Chem Lab Manual 0910

Department of Applied Chemistry Practical Manual

Shri Ramdeobaba Kamla Nehru Engineering College, Nagpur

1

SHRI RAMDEOBABA KAMLA NEHRU ENGINEERING COLLEGE,

NAGPUR

CHEMISTRY PRACTICAL MANUAL

INDEX

Expt. No Name of the Experiment. Page No.

1 To determine alkalinity of given water samples. 02-05

2 Determination of hardness of water by Complexometric

method

06-13

3 To estimate the amount of Copper in given solution by

Iodometry.

14-18

4 Determination of Chlorine in given water sample. 19-22

5 To estimate the amount of Ferrous and Ferric ions present in

the given solution.

23-26

6 Determination of viscosity of a given sample at different

temperatures.

27-32

7 Determination of Flash Point of given sample by Abel’s

Flash Point apparatus.

33-35

8 Determination of Flash Point of given sample by Pensky-

Marten’s flash point apparatus

36-38

9 Determination of Flash Point of given sample by Cleveland’s

Flash Point apparatus.

39-40

10 Determination of Chemical Oxygen Demand in a given

wastewater sample.

41-45

11 Determination of Acid Value of given sample. 46-48

12 Determination of Saponification Value of a given sample. 49-52

.



2

EXPERIMENT NO. 1

Aim: To estimate the Type and Extent of alkalinity present in the given water sample

Principle:

The alkalinity of a water sample is due to carbonates (CO32-

), bicarbonates (HCO3-)

and hydroxide (OH-) ions. Thus in a given water sample the possible combination of ions

causing alkalinity are as follows.

(i) OH - alone

(ii) CO3 2 - alone

(iii) HCO3 - alone

(iv) OH- and CO3

2 – together

(v) CO32-

and HCO3- together

The possibility of OH- and HCO3

-ions together in the same solution is ruled out as

they react as-

HCO3 - + OH

- CO3

2 -- + H2O

In the same way, possibility of co -existence of all the three OH- , CO3

2-and HCO3

-

is ruled out.

The determination of alkalinity involves following reactions.

OH- + H

+ H2 O ----------------1)

CO3 2-

+ H+ HCO3

- ---------------- 2)

HCO3- + H

+ H2CO3 H2O + CO2 ---------------- 3)

The phenolphthalein is pink in color above pH 10 and it is colorless below pH-8.

While, The methyl orange is yellow above pH 4.4 and it turns pink below pH 3.1



3

Thus titration of a given water sample in presence of phenolphthalein as an

indicator indicates completion of reaction 1 and 2 whereas the same water sample, if titrated in

presence of Methyl orange as an indicator indicates the completion of reaction 1,2 and 3.

The water sample, when titrated with an acid solution using phenolphthalein

indicator gives (End point = P) and with methyl orange indicator gives (End point = M). The

relation between P and M points gives the type and extent of alkalinity is established as follows:

Relation between

P&M point

Type of alkalinity Extent of alkalinity

OH- CO3

2- HCO3

-

P =M Only OH- M - -

P = 1/2 M Only CO3-- - 2P -

P = 0 Only HCO3- - - M

P>1/2 M OH-&CO3

-- (2P-M) 2(M-P) -

P<1/2 M CO3-- & HCO3

- - 2P (M-2P)

Procedure:

Part 1: - Determination of type of alkalinity.

Take 25-ml. water sample in a conical flask. Add 1-2 drops of phenolphthalein as an

indicator. The solution becomes pink. Titrate this solution with acid solution from the burette. At

the end point the pink color of the solution changes to colorless. Note this end point as P point.

Now At this point, add 2 drops of methyl orange indicator to the same solution. Solution

becomes yellow. Again titrate the solution with same acid solution from the same burette. The

end point may be obtained immediately or after addition of some amount of acid. In the former

case, obviously, P = M and in later P ≠ M.

The relationship between P point and M point denotes the type of alkalinity present in the given

water samples as shown in earlier table.



4

Calculations:-

Part 1 : Determination of type of alkalinity present in water sample ‘ A’

The given Normality of HCl solution is = N1 and

If P > 1 / 2 M

then,

The type of alkalinity present in given water sample is OH- and CO3

2-

Now, 1)Volume of acid corresponding to OH - = 2P- M

2) Volume of acid corresponding to CO32-

= 2 (M- P)

a) Normality of water sample due to OH-

HCl Water sample A

N1V1 = N2V2

N1 x 2P- M

N2 = -------------------

25

Wt/ lt for OH-

= Normality x Eq. Wt.of CaCO3

= N2 x 50

= g/l

= x 1000 mg/l

= ppm

b)Normality of water sample due to CO3 2-

.

HCl Water sample A

N1V1 = N3V3

N1V1 x 2 (M- P)

N3 = ------ --------------------

25

Wt./ lt for CO3 2-

. = Normality x Eq.. Wt. of CaCO3

= N3 x 50

= g/l

= x 1000 mg/l

= ppm

Same calculations are to be carried out for Water sample C



5

Result: -

The given water sample A contains:-

OH – Alkalinity = ------------ ppm

CO3 2–

Alkalinity = ------------- ppm

Total Alkalinity = ------------- ppm



6

EXPERIMENT NO. 2

Aim : Determination of hardness of water by comlexometric method.

Principle: Complexometry involves the interaction of an organic compound with a

complexable metal ion and results in the formation of compounds known as Werner’s

complexes. The organic compound is an anion or a Lewi’s base and is known as Ligand.

Several factors, such as pH, basicity of ligand, type of ring formed etc. govern the

successful formation of a complex. Thus:

X L + M n +

---------� (L) x M

Ligand Metal ion Complex

Ethylene diamine tetra acetic acid (EDTA) is a well known complexing agent which is

widely used in analytical work on account of its powerful complexing action and commercial

availability.

HOOC-H2C CH2-COOH

\ /

N-CH2-CH2-N

/ \

HOOC-H2C CH2COOH

EDTA

The spatial structure of its anion which has six donor atoms , enables it to satisfy the co-

ordination number of six, frequently encountered among the metal ions and to form strain less

five membered rings on chelation.

EDTA forms complexes with Ca 2+

, Mg 2+

, Zn 2 + as well as with many other metal

ions in an aqueous solution. The complexes have the general formula.



7

N

H2

C

H2

C N

CH2

CH2

H2C

H2C

O O

OO

OO

OO

M2+

Structure of Metal-EDTA Complex

EDTA is generally used in the form of disodium salt or tetra sodium salt on account of their

greater solubility. For simplicity of discussion, EDTA is assigned the formula H 4 Y, the

disodium salt will hence be Na 2 H2 Y which gives the complex forming ion H2Y 2-

in aqueous

solution. It reacts with metal ions in 1:1 ratio. The reaction with divalent cations, M 2+

(Such as

Mg 2+

or Ca 2+

) may be represented as,

M2+

+ H2 Y2 - -----------� MY

2- + 2H

+

Thus it is evident that 1 mole of the complex forming H2 Y2 - reacts in with one mole

of the metal ion and, two moles of hydrogen ions are generated. It is apparent from the equation

above that the dissociation of the complex will be governed by the pH of the solution. Lowering

of the pH will decrease the stability of the metal EDTA complex. .

