I04 THE JOURNAL OF PHARMACOLOGY. 



the contents of the crucible (Magnesium Pyrophosphate MgsPaOy), with 

 nitric acid, and heat cautiously over a small flame until this is again re- 

 moved. Finally heat to redness, and allow to cool in a dessicator, and 

 weigh, deducting the filter ash. 



ICO parts of MG2P2O7 = 63.97 parts of P2O5, 



2NH4MGPO4 = MG2P2O7 + 2NH3 + H2O, 

 MG2P2O7 = P2O5 + 2MGO, 



222 142 80 



222 : 142 : : 100 : x = 63.97. 



2. Volumetric: If phosphates or phosphoric acid are brought into 

 reaction in hot acetic acid solution, with uranium acetate or nitrate, a 

 precipitate of uranium phosphate is formed as follows: 



NA2HPO4 + UR02(C2H302)2 = URO2HPO4 + 2NAC2H3O2. 



Solution of uranium nitrate keeps its titre longer than uranium acetate 

 and is, therefore, employed preferably. During the reaction nitric acid 

 is liberated which will dissolve the precipitate of uranium phosphate. 

 This may be prevented by the addition of solution of sodium acetate. A 

 solution of potassium ferrocyanide is used as indicator, forming with 

 uranium salts a red-brown precipitate of uranium ferrocyanide. 



Reagents required: 



A. Solution of di-sodic hydric phosphate which is the equivalent of o. i 

 gramme of P2O5 in each 50 Cc. Prepared by dissolving 10.085 grammes 

 of chemically pure salt (NA2HPO4 + 12H2O), in one liter of water. The 

 strength of this solution must always be ascertained gravimetrically, as 

 the salt is of variable composition. If freshly recrystallized it is apt to be 

 contaminated by adhering water, or if old it has usually effloresced. 

 20 Cc. of the solution are, therefore, treated with 15 Cc. of Magnesia mix- 

 ture, and the precipitate treated as given under the gravimetric process 

 for phosphoric acid. If it is found that 50 Cc. of this solution are not the 

 exact equivalent of o.i gramme of P2O5, it must be either diluted concen- 

 trated as the case may be. Examples: 



a. 50 Cc. of solution were found to be the equivalent of o. 115 gramme 

 P2O5, consequently a dilution will be necessary as follows : 



.1 :50 : : .115 :^ = 57-5 

 or 500 Cc. must be diluted to 575 Cc. 



b. If it should be found that 50 Cc. of solution were the equivalent of 

 only 0.0S5 gramme of PsOg, concentration will be required as follows : 



.1 : 50 : : .085 :;f = 42.5 



or 500 Cc. must be concentrated to 425 Cc. 



After dilution or concentration, if either were necessary, it is advisable 

 to make a second determination. 



