Volumetric Analysis - Glycine, Acid NO, Iodine No etc

COMPILED BY Dr. A.LIYAKATH ALI & Mr. V. MAGENDIRA MANI

VOLUMETRIC ANALYSIS

CORE PRACTICAL – I

COMPILED BY

Dr. A.LIYAKATH ALI & Mr. V. MAGENDIRA MANI

DEPARTMENT OF BIOCHEMISTRY

ISLAMIAH COLLEGE (AUTONOMOUS)

VANIYAMBADI - 635751

FOR THE STUDENT OF IST

B. Sc BIOCHEMISTRY


ESTIMATION OF AMINO ACID (GLYCINE) BY SORENSON’S FORMAL TITRIMETRIC METHOD

CALCULATION

Weight of oxalic acid = 1.576 g

Titration – I

Standard oxalic acid Vs Sodium hydroxide

Indicator: phenolphthalein

S.NO Volume of standard oxalic

acid (ml)

Burette readings Volume of Sodium

hydroxide (ml) Initial (ml) Final (ml)

1 20 ml 0 ml

2 20 ml 0 ml

Volume of Standard oxalic acid solution (V 1) = 20 ml

Normality of Standard oxalic acid solution (N1 ) = 0.1 N

Volume of Sodium hydroxide solution (V2) = -------- ml

Normality of Sodium hydroxide solution (N2) = ?

We know that, V1N1 = V2N2

N2 = V1N1

V2

N2 = 20 x 0.1/---------

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

Normality of Sodium hydroxide solution (N2) = --------------------------- N


ESTIMATION OF AMINO ACID (GLYCINE) BY SORENSON’S FORMAL TITRIMETRIC METHOD

Aim

To estimate the amount of amino acid (Glycine) present in the whole of the given unknown

solution

Principle

Amino acid reacts with excess of formaldehyde to give free hydrogen ion and act as acidic

solution. This acidic solution is titrated against standard alkali (Sodium hydroxide) using phenolphthalein

as indicator.

Reagents required

i. Standard oxalic acid solution (0.1 N)

ii. Sodium hydroxide solution

iii. Phenolphthalein as indicator.

iv. Formaldehyde

v. Amino acid (Glycine)

Ex. No :

Date :


Titration – II (Test value)

Unknown amino acid (Glycine) Vs Standardized Sodium hydroxide solution

Indicator: phenolphthalein

S.NO Contents taken in the

conical flask (ml)



1

20 ml amino acid + 5 ml

formaldehyde

0 ml

2

20 ml amino acid + 5 ml

formaldehyde

0 ml

Volume of Sodium hydroxide consumed by amino acid (Glycine)

and formaldehyde (Test value) = ------------------ ml

Titration – III (Blank value)

Blank Vs Standardized Sodium hydroxide solution

S.NO Contents taken in the

conical flask (ml)



1

20 ml distilled water + 5 ml

formaldehyde

0 ml

2

20 ml distilled water + 5 ml

formaldehyde

0 ml

Volume of Sodium hydroxide consumed by formaldehyde (Blank value) = ------------------ ml

Volume of Sodium hydroxide consumed by amino acid alone = Test value – Blank value

= ------------------------ ml


Procedure

Titration – I

Standard oxalic acid Vs Sodium hydroxide solution

Weighed accurately 1.576 g of crystalline oxalic acid and transfer into a 250 ml of standard flask

then the volume is made up to 250 ml using distilled water. Pipette out exactly 20 ml of this solution

into a clean conical flask and two drops of phenolphthalein as indicator is added. This is titrated against

the Sodium hydroxide solution taken in the burette. The end point is the appearance of pale permanent

pink colour. The titrations are repeated for concordant values. From the titre value the normality of

Sodium hydroxide solution is calculated.

Titration – II (Test value)

Unknown amino acid (Glycine) Vs Standardized Sodium hydroxide solution

The given unknown amino acid (Glycine) solution is made up to 100 ml standard flask using

distilled water. Pipette out exactly 20 ml of this solution into a clean conical flask to this 5 ml of

formaldehyde and two drops of phenolphthalein as indicator is added and shaken well for 5 minutes.

This is titrated against the standardized Sodium hydroxide solution taken in the burette. The end point is

the appearance of pale permanent pink colour. The titrations are repeated for concordant values.

Titration – III (Blank value)

Blank Vs Standardized Sodium hydroxide solution

Pipette out exactly 20 ml of distilled water into a clean conical flask to this 5 ml of formaldehyde

and followed by two drops of phenolphthalein as indicator is added. The contents are mixed well for 5

minutes. This is titrated against the standardized Sodium hydroxide solution taken in the burette. The

end point is the appearance of pale permanent pink colour. The titrations are repeated for concordant

values.