INDICATOR:

Metal ion indicators are commonly used in EDTA titrations. Eriochrome black-

T, also known as EBT or Solochrome black is the metal ion indicator used for complexometric

titration. Eriochrome black-T is 3-hydroxy,4(1-hydroxy, 2 -azo napthyl), 7-nitro 1 sodium

napthyl sulphonate.



8

N NNaO3S

O2N

HOOH

Structure of EBT Indicator

This compound tends to polymerize in strongly acidic solutions to a red brown

product and hence this indicator is used in EDTA titrations with solution having pH greater than

6.5. This EBT posses three replacable hydrogen atoms, one from sulphonic acid group and two

form phenolic –OH groups. The sulphonic acid looses its proton to form Na salt, as it is more

acidic than the other groups. The dissociation of the two hydrogen atoms of phenolic groups only

should therefore be considered and hence EBT for the simplicity is represented asNaH2D which

gives H2D – ion which

exhibits different colours at different pH values.

H2D - -----------� H D

2- -------------� D

3-

pH 6.3 pH 10 pH 11.5

( Red) (Blue) (Yellowish Orange)

In the pH range 8-10, the blue form of the indicator HD 2-

gives a wine red

complex with Metal ion.

For ex.:

Mg 2+

+ HD2-

----------� Mg D- + H

+

Blue Wine red

Now, if EDTA is added to such a solution, Mg 2+

/ Zn2+

preferentially complexes

with EDTA (since the metal-EDTA complex is more stable than the metal indicator complex)



9

and liberates the free indicator HD2-

at the end point, thereby producing a sharp colour change

from wine red to blue.

The hardness producing constituents of water i.e. Ca2+

and Mg2+

can be

estimated by titrating with Ethylene diamine tetracetic acid (EDTA) which acts as the ligand,

using Ericromic Black-T as an indicator, which as explained above is also a ligand. In the pH

range 8-10, the indicator is blue and forms a wine red coloured complex with Ca2+

/ Mg2+

ions.

In this estimation, thus four types of complexes are possible

viz :

a) EBT – Mg2+

complex (Wine red)

b) EBT - Ca2+

complex (Wine red)

c) EDTA- Mg2+

complex (colourless)

d) EDTA- Ca2+

complex (colourless )

Their order of stability and consequently, their preference of formation are:

EDTA- Mg2+

> EDTA- Ca2+

> EBT – Mg2+

> EBT – Ca 2+

The colour and stability of these complexes is made use of in the estimation of

hardness of water using EBT and EDTA in the complexometric methods. To a known volume of

hard water(say 25 ml) containing Ca2+

and Mg2+

, are added 2-3 drops of EBT at a pH of 8-10,

maintained with the help of ammonia buffer. The EBT forms a wine red complex with Mg2+.

EDTA is then added when it first reacts with the free Mg2+

then with free Ca2+

and finally after

all the free metal ions are exhausted, it reacts with the EBT- Mg2+

complex, extracting the Mg2+

ions (as EDTA-Mg2+

complex is more stable than EBT-Mg2+

complex). In this way, the free

EBT ion (HD2-

) is generated which being blue in the pH range 8-10, imparts a blue colour to the

solution. Thus the colour change at the point is wine red to blue.



10

Reactions:

pH

EBT + Mg2+

--------� EBT-Mg2+

Complex

8-10 (Wine red colour obtained)

pH

EDTA + Mg2+

--------� EDTA- Mg2+

complex (Colourless)

8-10

pH

EDTA + Ca2+

--------� EDTA-Ca2+

complex (Colourless)

8-10

EDTA + EBT-Mg 2+

complex ----------� EDTA-Mg 2+

complex + EBT

(Wine red) PH 8-10 ( blue)

PROCEDURE:

Part 1: Standardisation of EDTA

Take 25 ml.of standard hard water sample in a clean conical flask. To it add 2 ml

of buffer solution (NH4Cl + NH4OH) and 3-4 drops of indicator EBT. The solution becomes

wine-red in colour. Titrate this solution against EDTA solution. A change of colour from wine-

red to blue indicates end point.

Part 2: Determination of Total hardness.

Pipette out 25 ml. of given hard water sample in a clean conical flask. Add 2 ml.

of buffer solution and 3-4 drops of EBT indicator. Titrate this solution against EDTA solution

until the colour changes from wine red to blue.. Note the titre Value, which corresponds to total

hardness.



11

Part 3: Determination of Permanent Hardness.

Pipette out 100ml of given hard water sample in 250 ml borosil conical flask, boil

gently to reduce to 1/3rd

of its volume. Filter it into 250 ml volumetric flask. Wash the funnel

along with the filter paper, 3-4 times (about 10ml each washing) with distilled water and collect

the washings in volumetric flask .Make up the volume using distilled water. Transfer the solution

to 250 ml clean beaker.

Now pipette out 25 ml of diluted hard water sample, add 2 ml of basic buffer

solution and 3-4 drops of EBT indicator. Titrate with EDTA solution until the wine red colour

changes into blue.This titre value (on multiplying by 2.5) corresponds to the permanent hardness.

Observation Table:

Part 1 : Standardization of EDTA

Control reading: ml

S.No Vol. of std. Hard

water Vol.of EDTA solution

Constant

Reading.

Part 2: Determination of Total Hardness of water.

Control reading: ml

S.No

Vol. of given.

Hard water

sample

Vol.of EDTA solution Constant

Reading.



12

Part 3: Determination of Permanent Hardness:

Control reading: ml

S.No

Vol. of boiled

Hard water

sample.

Vol.of EDTA solution Constant

Reading.

Calculations:

For part 1: Standardization of EDTA

Let 1.2 g of CaCO3 is dissolved in 1000 ml of water.

Wt. of the CaCO3

Molarity of std. Hard water = -------- -------------------------------------- X 1000

Mole. wt. of CaCO3 X Vol.

= M ( M 1 )

Std Hard water EDTA

M1 V1 = M2 V2

M1 V1

M 2 = ------------------

V2

= ( molarity of EDTA)

Part 2 : Determination of Total Hardness present in given water sample.

EDTA Water Sample

M2V2 = M3V3

M2V2

M3 = ------------------

V3

= (molarity of hard water)



13

Total Hardness

Wt/Lit = M3 x Mole. Wt. of CaCO3

= ------------- g/lit.

= ------------ x 1000 mg/ lit.

= -- ppm. (Total Hardness.)

Part 3 : Determination of Permanent Hardness present in given water sample.

EDTA Water Sample

M2V2 = M4V4

M2V2

M4 = ------------------

V4

= (molarity of diluted hard water)

Permanent Hardness

Wt/Lit = M4 x Mole. Wt. of CaCO3

= ------------- g/lit.

= ------------ x 1000 mg/ lit.

= -- ppm.(Permanent Hardness)

Temporary Hardness = Total hardness – Permanent hardness

Result: The given water sample was found to contain

Total hardness = ppm,

Permanent hardness = ppm and

Temporary hardness = ppm



14

EXPERIMENT NO. 3

AIM: To Estimate Copper in the given copper sulphate solution by Iodometry.

THEORY: -

The Iodometric method of volumetric analysis is based on oxidation reduction process (Redox)

and involves interconversion of elemental Iodine into and Iodide ions and vice-versa as-

Iodometry

2 I - -------------- ----------� I2 + 2e

-

Iodimetry

There are two types of Iodometric titrations

a) Indirect Iodometric titrations (Iodometry) involves titrations of iodine liberated from

potassium iodide (KI) in a chemical reaction.

b) Direct Iodometric titrations (Iodimetry) involves titrations with Std. solution of Iodine.

