LAr 
of ec. acid used in liberating the casein. Since a quantity of milk has been 
taken so that each ce. of acid used equals 1 per cent. casein, then each 
ce. represents 1 per cent. of casein in the sample of milk. For example, it 
required 9.4 cc. to neutralize 100 ce. of the filtrate, and since it represented 
12.5 ce. of the acid added to the 200 cc. of the diluted milk, we have 12.5- 
N 9.4 = 3.10 per cent. casein. 
Below are some of Van Slyke’s results obtained by this method in com- 
parison with the official method. 
PERCENT CASEIN. 
Voh metric Method 
(Van Slyke-Bosworth). Official Method. 
3.00 3.00 
3.40 3.36 
3.30 By PAN 
Se) 3.16 
2.90 2.95 
2.70 2.60 
The second volumetric method which I wish to consider is that of E. B. 
Hart, of the University of Wisconsin, published in Research Bulletin, No. 
10, 1910. For speed and accuracy this method offers no advantage over 
that of Van Slyke’s and Bosworth’s, just mentioned. However, the method 
is unique and sound in principle. The fact that free casein has the properties 
of an acid and can combine with an alkali in a definite proportion, it seems 
rational that if we dissolve casein in excess of alkali and the uncombined 
alkali is estimated by titration, using phenolpthalein as an indicator, we 
are in a position to calculate the casein equivalent per cc. of standard alkali 
used. This is true, and upon this principle rests Hart’s volumetric method. 
Hart found the casein equivalent for each 1 ce. X| KOH to be .108 grams. 
Therefore, if we titrate the casein obtained from 10.8 grams of milk, we see 
that each ce. of alkali used must represent 1 per cent. of casein. 
Details of the method. Measure 10.5 ce. or weigh 10.8 grams of milk 
into a 200 ec. Erlenmeyer flask, add 75 ce. of distilled water at room tem- 
perature and add to this 1 to 1.5 ce. of a 10 per cent. solution of acetic acid. 
The flask is given a quick rotary motion, usually 1.5 ec. of acetic acid gives 
5084—12 
