8 PUBLIC HEALTH CHEMISTRY 



make a standard solution of NaOH 40 grammes to I litre, 

 and one of HC1 36-5 grammes to i litre, then 



i c.c. stand, sol. NaOH = i c.c. stand, sol. HC1, 



and if we can build other solutions on the same plan, we 

 shall have a whole range of solutions which are chemically 

 equivalent c.c. for c.c. This will obviate having a strength 

 of acid for titrating soda, another strength for potash, 

 another for baryta, and so on, and reversely. This result 

 is obtained, or rather attained, by using normal solutions, 

 which are thus denned : 



A normal solution is one which contains in i litre of 

 distilled water at 16 C. the hydrogen equivalent of the 

 active reagent weighed in grammes, hydrogen being taken 

 as one gramme. Such a solution is usually indicated by 



N 

 the mark N or . If diluted, the degree of dilution is 



i 



indicated by a denominator, thus 



N N N N 



2 IO IOO IOOO 



which are respectively 

 seminormal decinormal centinormal millinormal 



The hydrogen equivalent of a substance is found by 

 taking its molecular weight, the number of atoms usually 

 replaced by hydrogen, and the valency of these atoms. 

 Divide the molecular weight by the product of the number 

 of atoms and their valency. Thus 



NaCl m.w. =58-5 no. of replaceable atoms=i valency =i ; 

 therefore 



N/i NaCl=58-5/ixi=58-5 grm. to i litre of water. 

 Similarly 



HC1 m.w. =36-5 n.r.a. = i v. = i; hence 



N /i =36-5 grm. per litre. 



H 2 S0 4 m.w. = 98 grm. N/i =49 grm. per litre. 

 HNO 3 m.w. = 63 N/i; = 63 

 NaOH m.w. = 40 N/i=4O 

 KOH m.w. = 56 N/i = 56 ,; 

 Na 2 C0 3 m.w. = 106 N/i = 53 

 K 2 C0 3 m.w. = 138 N/i =69 