PRINCIPLE: In this Experiment an oxidizing agent such as K2Cr2O7 or CuSO4 is treated in acid

solution (dil. H2SO4) with an excess of iodide ion produced by KI. The oxidizing agent oxidizes

the iodide ions into Iodine, which is then titrated with the std. solution of reducing agent such as

sodium thiosulphate solution.

Chemical Reactions:

PART A: Standardization of Sodium thiosulphate solution.

1) K2Cr2O7 + 6KI + 7H2SO4 3I2 + Cr2(SO4)3 + 4K2SO4 + 7H2O

In ionic form, the same reaction can be written as

Cr2O72-

+ 6I- + 14H

+ 3I2 + 2Cr

3+ + 7H2O

2) 2 Na 2S2O3 + I2 2Na I + Na 2 S4 O6

Sodium thiosulphate Sodium tetrathionate



15


2S2O32-

+ I2 2I- + S4O6

2-

PART B: Estimation of copper ions

2CuSO4 + 4KI Cu2I2 + I2 +2 K2SO4


2Cu2+

+ 4 I- 2CuI2 Cu2I2 + I2

The amount of iodine thus generated depends upon the concentration of Cu2+

ions and the

amount of I2 generated can be found out by titrating against sodium thiosulphate solution. .

Since the solubility of Iodine in water is low, a considerable excess of KI must be used

in iodometric determination of oxidizing agent. The iodine liberated by the reaction then

dissolves by forming the complex K (I3)

KI + I2

K (I3) OR I-+ I2 (I3)

-

Formation of this compound (I3)- does not interferes with in the titration of iodine with

thiosulphate because as free iodine is used with thiosulphate, the equilibrium between I2 and

(I3)- is disturbed and more I2 is produced by the backward reaction.

After addition of KI and /or acid the flask is kept in dark place for about 5 minutes

as rate of the reaction between oxidizing agent and I- is too low.. Secondly, light accelerates

side reaction

4I-+4H

+ + O2 2I2 + 2H2O

In which I- are oxidized to I2 by atmospheric oxygen and analytical results

may go wrong. It is for this purpose therefore before adding KI to the acidic solution of the

oxidizing agent, a pinch of Na2CO3 is added so as to generate a CO2 atmosphere which

would prevent access of the O2 to the solution.



16

Indicator:

Most commonly used indicator in iodometric titration is STARCH. It forms

deep blue complex with iodine (but not with iodide)

Starch is a polymer, consisting of two components Amylose and

Amylopectine. The active fraction amylose is a polymer of the sugar α –D- glucose. It has a

shape of coiled helix into which I2 combined with I-

can fit, thus reacting slowly with

thiosulphate. Due to this reason the starch solution must be added at the very end of the

titration (when iodine conc. is very less).

Starch solution is not very stable solution because various microorganisms

are degrading it and one of the degradation products, glucose, is a reducing agent. Thus an

old starch solution containing some glucose can cause an appreciable titration error. Hence

always fresh starch solution is used.

PROCEDURE: -

Part A: -Standardization of sodium thiosulphate solution.

Pipette out 25ml. std. K2Cr2O7 solution in a conical flask. Add 10 ml.dil. sulphuric acid,

followed by a pinch of sodium bicarbonate /sodium carbonate and finally 10 ml KI (10%). Keep

the flask covered for 5-6 minutes in dark. Appearance of dark brown color indicates the

formation of I2.

Now, titrate it against Na2S2O3 from burette until dark brown colour becomes faint brown. At

this stage, add 2 ml. starch solution, colour of the solution becomes intense blue. Continue

titration drop wise till blue color disappears. Note this point as an end point.

Part B: -Estimation of Cu2+

ions.

Pipette out 25 ml. given CuSO4 solution in a conical flask. Add few drops of dil.

Na2CO3 solution until a faint permanent precipitate remains. Add acetic acid solution dropwise

till the solution becomes clear. Now add 10 ml. KI (10%). Appearance of dark brown colour

indicates formation of I2. Titrate the liberated iodine with Na2S2O3 solution from burette till

brown colour of iodine fades, then add 2ml. of starch solution, and continue the addition of the



17

Na2S2O3 solution until the blue color commences to fade. Then add about 5-6ml of ammonium

thiocyanate solution into the flask due to which blue color become more intense. Continue the

titration till blue color disappears (there is formation of flesh coloured ppt in the flask). Note this

point as an end point.

Calculations:-

Given Normality of std. K2Cr2O7 is N1

Part A: -Standardization of Sodium thiosulphate solution.(Hypo.)

Std. K2Cr2O7 v/s Hypo

N1V1 = N2V2

N1x 25 = N2 X V2

N1x 25

N2 = -----------------

V2

= X

Part B: -Estimation of Cu2+

ions.

Hypo v/s given CuSO4

N2V’2 = N3V3

N2V’2

N3 = ------------

25

N3 = Y (Normality of CuSO4 solution.)

Wt./lt.of = Normality x Eq. Wt. Of Cu2 +

ions

Of Cu 2 +

= N3 X 63.546

= Z g/l

Result: The given solution of copper contains Z g/l of Cu2+

ions.



18

Precautions:

1) To suppress the volatility of Iodine, excess of KI is added.

2) Flask is to be kept in dark to avoid photo oxidation of Iodide ions to Iodine.

3) Freshly prepared starch solution only can be used as an indicator.



19

Experiment N0. 4

Aim: Estimation of chlorine in the given water sample.

Reagents required

1. Std K2Cr2O7 Solution

2. KI solution (10%)

3. Dilute H2SO4 solution

4. NaHCO3

5. Sodium thiosulphate solution

6. Starch solution

Theory:

Living organisms such as algae, fungi and bacteria are more abundant in surface

drainage waters while in deep well waters, the bacterial count is often low .The growth of these

organisms in water may have serious consequences .The slime surrounding these organisms

causes them to adhere to the metal surfaces and the film of slime thus formed serves as a base to

which suspended matter present in the water can adhere. This leads to a serious reduction in the

heat transfer rates and tube blockages. The growth of marine organisms such as mussels may

lead to serious reduction in the carrying capacity of pipelines and culverts. Thus measures to

prevent the development of the living organisms are often necessary. Organism growth generally

take place most readily in water at temperature range between 10 - 35 0C.

Chlorine is widely used for disinfecting water to remove bacteria, fungus and other

pathogenic microorganisms and for deodorization, as it is a powerful oxidizing agent and is

cheaply available.

Chlorination can be carried out by using bleaching powder or chlorine gas or

chlorine dissolved in water in the form of concentrated solution or with chloramines.The

sterilizing action of is supposed to be due to its reaction with water producing hypochlorus acid

and nascent oxygen both of which have powerful germicidal properties.



20

Reactions:

CaOCl2 + H2O �� Ca (OH) 2 + Cl2

Cl2 + H2O �� HOCl + HCl

HOCl �� HCl + [O]

Hypochlorus acid nascent oxygen

Principle:

The principle involved in the estimation of chlorine in water is that when a measured

quantity of water is treated with excess of potassium iodide, the free chlorine present in the water

oxidizes the corresponding amount of potassium iodide to Iodine. The liberated iodine is

estimated by titrating against sodium thiosulphate solution.

