ANALYTICAL cHEMISTRY - Iv

MSCM302

Experiment 3

Aim of the Experiment

Determination of saponification value of the the oil.

Principle

  • Saponification is the process of making soap from oil.

  • Saponification value is the number of mg of KOH required for complete hydrolysis of 1 gm of oil into potassium salt of fatty acid (known as soap).

  • This is hydrolysis reaction, chemically, fats are carboxylic acid ester derivatives, which are derived from single alcohol - glycerol, and is thus known as glyceride.

  • Oil and fats are long chains of fatty acids with ester groups. Saponification is process of converting esters into acids having alkali metal (in salt form) by hydrolysis. The saponification value is inversely proportional to relative molecular mass of fatty acid obtained from esters.

Saponification of oils

Apparatus required

Round bottom flask, Boiling chips, Condenser unit (air/water), Bent tube, Burette, Pipette, Funnel, Graduated pipette, etc.

Chemicals required

Concentrated HCl, 1 N HCl solution, 1 N alcoholic KOH solution, Oil samples, etc.

Preparation of reagents

  • Preparation of 1 N HCl solution from 35% concentrated HCl:

1000 mL 1 N HCl ≡ 36.5 gm HCl

100 gm of HCl ≡ 35 gm HCl (as concentrated HCl is 35% solution)

Therefore, 36.5 gm of HCl would be present in [36.5×100] / 35 = 104.28 gm of HCl solution.

1000 mL of 1 N HCl ≡ 104.28 gm

100 mL of 1N HCl = [104.28 × 1 ×100] / 1000 = 10.42 gm

Density of concentrated HCl = 1.18 g/mL, therefore,

1.18 gm HCl ≡ 1 mL HCl

10.42 gm HCl = [1 × 10.42] / 1.18 = 8.83 mL

Procedure

  • Wash the round bottom flask with acetone and then keep it in oven for drying. To this flask, add 1 gm of oil sample (weighed accurately) followed by 25 ml of 1 N alcoholic KOH solution. Drop some boiling chips and keep the flask for refluxing for about 90 minutes (attach appropriate condenser).

  • Meanwhile, prepare 1 N HCl solution as follows: transfer 8.9 ml concentrated HCl to a 100 ml volumetric flask already containing about 50 ml distilled water. Mix the solution and make up the volume using more distilled water. This gives 1 N HCl solution.

  • Perform a blank titration of 25 ml of 1 N alcoholic KOH solution against 1 N HCl from burette using phenolphthalein as indicator. Repeat the titration for concordant result. Let this reading be A ml.

  • Cool the refluxed solution to room temperature and titrate the solution against 1 N HCl from burette using phenolphthalein as indicator. Let this reading be B ml.

  • Subtracting blank reading from this reading gives the amount of KOH consumed for saponification, in terms of 1 N HCl.

1_Determination of saponification value of vegetables oil.mp4
2_Determination of saponification value of vegetables oil.mp4
3_Determination of saponification value of vegetables oil.mp4

Observations

Calculations

Result

Saponification value of the given oil is _________ mg/g of oil.

Reference Material

  1. G H Jeffery, J Bassett, J Mendham and R C Denney, Vogel's Textbook of Quantitative Chemical Analysis, 5th Edition

  2. S. Suzanne Nielsen, Food analysis, 4th Edition

Questions

  1. What is the relevance between saponification and hydrolysis reactions?

  2. Why we need to reflux the mixture of oil and alcoholic KOH solution to carry out saponification reaction?

  3. Why we need alcoholic KOH solution? Will aqueous KOH solution do the same job of alcoholic KOH solution?

  4. Give the examples of mono-, di-, and polyunsaturated fatty acids and discuss the effects of them on human body if they are present in any food source.

Developed by

Dr. Viraj Bhanvadia,

Assistant Professor, Chemistry,

viraj.bhanvadia@gsfcuniversity.ac.in