Example.—1.8976 grammes were taken, and required 5.2 c.c. of N/10 KOH solution for neutralisation.
| 5.2 × 0.0282 × 100 | ||
| ————————— | = | 7.72 per cent. free fatty acids, expressed as oleic acid. |
| 1.8976 |
The free acidity is sometimes expressed as acid value, which is the amount of KOH in milligrammes necessary to neutralise the free acid in 1 gramme of fat or oil.
In the above example:—
| 5.2 × 5.61 | ||
| ————— | = | 15.3 acid value. |
| 1.8976 |
The saponification equivalent is determined by weighing 2-4 grammes of fat or oil into a wide-necked flask (about 250 c.c. capacity), adding 30 c.c. neutral alcohol, and warming under a reflux condenser on a steam or water-bath. When boiling, the flask is disconnected, 50 c.c. of an approximately semi-normal alcoholic potash solution carefully added from a burette, together with a few drops of phenol-phthalein solution, and the boiling under a reflux condenser continued, with frequent agitation, until saponification is complete (usually from 30-60 minutes) which is indicated by the absence of fatty globules. The excess of alkali is titrated with N/1 hydrochloric or sulphuric acid.
The value of the approximately N/2 alkali solution is ascertained by taking 50 c.c. together with 30 c.c. neutral alcohol in a similar flask, boiling for the same length of time as the fat, and titrating with N/1 hydrochloric or sulphuric acid. The "saponification equivalent" is the amount of fat or oil in grammes saponified by 1 equivalent or 56.1 grammes of caustic potash.
Example.—1.8976 grammes fat required 18.95 c.c. N/1 acid to neutralise the unabsorbed alkali.
Fifty c.c. approximately N/2 alcoholic potash solution required 25.6 c.c. N/ acid..
25.6 - 18.95 = 6.65 c.c. N/1 KOH required by fat.
1.8976 × 1000 / 6.65 = 285.3 Saponification Equivalent.