Cl2 + 2KI ------�� 2KCl + I2

I2 + 2Na2S2O3 -----�� Na2S4O6 + 2NaI

Sodium thiosulphate sodium tetra thionate

Procedure:

Part A: - Standardization of Hypo solution

Pipette out 25ml. std. K2Cr2O7 solution in a conical flask. Add 10 ml.dil. sulphuric acid,

followed by a pinch of sodium bicarbonate /sodium carbonate and finally 10 ml KI (10%). Keep

the flask covered for 5-6 minutes in dark. Appearance of dark brown color indicates the

formation of I2.

Now, titrate it against Na2S2O3 from burette until dark brown colour becomes faint brown. At

this stage, add 2 ml. starch solution, colour of the solution becomes intense blue. Continue

titration drop wise till blue color disappears. Note this point as an end point.



21

Part B: - Determination of chlorine in the given water sample.

Take 25-ml.given water sample in a conical flask. Now add 10 ml. KI (10%).

Appearance of dark brown color indicates formation of I2. Titrate the liberated iodine with

Na2S2O3 solution from burette till brown color of iodine fades, then add 2-ml. of starch solution,

and continue the addition of the Thiosulphate solution until the blue color disappears completely.

Note this point as an end point.

Calculation:-

Given Normality of std. K2Cr2O7 is N1

Part A: -Standardization of Sodium thiosulphate solution.(Hypo.)

Std. K2Cr2O7 v/s Hypo

N1V1 = N2V2

N1x 25 = N2 X V2

N1x 25

N2 = -----------------

V2

= X

Part B: - Determination of chlorine in the given water sample.

Hypo v/s given water sample

N2V’2 = N3V3

N2V’2

N3 = ------------

25

N3 = Y (Normality of given water sample.)



22

Wt./lt. of Chlorine = Normality x Eq. Wt. of Cl 2

= N3 X 35.5

= Z g/l

Result: - The water sample contains --------------g/l of chlorine.



23

EXPERIMENT NO. 5

Aim: - To estimate the amount of Ferrous, Ferric and total Iron present in the given

solution.

Principle:

The amount of ferrous ions present in the given solution containing ferrous and

ferric salt is determined titrimetrically by titrating the solution with standard K2Cr2O7 solution.

Then the Ferric ions are reduced to ferrous with a reducing agent such as SnCl2 and then the

total iron (present now in the ferrous state only) is titrated with standard potassium dichromate.

1) Cr2O72- + 6Fe

2+ + 14H

+ 2Cr

3+ + 6Fe

3+ + 7H2O

2) 2Fe3+

+ Sn 2+

2Fe 2+

+ Sn 4+

In acid solution, when potassium dichromate oxidizes, it it itself is reduced which

may be represented as:

Cr2O7 2-

+ 14H +

+ 6e- 2Cr

3 + + 7H2O

From which it follows that the equivalent wt. of Potassium dichromate is one sixth of

molecular weight i.e ; 294.22/6 = 49.03 g.

Internal indicators used for redox titrations (barring starch used in-iodometric

titrations ) change colour when the oxidation potential of the titrated solution reaches a definite

value. Such colour changes do not depend on the specific properties of the oxidation reduction

or redox indicators. Such indicators include diphenylamine, neutral red, methylene blue,

diphenylaminebenzene sulphonic acid, N- phenylanthranilic acid etc.

Ferric salts are not oxidized either by dichromate or permanganate. They can be

determined in solution only after reduction of Fe3+

to Fe2+

ions. This reduction can be done by



24

reducing agents such as SnCl2. Ferric salt e.g. FeCl3 is reduced by the action of SnCl2 in presence

of hydrochloric acid.

2FeCl3 + SnCl2 2FeCl2 + SnCl4

The excess SnCl2 must be removed completely, as K2Cr2O7 oxidizes it. This is done by

adding mercuric chloride, which reacts with SnCl2 as follows:

SnCl2 + 2HgCl2 SnCl4 + Hg2Cl2

(Excess added) (silky white ppt)

Mecurous chloride (Hg2Cl2) is deposited in the form of silky white precipitate.

The solution is directly titrated with K2Cr2O7.

Following reaction takes place: -

Cl2 + K2Cr2O7 + 14HCl 6FeCl3 + 2KCl + 2CrCl3 + 7H2O

The following points must be taken into consideration in this analysis. The precipitated

Hg2Cl2 can also oxidize by dichromate. However, if only a small excess of SnCl2 is used in the

previous stage, the amount of HgCl2 precipitated is small. It reacts with the dichromate slowly

and no appreciable error is introduced into the result. On the other hand, if a large excess of

Hg2Cl2 is precipitated and, in particular, if it has a grey or dark colour as a result of further

reduction to metallic mercury by the reaction as follows:

Hg2Cl2 + SnCl2 2Hg + SnCl4

Then it is oxidized so rapidly in the titration that the result is quite erroneous.

It follows, therefore, that an important condition for accuracy in this method is the use

of a very slight excess of SnCl2. Stannous chloride is added drop by drop until the reddish



25

yellow/ yellow colour of FeCl3 has completely disappeared, after which one or two more drops

are added.

PROCEDURE:

Part-I: Determination of Ferrous ion

Pipette out 25 ml. of the given ferrous ferric solution in a conical flask, add about 10

ml. of 4N sulphuric acid and 4-5 drops of N- Phenylanthranilic acid indicator. Titrate slowly

with constant shaking of the flask against K2Cr2O7 solution. Appearance of a violet red or brick-

red colouration marks the end point. This titre value corresponds to the ferrous ions present in

the solution.

Part-II: Determination of Total Iron (Fe3+

& Fe2+

)

Pipette out 25 ml. of the given ferrous ferric solution in a conical flask. Add about 10 ml.

of concentrated HCl. Its colour changes colourless to yellow. Heat the contents of the flask to

boiling and to this hot solution add SnCl2 dropwise with continuous stirring till the solution

becomes colourless. Do not add excess of SnCl2 .

Cool the flask rapidly under tap water. Add 5 ml of mercuric chloride, a light silky white

precipitate of mercurous chloride is formed. (if black ppt occurs it indicates that too much SnCl2

was added. In this case discard it.

To the white ppt containing solution add 4-5 drops of N- Phenylanthranilic acid indicator.

Titrate slowly with constant shaking of the flask against Std K2Cr2O7 solution. At the end of

titration faint green colour develops which changes to violet-red or brick-red colour. This

marks the end point. This titre value corresponds to the total iron present in the solution.

Difference between these gives the iron present in ferric state.



26

CALCULATIONS:

Part-I: Determination of Ferrous ion

. Std K2Cr2O7 Vs Fe2+

/ Fe3+

mixture

N1V1 = N2V2

N1V1

N2 = ----------------

V2

Where, N1 and V1 are the normality and volume respectively of the

K2Cr2O7 solution and V2 is the volume of the Fe2

+ / Fe

3+ mixture used during part I.

N2 x 55.85 = Wt./ lit.of Fe2+

= x g.