Calculation

Volume of Sodium hydroxide solution (V 1) = -------------- ml

Normality of Sodium hydroxide solution (N1) = -------------- N

Volume of amino acid (Glycine) solution (V2) = 20 ml

Normality of amino acid (Glycine) solution (N2) = ?

We know that,

V1N1 = V2N2

N2 = V1N1

V2

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

Normality of amino acid (Glycine) (N2) = ------------------ N

Equivalent weight of Glycine - 75

The amount of amino acid (Glycine) present in the whole of the given unknown solution

= Normality of amino acid (Glycine) x Equivalent weight of Glycine x 100

1000

= ----------------------------- grams


Test value – Blank value

Test value – Blank value will give the actual amount of Sodium hydroxide consumed by the

amino acid solution. From this value the strength of amino acid is calculated, and from this strength the

amino acid present in the whole of the given unknown solution is calculated

Equivalent weight of Glycine = 75

Result

The amount amino acid (Glycine) present in the whole of the given unknown solution is

= ----------------------- grams


ESTIMATION OF SUGAR (GLUCOSE) BY BENEDICT’S TITRIMETRIC METHOD

CALCULATION

Standard Glucose = 250 mg of Glucose/250 ml of distilled water

Titration – I

Standard Glucose Vs Benedict’s Reagent

S.NO Volume of Benedict’s

Reagent (ml)

Burette readings Volume of Standard

Glucose (ml) Initial (ml) Final (ml)

1 5 ml 0 ml

2 5 ml 0 ml

5 ml of Benedict’s reagent consumes --------------------------- ml of Standard Glucose solution

Titration – II

Unknown Glucose Vs Benedict’s Reagent

S.NO Volume of Benedict’s

Reagent (ml)

Burette readings Volume of Unknown

Glucose (ml) Initial (ml) Final (ml)

1 5 ml 0 ml

2 5 ml 0 ml

5 ml of Benedict’s reagent consumes --------------------------- ml of unknown Glucose solution

Ex. No :

Date :


ESTIMATION OF SUGAR (GLUCOSE) BY BENEDICT’S TITRIMETRIC METHOD

Aim

To estimate the amount of Sugar (Gluose) present in the whole of the given unknown solution

Principle

When glucose is heated with an alkaline solution of copper (Cu2+

) ions, the copper (Cu2+

) ions is

reduced to cupric (Cu+) ion, which is precipitated as copper oxide CuO2. This is the basic for the

estimation of reducing sugar.

Reagent required

Benedict’s quantitative Reagent

Crystalline copper sulphate – 18 grams, anhydrous sodium carbonate – 200 grams, potassium

thio cyanate – 125 grams, sodium nitrite – 200 grams are added to 250 ml of distilled water and

dissolved with the aid of heat. The contents are finally made up to 800 ml using distilled water.

Standard glucose solution

Weighed accurately 250 mg of Glucose and transfer into a 250 ml of standard flask then the volume is

made up to 250 ml using distilled water. (concentration – 1 mg/1 ml)

Procedure

Titration – I

Standard Glucose solution Vs Benedict’s quantitative Reagent

Pipette out exactly 5 ml of Benedict’s quantitative reagent into a clean conical flask. To this add 1 gram

of sodium carbonate. The contents are mixed well and heated in boiling water bath till the first bubble

appearance. This is titrated against the Standard Glucose solution taken in the burette. The end point is

the disappearance of blue colour. The titrations are repeated for concordant values


X - Volume of standard Glucose

Y - Volume of unknown Glucose

ie. --------------------- ml of standard Glucose

--------------------- ml of unknown Glucose

100 ml of unknown solution contain = X

Y

= -------------------- mg

X 100


Titration – II

Unknown Glucose Vs Benedict’s Reagent

The given unknown Glucose solution is made up to 100 ml standard flask using distilled water.

The burette is rinsed with unknown Glucose solution and filled with the same unknown Glucose

solution. Pipette out exactly 5 ml of Benedict’s quantitative Reagent into a clean conical flask to this

1gram of sodium carbonate added. The contents are mixed well and heated till the first bubble

appearance. This is titrated against the unknown Glucose solution taken in the burette. The end point is

the disappearance of blue colour. The titrations are repeated for concordant values.

