THE TESTING OF MILK AND CREAM 



135 



The strength of the solution will vary according to the number 

 of tablets used and the number of cubic centimeters of water 

 in which they are dissolved. The indicator is added to the 

 tablets when 

 they are manu- 

 factured. Con- 

 sequently, when 

 using an alkali 

 tablet solution 

 no phenol- 

 phthalein is 

 needed. A con- 

 crete example 

 will show how 

 these tablets are 

 used. 



Suppose that 

 it required 15 

 c.c. of an alkali tablet solution to neutralize the acid in 20 

 grams of milk. The tablet solution was made by dissolving 

 five tablets in 100 c.c. of water. What is the per cent of acid 

 in the milk? .03492 = gram of lactic acid one tablet will 

 neutralize. .03492x5 = .1746 gram of lactic acid five tablets 

 will neutralize. Since the five tablets are dissolved in 100 c.c. 

 of water, .1746 is the amount of lactic acid 100 c.c. of the solu- 

 tion will neutalize. Then .1746 -v- 100 = .001746, the strength 

 of 1 c.c. of the solution ; in other words, the number of grams 

 of lactic acid 1 c.c. of the solution will neutralize. 



.001746 x 15 = .026190 



.026190 -h 20 = .0013095 X 100 = .13095 per cent 



If five of the alkali tablets are dissolved in 97 c.c. of water, 

 each cubic centimeter of the solution will neutralize .01 per cent 

 of acid when 18 grams of milk are used. This solution is often 



FIG. 36. Apparatus for the Farrington acid test. 