(Eq. Wt. Of Fe)

Part-II: Determination of Total Iron ( Fe3+

& Fe2+

)

K2Cr2O7 vs. Fe2+

(after reduction of Fe3+

by SnCl2)

N1V1’ = N3V3

N3 = N1V1’

---------------

V3

Where, N1 and V1’ are the normality and volume of the Std. dichromate

solution and V3 the volume of the Iron solution used during Part II.

N3 x 55.85 = Wt. / lit. of total iron = Y g

Wt/lit of Fe3+

= (y-x) g = z g.

Result: The amount of Fe2+

and Fe3+

present in the sample are: ------.



27

EXPERIMENT NO. 06

Aim: Determination of viscosity of a Lubricating Oils at different Temperatures by Red -Wood

Viscometer.

Apparatus: Redwood viscometer (1 or 2), thermometer, energy regulator, etc.

Theory and General Discussion: Viscosity is one of the most important properties of liquid.

Viscosity is a measure of the internal resistance to the motion of a liquid and is mainly due to the

forces of cohesion between the liquid molecules. Viscosity of the liquid may be measured in

several ways, one of which is determining time required for a definite amount of the liquid to

flow through a capillary. Such method includes the use of Saybolt, Engler and Redwood

Viscometers.

Viscosity may be expressed as dynamic (or absolute) viscosity, kinematic viscosity or the

viscosity coined after the name of the apparatus used for its determination.

Dynamic or Absolute Viscosity: Dynamic or absolute viscosity (often denoted by ‘η ’) is the

tangential force per unit area required to maintain unit velocity gradient between two parallel

planes, unit distance apart in the fluid. It can be also be defined as the ratio of shearing stress to

the rate of shearing strain.

If F is the force required to keep moving an particle of surface area A in contact with the

fluid, separated from stationary surface by a thickness D, and moving with velocity V, then

Shearing stress = Force acting on the surface area A F

A= (1)

and rate of shearing strain = V/D (2)

Absolute Viscosity (η ) = Shearing stress/ rate of shearing strain.

/

/

F A F D

V D V A

×=

× (3)

The numerical value of ‘η ’ depends upon the unit used in equation (3). In metric system

the unit is ‘poise’ (F = 1 dyne, A = 1 cm. and V = km/sec.). A smaller unit, the centipoise is more



28

often used. Poise is equal in one dyne.sec/cm2

(Dimensions of absolute viscosity are ML –1

T-1

).

Absolute viscosity is also referred to as ‘coefficient of viscosity’.

Kinematic Viscosity: -

The ratio of absolute viscosity to density for any liquid is known as absolute kinematics

viscosity. It is donated by µ. Its units is stoke. A smaller unit, centistokes (1/100 th of stoke) is

often used.

µ η ρ= (4)

Where,

µ = Absolute kinematic viscosity

η = absolute dynamic viscosity

ρ = density of liquid.

Dimension of µ is L2 T

1

Since the rate, at which a liquid will flow through an apparatus increases as the viscosity

of the liquid decreases the rate of flow through an orifice or short tube may be used as a means

for measuring viscosity. This is the principle involved in the Redwood viscometer, which is used

in UK and India whereas Saybolt’s Viscometer is used in U.S.A. Engler’s Viscometer is used in

Europe. In these viscometers a fixed volume of the liquid is allowed to flow (in case of Redwood

it is 50 ml, Engler-200 ml and Saybolt –60 ml.) through a capillary tube of specified dimension

under given set of conditions and the time of flow is measured at a particular temperature. The

result is usually expressed in terms of the time in seconds taken by fixed volume of the liquid to

flow through the standard orifice of the particular apparatus used.

e.g.: Viscosity of oil is 250 sec using Redwood ( no. 1 or 2) at 270C. This Viscosity so

determined is also called as Relative Viscosity.

Absolute and kinematic Viscosities can also be determined from the relative Viscosity (i.e.

Redwood values) from the equations:



29

µ = Ct (5)

(for liquid whose kinematic viscosity is more than 10 centistokes)

µ = Ct – β / t (6)

(for liquid whose Kinematics Viscosities less than or equal to 10 centistokes)

Where,

µ = kinematic viscosity in centistokes

t = time of flow in seconds

C = viscometer constant

β = coefficient of kinetic energy which may be determined experimentally or

eliminated by choosing long flow-times

Test Viscometer may be calibrated by solutions of known viscosity. The primary

standard used is freshly distilled water, whose kinematic viscosity is 1.0008 centistokes. Other

standards usually employed are-

40% sucrose solution µ = 4.390 cs at 25 0C

60% sucrose solution. µ = 33.66 cs. at 25 0C

For Redwood viscometer No.1, the values for the constants are as below:

Time of flow, t β C

40 to 85 seconds 190 0.264

85 to 2000 seconds 65 0.247

For Redwood viscometer No.2, the values for the constants are β =1120, C= 2.720

Redwood viscometer No. 2 is used for very viscous liquids which require 2000 or

more seconds for specified volume of the liquid to flow.



30

Brief mechanism of lubricants:

The formation of a fluid film of a lubricant between the moving friction surface and the

generation of the frictional heat under the particular condition of load, bearing speed and

lubricant supply is mainly depends upon the viscosity of the lubricant and to some extent on its

oiliness. If the viscosity of the oil is too low, the fluid lubricant film cannot be maintained

between the moving surfaces as a result of which excessive wear may take place. On the other

hand, if the viscosity of the oil is too high, excessive friction due to the shearing of oil itself

would result. Hence in hydrodynamic lubrication, the lubricant should posses the proper

viscosity. So, it is of vital importance to have knowledge of the viscosity of the lubricating oil.

Procedure:

1) Level the instrument with the help of spirit level.

2) Put the valve rod at its position (i.e. in the concavity at the bottom of

the cup) to close the opening of orifice.

3) Keep clean dry Kohlrausch’s flask centrally below the jet.

4) The test sample is poured into the oil cup and adjust the level of oil, with that of

the pointer.

5) Cover the oil cup with the lid and insert thermocouple / thermometer of desired

range into the oil cup from the thermometer bracket. Care

should be taken that it dose not touch the bottom of the cup.

6) Fill up the water bath with water and adjust the level of water with that

of the pointer in the oil cup. Let the oil in the cup to attain the room temperature.

7) Lift up the value rod and suspend it from the thermometer bracket and start the

stop watch simultaneously.

8) Receiving flask is so located that oil strikes the flared mouth and does not drop

directly into the opening, which would cause foaming.

9) When the level of the oil reaches 50 ml. mark in the neck of the flask,

stop the stop-watch.

10) At the same time close the passage of the orifice by keeping the valve

rod at its original position to prevent any overflow of oil.

11) The time elapsed in seconds is the relative viscosity of oil at room temperature.



31

12) Again, pour oil into the cup and adjust oil level a little lower than the pointer.

13) Check the oil level and draw or add a little oil so as to be exactly at the pointer.

14) Switch on the water bath adjust the knob of the so as to heat the water and raise

the temperature of the oil to say 85 oC or 80

oC.

15) When the temperature of sample has become steady at the desired value lift the

valve rod and suspend it from the thermometer bracket and start the stopwatch

simultaneously.

16) When the level of the oil touches the 50 ml. mark on the neck of the flask stop

the watch. The time elapsed in seconds in the relative viscosity of oil at elevated

temperature. (85 oC or 80

oC as the case may be)

17) Now take out some hot water from the water bath and add equal amount of cold

water to bring the temperature of the bath.