Precaution

i. Excess amount Sodium carbonate will cause frothing

ii. Benedict’s quantitative Reagent is kept at boiling throughout in the titration

iii. Porcelain bits are added to prevent bumping

Result

The amount of Sugar (Glucose) present in the whole of the given unknown

Solution ------------------------ grams


ESTIMATION OF ASCORBIC ACID USING 2, 6 DICHLORO PHENOL INDOPHENOL

Titration – I

Standard Ascorbic acid Vs Dye

S.NO Volume of Standard

Ascorbic acid (ml)

Burette readings Volume of Dye

(ml) Initial (ml) Final (ml)

1 10 ml 0 ml

2 10 ml 0 ml

Volume of the Dye = --------------------------- ml

10 ml of standard ascorbic acid consumes = ------------------------ ml of the dye

Titration – II

Unknown Ascorbic acid Vs Dye

S.NO Volume of Unknown

Ascorbic acid (ml)

Burette readings Volume of Dye

(ml) Initial (ml) Final (ml)

1 10 ml 0 ml

2 10 ml 0 ml

Volume of the Dye = --------------------------- ml

X - mg of ascorbic acid consumes = ------------------------ ml of the dye


ESTIMATION OF ASCORBIC ACID USING 2, 6 DICHLORO PHENOL INDOPHENOL

Aim

To estimate the amount of ascorbic acid present in the whole of the given unknown solution

Principle

Ascorbic acid is oxidized by the colourD dye 2, 6 – Dichloro phenol indophenol to dehydro

ascorbic acid. At the same time the dye is reduced to colourless compound, so that, the end point of the

reaction can be easily determined.

Ex. No :

Date :


Calculation

Working standard ascorbic acid = 10 mg ascorbic acid/100 ml of 0.6 % oxalic acid

[Concentration – 0.1 mg/1 ml]

Ie. 1 ml of standard contains 0.1 mg of ascorbic acid

10 ml standard contains [0.1 mg x 10] – 1 mg of ascorbic acid

10 ml standard ascorbic acid contains -------------------------- ml of Dye

Ie. 1 mg of ascorbic acid consumes ---------------------------- ml of Dye

But X of ascorbic acid consumes ------------------------ ml of Dye

The amount of ascorbic acid present in the 100 ml of unknown solution

=

= -------------------- mg

The amount of ascorbic acid present in the whole of the given unknown solution

= ------------------ mg x 10

= ------------------- mg

Volume of dye consumed by unknown solution

Volume of dye consumed by standard solution


Reagent required

2, 6 – Dichloro phenol indophenols

Dissolve 42 grams of Sodium bi carbonate and 52 grams of Dichloro phenol indophenols in 50 ml

of distilled water and finally dilute to 250 ml using distilled water.

Stock standard ascorbic acid solution

100 mg of ascorbic acid is weighed exactly and carefully transfer in to 100 ml standard flask and

made up to 100 ml using 0.6 % oxalic acid.

Working standard ascorbic acid solution

10 ml of Stock standard ascorbic acid solution is pipette out in to a 100 ml standard flask and

made up to 100 ml using 0.6 % oxalic acid.

Procedure

Titration – I

Standard Ascorbic acid Vs Dye

Pipette out exactly 10 ml of working standard ascorbic acid solution into a clean conical flask

and it is titrated against the dye taken in the burette. The end point is the appearance of pale

permanent pink colour. The titrations are repeated for concordant values.

Titration – II

Unknown Ascorbic acid Vs Dye

The given unknown ascorbic acid solution is made up to 100 ml standard flask using 0.6 % oxalic

acid. Pipette out exactly 10 ml of this unknown ascorbic acid solution into a clean conical flask and it is

titrated against the dye taken in the burette. The end point is the appearance of pale permanent pink

colour. The titrations are repeated for concordant values.


Result

The amount of ascorbic acid present in the whole of the given unknown

Solution ------------------------ mg


DETERMINATION OF ACID NO OF FAT

CALCULATION

Standard 0.1 N oxalic acid = 1.576 g of oxalic acid/250 ml of distilled water

Titration – I

Standard oxalic acid Vs Potassium hydroxide

S.NO Volume of standard oxalic

acid (ml)

Burette readings Volume of Potassium


1 20 ml 0 ml

2 20 ml 0 ml

Volume of Standard oxalic acid solution (V 1) = 20 ml

Normality of Standard oxalic acid solution (N1 ) = 0.1 N

Volume of Potassium hydroxide solution (V2) = -------- ml

Normality of Potassium hydroxide solution (N2) = ?

