Analysis of oils: Determination of acid value
Item Type book_section
Authors Low, Lai Kim; Ng, Cher Siang
Publisher Marine Fisheries Research Department, Southeast Asian
Fisheries Development Center
Download date 26/09/2024 05:46:25
Link to Item http://hdl.handle.net/1834/41018
DETERMINATION OF ACID VALUE
LOW L. K. & NG C. S.
INTRODUCTION
The acid value is a measure of the extent to which the glycerides in the oil have been
hydrolysed by lipase action. The glycerides are also hydrolysed with water in the presence of air
and possibly bacteria. The decomposition is accelerated by heat and light.
As rancidity is usually accompanied by free fatty acid formation, determination of acid
value is often used as a general indication of the condition and edibility of oils.
The acid value is the number of milligrams of potassium hydroxide required to neutralize
the free fatty acids in 1.0 g of fat or oil.
I APPARATUS
Microburette (2 ml with 0.01 ml intervals)
Conical flasks (100 ml)
5 ml pipettes
II REAGENTS
a) 0.02N KOH in ethyl alcohol
Weigh 5.6 g of potassium hydroxide, dissolve in distilled water and make up to 100 ml with
distilled water (1N solution). Dilute 50 times with ethyl alcohol when required.
b) n-Hexane
c) 1% phenolphthalein or thymolphthalein in ethyl alcohol
Dissolve 1 g of either indicator in 100 ml of ethyl alcohol.
C-5.1
III PROCEDURE
1. Take 0.1 -0.3 g of fat sample or A ml of the extract containing 0.1 -0.3 g of fat in a 100 ml
Erlenmeyer flask.
2. Add [10 — A] ml of n-Hexane and 1-2 drops of indicator.
3. Titrate the solution against 0.02N KOH solution. The end point is reached when pink
(phenolphthalein) or blue (thymolphthalein) colour persists for 30 seconds.
4. Carry out a blank test using A ml of C-M Mixture instead of the extract.
IV CALCULATION
56.11 x 0.02 x (Vs — V b) x F
Acid value (mg/g) =
W
where Vs = titration volume of sample (ml);
Vb = titration volume of blank (ml);
W = weight of fat in the volume of extract usd (g);
F = factor of 0.02 KOH solution, where
5
F = : Vf is the volume of 0.02N KOH required to neutralize 5 ml of the
Vf 0.02N H2SO 4 solution.
56.11 = Molecular weight of KOH
0.02 = Concentration of KOH
C- 5.2
