Chemical properties Testing of Leather revised PPT

methods used to test the chemical properties of leather

1. TESTING OF LEATHER
ANALYTICAL CHEMICALS

2. Chemical Testing in Leather Technology
Chemical Testing
In chemistry, a chemical test is a qualitative or quantitative
procedure designed to prove the existence of, or to quantify, a
chemical compound or chemical group with the aid of a specific
reagent.

3. Purposes of chemical testing
•Determine if, or verify that, the requirements of a specification,
regulation, or contract are met
•Decide if a new product development program is on track
•Demonstrate the utility of a proposed patent
•Determine the interactions of a sample with other known substances
•Determine the composition of a sample
•Provide standard data for other scientific, medical, and Quality
assurance functions
•Validate suitability for end-use
•Provide a basis for Technical communication
•Provide evidence in legal proceedings

4. Chemical testing can be divided into a few overlapping
broad fields, to name most:
• Inorganic
• Organic
• Qualitative
• Quantitative
• Biological
• Wet or "classical"
• Instrumental

5. Quantitative and Qualitative Analysis
By definition, analytical chemistry has two sides:
Qualitative analysis Quantitative analysis
Focuses on what is present in a samplefocuses on how much
Dealing with the identification of elements or
compounds in an unknown substance.
determining the quantity of a
particular chemical is in a substance
"Does the water contain benzene?" or "Does
it contain arsenic?"
A flame test, in which a drop of solution is
held over a flame, the colour of which is
examined to determine the predominant
metal ions in the solution.
Classical methods in the lab include
careful measurements of sample
weights or volumes before and after
reactions, titrations, and so on.
"spot tests". These experiments involve
placing one drop of a particular reagent with a
drop of an unknown sample and observing
any changes: color, bubbling, precipitation,
and the like.
modern methods use extremely
sophisticated techniques of ultra-
high-power microscopy,
spectroscopy, electro- and thermo-
chemistry, and so on.

6. Titrations
Main Idea: Titrations are an application
of acid-base neutralization reactions
that require the use of an indicator.
6

7. Titration
•Titration is a method for determining the
concentration of a solution by reacting a known
volume of that solution with a solution of known
concentration.
•If you wish to find the concentration of an acid
solution, you would titrate the acid solution with
a solution of a base of known concentration.
•You could also titrate a base of unknown
concentration with an acid of known
concentration.
7

9. ACID –BASE TITRATION
•An acid-base titration is the determination of the concentration
of the acid or base by exactly neutralizing the acid or base with
an base or acid respectively of known concentration.
•This is a quantitative analysis which involves neutralization
reaction.
•Before starting the titration suitable pH indicator must be
chosen. The equivalence point of the reaction ,the point at
which equal amount of the reactants have reacted will have a
pH dependent on the relative strengths of the acid and the base
used.

10. •The pH of the equivalce point can be estimated using
the following rules:



•The point at which the indicator changes colour is called
the end point.A suitable indicator must be chosen that
will experience a change in colour close to the endpoint.
Nature of
the acid
Nature of the
base
Nature of the
solution
pH Range of the
solution
Strong Strong Neutral 7
Strong Weak Acidic <7
Weak Strong Basic >7

11. Principle
An acid base titration is when we add a base to an
acid or an acid to the base ,until the equivalence
point is reached where the moles of acid equals the
moles of base .

12. PROCEDURE:
•Rinse the burette, pipette and conical flask with distilled
water.
•A known volume of the unknown solution is pipetted out into
the conical flask along with a small amount of the chosen
indicator.
•The burette should always be filled to the top of its scale with
the known solution for the ease of reading.
•The known solution should then be allowed into the conical
flask out of the burette
• The solution is let out of the burette until the colour changes
and the value on the burette is recorded.

13. •This titration is carried out similarly three
times and the burette readings are noted
each time and their average is taken to
calculate the final result using the following
formula:
According to the law of volumetric analysis,

V
1 = Volume of the base
N
1
= Normality of the base
V
2 = Volume of the acid
N
2= Normality of the acid
V
1N
1=V
2N
2

14. Chemicals in the beam house operations
•The steps in the production of leather between curing and
tanning are collectively referred to as beamhouse operations.
•They include soaking, liming, removal of extraneous tissues
(unhairing, scudding, and fleshing), deliming, bating and
pickling.
Pretanning Chemicals
•Water
•Salt
•Lime
•Sodium sulphid
•Ammonium salts
•Analysis of Deliming acids, bates, pickling acids and
neutralising agents

15. 1. Analysis of water
Physical parameters
•pH
•Hardness
•Conductivity (TDS)
•Turbidity
•Calcium, Magnesium, Sodium, Potassium, Iron, Chloride
•BOD
•COD

What is the effect of water quality on leather?

16. Turbidity
•Turbidity is the amount of particulate matter that is suspended in
water. Turbidity measures the scattering effect that suspended
solids have on light: the higher the intensity of scattered light, the
higher the turbidity.

17. Material that causes water to be turbid includes:
•Clay.
•Silt.
•Finely divided organic and inorganic matter.
•Soluble coloured organic compounds.
•Plankton.
•Microscopic organisms.
The turbidity of sample solution can be measured by using
Nephelometer. It is usually measured in nephelometric turbidity
units (NTU) or Jackson turbidity units (JTU).

18. Principle
This method is based on a comparison of the intensity of light
scattered by the sample under defined conditions with the
intensity of light scattered by a standard reference suspension
under the same conditions. The higher the intensity of scattered
light, the higher the turbidity. Commonly used primary standard
reference suspension is Formazin polymer. The turbidity of a
specified concentration of formazin suspension is defined as 400
NTU.

19. Equipment
Any approved model of turbidimeter can be used to test
turbidity.
Reagents
Ether of the following standards can be used:
• Stock primary standard formazin suspension
• Secondary standard.

20. 1.Remove the dust cover from the instrument. 
Turn on the instrument and allow it to warm up for at least 30
minutes.
2.Inspect and clean sample cells.
3.Rinse the sample cell three times with the sample water.
4.Fill the sample cell with approximately 25 mL of the test sample.
5.Hold the cell up to the light to check for small air bubbles on the
walls.
6.Select the range.
7.Insert the cell riser, if necessary.
8.Standardize the instrument each time the measurement range is
changed to ensure accurate results.
9.Place the sample cell into the sample compartment and cover with
the light shield.
10.The instrument can be left on when readings are performed
periodically.
Procedure

21. pH
•In chemistry, pH is a measure of the acidity or basicity of an
aqueous solution. pH is a logarithmic measure of hydrogen ion
concentration, originally defined by Danish biochemist Søren
Peter Lauritz Sørensen in 1909 .
pH = -log[H+]
•Solutions with a pH less than 7 are said to be acidic and solutions
with a pH greater than 7 are basic or alkaline. Pure water has a
pH very close to 7.
•Measured by pH meter.

22. Samples pH
Tap water 7
Sea water Varies from 7.5 -8.4
Well water Varies from 6.5-8.5
Distilled water 5.8
pH of various samples
Water molecule can be dissociate
H2O H+ + OH-

23. Procedure
1.Calibrate the instrument according to the manufacturer
instruction
2.Place about 200 ml of water sample in a clean a clean 250
ml beaker.
3.Rinse the electrode with distilled water, wipe dry with soft
clean tissue and immerse in the sample
4.Allow the reading to stabilize then read the pH
5.As temperature does affect pH reading the instrument
normally has a built in temperature compensation device.

24. Electrical Conductivity
•Electrical conductivity (EC) is a measure of how conductive the
water is to electrical current.
•an indirect measure for finding the total dissolved solids in a water
body.
•Pure Water (distilled water ) does not contain any solids and does
not conduct electricity.
•An electrical current will not flow in pure water due to infinite
resistance to electron movement.
•Conductivity increase as the dissolved salt concentration increase
that means the water resistance decreases.

25. •TDS (ppm) = Conductivity { (μS/cm) x Conversion Factor}
•The conversion factor is between 0.46 and 0.9 (depending on
the unique mixture of the dissolved materials).
•A widely accepted conversion factor is 0.67.
•A siemens unit is a reciprocal of ohm (the unit of resistance).
Type of salt Conversion factor
NaCl 5
NaSO4 8.5
River water 7.2

26. Water hardness
•The amount of calcium and magnesium compounds dissolved in the
water.
•Measured is the milligram equivalent per litre(mval/l).
1 mval/l = 1 mmol/Z.
(Z = absolute value of the electrochemical valence of alkaline
earth ions).
•Total hardness (TH)
•Temporary or carbonate hardness (CH)
Consists of: calcium- and magnesium hydrogen carbonates, (=
bicarbonates), -carbonates
•Permanent or non-carbonate hardness(NCH)
Consists of: calcium and magnesium chlorides, sulfates,
silicates, nitrates and humates.

27. Water softening
•Heating - Hardness due to carbonates is reduced to about 2
German degrees of hardness.
•By precipitation and with lime or caustic soda: hardness due to
separation carbonate is reduced to about 2 German degrees of
hardness. With soda: total hardness is reduced to1–2 German
degrees of hardness.
•By ion exchange with per mutates, phenolic resin bases:
complete desalting is achieved in most cases.
•By complexing With polyphosphates or organic polyacids,
e.g.agents Trilon types: total hardness is removed.

28. 2. ANALYSIS OF SALT
A.Moisture Content
B.Percent insoluble matter in water
C.Solubility
D.Sodium Chloride content
E.Na
+
AND Cl
-

29. Analysis of common and table salt
Common salt is a mineral substance composed primarily of
sodium chloride (NaCl), a chemical compound belonging to the
larger class of ionic salts; salt in its natural form as a crystalline
mineral is known as rock salt or halite.
Difference between Sea Salt and Table Salt
Sea salt and table salt differ from each other, right from the
method of extraction to the physical properties such as the
color, texture, taste, and flavor.