We know that,

V1N1 = V2N2

N2 = V1N1

V2

N2 = 20 x 0.1/---------

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

Normality of Potassium hydroxide solution (N2) = --------------------------- N


DETERMINATION OF ACID NO OF FAT

Aim

To estimate the amount of Acid no of the given Fat

Principle

During storage of fat become rancid. As a result the peroxide formation of the double bond by

atmospheric oxygen and or hydrolyzed by micro organism with liberation of free fatty acids. The amount

of acid present gives the indication of age and quality of the fat.

Acid value is the number of milligrams of KOH required to neutralize the free fatty acids in one gram of a

fat or oil. It is a measure of free fatty acid contents in a fat or oil.

Reagents required

i. Fat

ii. Fat solvent

iii. Standard 0.1 N oxalic acid

iv. Potassium hydroxide solution

v. Phenolphthalein as indicator.

vi. Methanol / Ethanol

Procedure

Titration – I

Standard oxalic acid Vs Potassium hydroxide solution

Weigh accurately 1.575 g of oxalic acid and transfer into a 250 ml of standard flask then the

volume is made up to 250 ml using distilled water. Pipette out exactly 20 ml of this solution into a clean

conical flask and two drops of phenolphthalein as indicator is added. This is titrated against the

Potassium hydroxide solution taken in the burette. The end point is the appearance of pale permanent

pink colour. The titrations are repeated for concordant values. From the titre value the normality of

Potassium hydroxide solution is calculated.

Ex. No :

Date :


Weight of the oil

Weight of the weighing bottle + Oil =

Weight of the weighing bottle = ( - )

Weight of the Oil transferred =

Titration – II

Test value

S.NO Contents taken in the conical

flask



1

-------------- gram of Oil + 50 ml

of alcohol + 3 drops of

Phenolphthalein

0 ml

Titration – III

Blank Value

S.NO

Contents taken in the conical

flask



1

50 ml of alcohol + 3 drops of

Phenolphthalein

0 ml

Volume of Potassium hydroxide consumed = Test value – Blank value

By Oil alone = ------------------ ml


Titration – II

Test Value

Weigh about one gram of edible oil and carefully transfer in to a clean dry conical flask. Then 50

ml of alcohol is added followed by two drops of phenolphthalein as indicator is added. The contents are

mixed well for 20 minutes. This is titrated against the standardized Potassium hydroxide solution taken

in the burette. The end point is the appearance of pale permanent pink colour and persisting up to

20 – 30 seconds.

Titration – III

Blank Value

50 ml of alcohol is taken in a conical flask and three drops of phenolphthalein as indicator is

added. The contents are mixed well. This is titrated against the standardized Potassium hydroxide

solution taken in the burette. The end point is the appearance of pale permanent pink colour and

persisting up to 20 – 30 seconds.


100 ml of 1 N Potassium hydroxide contain 56 grams of Potassium hydroxide

X = 56 x Strength of Potassium hydroxide x Test value – Blank value

1000

= --------------------- grams

Acid no of Fat = ---------------------- grams x 1000

Weight of Oil

= ----------------------------- Acid no of Fat (Oil)

Result

Acid No of the given Fat = ------------------------------


DETERMINATION OF IODINE NO OF FAT

CALCULATION

Standard 0.1 N potassium dichromate solution = 1.225 g of potassium dichromate solution /250

ml of distilled water.

Titration – I

Standard potassium dichromate solution Vs Sodium thio cyanate solution

S.NO Volume of potassium

dichromate solution (ml)

Burette readings Volume of Sodium thio

cyanate solution (ml) Initial (ml) Final (ml)

1 20 ml 0 ml

2 20 ml 0 ml

Volume of Standard potassium dichromate solution (V 1) = 20 ml

Normality of Standard potassium dichromate solution (N1) = 0.1 N

Volume of Sodium thio cyanate solution (V2) = -------- ml

Normality of Sodium thio cyanate solution (N2) = ?

We know that,

V1N1 = V2N2

N2 = V1N1

V2

N2 = 20 x 0.1/---------

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

Normality of Sodium thio cyanate solution (N2) = --------------------------- N


DETERMINATION OF IODINE NO OF FAT

Aim

To estimate the amount of Iodine no of the given Fat

Principle

Iodine no of fat is defined as the no of grams of iodine absorbed by 100 gram of fat or oil. It is

a measure of degree unsaturation of the fatty acids in a fat or oil. Unsaturated fatty acids, either free or

combined in lipids react with halogens like bromine and iodine which get decolorized. These halogens

add at the carbon – carbon double bond.

