FORMATION OF IODIC ACID. 361 



Ditte found - 2'24, and Thomsen - 217, at a slightly different 

 temperature. 



(c) Dilution of iodic acid. 



HI0 3 (1 equiv. = 1 litre) + its volume of water, at 13 ... - 0-30 

 HIO (1 equiv. = 2 litres) + ... - 0-08 



HI0 3 (1 equiv. = 4 litres) + ... - O'O 



(d) Solution of anhydrous iodic acid. This body was pre- 

 pared pure, and its composition ascertained by analysis. 



^[I 2 5 (1 part to 45 parts of water, at 12) 4- water, - 0'81. 



Ditte found 0*95, and Thomsen 0*89, at a slightly different 

 temperature. 



(e) Solution of semihydrated iodic acid. This body is 

 crystallised and well defined. The composition was ascertained. 



J[2HI0 3 I 2 5 (1 part 4- 45 part of water, at 12) 4- water, - 2-86. 



(/) It was thought necessary to ascertain whether the three 

 solutions formed by anhydrous, monohydrated, and semi-hydrated 

 acid contain the acid in the same molecular state. 



To this end, these solutions were treated, as soon as they 

 were made, with potash (1 equiv. == 2 litres). They all liberated 

 the same quantity of heat 



For[I 6 ] + 14-28 



ForHIO, +14-31 



i[2HI0 3 ,I 2 OJ ... +14.35 



(g) Solution of the potassium iodates, Neutral iodate (crystal- 

 lised) 



KI0 3 (crystallised) (1 part 4- 40 parts of water) 4- water, at 12, 



- 6-05. 

 Dilution. 



KI0 3 (1 equiv. = 2 litres) 4- its volume of water, at 13, - 0-36 

 KI0 3 (1 equiv. = 4 litres) -f - 0-0 



Acid iodate (crystallised) 



KI0 3 ,HIO 3 (crystallised) (1 part 4- 40 parts of water) 

 4- water -11-8 



(h) Formation of iodic acid from the elements. From the pre- 

 ceding data we deduce 



J[I 2 solid 4- 5 + water = H 2 0, I 2 5 diluted], + 22-6. 

 This number, obtained by the synthetical method, is con- 

 sistent with the value 4- 21'5 found by Thomsen, by means of 

 analytical processes. 

 We have, moreover 



_ + 6 = I 2 5 (anhydrous)] ... + 18-0 Cal. 



I 2 (gas) + 6 = I 2 5 (solid) +23-4 



+ I + 3 + water = HI0 3 (dissolved) ... + 57-1 



H + I + 3 = HI0 3 (crystallised) + 59-8 



J[H 2 (solid) + I 2 5 (solid) = 2HI0 3 (crystallised)] + 1-13 



