A DAIRY LABORATORY GUIDE 45 



neutralize 6X-OO9=.O54 grams of acid. .054-^-20 

 .0027. .oo27Xioo=.27% acid in the milk. Formu- 

 lated, the above example is expressed as follows: 



wjp x ioo=.27% 



If the milk for the acid test is measured in cubic 

 centimeters it should be reduced to grams by mul- 

 tiplying by the specific gravity of milk. The acid 

 is obtained in terms of grams and we cannot divide 

 grams by cc. and obtain per cent. 



Professor Farrington has devised some alkali 

 tablets, each one of which will neutralize .03492 

 grams of lactic acid. These tablets are dissolved 

 in water and an alkali solution made. The strength 

 of the solution per cc. will vary according to the 

 number of tablets used and the number of cc. of 

 water in which they are dissolved. The indicator 

 is added to the tablets when they are manufactured. 

 Consequently, when using an alkali tablet solution 

 no phenolphtholein is needed. A concrete example 

 will show how these tablets are used. 



Suppose that it required 15 cc. of an alkali tablet 

 solution to neutralize the acid in 20 grams of milk. 

 The tablet solution was made by dissolving five 

 tablets in TOO cc. of water. What is the per cent 

 of acid in the milk? 



.O3492=grams of lactic acid one tablet will neu- 

 tralize. .03492X5=- 1 746 grams of lactic acid five 

 tablets will neutralize. Since the five tablets are 

 dissolved in 100 cc. of water .1746 is the amount of 

 lactic acid 100 cc. of the solution will neutralize. 

 Then . 1 746-^- ioo=.oo 1746, the strength of I cc. of 

 the solution ; in other words, the number of grams 

 of lactic acid i cc. of the solution will neutralize. 



.026i9o-=-20=.ooi 3095 X i oo=. 1 3095 % 



