262 THE BOOK OF BUTTER 



another. In 1 c.c. of a normal solution of lactic acid 

 there is .09 gram of lactic acid. According to the above 

 rule 1 c.c. of any normal alkali solution would just 

 neutralize .09 gram of lactic acid. 



" In actual practice a solution weaker than a normal 

 solution is usually employed, because a normal solution is 

 so strong that any small variation in the amount used 

 makes a big variation in results. A common solution used 

 is THT normal (expressed N/lO). One c. c. of N/10 alkali 

 solution would neutralize .009 gram of lactic acid. An 

 example will illustrate how the percentage of acid in 

 milk is calculated. Suppose it took 6 c.c. of N/lO alkali 

 solution to neutralize the acid in 20 grams of milk. What 

 is the per cent of acid? One c.c. of N/lO alkali will 

 neutralize .009 gram of lactic acid. Six c.c. will neutralize 

 6 X .009 = .054 gram of acid. .054 -h 20 = .0027. 

 .0027 X 100 = .27 per cent acid in the milk. Formulated, 

 the above example is expressed as follows : 



^^ 100 =.27%. 



If the milk for the acid test is measured in cubic centi- 

 meters, it should be reduced to grams by multiplying 

 by the specific gravity of milk. The acjd is obtained in 

 terms of grams and we cannot divide grams by cubic centi- 

 meters and obtain per cent." 



Usually the milk or cream is measured, since the acidity 

 does not have to be determined exactly. For this reason 

 one Babcock pipetteful is considered to be 18 grams, which 

 is true of whole milk but not quite accurate in case of cream. 

 The computation is as follows : Suppose that it took 8 

 c.c. of N/10 alkali solution to neutralize the acidity in 

 one pipetteful or approximately 18 grams of cream. 



