink colour

appears indicating pH of 4.5.

5. Note the volume of H2SO4 consumed (V2 ml).




7. Calculate total alkalinity by using the formula given below.

Total alkalinity = ppmV

20

10005002.02

RESULT:

Phenolphthalein alkalinity of water sample in terms of CaCO3 equivalents = ____________ppm.

Total alkalinity of water sample in terms of CaCO3 equivalents = _____________________ppm.

OBSERVATIONS and CALCULATIONS:

Part A: Phenolphthalein Alkalinity: Burette = 0.02 N H2SO4 solution

Conical flask = 20 ml of sample

Indicator = Drop of Phenolphthalein

End point = Pink to colourless.

S.No.

Volume of Sample

(ml )

Burette Reading (ml) Volume of H2SO4 (V1 ml )

Initial Final

1.

2.

3.

Phenolphthalein alkalinity in terms of CaCO3 equivalents

ppmV

20

10005002.01 = ______________ppm(or) mg/litre.

Phenolphthalein alkalinity = ___________ ppm.

Part B: Total alkalinity:

Burette = 0.02 N H2SO4 solution

Conical flask = 20 ml of sample

Indicator = 2 to 3 drops of methyl orange

End point = Yellow to pink

S.No.

Volume of Water

Sample (ml )

Burette Reading (ml) Volume of NaOH (V2 ml )

Initial Final

1.

2.

3.

Total alkalinity in terms of CaCO3 equivalents

ppmV

20

10005002.02

= _____________ppm (or) mg/litre.

Total alkalinity ___________ ppm.



Expt. No. : 07

Date :

DETERMINATION OF ACIDITY OF WATER SAMPLE

AIM: To determine the acidity of the given water sample.


CHEMICALS: Standard sodium hydroxide (0.02 N), Phenolphthalein indicator, Methyl orange

indicator, Sodium thiosulphate (0.1 N), Water sample.

PRINCIPLE: The acidity of a solution is a measure of its capacity to neutralize bases.

Acidity is due to the presence of mineral acids like H2SO4, HCl, HNO3 and dissolved CO2 in the

form of H2CO3. These acids can be estimated by titration against standard sodium hydroxide

using methyl orange and phenolphthalein.

Titration to pH 4.5 or a sharp change from faint orange to yellow of methyl orange indicator will

indicate the neutralization of mineral acids.

H+ + OH

- H2O

Titration to pH 8.5 or the appearance of faint pink colour, by the use of phenolphthalein will

indicate total acidity. (Complete neutralization of mineral acids and dissolved CO2).

Interference due to to presence of residual Chlorine can be removed by the addition of one drop

of Sodium thiosulphate (Hypo) solution to the water sample.

The acidity is measured interms of CaCO3 equivalents and expressed in ppm or mg/litre.

FORMULA:

Acidity of water sample interms of CaCO3 equivalents

ppmsampleofVolume

NaOHofConcvalueTitre 100050.

Where 50 is the equivalent weight of CaCO3.

PROCEDURE:

Part A: Methyl Orange Acididty:

1. Rinse and fill the burette with 0.02 N NaOH solution.


3. Add a drop of hypo solution and 2 drops of methyl orange indicator. The solution colour

changes to orange.

4. Titrate the solution against 0.02 N NaOH solution till the colour changes from faint

orange to yellow indicating pH of 4.5.

5. Note the volume of NaOH consumed (V1 ml).


7. Calculate methyl orange acidity by using the formula given below.

Methyl orange acidity ppmV

20

10005002.01

Part B: Phenolphthalein Acidity (Total Acidity):

1. Rinse and fill the burette with 0.02 N NaOH solution.


3. Add a drop of hypo solution and 2 drops of phenolphthalein indicator.



4. Titrate the solution against 0.02 N NaOH solution till the appearance of faint pink colour

indicating pH of 8.5.

5. Note the volume of NaOH consumed (V2 ml).


7. Calculate total acidity by using the formula given below.

Total acidity ppmV

20

10005002.02

RESULT:

Methyl orange acidity or Mineral acidity = _____________ ppm.

Phenolphthalein acidity or Total acidity = ______________ppm.

Carbonic acid acidity of water sample =______________ppm.


Part A: Methyl orange acidity (Mineral acidity)

Burette = 0.02 N NaOH solution

Conical flask = 20 ml water sample + Drop of Na2S2O3

Indicator = Methyl orange; End point = Faint orange to yellow.

S.No.

Volume of Water

Sample (ml )


Initial Final

1.

2.

3.

Methyl orange acidity = Titre value x conc. of NaOH x 50 x 1000 ppm

Volume of sample


20

10005002.01

= ______________ppm (or) mg/litre.