30. Sea salt
Source of Extraction
Extracted from the seas and
oceans.
Chemical Properties
Extracted from the rock salt (halite),
that is formed on the mineral beds due
to drying of water bodies.
Sodium Chloride: 97%
Potassium Chloride: 2%
Trace Minerals: 1%
Sodium Chloride: 97.5 - 99%
Anti-caking Agents: 1 - 2.5%
Texture
Has a lot of coarseness in the
texture
It has a fine texture, which makes
it easier to mix.
Table salt

31. Determination of matter insoluble in water
Apparatus
Petri dish, analytical balance, glass crucible, graduated flask
Procedure
About 0.585 gm of the sample shall be accurately weighed and
dissolved in 200 ml of water in a beaker it shall be heated to boil
and cooled.
 Filter the solution by filter paper and washed the residue till it
is free from soluble salt.
The filtrate and washing shall be cooled in a one liter
graduated flask and dillute to the mark
Preserve the solution for subsequent tests
The crucible filter paper along with the insoluble residue shall
be dried to constant mass

32. Moisture Content of Salt
Scope
This standard specifies the method for the determination of
moisture in a salt.
Apparatus
Moisture analyzer at 140
0
c and mortar and pestle
Procedure
About 100 g of the common salt shall be rapidly ground to a
size of 2.8mm.
Place the powder in airtight container
About 20 gram of the material shall be loaded 1n moiture
analyzer at 140 -150
0
c

33. A. SODIUM CHLORIDE CONTENT
•AIM :
To detrmine the purity of the common salt used for curing of
hides and skins .
•REAGENTS REQUIRED :
o0.1 N AgNO
3
Solution
o1 g of comman salt
oPottasium chromate indicator
•PRINCIPLE :
The salt is made into a solution and is titrated with silver
nitrate solution using pottasium chromate indicator .The
experiment involves the principles of Argentometric titrations.
The following reaction takes place :

34. In this reaction silver ions react with the chloride ions and forms
silver chloride which is a white precipitate. Potassium chromate
is used as an indicator in this reaction. Silver ions can combine
with chromate (VI) ions to produce a red precipitate of silver
chromate.

When both chloride and chromate ions are present ,the white
silver chloride appears first .When all the chloride ions have
reacted ,the chromate ions react and a red precipitate appears.
Thus the appearance of the red silver chromate can be used to
indicate the end point of the titration.
Ag
+
+ CrO
2-
Ag
2
CrO
4
(Red Precipitate )
AgNO
3
+ NaCl AgCl + NaNO
3

35. PROCEDURE :
o1 g of the common salt is exactly weighed and is made upto
250 ml in a std flask.
o20 ml of the solution is pipetted out into a clean ,dry conical
flask.
oA few drops of potassium chromate indicator is added to the
conical flask.
oThe titration is carried out with the 0.1 N AgNO
3 taken in the
burette.
oThe end point is the appearance of the brick red colour
precipitate .
oNote the titre value ( x ml )

36. CALCULATION :
1000 ml of 1 N AgNO
3
= 1 equivalent weight of NaCl = 58.45 g
V ml of 0.1 N AgNO
3 = 58.45
x 0.1 V = y (g) of NaCl
1000 x 1
20 ml contains = y (g) of NaCl
250 ml contains = y (g) x 250 = z (g)
20
y (g) x V total = z (g)
V aliquote
% of NaCl in the given Salt Sample = z (g) x 100
1 (g)

37. B. Na
+
AND Cl
-

•AIM :
To determine the amount of Na
+
and Cl
-
in the given salt.
•PRINCIPLE :
The procedure to calculate the amount of Na
+
and Cl
-
is the
same as that of the determination of the purity of the salt
•CALCULATION :
1000 ml of 1 N AgNO
3
= 1 equivalent weight of Na
+
= 23 (g)
1000 ml of 1 N AgNO
3
= 1 equivalent weight of Cl
-
= 35.45 (g)

38. 3.ANALYSIS OF LIME
A . PURITY OF LIME MIN. 70% CaO
B . CHECK FOR AVIALABILITY OF LIME DURING STORAGE
CaO MAY HAVE BECOME CONVERTED TO CaCO3 ON EXPOSURE TO AIR
C . TOTAL BASES

39. The name "lime" is used for both calcium oxide (quicklime), and
calcium hydroxide (slaked lime).
Quicklime reacts with water to form slaked lime (calcium hydroxide).
The reaction is highly exothermic.
calcium oxide + water calcium hydroxide.
CaO(s) + H
2
O(l) Ca(OH)
2
(s)
When limestone (calcium carbonate) is heated, at about 1000 °C
it undergoes thermal decomposition.
It loses carbon dioxide and turns into quicklime (calcium oxide).
calcium carbonate calcium oxide + carbon dioxide.
CaCO
3(s) CaO(s) + CO
2(g)
ANALYSIS OF LIME

40. The lime cycle for high-calcium lime.

41. •Slaked Lime Ca(OH)
2 is the most widely used alkali for
liming .
•pH of the saturated solution of lime is 12.4 which is the
most ideal for liming .
•Lime is sparingly soluble in water(0.185g/100cc of water).
•Analysis of lime involves :
A. Acid soluble quick lime détermination
B. Active Lime Alkalinity Determination
C. Total Lime Alkalinity

42. A . Acid soluble quick lime détermination
by EDTA method
•AIM:
To estimate the percentage of
available lime in the given sample
of lime.
•PRINCIPLE:
Commercial lime often contains
insoluble CaCO
3.Thus the purity of
lime is determined by estimating
the percentage of lime in the given
sample by titrating against
standard HCl . Acid –Base Titration
is the principle behind this
experiment.

43. Titration of solutions of Ca2+ With EDTA has an optimal pH
above 8, which we will maintain with an Ammonia /Ammonium
Ion buffer. Under these conditions dibasic
EDTA (H
2
EDTA
2-
)is the active form of the chelating agent.
Thus, the Titration Reaction can be taken as:
Reaction 1:Ca(OH)
2(s)
⇌ Ca
2+
(aq) + 2OH¯
(aq)
Reaction 2:Ca
2+

(aq)
+ EDTA
4-
(aq)

→
CaEDTA
2-
(aq)

(chelate)
Structure of EDTA

44.  REAGENTS REQUIRED :
1.0.5 N HCl
2.Phenolphthalein Indicator
PROCEDURE :
1.370 mg of the lime sample is taken in a conical flask
2.Add 0.5N HCl slowly till the sample dissolves completely
3.Add 200 ml distilled water and boil for 10 minutes with string
4.Cool solution and add 3-5 drops of methyl red indicator adjust
color to orange by adding o.5 N HCl.
5.Transfer solution to 500 ml volumetric flask and make up volume
to the mark.
6.Pipettee 20m ml solution into 500 ml titration flask
7.Adjust pH 12-13 by adding 0.5 N NaOH
8.Add 0.1 g indicator mixture ( 200 g mureside + 100 g NaCl )
9.Titrate with 0.01M EDTA to purple color.

45. Expression of Results
% CaO = V
Edta
x N x 56.08 x 500 x 100
Initial Mass x 20
B. Active Lime Alkalinity Determination
Procédure
Weigh 500 mg lime into 500 ml conikal flask
Dissolve by 100 ml hot water and boil for 5 minutes.
Add 3 drops of phenolphtalien indicator
Titrate with 0.5 N HCl till red colour disapear forever
Expression of Results
% Active CaO = V
HCL
x 0.5N x 28.04 x 100
500mg
Ca(OH)
2 +2HCl CaCl
2 + 2H
2O

46. C . Total Lime Alkalinity
AIM :
To determine the percentage of total base in the given lime sample .
PRINCIPLE :
Lime contains all forms of alkalinity namely oxides and carbonates of
Ca and Mg, decomposable silicates . The purpose of the experiment is
to find out the total alkalinity of lime. All forms of alkalinity are
neutralized by using known amount of HCl .The excess HCl
remaining after neutralization with the bases is back titrated against
std NaOH solution . Also blank titration is carried out using the same
amount of HCl that was used for the experiment .Thus the difference
in both the titre values (NaOH consumed ) to neutralize the acid
gives the amount of acid consumed by the bases ,from which the
total bases present in the given sample can be calculated.

47. •REAGENTS REQUIRED :
o0.5 N HCl
o0.5 N NaOH
oPhenolphthalein indicator
•PROCEDURE :
oWeigh 500 mg of lime and transferred into a conical flask.
oAdd 50 ml of 0.5N HCl and dissolve by heating.
oAdd 3 drops phenolphthalein indicator
oTitrate the solution against 0.5 N NaOH
standard NaOH using methyl orange indicator.
oThe end point is the change of color to purple.
oA blank titration using 50 ml of acid is conducted against the
alkali.

48. Expression of Results
% Total Alkalinity = (V Blank – V sample) x 0.5N x 28.04 x 100
500 mg

49. 3.ANALYSIS OF SODIUM SULPHIDE
Check for purity, minimum 50 %.
Check for Iron contamination
Avoid usage if the iron contamination is high
Na
2
S is used to remove the hair of
the skin/hide.

