PHENOLS 71 



The preparation is carried out as follows: Make a phenol solution 

 in N/10 hydrochloric acid, which contains approximately 1 mg. 

 of crystallized phenol per cubic centimeter. Transfer 25 c.c. of 

 this solution to a 250 c.c. flask, add 50 c.c. of N/10 sodium hydrox- 

 ide, heat to 65 C., add 25 c.c. of N/10 iodine solution, stopper 

 the flask, and let stand at room temperature thirty or forty 

 minutes. Add 5 c.c. of concentrated hydrochloric acid and titrate 

 the excess of iodine with N/10 thiosulphate solution. Each cubic 

 centimeter of N/10 iodine solution corresponds to 1.567 mg. of 

 phenol. On the basis of the result dilute the phenol solution so 

 that 10 c.c. contain 1 mg. of phenol. Five c.c. of this solution 

 (equivalent to 0.5 mg. of phenol), when 10 c.c. of the phospho- 

 tungstic-phosphomolybdic reagent and 25 c.c. of saturated sodium 

 carbonate solution are added, and the whole made up with water 

 at about 30 C. to 100 c.c. give when set in the colorimeter at 

 20 mm. a convenient standard. 



Calculation. The filtrate used for the determination of free 

 and total phenols contains the phenols from one-half the amount 

 of urine analyzed. The actual determination of phenols, both 

 free and total, is made upon a two-fifths portion of this filtrate, 

 and this amount of filtrate contains the phenols from one-fifth of 

 the amount of urine analyzed. In the determination of free phenols 

 the colored solution is diluted to only half that of the standard 

 while in the determination of total phenols the dilution is the same 

 as that of the standard. 



Hence, 



r> 



-T = milligrams of free phenol 

 /faX 4 



and 



r> 



^- = milligrams of total phenol 

 112X2 



in 2 c.c. or 4 c.c. of urine according to whether 10 e.c. or 20 c.c. of 

 urine was taken for analysis, when R\ is taken as the reading 

 obtained with the standard solution, and R2 is taken as the reading 

 obtained with the unknown. 