Part B: Phenolphthalein acidity (Total acidity)

Burette = 0.02 N NaOH solution

Conical flask = 20 ml water sample + Drop of Na2S2O3

Indicator = Phenolphthalein; End point = Colourless to pink

S.No.

Volume of Water

Sample (ml )


Initial Final

1.

2.

3.

Total acidity = Titre value x conc. of NaOH x 50 x 1000 ppm

Volume of sample


20

10005002.02

= ______________ppm (or) mg/litre.

Acidity due to carbonic acid = Total acidity – mineral acidity= ______ ppm (or) mg/litre.



Expt. No. : 08

Date :

PREPARATION OF PHENOL – FORMALDEHYDE RESIN

AIM: To prepare Phenol-Formaldehyde resin.

APPARATUS: Glass rod, beakers, funnel, filter paper.

CHEMICALS: Glacial acetic acid, 40% formaldehyde, phenol, Conc.HCl.

PRINCIPLE:

PHENOLIC RESINS:

In phenol formaldehyde resins one of the monomer is always a phenol molecule. One of the most

important numbers of this class is phenol formaldehyde resin. The phenol formaldehyde is

catalyzed by acids and alkalies. The nature of the product depends on several factors. The major

one is being the nature of catalyst and proportion of reactants. The formation of phenol

formaldehyde resin comprises of the following steps:

METHYLOLLATION:

When phenol and formaldehyde react together the first step is the entry of methylol

(CH2OH).Groups in ortho and para position to the hydroxyl group the reaction takes place in the

presence of acid or alkali. Depending upon phenol formaldehyde ratio various phenol alcohols

may be formed.

NOVOLAC FORMATION:

Depending upon the ratio of phenol formaldehyde different resins namely novolac and resole

resins are obtained. In presence of acid catalyst, when P/F ratio is greater than unity the methylol

derivatives condense with phenol to form a linear polymer with little methylol groups. The

product is thermo plastic in nature and is known as novolac.

PROCEDURE:

Place 5ml of glacial acetic acid and 2.5 ml of 40% formaldehyde solution in a 500ml

beaker

Add 2gm of phenol to the above solution.

Wrap a cloth loosely round the beaker

Add a few ml of Conc.HCl in to the mixture carefully and heat it slightly

A large mass of plastic pink in colour is formed

The residue is washed with water and is filtered.

The product is dried and yield is calculated.

RESULT:

The yield of the product is_______________________



METHYLOLLATION:

H+

CH2OHCH H

O

OH OH

CH2OH

CH2OH

+

OH

CH2OH

2-(hydroxymethyl)phenol 4-(hydroxymethyl)phenol

NOVOLAC FORMATION:

OH

CH2OH H

OH

H

OH

HHOH2CH

OH

HHOH2C

+ + +

OH OH OH OH

CH2 CH2 CH2 H2C



Expt. No. : 09

Date :

CONDUCTOMETRIC TITRATION OF STRONG ACID AGAINST STRONG BASE

AIM: To estimate the strength of strong acid against strong base by conductometrically.

APPARATUS: Conductivity Meter, Conductivity Cell, Micro Burette, Beaker, Glass rod

CHEMICALS: HCl, NaOH

PRINCIPLE: Conductometric titration is the volumetric analysis based upon the measurement

of conductance during the course of titration. The number of free ions, charge on the free ions

and mobility of the ions affect the conductance of an aqueous solution. When one electrolyte is

added to another electrolyte, the change in number of free ions causes a change in the

conductance.

When a strong acid (HCl) is titrated against a strong base (NaOH), before NaOH solution

is added from the burette, the acid solution has high conductivity due to highly mobile H+ ions.

When NaOH is added to the acid, the conductivity of the acid solution decreases due to

neutralization of highly mobile H ions of the acid with OH ions of the base.

OHClNaNaOHClH 2

Thus the conductance of the solution continues to decrease until the equivalent point is reached.

Further, the addition of NaOH solution will increase the conductance by highly mobile hydroxyl

ions. The point of intersection of the graph plotted by conductance of the solution against volume

of alkali added correspond the endpoint of the titration.

PROCEDURE:

CONDUCTOMETRIC TITRATION:

Fill the burette with standard 1N NaOH solution. Take 20ml of the given HCl solution in

a 100ml beaker and dip the conductivity cell in it and measure the conductance initially. Now

add NaOH from burette drop wise i.e.; 0.5ml for each addition. After each addition, stir the

solution gently by shaking and note down the change in conductance. The measured

conductance are recorded and tabulated in the table. Plot the graph between corrected

conductance against volume of base added, the intersection of two straight lines gives the

end point. Calculate the strength of the given strong acid (HCl) from the known strength of

the NaOH solution.

