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Investigatory Project on Determination of Saponification Value of Oil

Investigatory Project on Determination of Saponification Value of Coconut Oil Experiment (Practical) 

Aim of the Investigatory Project

 

To determine the saponification value of coconut oil

 

Theory

 

Saponification value of an oil or fat is the number of milligrams of potassium hydroxide required to hydrolyse completely 1g of oil or fat. Oils and fats are glycerides of higher fatty acids. On heating with NaOH or KOH, they are hydrolysed to form glycerol and sodium or potassium salt of higher fatty acids.

By knowing the amount of KOH required to hydrolyse a definite weight of oil or fat, we can calculate its saponification value.

 

Apparatus and Chemicals

 

1. 250 mL round bottomed flasks and reflux condensers (quick fit type).

2. Chemical balance

3. Potassium hydroxide

4. Alcohol (rectified spirit or methylated spirit)

5. 0.5N hydrochloric acid solution

6. Pure anhydrous sodium carbonate

7. Coconut oil

8. Phenolphthalein and methyl orange indicators

9. Pipettes and burette

 

Procedure

 

First prepare 0.5N alcoholic potash as follows. Dissolve about 2.8g of KOH in 100mL of pure rectified spirit or methylated spirit. (The weight of KOH need not be very accurate. The alcohol used should be colourless).

 

Weigh accurately about 2g of pure coconut oil and transfer it into a 250mL round bottomed flask. (Take about 2 mL of coconut oil in a dry weighing bottle and find its mass using a chemical balance. Pour the oil into the R.B. flask as much as possible and then take the mass of the weighing bottle again. The difference in mass will give the exact mass of oil transferred to the R.B flask). Add exactly 25 mL of 0.5N alcoholic KOH to it. The flask is fitted with reflux condenser and fixed on a stand. Connect running water to the condenser for cooling it. The flask is heated on a low flame for about one hour (i.e., till the reaction is complete and the liquid is clear).

 

A blank experiment is simultaneously conducted as follows. Take exactly 25 mL of 0.5N alcoholic KOH alone in another R.B. flask. Fit the reflux condenser and reflux it also for one hour. Both the flasks are then cooled. Wash down the inner side of the reflux condenser into the respective R.B. flasks with minimum quantity of water. Add 1 or 2 drops of phenolphthalein to each R.B. flask and titrate against 0.5N HCl taken in the burette.

 

The hydrochloric acid is then standardised using standard sodium carbonate solution. Weigh accurately about 2.7 g of pure anhydrous sodium carbonate and transfer it to a 100 mL standard flask. Dissolve it in water and make up to 100 mL. Pipette out 20 mL of this sodium carbonate solution into a conical flask and titrate against the hydrochloric acid using methyl orange as indicator. Repeat to concordant titre values.

 

OBSERVATION AND CALCULATIONS

 

Article weighed

Weights added

Total weight

gram

mg

Rider weight

Weighing bottle + oil

 

 

 

 

Weighing bottle after transferring oil

 

 

 

 

 

Therefore, Mass of coconut oil taken - _____g (Wg)

 

Article weighed

Weights added

Total weight

gram

mg

Rider weight

Weighing bottle + Na2CO3

 

 

 

 

Weighing bottle alone

 

 

 

 

 

 

Therefore, Mass of Na2CO3 in 100 mL = _____g(w1g)


Mass of coconut oil taken = Wg


Mass of sodium carbonate in 100 mL = w1g


Equivalent mass of sodium carbonate = 53


Therefore, Normality of sodium carbonate solution = w1x10/53 = N1


Volume of Na2CO3 solution pipette out = 20 mL


Volume of HCl required to neutralize 20 mL of sodium carbonate solution = v1mL


Volume of HCl x Normality of HCl = Volume of Na2CO3 x Normality of Na2CO3


v1 x Normality of HCl = 20 x N1


Therefore, Normality of HCl = 20 x N1/v1 = N2


Volume of HCl required to neutralize the alkali in the blank experiment = v2 mL


Volume of HCl required to neutralize the excess alkali in the R.B. flask containing the oil = v3 mL


Therefore, Volume of HCl equivalent to the alkali used up by the oil = (v2 – v3) mL


Equivalent mass of KOH = 56.1


Therefore, Saponification value of the oil = [(v2 – v3) x N2 x 56.1]/W


Note: Saponification value of coconut oil = 253 – 262

Saponification value of olive oil = 185 – 196

Saponification value of palm oil = 200 – 205

 

Model Viva Questions and Answers

 

1. What are oils and fats chemically ?

Ans: Oils and fats are glycerides of higher fatty acids.

 

2. Give the name and formulae of three important fatty acids found in oils and fats.

Ans: Palmitic acid - C15H31COOH, stearic acid - C17H35COOH, Oleic acid - C17H33COOH

 

3. What is saponification ?

Ans: Alkaline hydrolysis of esters is known as saponification.

 

4. What are the products obtained when oils are heated with caustic soda ?

Ans: Glycerol and sodium salt of higher fatty acids (soap).

 

5. What is soap ?

Ans: Soap is the sodium or potassium salt of higher fatty acids.

 

6. Define saponification value of an oil or fat

Ans: See the 'theory' of the project.

 

7. Define iodine value of an oil. What does this indicate ?

Ans: Iodine value of an oil is defined as the mass in grams of iodine that react with 100 grams of oil. It is a measure of the degree of unsaturation of an oil.

 

8. What is the main structural difference between an oil and a fat ?

Ans: Fats are glycerides of saturated fatty acid while oils are glycerides of unsaturated fatty acids.

 

9. Which indicator is used for the titration of Na2CO3 with HCI?

Ans: Methyl Orange.

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