Hane’s method is used for the determination of Iodine number. About 1 gram of the fat is taken in a well

cleaned dry iodine flask. To this 20 ml of chloroform is added to dissolve the fat. The contents are

shaken well and kept for 30 minutes. Then 20 ml of potassium iodide is added to liberate the iodine and

it is titrated against standard sodium thio cyanate solution. From this titration iodine number of fat is

calculated.

Reagents required

i. Hane’s solution

ii. Fat

iii. Standard 0.1 N potassium dichromate solution

iv. Sodium thio cyanate solution

v. 10 % potassium iodide

vi. 1 % Starch

vii. Chloroform

Preparation of Hane’s solution

3.3 grams of iodine is dissolved in 200 ml of acetic acid by constant shaking and heating. It is cooled to

room temperature, to this 50 ml of glacial acetic acid containing 0.75 grams of Bromine is added and

mixed well and stored in brown bottle.

Ex. No :

Date :


Titration – I

Standard potassium dichromate solution Vs Sodium thio cyanate solution

Weighed accurately 1.225 g of potassium dichromate solution and transfer into a 250 ml of

standard flask then the volume is made up to 250 ml using distilled water. Pipette out exactly 20 ml of

this solution into a clean conical flask to this 5 ml of Conc. Hydrochloric acid is added, followed by 10 ml

of 10 % potassium iodide is added. This contents are mixed well and titrated against the Sodium thio

cyanate solution taken in the burette, the titration is continued until a pale brown colour is appears. At

the time 1 ml of 1 % Starch solution is added. And the titration is continued till to get the end point

appearance of emerald green colour, it is the end point. The titrations are repeated for concordant

values. From the titre value the normality of Sodium thio cyanate solution is calculated.

Titration – II

Determination of iodine no of fat (Test value)

Weigh about one gram of edible oil and carefully transfer in to a clean dry iodine flask. Then

20 ml of Chloroform is added, the contents are mixed well to dissolve the oil. To this 20 ml of Hane’s

solution is added, shaken well and kept in dark for 30 minutes with occasional shaking. Then the flask is

taken out to this 20 ml of 10 % potassium iodide is added to liberate iodine. Except the iodine that is

absorbed by the oil. To this mixture 100 ml of distilled water is added, so the liberated iodine is nicely

disturbed in the solvent then it is titrated against the Sodium thio cyanate solution taken in the burette,

the titration is continued until a pale brown colour is appears. At the time 1 ml of 1 % Starch solution is

added. And the titration is continued till to get the end point disappearance of blue colour it is the end

point.

Blank value

Blank value is also done without oil

Equivalent weight of iodine = 127


Titration – II

Determination of iodine no of fat (Test value)

Titration Contents taken in the conical flask


thio cyanate solution

(ml)

Initial (ml) Final (ml)

Test

-------------- gram of Oil + 20 ml of

chloroform + 20 ml of Hane’s Solution

+ 100 ml of distilled water + 20 ml of

10 % Potassium iodide + 1 ml of 1 %

Starch

0 ml

Blank

20 ml of chloroform + 20 ml of Hane’s

Solution + 100 ml of distilled water +

20 ml of 10 % Potassium iodide + 1 ml

of 1 % Starch

0 ml

Weight of the oil

Weight of the weighing bottle + Oil =

Weight of the weighing bottle = ( - )

Weight of the Oil transferred =

Blank value - Test value = -------------------- ml


Determination of iodine no of fat =

Equivalent weight of iodine x Blank value - Test value x Normality of Sodium thio cyanate solution x 100

1000 x Weight of the Oil

= ---------------------------------- Iodine no of fat (oil)

Result

Iodine No of the given Fat (Oil) = ------------------------------


Analytical contents of oils and fats

OILS

ACID VALUE

SAPONIFICATION VALUE

IODINE VALUE

Coconut oil 5 – 13 250 – 264 8.0 – 9.5

Sesame oil 1 – 10 188 – 193 103 – 117

Castor oil 0.3 – 4 178 – 188 80 – 88

Linseed oil 1 – 8 190 – 196 170 – 203

Ground nut oil 2 – 6 186 – 196 83 – 105

Butter 0.4 – 2 220 – 241 26 – 38

Ghee (cow) - 225.5 – 236 31.5 – 45

Ghee (Buffalo) - 228.5 – 236 26.5 - 44

Volumetric Analysis - Glycine, Acid NO, Iodine No etc

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Transcript of Volumetric Analysis - Glycine, Acid NO, Iodine No etc