The corrected conductance is calculated by applying volume correction which is given by

V

VUCC |

Where |C = Corrected Conductance

C = Measured Conductance

V = Volume of Acid

U = Volume of Base

PRECAUTIONS:

1. The conductivity cell should be never kept dry.

2. The electrode must be washed.

RESULT:

The strength of given hydrochloric acid solution is_______ N.



OBSERVATIONS AND CALCULATIONS:

S.No. VOLUME OF BASE

ADDED (ml)

MEASURED

CONDUCTANCE (Siemens)

CORRECTED CONDUCTANCE

V

VUCC |

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

The titre value corresponding to the point of inflection in the end point graph is_______ ml.

Therefore,

kerbeaintakenHClofVolume

)A(NaOHofNormalityvaluetitreintPoEndHClofStrength

=.ml25

)A().ml(volumeintPoEnd =________________N.

Rough Graph:

Conductance

End Point

Volume of Base (ml)



Expt. No. : 10

Date :

ESTIMATION OF CHLORIDE CONTENT PRESENT IN WATER SAMPLE

AIM: To estimate the amount of chloride present in the given water sample.


CHEMICALS: Standard Silver nitrate solution (0.0141 N), Potassium chromate indicator,

Water Sample, Distilled water.

PRINCIPLE: Chloride ions are present in water in the form of compounds like NaCl, CaCl2,

MgCl2, etc. These chlorides are estimated by titrating with standard silver nitrate solution using

potassium chromate indicator in the pH range of 7 to 8. This can be maintained by adding 1 N

NaOH if the pH of the sample is less than 7 or by adding 1 N H2SO4 if it is greater than 10.

Silver chloride is precipitated quantitatively before red silver chromate is formed.

33 NOAgClAgNOCl

)( pptWhite

342342 22 KNOCrOAgAgNOCrOK

)( pptredBlack

FORMULA:

Chloride content in water sample litmgsamplewaterofVoluime

AgNOofConcvalueTitre/

10005.35. 3

Where 35.5 is the equivalent weight of chlorine.

PROCEDURE:

Part A – Titration of Water Sample with Standard AgNO3 Solution:

1. Rinse and fill the burette with 0.0141 N AgNO3 solution.

2. Pipette out 20 ml water sample into a clean conical flask.

3. Add 2 or 3 drops of potassium chromate indicator. The solution turns yellow colour.

Adjust to pH 7.0 to 8.0.

4. Titrate with 0.0141 N AgNO3 solution until a faint distinct reddish brown colour is

formed after brisk shaking.

5. Note the volume of AgNO3 used (V1 ml).

6. Repeat the titration till the concordant readings are obtained.

Part B – Titration of Distilled Water with Standard AgNO3 solution:

1. Rinse and fill the burette with 0.0141 N AgNO3 solution.

2. Pipette out 20 ml distilled water into a clean conical flask.

3. Add 2 or 3 drops of potassium chromate indicator. The solution turns yellow colour.

4. Titrate with 0.0141 N AgNO3 solution until a faint distinct reddish brown colour is

formed.

5. Note the volume of AgNO3 used (V2 ml).

6. Repeat the titration till the concordant readings are obtained.

Chloride content in the water sample

litmgVV

/20

10005.350141.0)( 12



= ______________mg/litre

RESULT: The amount of chloride present in the water sample =_______________mg/litre.


Part A – Titration of Water Sample with Standard AgNO3 Solution:

Burette = 0.0141 N AgNO3 solution

Conical flask = 20 ml water sample

Indicator = 2 or 3 drops of K2CrO4

End point = Yellow to Brick red

S.No. Volume of Water Sample (ml) Burette readings (ml) Volume of AgNO3 Used

(V1 ml) Initial Final

1.

2.

3.

Part B – Titration of Distilled Water with Standard AgNO3 solution:

Burette = 0.0141 N AgNO3 solution

Conical flask = 20 ml Distilled water

Indicator = 2 or 3 drops of K2CrO4

End point = Yellow to Brick red

S.No. Volume of Distilled Water (ml) Burette readings (ml) Volume of AgNO3 Used

(V2 ml) Initial Final

1.

2.

3.

Chloride content in the water sample

litmgVV

/20

10005.350141.0)( 12

= ______________mg/litre

Chloride content in the water sample =_____________________mg/litre.

Date post:	06-Mar-2018
Category:	Documents
Upload:	phamkhuong
View:	225 times
Download:	0 times

ENGINEERING CHEMISTRY LAB MANUAL I B - T K...

Documents