50. 3.ANALYSIS OF SODIUM SULPHIDE
AIM :
To determine the percentage of sodium sulphide
content in the sulphide flakes
REAGENTS REQUIRED :
oDistilled water
oAmmonia Buffer solution –pH 10
oDimethyl Glyoxime Indicator
oBarium chloride solution
o0.1N Potassium ferriciyanide solution

51. Procedure
Weigh 5 gram of the sulphide flakes and determine its moisture
content
Add the dried sample to 500ml volumetric flask
Dissolve the sample and dilute with distilled water up to the mark
Add 20 ml of ammonium buffer solution and 1 ml dimethyl glyoxime
indicator
Add 25 ml of barium chloride solution
Pipette out 10 ml of the prepared sample solution
Titrate the sample with 0.1N potassium ferricyanide solution until
pink coloration disappears for 30 seconds.
Record the amount of ferricyanide and calculate the % Na2S.
The sulphide is oxidized to sulphur. Sulphite interferes and must be
precipitated with barium chloride.

52. Expression of Results
%Na
2S = gm Na
2S (pure)
Sample weight (dry weight)
= V x N x Eq.weight x V
total x100
Dry weight of sample x V aliquot
= V x 0.39 x 50
Sample weight (dry weight
Where V – Volume in ml of K
3Fe(CN)
6

N - Normality of K
3
Fe(CN)
6

53. 3.Liming production sulphide analysis Méthod
AIM :
To determine the percentage of sodium sulphide in the
lime liquor which is used as paint or unhairing solution
REAGENTS REQUIRED :
oDistilled water
oAmmonia Buffer solution –pH 10
oDimethyl Glyoxime Indicator
oBarium chloride solution
o0.1N Potassium ferriciyanide solution
oLime liquor

54. Procedure
Filter sample through wire gauze to remove any hair.
Add 20 ml of ammonium buffer solution and 1 ml dimethyl
glyoxime indicator
Add 25 ml of barium chloride solution
Pipette out 10 ml of the prepared sample solution
Cork the flask, mix the reagents and stand for 1 min.
Titrate the sample with 0.1N potassium ferricyanide solution until
pink coloration disappears for 30 seconds.
Record the amount of ferricyanide and calculate the % Na2S.
Calculate the amount of sodium sulphide to be added if necessary
Using the liming chemical addition chart read the amount of
sodium sulphide to be added to the load size given for the test
result obtained.

55. Expression of Results
%Na
2S = gm Na
2S (pure)
Sample weight
= V x N x Eq.weight x 100
weight of sample
= V x 0.39
Sample weight
Where V – Volume in ml of K
3Fe(CN)
6

N - Normality of K
3
Fe(CN)
6
1 ml of 0.1 N K
3Fe(CN)
6 = 0.0039 g of Na
2S

56. 4.ANALYSIS OF DELIMING AGENT
A.ANALYSIS OF AMMONIUM SALTS
B. ANALYSIS OF BORIC ACID

57. A. ANALYSIS OF AMMONIUM SALTS
•Ammonium chloride or ammonium sulphate is used during
the deliming process and helps remove lime from the hides
or skins and used to reduce the pH.
Two analysis methods
•back titration Method
•substitution Méthod

58. i. ANALYSIS OF AMMONIUM SALT BY BACK TITRATION
MÉTHOD
•AIM :
To estimate the amount of NH
4
Cl / ( NH
4
)
2
SO
4
in the given deliming
agent.
•REAGENTS REQUIRED :
o0.1N NaOH
o0.1N HCl
o Methyl orange indicator
•PRINCIPLE :
In this determination excess amount of NaOH is added to the
ammonium sulphate sample and the excess NaOH is titrated with HCl.
The amount of gram equivalents of NaOH consumed by the sample is
equivalent to the gram equivalent of NH
3
produced in the reaction

59. (NH
4
)
2
SO
4
+ 2NaOH

NH
3


+ 2Na
2
SO
4
+ 2H
2
O
Chemical Reaction
Procedure
Weigh 0.15gm (NH
4
)
2
SO
4
sample in the flask
Dissolve in 60ml distilled water
Add 30ml of 0.1N NaOH using pipette
Heat until the NH
3
formed is removed completely
Continue heating until a piece of filter paper moistened with
Hg(NO
3)
2 solution no longer turns black in the vapor from
the liquid
Cool the solution
Titrate the excess alkali with 0.1N HCl using methyl orange
indicator

60. Expression of Results
V1N1 = V2N1 V1 = V2N2
N1
NH3 = N1 x eq.wt x (30 – V1)
1000
% NH3 = N1 x eq. wt x (30 – V1)
1.5
Where V1 & N1 Volume and normality of excess NaOH
V2 & N2 – Volume and Normality of HCl consumed by excess
NaOH

61. ii. Analysis of Ammonium Salt by substitution Méthod
Ammonium sulphate which can not be directly titrated with a
particular standard solution (NaOH) is replaced by an equivalent
quantity of another substances (H
2SO
4) which can be so titrated. This
is effected by addition of a solution of formaldehyde CH
2O (formalin)
to the (NH
4)
2SO
4 solution.
REAGENTS REQUIRED :
oNeutral 20 % HCHO
o0.2 N NaOH
o 1% Phenolphthalein
•PRINCIPLE :
When formaldehyde is added to the deliming agent , free acid is
liberated which can be estimated by titrating with an alkali.

62. PROCEDURE :
oPROCEDURE :
oAbout 1g of the sample is weighed accurately ,dissolved in
distilled water and is made upto 100 ml in a standard flask.
o25 ml of the solution is pipetted out into a clean conical flask.
o20 ml of neutralized formaldehyde is added and the liberated
acid is titrated with 0.2N NaOH using phenolphthalein
indicator.
6HCHO + 4NH
4
Cl ( CH
2
)
6
N
4
+ 4HCl + 6H
2
O
6HCHO + 2(NH
4
)
2
SO
4
( CH
2
)
6
N
4
+ 2H
2
SO
4
+ 6H
2
O

63. Expression of Results
QNH
3 = N
1 x eq.wt x V
1000
% NH
3 = N
1 x eq. wt x V
1.5
Where: -
V
1
& N
1
Volume and normality of excess NaOH
V
2
& N
2
– Volume and Normality of HCl consumed by excess NaOH
QNH
3 – Quantity of Ammonia in the sample
1 ml of 0.2 N NaOH = 0.0107 g of NH
4Cl
1 ml of 0.2 N NaOH = 0.0132
Expression of Results

64. B. ANALYSIS OF BORIC ACID
•AIM :
To determine the purity of the given deliming agent(Boric Acid).
•REAGENTS REQUIRED :
oBoric Acid
oGlycerol
o0.1 N NaOH
oPhenolphthalein indicator
•PRINCIPLE :
Boric acid is a weak acid and thus does not ionize easily. Thus
pretreatment of boric acid is necessary, where boric acid can be
treated with polyols like –glcerol,glycol etc.. When boric acid

65. reacts with glycerol it donates a proton and forms glyceroboric
acid ,which is a strong acid that ionizes easily on titrating with a
strong alkali.
• PROCEDURE :
oThe stock solution of the boric acid is first prepared ,by
dissolving 1 g of boric acid in 1000 ml of distilled water.
oFrom that 20 ml is taken in a conical flask along with 2 ml
glycerol.
oThis is then titrated against 0.1 N NaOH using phenolphthalein
indicator.
oEnd point is the appearance of pink colour .
oNote the titre value.

Glycerol + H
3
BO
3
H
2
BO
3

66. •CALCULATION :
1 equivalent of boric acid = 1 equivalent of polyboric acid
1 equivalent of polyboric acid = 1 equivalent of NaOH

67. 5. ANALYSIS OF BATE

68. ANALYSIS OF BATE
•AIM :
To analyze the given bate sample .
•REAGENTS REQUIRED :
o0.2 N H
2
SO
4
o40% solution of NaOH
o4% boric acid solution
oMixed indicator (1 part Methyl red + 5 parts bromocresol
green-0.1% solution in ethanol )
o0.2 N HCl

69. •PRINCIPLE :
Bates used are enzymes (proteins ), thus for the estimation of
proteins we need to estimate the nitrogen content . The
nitrogen present in the bate is converted into ammonium
sulphate by acid digestion. It is distilled ,on the addition of
sodium hydroxide, ammonia is liberated which is absorbed in a
weak acid like boric acid .The ammonium borate formed is a
strong base which can be titrated against a strong acid.
•PROCEDURE :
o0.5 g of the bate is accurately weighed and transferred to a
dry, clean kjeldahl flask along with concentrated H
2SO
4 .
oHere all the ammonium compounds are converted into
ammonium sulphate.
oInitially the bate solution is dirty brown in colour and after
digestion clear solution is obtained which is the end point of the
acid digestion.

70. oThen 200 ml of water is added to this digested sample ,50 ml
of 40% NaOH is added and the contents are distilled for 40
minutes.
oThe liberated ammonia is absorbed in 4% boric acid solution
and forms ammonium borate.
oThe boric acid solution is red in colour before absorbing
ammonia ,after distillation the contents of the flask turns green.
o Thus after the formation of ammonium borate ,2-3 drops of
the mixed indicator is added to the contents of the flask .
oThis is then titrated with strong acid 0.2 N HCl . The end point is
the appearance of original red colour.
•CALCULATION :
1000 ml of 1 N acid = 14 g of nitrogen = 1 equivalent of bate

71. 6. ANALYSIS OF VEGETABLE TANNING
MATERIALS
A.QUALITATIVE ANALYSIS
B.QUANTITATIVE ANALYSIS
C.ACIDS AND SALTS IN VEGETABLE TANNIN
EXTRACTS BY DIFFERENT METHODS

72. A. QUALITATIVE ANALYSIS
•Vegetable tannins are classified chemically into
two groups namely, pyrogallol and catechol.
•Pyrogallol tannins tend to be yellow to tan
brown ,high in acidity.eg: Oakwood, chestnut,
myrobalans, valonia, divi-divi, sumac,
algarobilla .
•The catechol tannins tend to be reddish ,low in
acidity, sugar and salt. eg: cutch, quebracho,
goran, gambier.
Qualitative analysis is required to identify if the
given tannin is a pyrogallol or catechol tannin.

