_Free acidity_ is estimated by weighing out from 2 to 5 grammes of the fat or oil, dissolving in neutral alcohol (purified methylated spirit) with gentle heat, and titrating with a standard aqueous or alcoholic solution of caustic soda or potash, using phenol-phthalein as indicator. The contents of the flask are well shaken after each addition of alkali, and the reaction is complete when the slight excess of alkali causes a permanent pink coloration with the indicator. The standard alkali may be N/2, N/5, or N/10. It is usual to calculate the result in terms of oleic acid (1 c.c. N/10 alkali = 0.0282 gramme oleic acid), and express in percentage on the fat or oil. _Example._--1.8976 grammes were taken, and required 5.2 c.c. of N/10 KOH solution for neutralisation. 5.2 x 0.0282 x 100 ------------------ = 7.72 per cent. free fatty acids, 1.8976 expressed as oleic acid. 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 x 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 x 1000 / 6.65 = 285.3 Saponification Equivalent. The result of this test is often expressed as the "Saponification Value," which is the number of milligrammes of KOH required for the saponification of 1 gramme of fat. This may be found by dividing 56,100 by the saponification equivalent or by multiplying the number of c.c. of N/1 alkali absorbed, by 56.1 and dividing by the quantity of fat taken. Thus, in the above example:--