102 PROCEEDINGS OF THE AMERICAN ACADEMY 



Analytical Methods. — Arsenic acid may be directly determined in 

 the soluble salts of this series by means of magnesia-mixture, but it is 

 best to redissolve the precipitated ammonio-maguesic arsenate in 

 chlorhydric acid, and precipitata a second time by ammonia. Arse- 

 nous oxide may be determined by titrition with iodine after adding an 

 excess of an alkaline dicarbonate. As a check upon the determination 

 of the two oxides of arsenic, it is well to oxidize a portion of the salt 

 directly by means of nitric acid, and then determine the total arsenic 

 oxide by means of magnesia-mixture. A further check may also be 

 obtained by reducing the arsenic oxide by boiling with sulphurous 

 acid, and then titrating the total ai'senous oxide by means of iodine. 

 In solutions from whicli the arsenic has been completely removed as 

 AsgS. , tungsten can be determined by means of mercurous nitrate, 

 provided that no chlorhydric acid or chloride is present. 



21 : 1 : 4 : 10 Potassic Arsenoso-phospho-tiinf/state.-'—'When potassic 

 bromide is added to a solution of arsenoso-tungstate of soda, and a 

 solution of potassic arsenate, AsO^K^II, is then poured into the clear 

 filtrate, a white crystalline precipitate is formed. This salt requires a 

 very large quantity of hot water for solution. Baric chloride boiled 

 with it gives a well-defined flocky-crystalline salt. Argentic nitrate 

 gives a similar salt, but this has a faint tinge of fawn-color, perhaps due 

 to the presence of a trace of potassic arsenate not completely removed 

 by washing. Mercurous nitrate gives on boiling a clear yellow flocky- 

 crystalline salt. In the i^otassic salt, 



1.9985 gr. gave 0.34G8 gr. As.O.Mg., = 12.36<% As,Og 



2.0085 gr. '• 0.0527 gr. Aspg by iodine = 2.62% AsgOg 



I 



After complete oxidation by nitric acid, 



1.5004 gr. gave 0.3 1G2 gr. As,0-Mg2 = 15.03% As.fis 

 1.9G84 gr. " 1.2935 gr. WOg = 65.72% 



1.5825 gr. " 1.0346 gr. " =65.37% 



1.3890 gr. '• 0.9382 gr. PtCl.K, = 13.09% K^O 



1.5765 gr. lost over a radiator 0.0955 gr. = 6.06% 



The analyses lead to the formula 



21 WO. . As.Og . 4 AsoO^ . 10 K,0 + 26 aq, 



which requires : 