73. EXPERIMENT OBSERVATION INFERENCE
To a few ml of the given
substance add 3-4 ml of 1%
gelatin
Precipitation
Occurs
May be hydrolysable or
condensed tannins
To a few drops of the tannins add
2 ml of lead acetate solution
Precipitation
Occurs
May be hydrolysable or
condensed tannins
To a few drops of the tannin
solution add 3-5 drops of 1% Iron-
Alum solution
1.Bluish violet
2.Green colour
1. Hydrolysable
Tannins
2. Condensed Tannins
Take 50 ml of tannin solution
along with 20 ml of HCHO and 10
ml of HCl in a round bottomed
flask and heated in a reflux
condenser and cool it
1. No
Precipitate
2. Precipitate
1. Hydrolysable
tannins
2. Condensed tannins
Then the precipitate is filtered in
a filter paper and the filtrate is
added with 1%
iron-alum
1. Bluish Violet
2. Green
colour
1.Hydrolysable
Tannins
2.Condensed tannins

74. B. QUANTITATIVE ANALYSIS
•AIM :
To determine the total tannin and non tannin content in the
given sample of tan liquor.
Firstly the approximate tan content in the given material
may be ascertained. The test also involves the determination
of the total solids, total soluble, non-tans ,tans and colour.
•PROCEDURE :
Preparation of the tan liquor for ananlysis :
oSoak about 25 – 50 gms of the ground tanning material
over night in a proctor extractor in cold distilled water.

75. Next morning draw off the extracted liquid at uniform rate of 2l
in 4 hours .After collecting the first 150 ml the bath temperature
is raised and maintained at 50˚C till 750 ml are collected.
Raise the temperature of the bath to boiling temperature and
then collect the remaining quantity of the 2l.
Then cool the flask and make up the volume accurately to 2l.
Shake the contents well and keep for analysis.
1. Total Solids :
oWeigh accurately a porcelain evaporating dish.
oPipette out 50ml of the prepared analytical solution into the
porcelain basin.
oEvaporate the contents till it becomes dry.
oNote the weight of the evaporating dish after drying.

76. •CALCULATION :
Initial weight of the porcelain dish = x gm.
Final weight of the porcelain dish = y gm.
Weight of total solids in 50 ml = (y-x) gms.
Weight of total solids in 2 l = 40 (y-x)
% of total solids = 40 ×(y-x) ×100
Weight of the tannin taken
for preparing the tan liquor .

77. 2. Determination of the total soluble :
oFilter the stock solution through no1 whatmann filter paper.
oWeigh the empty porcelain dish.
oPipette 50 ml of the filtrate into the porcelain dish.
oEvaporate the content till it becomes dry. Then weigh it.
•CALCULATION :
% Total soluble = 40 × difference in weight ×100
Weight of the tanning materials taken

78. 3. Determination of non- tans :
•Add 100 ml of the tan stock solution into the bottle containing
the hide powder .
•Shake thoroughly and place in a mechanical shaker for 10
minutes.
•Then filter through a dry filter cloth into a dry beaker .
•Add 1 gm of chrome alum to the filtrate and filter through no1
whatmann filter paper till a clean solution is obtained .
•Pipette out 50 ml of the filtrate into an evaporating basin and
evaporate to a constant weight .
•The non- tans in the 100 ml of the stock solution is given by
multiplying the weight of the dry residue with 1.2 since 100 ml is
diluted to 120 ml by the moisture present in the hide powder.
% of non – tan = 40 ×1.2 × dry weight of residue × 100
weight of the tan material taken

79. 4. Determination of tans :
oThe matter absorbable by the hide powder = Tans = % Total of
soluble - % total of non –tan.

80. 9. ANALYIS OF CHROME
TANNING AGENT
A.MOISTURE
B.Cr
2
O
3
CONTENT
C.ACID COMBINED WITH CHROMIUM
D.BASICITY : PROCTOR AND LEHIGH BASICITIES
E.DEGREE OF OLATION

81. A. MOISTURE
•Determination by wet oxidation method
•AIM :
To determine the moisture content in the given chrome tanning
agent.
•PROCEDURE :
oWeigh 1 g of basic chromium sulphate in a wide mouthed
weighing bottle .
oDry it in an air oven for 3-4 hours.
oCool in a desiccator and weigh again.
•CALCULATION :
Final weight – Initial weight x 100
Weight of the sample taken

82. B. Cr
2O
3 CONTENT
•AIM :
To determine the Cr
2
O
3
content in the given sample of the
chrome tanning agent.
•REAGENTS REQUIRED :
oPer-chloric acid
oConc. Sulphuric acid
oConc. Nitric acid
o10 % Potassium iodide solution
oStandard Sodium thio sulphate solution
oStarch indicator

83. •PRINCIPLE :
As we are not sure if all the Cr in the given sample of the
chrome tanning agent is in the trivalent state, the
determination depends upon the oxidation of the trivalent
chromium into hexavalent chromium by acid digestion .When
all the Cr(III) has been converted into Cr(VI) the green color
solution turns into orange color solution. Here Chrome acts as
the self indicator. Thus all the Cr in the taken sample are now
converted into Cr(VI). This is now reduced back to Cr(III) by
adding a reducing agent like KI . The liberated iodine is
titrated against standard sodium thio sulphate solution using
starch as the indicator. Solutions containing 1% Cr (III) are found
to be desirable.
Cr
2
O
7
2-

+ 6 I
-
+ 14 H
+
2 Cr
3+
+ 3 I
2
+ 7 H
2
O

84. •PROCEDURE :
oWeigh 300 mg of sample and dissolved in distilled water and
made upto 500 ml in heat resistant conical flask .
oAdd 20ml HNO3 + 10 ml (3 parts HClO4 + 2 Part H2SO4)
oPlace small funnel on a conical flask and reflux to reddish
orange color.
oAfter changing of the color heat for another 2 minutes.
oCool solution immediately on air and then on tap water.
oDilute the solution with distilled water to 100 ml.
oBoil solution for 10 minutes to expel chrome if any.
oAdd 10 ml H3PO4 or 10 ml HCl to mask iron
oAdd 10 ml of 10% KI solution and stand in a dark place covered
with a glass stopperd.

85. oAfter 10 minutes the liberated iodine is titrated against
standard sodium thio sulphate solution using starch as the
indicator.
oOn adding starch the solution turns blue. The end point is
marked by the color change from blue to colorless.
•CALCULATIONS :
oPreparations of Cr liquor
%Cr
2O
3 = 0.1N x V x 2.533
Initial Weight

86. C. Determination of acid combined with
chromium
•AIM :
To determine the acid combined with the chromium in the given
chrome tanning agent in order to determine the basicity
•REAGENTS REQUIRED :
o0.5 or 0.2 N NaOH
o1 % Phenolphthalein indicator
Procedure:
•prepare 1% chromium in 250 ml volumetric flask (usually 15 g for
solid chrome liquor)
•Pipette 25 ml of analytical solution and dilute with 300ml
distilled water

87. oAdd 3-4 ml of 1 % Phenolphthalein indicator
oTitrate with standard caustic soda until pink color develops
oBoil the contents with constant stirring
oContinue the titration until a grayish violate color appear to
be pink
oBoil for another minutes and titrate until the pink color
persists.
•CALCULATION :
Basicity can be calculated as:
Basicity = (chrome combined with OH x 100)
Total Cr

88. D. CALCULATION OF BASICITY OF
CHROME LIQUOR
•AIM :
To determine the basicity of the given chrome liquor.
•REAGENTS REQUIRED :
o 2.5% sodium oxalate solution adjusted to pH 8.3. (Use Ab
Na
2
(COO)
2
(to remove salts of acetates, citrates, etc.)
o0.5N H
2SO
4 standardized
o0.25N NaOH
•PRINCIPLE :
SO
4
Cr

OH ,Here sodium oxalate is used to break the Cr-
OH bond ,but it is still difficult to break ,thus we use standard
sulphuric acid along with the sodium oxalate.

89. Here sodium oxalate masks the indicator phenolphthalein .Thus
we use a pH meter .
•PROCEDURE :
oA sample of Cr liquor containing 0.05 to 0.1 g of Cr
2O
3 is taken
in a 250 ml round bottomed flask.
o 25 ml of 0.5N H
2
SO
4
is added with 50 ml of sodium oxalate
solution to the contents of the round bottomed flask.
o The flask is refluxed for an hour and then cooled.
oThe contents are then transferred to a beaker and the cold
solution is back titrated with 0.25N NaOH to pH 6.3 by using a
pH meter.
oA blank containing all reagents is run along with the sample.

90. •CALCULATION :
% Basicity = (V
B – V
E)
NaoH x N
NaoH x 100 x meq. Of Cr.
x 100
Chrome content in the stock taken
It can be calculated as follows
If a ml of sodium thiosalphate are required in the
chromium titration and b ml of 0.1 N sodium hydroxide for
a similar volume in the acid titration then
Basicity = (a-b) x 100
a
This method is not usually accurate in the presence of
organic salt.

91. E. DEGREE OF OLATION
•AIM :
To determine the degree of olation in the
given chrome tanning liquor .
•REAGENTS REQUIRED :
o0.1 N H
2SO
4
o0.1 N NaOH
•PRINCIPLE :
When chrome liquors are boiled, hydrolysis takes place with
the formation of a basic salt and free acid.

