430 H. L. Ward — Oxalate- Permanganate Process. 



alcohol, water, and acetic acid in equal parts. When very large 

 amounts of iron are present it is more satisfactory to increase 

 the dilution on precipitation to 100 cra3 . It is necessary in all 

 cases to have present a very large excess of oxalic acid to secure 

 the complete insolubility of the copper. 



Peters found that when potassium nitrate was present in the 

 water solution of a copper salt, all the copper was not thrown 

 down by oxalic acid. It becomes desirable, therefore, to ascer- 

 tain whether copper oxalate is completely insoluble in the 

 presence of commonly occurring salts, when one-half the solu- 

 tion consists of acetic acid. The results of experiments shown 

 in Table IX show clearly that the separation of the oxalate 

 is complete even when very small amounts of oxalic acid are 

 used. The potassium salts were chosen in preference to the 

 sodium salts because potassium oxalate is much more soluble 

 in water and is therefore less likely to crystallize out in the 

 course of an analysis. Ammonium salts may not be present, 

 as a soluble double oxalate is formed, which is stable in the 

 presence of a'large amount of free acetic acid. 



In the experiments detailed in the last division of the table 

 concentrated hydrochloric acid was neutralized with potassium 

 hydroxide and acetic acid added before precipitation. 



Table IX. 



Effect of Salts on the Precipitation of Copper Oxalate in the Presence of 



Acetic Acid. 

 Volume 



Copper 



present 



grm. 



Salt 



present 



grm. 



at pre- Oxalic 

 cipitation acid 

 cm 3 grm. 



Acetic 

 acid 

 cm 3 



Copper 



found 



grm. 



Error 

 grm. 



0*0501 

 0-0501 



1-0 



3-0 



100 

 100 



KN0 3 present. 



1 50 

 1 50 



0-0504 

 0-0504 



4-0-0003 

 + 0-0003 



0-0501 



1-0 



100 



K2SO4 present 



1 50 



0-0500 



—o-oooi 



0-0050 

 0-0250 

 0-0501 

 0*0501 



2-0 

 2-0 

 1-0 

 3-0 



100 

 100 

 100 

 100 



KC1 present 



1 50 

 1 50 

 1 50 

 1 50 



0-0045 

 0'0246 

 0-0501 

 0-0501 



-0-0005 

 — 0-0004 

 ±0-0000 

 ±0-0000 





HC1 

 cm 3 



HC1 neutralized with KOH 







0-0511 

 0-0511 

 0511 

 0-0511 

 0-0511 

 0-1002 



1 

 9 

 3 

 5 

 3 

 5 



100 

 100 

 100 

 100 

 100 

 150 



1 

 1 

 1 

 1 

 1 

 1 



50 

 50 

 50 

 50 

 50 

 100 



0-0513 

 0-0510 

 0-0511 

 00501 

 0-0511 

 0-1001 



+ 0-0002 



—o-oooi 



±0-0000 



—o-ooio 

 ±0-0000 

 —o-oooi 



