Roberts — Action of Reducing Agents on Iodic Acid. 155 



experiment showed plainly that the essential product was 

 iodine monochloride. 



In three different ways I have tried to estimate the propor- 

 tions in which iodic acid and potassium iodide unite in pres- 

 ence of an excess of the former and hydrochloric acid. The 

 direct method of letting potassium iodide of known strength 

 slowly into a mixture of the two acids and determining the 

 first appearance of free iodine is difficult on account of the 

 action of the iodine chloride on the blue iodide of starch 

 which one would naturally use as an indicator. I have suc- 

 ceeded, however, in getting an approximate result, by taking 

 out a drop of the liquid from time to time and testing on 

 starch paper. With 5 cubic centimeters of iodic acid, 9"5 

 cubic centimeters of potassium iodide were added before there 

 was any effect on the starch paper. This would indicate 

 molecular proportions of iodic acid to potassium iodide as 

 1:1-9. 



In the second method, 20 cubic centimeters of iodic acid, 10 

 of hydrochloric acid, and .4 of potassium iodide were put to- 

 gether, the resulting iodine chloride was extracted with ether 

 and separated with a separating funnel. The aqueous solution 

 containing free hydrochloric acid was then treated with an 

 excess of potassium iodide and the amount of iodine set free 

 determined, after neutralization, by alkaline arsenious acid. 

 From this the iodic acid remaining in the solution after the 

 extraction of the iodine chloride could be readily estimated, 

 and the difference between this and the amount taken gave the 

 amount used in the reaction. Three trials of this experiment 

 gave the following results for the molecular proportions in 

 which the iodic acid and potassium iodide unite in presence of 

 hydrochloric acid while the iodic acid is in excess : 



HI0 3 : KI 1 



1 : 2-06 I . 1 , no 



„„ ^Average = 1 : T98 



1:1-9 J 



The third method consisted in putting together the reagents 

 in the same proportions as above and then decomposing the 

 iodine chloride with acid potassium carbonate and estimating 

 the iodine set free by alkaline arsenious acid. According to 

 Gay-Lussac, the reaction between iodine chloride and an alkali 

 is represented thus : 



6KOH + 5IC1 = 5KC1 + KIO, + 3 H 2 + 2l 2 



The iodine shown by the arsenious acid is then only -| of that 

 in the iodine chloride. Making this correction and knowing 

 the amount of iodine in the potassium iodide used, it becomes 