92. The hydroxyl groups in the basic salt, are not, however, in
hydrolytic equilibrium with the acid so formed because the
hydroxyl groups attached to one chromium atom coordinates
with another chromium atom. With basic chromium chlorides,
two such ‘olated’ OH groups lead to a four member ring
consisting of two chromium and two oxygen atoms. Such
hydroxyl groups are termed, ‘olated’ and they are resistant to a
large degree of mineral acids in the cold.

OH
Cr(OH)SO
4
+ H
2
SO
4
Cr
2
( SO
4
)
3
+ Cr Cr

OH

93. •PROCEDURE :
o50 ml of chrome liquor which must have been dissolved with
cold distilled water is treated with 25 ml of 0.1N H
2
SO
4
and
immediately titrated with O.1N NaOH using pH meter to a pH
of 3.3.
•CALCULATION :
The difference between 25 ml of 0.1N acid and the volume of
alkali (in ml) used corresponds to the acid used by unolated
hydroxyl groups. Let this value be ‘c’. If ‘a’ represents the
volume of 0.1N thio for chromium determination and ‘b’ the
volume of 0.1N NaOH equivalent to the Procter-McClandish
titration for the same amount of chrome liquor then the %
olation is
a-b x 100
a

94. As a rule, the percentage olation with basic chromium sulphate
liquors is 100%.
oa = meq of thio
ob = meq of sulfates
oc = meq of free –OH
oa-b = meq of total -OH
o(a-b)-c = meq of olated -OH
 

96. AIM:
To determine the TiO
2
content in the given titanium tanning
agent.
REAGENTS REQUIRED:
oSulphuric acid
oHydrochloric acid
oHydrogen peroxide
oAmmonium sulfate
oRecrystallized titanium potassium oxalate
PRINCIPLE:
Titanium is determined by measuring the intensity of the yellow
color produced when a Ti compound in H
2SO
4 solution is oxidized
with H
2
O
2
. Halogens interfere, hence HCl if present must be
expelled by evaporating to fumes with H
2
SO
4
, or else Ti present
must be precipitated by ammonia and redissolved in sulphuric
acid.

97. To prepare the standard Ti solution, treat 2.214g of recrystallized
Titanium potassium oxalate, K
2TiO(C
2O
4)
2.2H
2O with 8g of (NH
4)
2
SO
4
and 100ml of concentrated sulphuric acid in a kjeldahl flask.
Heat to boil for 5 to 10 minutes, cool, dilute to about 300ml,
transfer to a 1000ml volumetric flask and dilute to volume.
1ml=0.3 mg Ti or 0.5 mg TiO
2
.
•PROCEDURE:
oPipette an aliquot of solution B, containing not more than 10mg
TiO
2
in to a 250ml beaker. For a 5 g leather specimen containing
not over 1 percent TiO
2, a 100ml aliquot will be satisfactory.
oAdd 5 to 6ml of concentrated sulphuric acid and evaporate to
fumes. Cool, add about 25ml water, transfer to a 100ml Nessler
tube, and dilute to about 90ml.
oPrepare a series of standards containing various amounts of
standard Ti solution measured from a burette, and 5ml
concentrated sulphuric acid. Add 3ml of 3% H
2O
2 to the sample

98. and each standard, bring all to the same temperature and dilute
to the mark.
oMatch the color of the sample with the standards. If a
colorimeter is available, transfer the sample, after expelling HCl,
to a 100ml volumetric flask, make to about 90ml, and add 3ml
of 3% H
2O
2.
oTo a second flask add standard Ti solution containing 10mg TiO
2
,
5ml concentrated H
2
SO
4
, water to make about 90ml, add 3ml
of 3% H
2O
2. If the color of the sample is weaker than that of the
standard, make both to 100ml, bring them to the same
temperature, and compare in the usual way.
oIf the color of the sample is stronger than that of the standard,
transfer the sample to a larger volumetric flask, and make to
volume with (1:9) sulphuric acid (by weight) before making the
comparison.

99. o If a spectrophotometer is available transfer an aliquot of
solution B containing about 0.1 to 0.25mg TiO
2
to a 250ml
volumetric flask, dilute to 200ml, and 8.5ml of 30% H
2
O
2
and
25ml of concentrated sulphuric acid and make to volume. Read
the color at 390 m
µ against a blank containing the same
concentrations of reagents.
Titanium di oxide(TiO
2
)=g TiO
2
in aliquot * (500/M) * 100
g leather specimen
Where M=volume of aliquot in ml

101. •AIM:
To determine the Al
2
O
3
content in the given aluminum tanning
agent.
•REAGENTS REQUIRED:
oConcentrated sulphuric acid
oThymol blue and cresol red indicator
oPotassium hydroxide solution
oPotassium tartrate solution
oPotassium fluoride solution
oStandard hydrochloric acid solution
•PRINCIPLE:
In this method an aluminum tartrate complex is formed. When
treated in slightly basic solution with potassium fluoride this
liberates an amount of base equivalent to the aluminum
present, which can then be titrated.

102. •PROCEDURE:
oAsh 1 to 2 g of accurately weighed ground leather in a
crucible at 600˚C, in a muffle furnace.
oLixiviate the ash with 5 to 10 ml of concentrated HCl and
then with water. Transfer the contents to a conical flask, and
heat on a hot plate in a fume chamber till the volume is
reduced to 5ml.
oAdd few drops of the mixed thymol blue-cresol red indicator
solution and neutralize the excess of acid with potassium
hydroxide solution indicated by disappearance of red color.
oCool the solution and add 20ml of potassium tartrate
solution and complete the neutralization of the acid with
potassium hydroxide solution, indicated by the appearance
of pink color.
oSubsequently add 20ml of potassium fluoride solution and
titrate with standard HCl.

103. oThe end point is reached when the solution assumes the same
pink color which goes over to orange-yellow with further
addition of acid.
•CALCULATION:
Aluminium(as Al
2
O
3
), percent by weight=1.7 * V * N
W
where,
V=volume in ml of standard HCl required for the titration.
N=normality in N of standard HCl.
W=weight in gram of the sample taken for test.

104. ANALYSIS OF ZIRCONIUM
TANNING AGENT

105. •AIM:
To determine the ZrO
2
content in the given zirconium tanning
agent.
•REAGENTS REQUIRED:
oFusion mixture(equal parts of sodium carbonate anhydrous,
potassium carbonate and borax glass (anhydrous borax)
oHydrochloric acid solution
oBromophenol blue
•PRINCIPLE:
A known quantity of leather is ashed. Ash is dissolved cut and
zirconium is precipitated as zirconium hydroxide. It is then
ignited and weighed as zirconium oxide.

106. •PROCEDURE:
o Fuse the ash from 2g of ground leather with 3-4 times its own
weight of fusion mixture in a platinum crucible and fuse for 30
minutes in a muffle furnace at 800-850°C.
oCover and cool, dissolve the fused mixture in boiling water and
filter through filter paper. Wash the residue on the filter paper
with boiling water.
oDissolve the residue left on the filter paper with hot, dilute
hydrochloric acid. Add few drops of bromophenol blue indicator.
oPrecipitate out zirconium as Zr(OH)
4 by ammonium solution at
pH 3.8 judged by change in the color of indicator from yellow to
blue.
oFilter through the qualitative filter paper and dry the residue in
an oven, ash the filter paper along with the precipitate till
particles are burnt off. Ignite the residue at 900°C, cool and
weigh.

107. o Repeat heating, cooling and weighing until constant weight is
obtained.
• CALCULATION:
Calculate zirconium as zirconium oxide(ZrO
2
) as follows:
Zirconium(as ZrO
2), percent by weight=50W
Where W=final weight of the ash.

108. 10. ANALYSIS OF FAT LIQUOR
A.Moisture
B.Acid value
C.Iodine value
D.Unsaponifiables

E. Saponification value
F. Free fatty acids
G. Total Alkalinity

109. A. MOISTURE
•AIM:
To determine the moisture content present in the fat liquor.
•PRINCIPLE:
The initial weight of the fat liquor along with the moisture
content is taken. It is then evaporated and the weight
excluding the moisture content is taken. the difference
between the weights give the moisture content present.
•PROCEDURE:
oweigh accurately 20 gms of oil in a cleaned and dried
porcelain dish, say it is about w1 gms.
oDry the same by keeping in an air oven at 105 degree Celsius
to 110 degree Celsius for 3 hours and weigh.

110. oContinue this drying and weighing until a constant weight (w2) is
obtained.
•CALCULATION:
o% moisture= (w2-w1) x 100
Weight of the fatliquor taken

111. ESTIMATION OF MOISTURE
•AIM:
To estimate the moisture present in the oils and fats.
•APPARATUS REQUIRED:
oDean and stark apparatus
•REAGENTS REQUIRED:
oToluene
oXylene
oPetroleum ether
•PROCEDURE:
oWeigh about 20 g of sample (fatliquor) into 500 ml of round
bottom flask.
oAdd 150 ml of solvent chosen (here xylene is used).
oAdd a few glass beads / porcelain bits.