20) Pour oil into the cup and adjust oil level as described earlier.

21) Stir the contents of the bath and cup regularly.

22) When the temperature of sample has become steady at the desired temperature 75

oC or 70

oC. Find out the relative viscosity of test sample at this temperature as

described earlier.

23) Repeat the procedure to find out relative viscosities of oil at

65 0C, 55

0C, 45

0C, respectively.

24) Record all the results in the tabular form.

25) Plot graph corelatively (a) Viscosity and temperature, (b) Log of

viscosity and temperature and (c) Kinematic viscosity and temperature

. (d) Log of Kinematic viscosity and temperature, (e) Find rate of change of

viscosity with respect to temperature.



32

Observation table: Redwood viscometer no.:

=

=

c

β

Sr.

No.

Temperature

(T) in 0C

Time in second

(Relative

Viscosity)

(r

η )

Kinematic

Viscosity ( )µ

in centistroke

Log (r

η ) Log ( )µ

1 Room Temp.:

2

3

4

5

6

7

8

Calculation:

Plot the following graphs

1. Relative viscosity (r

η ) against the temperature of sample (T)

2. Log of relative viscosity (r

η ) against the temperature of sample (T)

3. Kinematic viscosity ( )µ against the temperature of sample (T)

4. Log of kinematic viscosity ( )µ against the temperature of sample (T)

and also determine the slope of the graph 1

Result :-The rate of change of relative viscosity with respect to temperature was found to be

equal to ------------



33

EXPERIMENT NO. 07

Aim: Determination of Flash Point of given sample by using flash point apparatus.

Theory:. The Flash Point of sample may be defined as the minimum temperature to which it

must be heated to give off sufficient vapors to ignite momentarily when a flame of standard

dimensions is brought near the surface of the sample for a prescribed rate in an apparatus of

specified dimensions. This is detected by the appearance of momentary flash upon the

application of standard flame over the surface of sample. The Flash Point is defined as closed

cup or open cup flash point accordingly as the apparatus for the determination of flash point of

sample is provided with a cover the sample cup or not. This parameter is not related to the

temperatures of the ignition source or of the burning liquid, which are much higher.

The mechanism of the appearance of the flash can be explained in the following manner.

Every flammable liquid has a vapor pressure, which is a function of that liquid's temperature. As

the temperature increases, the vapor pressure increases. As the vapor pressure increases, the

concentration of evaporated flammable liquid in the air increases. Hence, temperature determines

the concentration of evaporated flammable liquid in the air. Each flammable liquid requires a

different concentration of its vapor in air to sustain combustion. The flash point of a flammable

liquid is the lowest temperature at which there can be enough flammable vapour to ignite, when

an ignition source i.e. flame of suitable dimension is applied.

Importance of flash point from view of lubricants:

A good lubricating oil should not volatilize under the working temperatures. Even if

some volatilization takes place, the vapors formed should not form inflammable mixture with air

under the condition of lubrication. From this point of view, the flash point of a lubricating oil are

of vital importance.

Lubricating oil selected for a job should have a flash point which is reasonably above its

working temperature. This ensures the safety against the fire hazards during the storage,

transport and use of the lubricating oil. This test is immense importance for illuminating and

Lubricating Oils. This helps in detecting the highly volatile constituents of the oils. If they are

highly volatile at ordinary temperature, the issuing vapors may cause fire hazards. So to ensure



34

safety, certain minimum temperatures are laid down for fuels and Lubricating Oils below which

they should not give off adequate vapors to make them burn.

Outline of the Method:

The sample is placed in the cup of apparatus and heated at a slow uniform rate about

2 oC . A small test flame is directed into the cup at regular intervals, and the lowest temperature

at which application of the test flame causes the vapor to ignite momentary, with a distinct flash

inside the cup, is recorded as Abel’s Flash Point.

Description of the apparatus:

The Abel’s apparatus consists of the following parts: -

a. SAMPLE CUP: This is a cylindrical vessel with a lid. Within the cup, near the top; there

is a sample level mark.

b. COVER: The cup is provided with a close fitting cover. The cover is provided with a

thermometer socket, to support an oil test lamp, a movable metal bead. The top of the cover has

three rectangular holes which are covered on uncovered by moving the slide which has two

perforations to allow air to enter the sample up and bring contact between vapours of the sample

and the test-lamp when in open position.

A metal bead, the dimension (4mm) of which represents the size of test flame, is mounted in the

cover. The apparatus is also provided with a stirrer.

(C) HEATING VESSEL: - The heating vessel consists of copper vessel and placed

coaxially, which serves as a water jacket. When the sample cup is placed into the hole at the top

of the vessel, it fits into it and leaves an air gap between itself and the outer copper water vessel.

The water can be heated electrically or with a burner or spirit lamp. Thermometer is

provided with the apparatus for the measuring the temperature of the sample.

Procedure: Fill the given sample in such a way that the sample level is exactly upto the mark in

the cup. Fix the cup into the apparatus and cover with lid. Insert the thermometer in the

thermometer holder given in the cup in such a manner that it will not directly touch the lower

bottom of the cup and the paddle stirrer inside the cup. Fill the water bath with the cold water.

Close the sliding shutter and light the standard flame. Adjust the size of flame (4 mm diameter)

with respect to the metal beed. Stir the oil using paddle stirrer. Introduce the flame by opening



35

the shutters and check the appearance of the flash. Now heat the apparatus and set the rate of

temperature increase at the rate of 1 to 2 oC per minute. Check the flash point of given sample at

the interval of 3 oC rise in the temperature. Discontinue the stirring the sample during the

introduction of the test flame. On observing a flash, stop the heating process and allow the

temperature to decrease. Check the occurrence of a flash at every 1 oC drop in temperature.

Record the lowest temperature at which the flash is observed as the flash point of the sample.

Observation:

Table 1

============================================

S. No. Increasing Inference

Temperature (0C) (No flash or flash observed)

============================================

1. Room temp

2.

3.

4.

5.

===============================================

Table 2

=================================================

S. No. Decreasing Inference

Temperature (0C) (Flash observed or no flash)

=================================================

1.

2.

3.

4.

5.

================================================

Result: The Flash Point of given sample determined by Able’s Flash Point apparatus is found to

be ------0C



36

EXPERIMENT NO. 08

Aim: - Determination of Flash Point of a sample by using Pensky Martin’s ‘closed’ cup

apparatus.

Description of the Apparatus: A Pensky Martin’s apparatus consists of the following parts:

SAMPLE CUP: Sample Cup is a cylindrical vessel, made of brass with a filling mark grooved

inside near the top. It is provided with a lid.

LID: The lid is equipped with the following parts:

i) Stirrer: The stirring device consists of a vertical steel shaft mounted in the center of the

cup and carrying two-bladed brass propellers.

ii) Cover: It has four openings: one for thermometer and the rest for the oxygen entry and

exposure of vapors to test flame.

iii) Shutter: The lid is equipped with a brass shutter operating on the plane of the upper

surface of the cover proper. The shutter is so shaped and mounted on the lid that when in one

position, the holes are completely closed and when in the other, these orifices are completely

opened.

iv) The Flame Exposure Device: The lid is equipped with a lamp with such a mechanism that

its flame operates simultaneously with the shutter. When the shutter is in the ‘open’ position, the

tip is lowered down in the center of the central orifice.

v) Heater: The cup is heated by means burner or it is electrically heated. The air bath has

cylindrical interior about four cm. Deep and can be heated by a direct flame or an electric

resistance element. The top-plate is also made of metal and mounted with an air gap between it

and the air bath.



