A. A. Ramsay 297 



Ca3P20g was prepared by pouring pure CaClg into a very dilute 

 solution of Na2HP04 plus 1 equivalent of ammonia. The alkaline 

 phosphate was kept in excess during precipitation. The triphosphate 

 was washed by decantation till no chlorine was found in the washings. 

 All analyses were made unless the contrary is stated by dissolving 

 an unweighed quantity of moist phosphate in hydrochloric acid, pre- 

 cipitating the lime with ammonium oxalate in acetic acid solution and 

 afterwards the phosphoric acid with magnesia mixture. 



The following figures are then given as representing the CaO and 



P2O5 in the phosphates prepared. 



(a) (6) (c) id) 



CaO 54-37 54-78 54-95 54-93 



P,0= ... 45-63 45-22 45-05 45-07 



100-00 100-00 100-00 100-00 



Warington has apparently assumed that because the ratio of lime 

 to phosphoric acid approximates that of 3CaO : P2O5, the compounds 

 obtained were necessarily tricalcium phosphates, and the possibility of 

 these products being mixtures of di- and tricalcic phosphate, or of these 

 with free lime, appears to have been overlooked. 



Let us examine for a moment the figures obtained by Warington. 

 (a) 54-37CaO with 45-63P205 . Now while theory requires 53-98CaO 

 to be associated with 45'63P205 to form tricalcic phosphate, 54:*37CaO is 

 found present. Therefore we have either 



(1) tricalcic phosphate plus free Ume, or 



(2) tricalcic phosphate plus dicalcic phosphate plus free hme. 



(2) will obtain if the amount of free hme is at all greater than that 

 required for supposition (1). 



(6) 54-78CaO with 45-22P205. Now theoretically the 45-22P205 

 present would require 53-05CaO to form CagPaOg, whereas there is 

 54'78CaO present. Again therefore we have either a mixture of tri- 

 calcic phosphate and free lime or of di-, tricalcic phosphate and free lime. 



(c) 54-95CaO with 45-05P2O5 . To produce CagPaOg from 45-05P2O5 

 would require 53-30CaO, and there is actually 54-95CaO present. Here 

 also we have either (1) a mixture of tricalcic phosphate and hme, or 

 (2) a mixture of di- and tricalcic phosphate and lime. 



In the first case a mixture of 98-35 % tricalcic phosphate (contain- 

 ing 53-30CaO combined with 45-05P2O5) with 1-65 % hme (CaO) would 

 agree with Warington's figures found, viz. 54-95CaO and 45-05P2O5, 

 Now if the free lime actually present exceeds 1-65 then di- and tricalcic 

 phosphate must be present and their amounts can be calculated but 