112. oAttach dean and stark apparatus with condenser.
oBring the round bottom flask to boiling and continue boiling
vigorously.
oThe boiling is continued until the liquid falling from the
condenser is clean and the volume of solvent is constant read
the volume.
•CALCULATION:
Percentage of water= T/W ×100
Where, T = water collected in the side tube
W = Weight of the sample

113. B. ACID VALUE
•AIM: To determine the acid value of the given fat liquor.
•REAGENTS REQUIRED:
oMixture of 1:1 benzene and alcohol (neutralized to
phenolphthalein)
oPhenolphthalein
oAlcoholic potassium hydroxide solution
•PROCEDURE:
oWeigh accurately 1 or 2 gm of the oil in 250 ml conical flask.
oAdd 40 ml of mixture of 1:1 benzene and alcohol

114. o(neutralized to phenolphthalein) and dissolve the fat- if
necessary heat the content to facilitate dissolving.
oCool the contents and titrate against N/10 alcoholic potassium
hydroxide solution using phenolphthalein indicator, until a
distinct pink color is obtained.(say litre value is a).
Conduct a blank titration using 1:1 benzene and alcohol mixture
(40 ml) against KOH using the same indicator (say titre value is
b)
•CALCULATION:
oAcid value= (a-b) X strength of alcoholic KOH X 56.1
weight of fatliquor taken

115. C. IODINE VALUE
•AIM:
To determine the iodine value of the given fat liquor.
•REAGENTS REQUIRED:
oChloroform
oPotassium iodine solution
oSodium thio sulfate solution
oStarch solution
oIodine solution
•PROCEDURE:
oThe amount of fatty matter to be taken for analysis:

116. •0.1 to 0.2 gms for oils having iodine value above 120.
•0.2 to 0.4 gms for oils having iodine value between 60-120.
•0.4 to 0.8 gms for oils having iodine value less than 60.
oWeigh accurately the required quantity of oil in a glass capsule
and transfer the same in to a 300 cc iodine value flask.
oDissolve the oil in 10 ml chloroform and add 25cc of the iodine
solution, and stopper the same. Keep in a dark place for 30
minutes to 1 hour.
oConduct a blank using 25 ml of iodine solution in an iodine value
flask, keeping for the same time as in (2).
oAdd 15 ml of 10% of potassium iodide solution and add 50 ml of
water to both the flasks and titrate against 0.1N sodium thio
sulphate solution, using starch solution as indicator. The end
pointy is indicated by the disappearance of blue color.

117. •CALCULATION:
oLet ‘a’ be the titre value for blank and let ‘b’ be the titer value
for the experiment.
oIodine value= (a-b) * strength of hypo * 12.69/weight of oil taken.

118. D. UNSAPONIFIABLES
•AIM:
To determine the amount of unsaponifiables present in the
fatliquor.
•REAGENTS REQUIRED:
oMethyl orange indicator
oEthyl ether
oEther solution
oAlcohol
oN/10 HCl

119. •PROCEDURE:
oTake 5 gms of the oil sample in a dish.
oAdd 12 to 15 ml of 0.2N alcoholic KOH.
oHeat the same over a sand bath with stirring till it becomes dry.
oDissolve the soap formed by washing repeatedly with warm
water and transfer to a separating funnel. Add 10 ml of alcohol,
cool the solution.
oShake the contents first with 50 ml of ethyl ether, and then twice
with 25 ml of ether.
oAllow the liquid to stand and separate in the separating funnel.
oWash the combined ether extracts with 8 ml of water acidified
with 1 to 2 ml of 0.1N HCl using methyl orange indicator.
oWithdraw the acidic layer from the funnel. Neutralize the other
layer by adding 3 ml of 0.2 N alcoholic KOH and 7 ml of water-
keep this for sometime.

120. oRemove the aqueous layer, and distill off the ether solution.
oCollect the ether extract, and dry at 100 degree Celsius, till
constant weight is obtained.
oThis weight represents the weight of unsaponifiables in the
sample. From this calculate the percentage.

121. E. SAPONIFICATION VALUE
•AIM:
To determine the saponification value of the given fat liquor.
•REAGENTS REQUIRED:
oAlcoholic potassium hydroxide
oBenzol
ophenolphthalein
•PROCEDURE:
oWeigh out accurately 2 to 5 gm of the oil or fat in a 300 ccs
conical flask.

122. o Add 25 ml of 0.5N alcoholic potassium hydroxide using pipette.
Reflux the whole contents for 1 hour over a burner.
oNOTE: Add 20 ml of benzol before refluxing in the case of waxes.
oAdd a few pieces of porcelain to prevent bumping.
oCool the contents and titrate against 0.5 HCl using
phenolphthalein indicator, till the pink color disappears. Let the
titer value be (a) ml.
oConduct a blank experiment using 25 ml N/10 alcoholic KOH
against N/2 HCl using the same indicator-let this titer value be
(b) ml.
•CALCULATION:
oSaponofication value= (b-a) * strength of HCl * 56.1
weight of fat liquor taken

123. F. pH
•AIM:
To determine the pH of the given fat liquor.
•PROCEDURE:
This method can only be approximate and placing drops of a
solution of the oil in water (at the usual concentration at
which it will be used in practice) on a white tile and testing
with indicators is a more reliable method. It is always possible
for the dilution of a sample of sulphated oil in water to cause
hydrolytic changes, and a solution may apparently become
more acid upon further dilution. The indicator may be added

124. to the sulphated oil in a test tube. The choice of indicators will
depend upon approximate pH, but B.D.H. universal indicator
will afford guidance. The pH value of the commercial sulphated
oils generally lie between 6 and 8.5 but both higher and lower
figures are sometimes obtained.

125. •AIM:
To determine the amount of free fatty acids present in the
given fat liquor.
•REAGENTS REQUIRED:
oAbsolute alcohol
oPhenolphthalein
•PRINCIPLE:
Acid value is the number of milligrams of KOH needed to
neutralize the free fatty acid in 1 gram of fat. It is a very
important indication of the quality since certain free fatty
acids in cod oils oxidize very rapidly and may give gummy
spue on the leather and oils containing much free acids
corrode metals. Commercially the result is expressed
generally
G. FREE FATTY ACIDS

126. as percentage of oleic acid. Therefore percentage of acids is equal
to :
% FFA = Acid value * molecular weight of oleic acid
561(equivalent of KOH)
= Acid value * 282
561
= Acid value (approx)
2
•PROCEDURE:
50cc of absolute alcohol is pipetted in to a conical flask, 1cc of 1%
phenolphthalein solution is added, the N/10 NaOH is dropped in
to the alcohol until a very faint color is obtained. To the alcohol

127. thus rendered neutral 1-5 gram of oil is added and the flask is then
slightly warmed over the water bath with a reflux condenser. The
dissolved fat is then titrated with N/10 KOH,
Preferably an alcohol solution. The acid value according to the
above definition is calculated as follows:
Acid value = cc of N/10 alkali used * 5.61
weight of oil taken

128. •AIM:
To determine the total alkalinity present in the given fat liquor.
•REAGENTS REQUIRED:
oSodium hydroxide
oMethyl orange or bromophenol blue
•PROCEDURE:
The aqueous solution from the total fatty acid determination is
warmed to remove the ether and filtered if necessary. It is then
cooled and titrated with 0.1N NaOH using methyl orange or
bromophenol blue as indicator. The difference between this
titration and the acid originally added gives the cc of N/1 acid
H. TOTAL ALKALINITY

129. required to neutralize the total alkali not only as soap but as
hydroxide, carbonate, borate or salts of other weak acids.
Including CaCO
3
-
if present. The total alkali is returned as
percentage of Na
2O in hard soap or K
2O in soft soap.
0.031 * cc N/1 HCl used to neutralize in case of hard soap.
0.047 * cc N/1 HCl used to neutralize in case of soft soap

and equals the base present * 100
weight of soap

130. 13. ANALYSIS OF PRETANNED
PELTS AND TANNED LEATHERS
A.LIMED PELT
B.CHROME TANNED LEATHER
C.VEGETABLE TANNED LEATHER

131. A. ANALYSIS OF LIMED PELT
•AIM :
To estimate the amount of caustic lime (Slaked lime ) in the
given limed pelt.
•REAGENTS REQUIRED :
o0.1 N HCl
oPhenolphthalein
•PRINCIPLE :
Acid –base titration is the principle behind this experiment.
Here the color of the phenolphthalein is returned because of
the diffusion of the lime . Thus the experiment is carried out
until the pink color disappears totally.

132. •PROCEDURE :
oThe limed pelt is cut into small pieces .
oAbout 5 g is weighed accurately in a bottle and about 150 ml of
distilled water is added and the contents are titrated against
0.1N HCl using phenolphthalein indicator .
oThe bottle of limed pelt is shaked well, and allowed to stand.
oThe color of phenolphthalein is returned because of the diffusion
of the lime.
oThe titration is continued till the pink color disappears
completely.
oThe shaking and the titration are continued until the pink color
fails to appear from the pelt bits on standing.

133. •CALCULATION :
% lime in limed pelt = V
HCl
x N
HCl
x 0.037 x 100
Weight of the pelt

134. B.CHROME TANNED LEATHER
•AIM :
To determine the given sample of the chrome tanned leather.
•SAMPLING :
oPositioning of sampling –Each sample consists of a
rectangular piece of leather measuring approximately 6.5
inches x 2.5 inches in the butt ,belly and the shoulder
positions.
oThe samples are cut up separately and mixed thoroughly .
oThe pieces ,so obtained ,are then put into glass stoppered
bottles ,about 100 gm being set aside for each analysis.

135. •MOISTURE :
o5 g of the prepared sample is weighed out in a weighing bottle
(which) was previously weighed in the oven and made constant
in weight and then dried to a constant weight in the oven at a
temperature of 100-110˚C .The percentage is calculated from
the loss in weight .
•DETERMINATION OF FAT AND SULPHUR :
oWeigh out accurately 20 gms of the leather cut samples (which
are free from moisture) and place in a soxhlet condenser
distillation unit .
oWeigh accurately the dry flask and connect it to distillation
condenser unit .
oAdd petroleum ether (boiling point -60˚C) and reflux
thoroughly for a few hours ,until the leather sample is free from
grease.