37

Procedure: The cup and its accessories are well cleaned and dried before the test is started. Now

the cup is filled with the sample to be tested up to the level indicated by the filling mark and

covered with the lid. The cover incorporating the stirring device, the thermometer and flame

exposure device is fixed on the top. The cup is now set in the apparatus properly and the

thermometer inserted. The test flame is lighted and adjusted until it is the size of a bead

(approximately 4mm in diameter). The apparatus is heated so that the heating rate is maintained,

with the help of a rheostat, at about 5 0C per minute and stirring rate at one-two rps.

Once the heating is started, the test flame is applied after each 2 0C rise of temperature

nearer to the sample in the cup by opening the shutter and check the appearance of the flash.

On observing a flash, stop the heating process and allow the temperature to decrease.

Check the occurrence of a flash at every 1 oC drop in temperature. Record the lowest

temperature at which the flash is observed as the flash point of the sample.

Observation:

Table 1

Sr. No.

Increasing Temperature

in o

C

Observation

(No flash or flash

observed)

1

2

3

4

5

6

7

8



38

Table 2

S. No. Decreasing Temperature

in o C

Observation

(Flash observed or No

flash)

1

2

3

4

5

6

RESULT: The Flash Point of given sample determined by Pensky Marten’s Flash Point

apparatus is found to be ---------0C



39

EXPERIMENT NO. 09

Aim: Determination of Flash Point and fire point of the given sample by Cleveland’s open cup

apparatus.

Definitions:

I) Flash Point: Defined earlier.

II) Fire point: It is defined as the minimum temperature at which the oil gives

off sufficient vapor which causes it to burn for a period of five seconds or more when a test

flame is passed over surface of the oil.

Description of Apparatus:

The apparatus consists of a test cup made of without any lid and is equipped with a

handle. The cup is supported by a metal plate known as heating plate. The cup may be heated by

an electric heater mounted below the cup in the apparatus itself.

The metal plate has an extension for mounting the test flame and the thermometer

support. The test flame is mounted in such a manner as to permit automatic duplication of the

sweep of the test flame over the sample cup. The size of the flame can be adjusted with respect to

the dimension of metal beed. (approximately 4 mm)

Procedure:

The apparatus is thoroughly cleaned and the thermometer is suspended in such a way so

that the bottom of the thermometer bulb just above the bottom of sample cup.

The cup is now filled with sample up to the filling mark grooved on the inner side of the

The test flame is adjusted to have a flame diameter approximately to the dimension of metal

beed. Now move the test flame over the sample cup and check the appearance of flash over the

sample inside the cup.

If no flash observed, increase the temperature of the sample taken and take a flame over

the cup after every 2 0C increase. On observing a flash, stop the heating process and allow the



40

temperature to decrease. Check the occurrence of a flash at every 1 oC drop in temperature.

Record the lowest temperature at which the flash is observed as the flash point of the sample.

Observation:

Table 1

Sr.

No.

Increasing Temperature

in o

C

Observation

(No flash or flash

observed)

1

2

3

4

5

6

7

8

Table 2

S. No. Decreasing Temperature

in o C

Observation

(Flash observed or No

flash)

1

2

3

4

5

6

Result: Flash Point of given sample determined by Cleveland’s apparatus is found to be

----- ----0C.



41

EXPERIMENT NO. 10

Aim : Determination of Chemical Oxygen Demand in a waste water sample.

Apparatus:

Reflux apparatus, consisting of a 250 ml.-flat bottomed Borosil flask with ground glass joint

and a condenser.

Regents required:

1. Standard 0.25 N K2Cr2O7 solution

2. Sulfuric acid-silver sulphate reagent:

3. Standard 0.1 N Fe (NH 4) 2SO4. 7H2O solution:

4. Ferroin indicator

5. HgSO4

N

N

Fe

+2

. SO-4-2

Figure: Structure of ferroin indicator

Theory: In Environmental Chemistry, the Chemical Oxygen Demand (COD) test is commonly

used to measure the amount of organic compounds that is susceptible to oxidation by a strong

oxidant present in water. Any organic matter such as hydrocarons when brought in contact with

oxygen is oxidised to get CO2 and H2O. If the organic matter contains hydrogen, nitrogen,

sulphur, etc. in addition to carbon, then the chemical oxidation leads to formation of CO2, H2O,

NO2, SO2 and etc. To get thos oxidation, oxygen is required and demanded by this organic



42

material. The possibility of meeting such demand is through dissolved oxygen in the water. Thus

dissolved oxygen in water gets depleted and survival of the bio-organisms in such water is

difficult and sometimes impossible. It is therefore, essential to check COD of effluent water

before letting it out from industr. It is expressed in milligrams per liter (mg/L), which indicates

the mass of oxygen consumed per liter of solution. Older references may express the units as

parts per million (ppm).

The basis for the COD test is that nearly all organic compounds can be fully oxidized to

carbon dioxide with a strong oxidizing agent under acidic conditions. The amount of oxygen

required to oxidize an organic compound to carbon dioxide, ammonia, and water is given by:

This expression does not include the oxygen demand caused by the oxidation of ammonia

into nitrate. The process of ammonia being converted into nitrate is referred to as nitrification.

The following is the correct equation for the oxidation of ammonia into nitrate.

The second equation should be applied after the first one to include oxidation due to

nitrification if the oxygen demand from nitrification must be known. Dichromate does not

oxidize ammonia into nitrate, so this nitrification can be safely ignored in the standard chemical

oxygen demand test.

Using potassium dichromate

Potassium dichromate is a strong oxidizing agent under acidic conditions. (Acidity is

usually achieved by the addition of sulfuric acid.) The reaction of potassium dichromate with

organic compounds is given by:



43

where d = 2n/3 + a/6 - b/3 - c/2. Most commonly, a 0.25 N solution of potassium dichromate is

used for COD determination, although for samples with COD below 50 mg/L, a lower

concentration of potassium dichromate is preferred.

In the process of oxidizing the organic substances found in the water sample, potassium

dichromate is reduced (since in all redox reactions, one reagent is oxidized and the other is

reduced), forming Cr3+

. The amount of Cr3+

is determined after oxidization is complete, and is

used as an indirect measure of the organic contents of the water sample.

Importance of Blank Titration

Because COD measures the oxygen demand of organic compounds in a sample of water,

it is important that no outside organic material be accidentally added to the sample to be

measured. To control for this, a so-called blank sample is required in the determination of COD

(and BOD, for that matter). A blank sample is created by adding all reagents (e.g. acid and

oxidizing agent) to a volume of distilled water. COD is measured for both the water and blank

samples, and the two are compared. The oxygen demand in the blank sample is subtracted from

the COD for the original sample to ensure a true measurement of organic matter.

Interference due to Chlorides in waste water

Some samples of water contain high levels of oxidizable inorganic materials which may

interfere with the determination of COD. Because of its high concentration in most wastewater,

chloride is often the most serious source of interference. Its reaction with potassium dichromate

follows the equation:

Prior to the addition of other reagents, mercuric sulfate can be added to the sample to

eliminate chloride interference.



