175 
N 
10 
casein is capable of neutralizing as much alkali as 1 ce. of a acid, so if we take 
1 ce. acid = .11315 grams casein, or in other words .11315 grams of 
11.315 grams of milk we see from the relation above that every ce. of 
N 
10 
we need only change the normality of our acid. 
acid used equals 1 per cent. casein. By using different quantities of milk 
If by using 11.315 grams of milk (or 11 ce.) where each ce. of Be acid 
corresponds to 1 per cent., by using a greater or larger quantity of milk the 
normality would have to be correspondingly less or greater. When we use 
8.75 cc. or 9 grams of milk the normality would not be X but 795 ce. 
AN acid plus water to make 1,000 ec. which equals Upon the above 
< 
12.56 +. 
facts the volumetric method of Van Slyke and Bosworth is based. 
Procedure in carrying out in detail the volumetric estimation of casein: 
‘““A given amount of milk, diluted with water, is made neutral to phenolpthalein 
by the addition of a solution of sodium hydroxide. The casein is then completely 
precipitated by the addition of the standardized acetic acid; the volume of the 
mixture is then made up to 200 cc. by the addition of water, thoroughly shaken 
and then filtered. Into 100 ce. of the filtrate a standard solution of sodium 
hydroxide is run until neutral to phenolpthalein. The solutions are so stand- 
ardized that 1 ce. is equivalent to 1 per cent. of casein when a definite amount of 
milk is used. The number of ce. standard acid used, divided by two (since 
only 100 ce. of the 200 ce. is used), less the standard alkali used in the last 
titration gives the percentage of casein in the milk examined.” When 17.5 
or 18 grams of milk are used the strength of acetic acid and alkali are made 
by diluting 795 ce. of N to 1,000 ce. The same normality as was derived 
above. Since only 100 ce. of the 200 cc. were titrated this then represents 
the acid required to liberate the casein in 8.75 ce. or 9 grams of milk. Like- 
wise by using 22 ce. cr’*22.6 grams of milk treated as above, then 1 ce. of 
cat acid equals 1 per cent of casein. By the use of a factor any con- 
venient quantity can be used. Example, by the use of 20 ce. of milk and 
- solution, adjustment is made by multiplying the final result by 1.0964. 
Apparatus and reagents necessary to carry on the volumetric estimation 
of casein in milk are, first, two 50 ec. burettes, graduated to 1/10 ce. or better 
1/20 ee., these must be accurate. One of the burettes should be supplied 
with a glass stop cock for the acid, and one with a pinch cock for the alkaline 
solution. Second, flasks, volumetric, holding 200 ec. At least two of these 
are needed and where a number of estimations are to be made more are 
required to do rapid work; ten to twelve are necessary for rapid work. The 