136. oAfter complete extraction, distill off the solvent and dry the flask
in an air oven at 100˚C for 5 hours. Then cool in a desiccators.
Finally weigh the flask ,let its weight be say ( a ) gms.
oDry the extracted leather by spreading out on a clean surface
exposed to air .
•DETERMINATION OF SULPHUR :
•The residue is added into the flask ,and this contains fat and
sulphur .
•Treat this mixture with ether ,which dissolves the fat and grease.
Then decant the liquid from the residue.
•The decanted liquid is then transferred to another flask,
evaporate the ether and find out the weight of the fat and oil.

137. •Treat the residue in the first flask with fuming nitric acid.
Then allow it to stand overnight .
•Pour the contents into a porcelain basin and evaporate.
•Repeat the above washings and evaporate to dryness in a
fume chamber.
•Dissolve the above residue in hot water and filter .
•Acidify with few ml of HCl. Then add 10 ml of 10 % barium
chloride. The sulphur gets oxidized to sulphate and is then
precipitated as barium sulphate.
•Filter over a ash less filter paper and ignite to completion
over a previously weighed crucible.
•Weight of sulphur = Weight of barium sulphate × 32/233
•Let this be (b)g.

138. •DETERMINATION OF FAT :
oWeight of fat = (weight of sulphur + fat ) – Weight of sulphur
= a-b
o% of fat = weight of fat × 100 / Weight of the leather taken
o% of sulphur = Weight of sulphur × 100 / Weight of the
leather taken
•DETERMINATION OF CHROMIUM :
•PRINCIPLE :
oThe determination of chromic oxide depends upon the
oxidation of trivalent chromium into the hexavalent state
using strong oxidizing agent like the acid mixture and the
subsequent liberation of iodine from KI, which is titrated with
standard sodium thio sulphate leading to the estimation of
chromium.

139. •REAGENTS USED :
o 10% KI solution
o0.1 N Sodium thiosulfate
o20 ml acid mixture (conc. sulphuric acid, conc. nitric acid,
perchloric acid 3.5:5:11.5).
•PROCEDURE :
oA piece of tanned leather is taken and is cut into small pieces.
oKnown weight was taken and 20 ml of acid mixture is added in
a conical flask.
oThe mixture is digested till the color change from green to
orange.
o20 ml distilled water is added to wash away the chlorine
liberated, it is boiled and cooled, and made up to 250 ml.
o20 ml of this digested solution is taken in a iodine flask and

140. 15 ml KI solution is added to it and immediately kept in dark
place for 2-3 min and titrated against 0.1N Sodium thio
sulphate till brown color is converted into straw yellow.
o2 ml of starch is added to the iodine flask to get blue color
solution.
oIt is titrated against 0.1N Sodium thio sulphate till the color
disappears and titre value is noted.
•CALCULATION :
% Cr
2O
3 in leather = Titre value x Eq.weight of chromium x 100
Weight of the leather sample taken

141. C. VEGETABLE TANNED LEATHER
1. MOISTURE CONTENT:
oA piece of tanned leather is taken and cut into small pieces
in a crucible.
oInitial weight of crucible and weight of crucible and leather is
noted.
oThe crucible is covered with a filter paper to prevent it from
other contaminants and dried in air oven at 100–105
o
C for a
period of 5 hrs, cooled and weighed.
oThe process of drying, cooling and weighing is repeated until
constant weight is obtained.
oPercentage of moisture is calculated from the difference in
weights.

142. oWeight of empty crucible = a gm
oWeight of crucible + leather = b gm
oWeight of crucible + leather after drying = c gm
•CALCUALTION :
% Moisture = b-a x 100
b-c
2. DETERMINATION OF OILS AND FATS :
o10gm of the leather sample is weighed accurately and put into
soxhlet apparatus by means of thimble which is attached to a
previously weighed round bottomed flask.
oOils and fats are extracted with petroleum ether (B.P 40-60
o
C),
in a soxhlet extraction unit for not less than 4 hrs.
oThe round flat bottomed flask is dried and weighed till a
constant weight is obtained.
 

143. oAfter distilling the petroleum ether, the oils and fats in the flask
were dried at 98-100
o
C for 3 hours, cooled and weighed.
o The process of drying, cooling and weighing is repeated until
constant weight is obtained. The amount of oils and fats
expressed as % of the weight of the sample.
•CALCULATION :
oWeight of the leather sample = w
1
oWeight of the empty round bottomed flask =

w
2
oWeight of the round bottomed flask after extraction and
drying = w
3
% oil and fat content = W
3 – W
2 x 100
W
1

144. 3. METHOD OF DETERMINING THE TOTAL ASH :
oWeigh a silica crucible accurately.
oAdd about 5 g of leather and find out the weight accurately
oIncinerate the same over a burner and then keep ina muffle
furnace at about 500˚C ,until all the carbon is consumed and
the leather is fully ashed .Then take out and keep in a
descicator to cool.
oFind out the final weight accurately and calculate theweight of
ash.Let this be “a” g.
•CALCULATION :
% of total ash = a x 100
5

145. 4. DETERMINATION OF WATER SOLUBLE MATTER :
oKeep about 20 g of the original sample of leather (extraction of fat )
in a beaker in a Procter extractor and cover with distilled water.
Allow it to stand for 12 to 18 hours .
oKeep the temperature of the water bath at 45ºC and extract by
continuous extraction .Collect about 1 litre in 3-3.5 hours and make up
to the mark.
oFilter this extracted solution .
oWeigh accurately an evaporating basin .Then add 50 ml of the
filtrate from that used in oils and fats determination by pipetting.
Evaporate over a water bath to almost dryness .Then place in an air
oven at 100˚C till constant weight is obtained .
oWeigh the evaporating basin and the contents accurately and find
out the weight of the water soluble matter .let this be “x”.
CALCULATION :
% of water soluble = x * 1000 * 100
50 * 20

146. 5. ASH OF WATER SOLUBLES :
oWeigh accurately a silica basin or big crucible .
oPipette out 100 ml of the water soluble extract from the filtered
solution of the previous experiment . Evaporate to dryness and
ignite and ash in a muffle furnace .
oCool, and find out the accurate weight of the ash of the water
soluble ,let this be “y” g.
•CALCULATION :
% of ash = y * 1000 * 100
100 * 20

147. 6. DETERMINATION OF NITROGEN AND HIDE
SUBSTANCES :
•REAGENTS REQUIRED :
oStandard 0.5N HCl or H
2SO
4
oStandard 0.5N NaOH, pure anhydrous potassium sulphate
powder and anhydrous copper sulphate .
•PROCEDURE :
oWeigh accurately 1.5 g of the leather sample and place in a
700 ml kjeldahl flask. Add 15 –20 ml of concentrated sulphuric
acid to cover the leather.
oHeat this flask by keeping it in a slightly inclined position, till
the leather is disintegrated. Then add 10 g of potassium
sulphate and 0.5 g of copper sulphate .Raise the temperature
and boil continuously until the liquid becomes clear and nearly
colorless or bluish green .Allow the flask and the contents to
cool .
oAfter cooling add 250 ml of distilled water and 1 or 2 pieces of

148. glass beads or broken porcelain bits to avoid bumping during
heating, then add 1 ml of 1 % phenolphthalein indicator
solution, then pour 80 ml of 40 % NaOH solution through the
sides of the flask to avoid violent mixing .
oConnect the flask to a vertical condenser through a spray
trap.
oKeep at the bottom end of the condenser a 400 ml beaker
containing 80 ml of 0.5N HCl to which 2 drops of methyl red
indicator is added .
oHeat and boil the flask continuously to distill off the evolved
ammonia into the beaker .
oWhen the distillation is complete ,take out the beaker and
titrate against 0.5N NaOH solution ,until the colour changes
to yellow, note the titre value, let that be “x”.
oConduct a blank titration using 20 ml of 0.5N HCl against
0.5N NaOH . Let this value be “y”.

149. •CALCUALTION :
Volume of acid consumed
by evolved ammonia = (4y - x ) = z
1 ml of N/5 HCl = 0.0028 g of nitrogen.
% of Hide substance = (4y-x) * 0.0028 * 5.62 * 100/1.5
To find out the hide substance multiply the nitrogen content by
5.62, the factor which represents the pure collagen content.

150. 14. ANALYSIS OF SOAK LIQUOR
AIM:
To estimate the amount of sodium chloride present in the
soak liquor.
REAGENTS REQUIRED:
o0.1N silver nitrate
oPotassium chromate indicator
PRINCIPLE:
This analysis follows the principle of argentometric titration
where the salt in the soak liquor reacts with silver nitrate to
form silver chloride. When all the chloride ions have reacted,
the chromate ion reacts with Ag
+
and a red precipitate
appears which is the end point.

151. •PROCEDURE:
The given soak liquor is filtered to remove the suspended
impurities. 10ml of the filtrate is titrated against standard silver
nitrate using potassium chromate indicator. The end point is the
appearance of brick red precipitate.
•CALCULATION:
1ml of 0.1N AgNO
3=0.005845g of NaCl

152. 15. ANALYSIS OF USED LIME LIQUOR
AIM:
To estimate the amount of lime in the used lime liquor.
REAGENTS REQUIRED:
oSaturated solution of ammonium chloride(10%)
oAmmonium oxalate 10% solution
o10% sulphuric acid
oDecinormal solution of potassium permanganate
PRINCIPLE:
The calcium in the used lime liquor is precipitated as calcium
oxalate and dissolved in hot sulphuric acid. It is then titrated
against standard potassium permanganate.

