. METHODS FOR QUANTITATIVE DETERMINATIONS. 415 



normal alkali, it is necessary to know how much of the acid in 

 question is equivalent to i. e., is required to neutralize one c.c. of 

 the alkali. Knowing this, it is easy to find how much acid is equiva- 

 lent to a certain number of c.c. of normal alkali used in titration. 

 These alkali equivalents of normal acid, or acid equivalents of a 

 normal alkali, are easily found by the student from the equation 

 of reaction. 



Thus, to find how much HC1 is equivalent to one c.c. of normal alkali, we 

 use the equation : 



HC1 + KOH == KC1 + H 2 O, 

 36.18 55.74 

 from which we see that 



36.18 gm. HC1 = 55.74 gm. KOH == 1000 c.c. normal KOH, 

 or, 0.03618 gm. HC1 1 " 



2KOH + H 2 S0 4 = K 2 S0 4 + 2H 2 O. 

 2 X 55.74 97.35 



97.35 gm. H 2 SO 4 = 2 X 55.74 gm. KOH = 2000 c.c. normal KOH. 

 48.675 " " 55.74 " " = 1000 " 



0.048675 " 1 " 



Phosphoric acid presents a peculiar case. When tropaeolin is used as an 

 indicator the change in color takes place when this reaction is completed : 



H 3 P0 4 + KOH = KH 2 P0 4 + H 2 0. 

 97.29 55.74 

 Hence, 



97.29 gm. H 3 PO 4 =r 55.74 gm. KOH == 1000 c.c. normal KOH. 



0.09729 " " = 1 " 



With phenolphthalein, the indicator changes color when the subjoined reac- 

 tion is completed : 



H 3 P0 4 + 2KOH = K 2 HPO 4 + 2H 2 O. 

 Hence, 



97.29 gm. H 3 P0 4 = 2 X 55.74 gr. KOH ~ 2000 c.c. normal KOH. 

 or 0.04864 " " 1 " 



The calculation for the amount of acid or alkali is made in this way. Sup- 

 pose we weigh off 10 grammes of dilute sulphuric acid. On titrating with 

 normal KOH it is found that 20 c.c. are required to cause the change in the 

 indicator (litmus) i. e., to completely neutralize the acid. We know from the 

 above that 1 c.c. normal KOH requires 0.04867 gramme H 2 SO 4 for neutraliza- 

 tion, hence 20 c.c. normal KOH require 0.04867 X 20 = 0.9735 gramme H 2 SO 4 , 

 That is, in the 10 grammes dilute acid weighed off there are 0.9735 gramme 

 H 2 SO 4 , or in 100 grammes there are 9.735 grammes, or 9.73 per cent. This 

 instance will serve as a type for all calculations of percentages. 



Use of empirical solutions. The primary advantage in using normal, deci- 

 normal, etc., solutions is the fact that the calculations of results is very much 

 simplified in a system which involves molecular and atomic weights, or simple 

 fractions thereof. But any solution of definite strength can be employed. All 



