408 ANALYTICAL CHEMISTRY. 



one-half the gram-molecule must be taken, since this quantity contains one 

 atom of hydrogen, and so on. 



A normal alkali solution may be defined as one containing that quantity of 

 alkali in a liter which is chemically equivalent to i. e., neutralizes^ one gram- 

 atom of acid hydrogen. For such compounds as KOH, NaOH, NH 4 OH, the 

 whole gram-molecule must be taken to make a liter of normal solution. In 

 case of Ba(OH) 2 or Ca(OH) 2 , one half the gram-molecule is taken. 



It will easily be seen that all the acid solutions made as described are of 

 equivalent strength and are exactly equivalent to the solutions of alkali i. e., one 

 liter normal HC1 will exactly neutralize one liter of normal KOH, or NaOH, 

 or Ba(OH) 2 . That this is so will appear at once on writing the equations 

 which express the reactions between these alkalies and acids, thus: 



HC1 + KOH = KC1 + H 2 O. 

 36.18 55.74 



2HC1 + Ba(OH) 2 = BaCl 2 + 2H 2 O. 



36.18 



L 



KOH + H 2 SO * = KO*- + H 2 O. 

 55,4 ^ 



The above equations show that 36.18 gramme of HC1 are equivalent to 

 55.74 gramme of KOH, but these are the quantities taken for a liter of normal 

 solution respectively, hence these normal solutions must be equivalent. Simi- 

 larly for HC1 and Ba(OH) 2 , for KOH and H 2 SO 4 , etc. 



Conversely, if a solution of unknown strength be compared by titration with 

 a second solution known to be normal, and then be properly diluted so that the 

 two are equivalent, volume for volume, the first solution will also be normal, 

 and from the definition of normal solutions the quantity of reagent in a liter 

 of the solution becomes at once known. For example, if a solution of sul- 

 phuric acid be made equivalent to a normal solution of caustic alkali it will 

 then contain 48.675 grammes of absolute sulphuric acid per liter. It is thus 

 an easy matter to prepare normal solutions, although it may be impossible to 

 weigh exactly the amount of reagent necessary for a liter of such solutions. 

 All that is required is one normal solution as a starting-point. 



Sodium carbonate, Na 2 C0 3 , may be used as an alkali, just as KOH or NaOH, 

 because it neutralizes acids in precisely the same manner, for the carbonic acid 

 has no effect, being volatile and escaping into the air. The decomposition 

 taking place thus: 



Na 2 CO 3 + 2HC1 = 2NaCl + H 2 O + CO 2 , 



shows that one molecule of sodium carbonate neutralizes two molecules of 

 hydrochloric acid, consequently one-half gram-molecule of sodium carbonate 

 must be taken to make a liter of normal solution. A similar consideration 

 will show that of sodium bicarbonate the whole gram-molecule should be 