153. •PROCEDURE:
50ml of clear filtered lime liquor is pipetted out in to a beaker. It
is diluted to about 250ml with distilled water. 25ml of
ammonium chloride is added and boiled. To precipitate calcium,
hot solution of ammonium oxalate is added and allowed to
stand on water bath for 2 hours. The precipitate is then filtered
and washed with dilute ammonia solution till it is free of
oxalates. (About 25ml of the filtrate is acidified with dilute
sulphuric acid).
The calcium oxalate precipitated on the filter paper should be
dissolved in to a 500ml beaker using hot sulphuric acid. This hot
solution is titrated against standard permanganate.
The end point is the appearance of pale pink color.

154. •CALCULATION:
Lime present in 50ml of used lime liquor as calcium =V
KMnO4
*N
KMnO4

*0.02

5C
2
O
4
2-
+ 2 MnO
4
-
10 CO
2
+ 2 Mn
2+

155. 16. ANALYSIS OF PICKLE LIQUOR
•AIM :
To estimate the amount of NaCl and HCl in the given sample
of pickle liquor .
•PRINCIPLE :
The pickle liquor is titrated against standard NaOH to find
out the acid content and the neutralized pickle liquor is
titrated against standard AgNO
3
to find out the salt content
HCl + NaOH  NaCl + H
2O
NaCl + AgNO
3 AgCl + NaNO
3
2 AgNO
3 + K
2CrO
4  Ag
2CrO
4 + 2KNO
3

156. •PROCEDURE :
o10 ml of suitably diluted pickle liquor is taken in a conical flask
and titrated against 0.1N NaOH using phenolphthalein
indicator. The end point is the appearance of pale pink color.
oIn another flask, 10 ml of pickle liquor is pipetted out. The acid
present is neutralized by adding the same amount of alkali
consumed in the above experiment. Then it is titrated against
standard AgNO
3
using potassium chromate indicator. The end
point is the appearance of brick red color precipitate.
•CALCULATION :
Pickle liquor Vs NaOH= a ml
Pickle liquorVs AgNO3 = b ml
1 ml of 0.1N NaOH = 0.00365 g HCl
1 ml of 0.1N AgNO
3 = 0.005845 g NaCl

157. a ml of the N NaOH = 0.0365 x a x N
NaOH
= HCl g




10 ml of sample contains = HCl g
  100 ml of sample contains = 10 x HCl g
Total chlorides - chlorides in reaction with AgNO
3
= Salt content
Mill equivalents of
total chloride = V
AgNO3 x N
AgNO3 = (A)
 
Meq. of HCl = V
NaOH
x N
NaOH
= (B)

158. Meq. of NaCl = (A) - (B)
Amount of NaCl in 10 ml = (A-B) x 0.05845 = C (g)

  Amount of NaCl in 100 ml= C (g) x 10

159. ANALYSIS OF FINISHING
CHEMICALS
A.CAESIN
B.GELATIN
C.WAXES AND SULPHONATED OILS
D.WATER PIGMENT FINISHES
E.LACQUER FINISHES
NITROCELLULOSE FINISHES
VINYL RESIN SOLUTION FINISHES

160. A. CASEIN
oThe fat content and alkalinity are important properties and
should be known.
oThe fat content is obtained by extracting about 10gm of the
casein with chloroform in a soxhlet apparatus, using a
thimble to hold the casein.
oThe acidity to phenolphthalein is best measured by weighing
out 1gm of the casein in to a conical flask, adding 25cc of
N/10 sodium hydroxide solution and shaking until solution is
complete.
oThe excess sodium hydroxide is titrated with N/10 HCl until
the pink color of phenolphthalein is discharged. The result is
best stated as cc N/10 alkali per gram casein.

161. B. GELATIN
Gelatin should be completely soluble in water. With all the
proteins that yield clear solutions a determination of the pH of
say a 1% solution by means of indicators or other suitable means
would be useful.
•PROCEDURE:
oThe protein solution is brought to pH 8.3. To this solution is then
added an excess of neutral formaldehyde and the solution is
then titrated with 0.1N sodium hydroxide solution.
oFor a given volume of protein solution the ratio “cc 0.1N
NaOH/mg total nitrogen by kjeldahl” is at a minimum value
when the protein is in its purest form.

162. oAs degradation takes place and consequently the long poly-
peptide chains are broken the formaldehyde titration increases,
and this of course increases the above ratio.
oFor control purposes it would, therefore, be advisable to
determine for the pure proteins the minimum values of such
ratio.

163. D. WATER PIGMENT FINISHES
oWater pigmented finishes are analyzed by diluting the
concentrated pigment finish ten or twenty fold with water,
and allowing the solution to settle in long tubes or better still
by centrifuging, it is often possible to obtain a good
knowledge of the pigments used.
oBy hanging strips of filter paper in the diluted pigment finish
valuable information as to the nature of the dyestuffs
employed may be obtained.
oAs a rule, the more soluble and the smaller molecular weight
dyes show the greatest tendency to creep up the filter
papers.
o Moisture and total solids are determined by drying a
weighed sample in a porcelain dish in an oven at

164. about 105°C, cooling in a desiccator, and weighing to constant
weight. The loss of weight corresponds to water and volatile
matter.
oThe residue total solids, and the percentages are calculated on
the basis of the original sample of pigment finish.
o The total solids are transferred to a soxhlet thimble, and
extracted first with ether, which removes a large part of greases,
then with chloroform, which removes any retaining shellac.
oAfter evaporation of the solvents the residues are dried to
constant weight.
oThe residue in the soxhlet thimble will contain the pigments and
the binding materials.
oIf dyes and dye lakes are absent the determination of total
nitrogen by kjeldahl will give a measure of the protein content.
oThe ash, as a rule, is a good guide to the total amount of
pigment when lakes are absent.

165. o Some pigment finishes contain ammonia, so before
commencing any analysis this should be tested for, and if found
present, its amount should be determined by distilling with
alkali and absorbing the ammonia in an excess of standard
acid.
oThe excess of acid should then be titrated with sodium
hydroxide, using methyl red or bromophenol blue as indicator.
oThe alkalinity of the finish should also be determined by
titrating with standard acid to pH 8.3(neutrality to
phenolphthalein), and also to pH 5.0 approximate isoelectric
point of gelatin and casein.
oBromocresol green in comparison with a solution at pH 5.0 is a
suitable indicator to use in such cases.

166. oThe following are some of the common pigments used in water
finishes.
White pigments
Black pigments
Red pigments
Yellow pigments
Blue pigments
oFrom the list of pigments and with the aid of qualitative
analysis it should be possible to identify the pigments employed
except where these are the dyestuff lakes, though here the
bright shade is often an indication, especially if albumins and
barium sulphate are found to be present in the ash.
oDetailed analysis of the matter extracted by ether, chloroform
and alcohol will afford further information.
oSulphonated oils are used in the manufacture of pigment
finishes.

167. oThe un sulphonated portions are likely to be found in ether
or chloroform extract along with the waxes, whilst the
sulphonated part is likely to be present in the alcoholic
extract.

168. E. LACQUER FINISHES
NITRO CELLULOSE FINISH:
oThese are composed of cellulose nitrate(generally called nitro
cellulose), plasticizers, solvents and diluents.
oFor colored lacquer, pigments are dispersed in this system or
dyestuffs are dissolved in it. In addition to cellulose nitrate
other esters of cellulose such as cellulose acetate, cellulose
acetate butyrate or cellulose acetate proprionate may be
used and some use has been made of cellulose ethers in the
lacquer finish, but the most important cellulose derivative
used is still cellulose nitrate.

169. VINYL RESIN SOLUTION FINISHES:
oIn this type of solvent finish vinyl resins as poly vinyl chloride or
copolymers of vinyl chloride and vinyl acetate are used as the
film forming ingredients in place of nitro cellulose.
oThese products are produced by polymerization of vinyl chloride
(CH
2
=CHCl) or mixtures of vinyl chloride and vinyl acetate
(CH
2
=CH-O-COCH
3
). High polymers of vinyl chloride alone show
very little

stability in the common organic solvents but presence
of 10% to 15% vinyl acetate increases degree of solubility.
oFor solution coatings, therefore, vinyl resins used in general
consists of co-polymers of vinyl chloride and vinyl acetate.
oIn leather finishes, resins consisting of polyvinyl acetate alone do
not seem to be generally used because of their low softening
points which create problems in hot processing such as embossing
and hot plating.

170. oThe vinyl resins give tough, strong films with, however, a rather
low degree of extensibility and so it is necessary to use
plasticizers to produce satisfactory leather finish keeping an eye
on avoidance of brittlement caused by the migration of
plasticizer.
oThe most active solvents for the polyvinyl chloride resins are
ketones such as acetone, methyl-ethyl ketone and methyl iso
butyl ketone.

171. •DETERMINATION OF NITROCELLULOSE AND
PIGMENT:
o Weigh out accurately about 50gm of the lacquer in to a
tared beaker and add slowly, with vigorous stirring, a large
excess of benzene which precipitates the nitrocellulose and of
course, the pigment.
oAllow the precipitate to settle and pour off the clear liquid in
to a clean, dry, stoppered bottle or flask.
oRedissolve the nitrocellulose of course in a mixture of equal
volumes of acetone and alcohol, using the minimum of the
mixed liquids to bring about solution.
oReprecipitate with benzene and add the clear liquid to the first
decantation in the stoppered bottle. Repeat the above
operations once more.
oSuspend in benzene the precipitated nitrocellulose and
pigment, and transfer to a tared filter paper. This is then dried
and weighed. In the absence of lakes the pigment may usually
be determined by ashing and weighing.

172. oThe difference between the two weights corresponds to the
nitrocellulose.
oIf the presence of a lake is suspected(very bright red and
blue shades) a total nitrogen content determination by
kjeldahl should be carried out.
oThe nitrocellulose content may then be calculated on the
assumption that it contains 11.5% of nitrogen, though this is
only an average figure.