44

Procedure:

Part –I Determination of COD value of the given waste water sample

Pipette out 10 ml of the wastewater sample in a reflux flask, dilute the sample by 20

ml with distilled water. Mix well and add 10 ml.of 0.25 N K2Cr2O7 solutions. Drop some

pumice stones and slowly add 30 ml. of H2SO4—AgSO4 reagent while continuously swirling the

flask. If the colour changes to green, add more K2Cr2O7 and or alternatively discard the solution

and take a fresh sample with lesser aliquot.. Mix the contents of the flask thoroughly. Connect

the flask to the condenser and slowly heat flask. Reflux for at least 2 hours. Cool and wash down

the condenser with distilled water such that the washings fall in to the flask. Add few drops of

Ferro in indicator and titrate the content of the flask (un reacted K2Cr2O7) with the 0.1 N FAS

solution. Near the end point the yellow colour of the solution gradually fades up to bluish green,

continue the titration till bluish green colour changes to wine red color.

Part-II Blank Titration

Repeat the above procedure using 10 ml of the distilled water instead of the

Waste water sample for about half hour.

Part-III Standardization of the FAS solution

Pipette out 10 ml of the FAS solution in a conical flask add few drops of the Ferroin

indicator and titrate with the standard K2Cr2O7 solution till solution changes color from yellow to

wine red colour.

Calculation:

Part A] Standardization of FAS solution:

Std. K2Cr2O7 = FAS solution

N1V1 = N2V2

1 12

2

N VN

V=



45

Part B] Determination of COD of given water sample:

The COD value of the sample is calculated as follows:

4 3 2(V -V ) N 8COD in mg/l = 1000

V

× ××

Where,

V4= Volume of the FAS run down in the blank Experiment.

V3 = Volume of the FAS run down in the test Experiment.

N2 = Normality of FAS solution

V = Volume of the test sample taken.

Result:

The Chemical oxygen demand of the given waste water sample was found to be equal to

__________ppm.



46

Experiment No. 11

Aim: - To determine Acid Value or Neutralization number of given oil.

Introduction:

The Acid Value of an oil is defined as the number of milligrams of potassium

hydroxide required to neutralize the free acid present in 1 gram of the oil.

The presence of mineral acids in oil is so rare that it is almost unnecessary to look for it,

unless the oil is refined in a faulty manner. Free organic acids or acidic bodies are always found

in Oil, whether they be pure mineral oils or compounded oils with fatty oils. In unused refined

petroleum oils, the quantity is invariably negligible.

In good oils, the Acid Value should be very low (<0.1). Increase in Acid Value should be

taken as an indicator of oxidation of the oil, which may lead to gum and sludge formation

besides corrosion.

The Acid Value of fatty oils may vary from 0.2 to 50 and it shows the extent of hydrolysis

of glycerol ester of the oil.

Regents required:

(1) KOH solution (2) Neutral ethyl alcohol

(3) Phenolphthalein indicator.

Procedure:

Part-I: Standardization of KOH Solution

Take 10ml of oxalic acid solution. Add few drops of phenolphthalein indicator.

Titrate this with KOH solution from burette till there is a colour change. Note

the end point as V2 .



47

Part-II: - Determination of Acid value of Solution

Weight out accurately about 5 g. of the oil under test into a 250 ml conical flask and add

50 ml of neutral alcohol. Heat the flask over a water bath for about 30 minutes. Cool the flask

and the contents to room temperature and add a few drops of phenolphthalein indicator. Titrate

with the standard N/10 KOH solution until a faint permanent pink color appears at the end point

as V2’.

Part-III: - Blank titration

Repeat the same procedure without taking lubricating oil.

Reaction:

C17H35COOH+KOH ---------------- > C17H35COOK+H2O

Calculations:

Part-I: Standardization of KOH Solution

Std Oxalic acid Vs KOH

N1V1 = N2V2

N1V1

Normality of KOH i e; N2 = ----------------

V2



48

Part-II: - Determination of Acid value of Solution

Normality of KOH x V2’ x eq. wt. of KOH

Acid value = ---------------------------------------------------------

Weight of the oil taken in g

Result: The Acid Value of the given oil is found to be_________.



49

Experiment No.12

Aim: - Determination of saponification Value of Oil.

Introduction:

The saponification value of oil is the number of milligrams of potassium hydroxide required to

saponify one gram of the oil. The vegetables or animal oils are esters of fatty acids and glycerol.

They react with KOH to form the potassium salts of fatty acids. The saponification value of oil is

determined by refluxing a known quantity of the sample with a known excess of standard KOH

solution and determining the alkali consumed by titrating the unreacted alkali.

Reactions:

CH2OOCC17H35 CH2OH

| |

CHOOCC17H35 + 3KOH --------�� CHOH + 3C17H35COOK

| |

CH2OOC17H35

CH2OH

Fat

Stearic acid glycerol potassium steareate

Reagents required

(1) N/2 std Na2CO3 solution

(2) alcoholic KOH solution

(3) HCl solution



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Procedure:

Part-I: Standardization of HCl solution

Pipette out 10 ml of std. borax solution in a conical flask. Add phenolphthelin as an indicator.

Titrate it with HCl solution till pink colour turns to colourless.

Part-II: - Saponification Value of oil

Weight out accurately about 5 g. of the oil into a 250 ml conical flask and add

50ml of N/2 alcoholic KOH solution to the flask. Reflux the contents over a water bath for

about 1.30 hr using water condenser. Wash the inner wall of the condenser with a small amount

of distilled water and collect the washing into the same conical flask. Remove the flask from

the water bath cool the flask and the contents to room temperature and add a few drops of

phenolphthalein indicator. Titrate with the standardised N/2 HCl solution until solution

becomes colourless.

Note the end point as V2

Part-III: - Blank titration

Perform the blank experiment by repeating the entire procedure similarly but without using the

oil and titrate with the same standardized N/2 HCl as described above.

Note the end point as V2 ’.



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Calculation:

Part-I: Standardization of HCl solution

Standard Borax solution Vs HCl

N1V1 = N2V2

N1V1

Normality of HCl i e; N2 = ----------------

V2

Part-II: - Determination of Saponification value

Actual volume of KOH Consumed

A] Actual volume of KOH required can be found by:

HCl required for the titration of oil and alc. KOH is refluxed is------- V2

HCl required for the titration of only alc. KOH is (blank) ----------------V2’

HCl used corresponding to used KOH is = (V2’- V2) ml

= V ml.

B] Calculation of normality of KOH:

KOH solution Vs HCl solution

NKOH X VKOH = NHCl X VHcl

NHCl X VHcl

NKOH = ------------------------------------

VKOH

C] Calculation of volume of HCl:



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HCl Vs used KOH

N2 x V = N3 x V3 (?)

V3 = N2 x V / N3

V3 = ---------ml.

V3 ml of KOH is used for saponification for x g of oil

Say Normality of KOH =----- N

Wt / litre = N KOH x 56

= z g

1000 ml of KOH = z g of

V3 ml of KOH = z x V3

---------

1000

= b g of KOH

b

Normality per gram of oil is = --- g

x

b

= ---- x 103

mg

x

Result: Therefore the Saponification value of the given oil is found to be_________.

found to be_________.

Chem Lab Manual 0910

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Transcript of Chem Lab Manual 0910