22 CHEMISTRY OF FARM PRACTICE 



In the reaction between phosphoric acid, HaPCU, and 

 sodium hydroxide, NaOH, if there were present enough of 

 the base to replace one only of the three acid hydrogens, 

 we could represent the action by the equation, 



I NaOH+H 3 P0 4 = NaH 2 P0 4 +H 2 0. 



In the formula of the sodium dihydrogen phosphate, it 

 will be noticed that the three bonds of the acid radical, 

 PO4, are held by a combination of three bonds, one from 

 the sodium atom and two from the two hydrogen atoms. 



Should there be sufficient sodium hydroxide to replace 

 two of the three hydrogen atoms of phosphoric acid, the 

 trial equation would be, 



NaOH+H 3 P0 4 = Na 2 HPO 4 +H 2 O. 



It will be seen that the total number of sodium, oxygen, 

 or hydrogen atoms on one side of the equation does not 

 correspond or balance the atoms of these elements on the 

 other side. Starting with the disodium hydrogen phos- 

 phate, which is the most complicated formula of the equa- 

 tion, its two sodium atoms require two sodium atoms in 

 the left-hand or factor side of the equation, therefore, we 

 take two molecules of sodium hydroxide. The equation 

 then will be balanced, except its hydrogen and oxygen, 

 which may be made equal on each side by making two 

 molecules of water on the product side. 



II. 2NaOH+H 3 P0 4 = Na 2 HP0 4 +2H 2 0. 



When there is excess of sodium hydroxide, Na 3 P0 4 , would 

 be formed and the equation would be, 



III. 3NaOH+H 3 P0 4 = Na 3 P0 4 +3H 2 0. 



Thus we find three different reactions represented by the 

 three equations which are possible when sodium hydroxide 

 is treated with phosphoric acid. 



