166 Proceedings of Royal Society of Edinburgh. [sess. 
Instead of that, the limit was clearly considerably higher, about 12 
or 13. (The number actually obtained was 12*75.) Further, at 
no stage was there any evolution of gas observable, even on shaking 
the liquid, notwithstanding the large quantity of salt decomposed. 
The only reasonable assumption to he made was that the oxygen, 
instead of being liberated, was being used up to oxidise the hydro* 
gen of ammonium, probably also the nitrogen — otherwise there 
should still have been a considerable evolution of gas, as shown by 
the equation — 
3(NH 4 ) 2 S 2 0 8 = 2(NH 4 ) 2 S0 4 + 4H 2 S0 4 + N 2 . 
This would give an increased acidity of one third, making the 
limit about 13*5. On the other hand, assuming nitric acid to be 
the oxidation product, as shown by the equation 
8(NII 4 ) 2 S 2 0 8 + 6H 2 0 = 7(NH 4 ) 2 SQ 4 4- 9H 2 S0 4 + 2HN0 3 , 
the increased acidity would be only one fourth, giving a limit of 
slightly over 12*5.* As a matter of fact, the liquid was found to 
give a very well marked nitric acid reaction, although the pro- 
portion of silver nitrate originally added was far too small to be 
appreciable in a small quantity of the solution by means of the 
usual test for nitric acid. The matter was put beyond all doubt 
by heating about a gram of ammonium persulphate with solution 
of silver sulphate. There was only a slight evolution of gas, 
although the liquid was heated nearly to boiling, and the resulting 
liquid contained so much nitric acid that there was a quite con- 
siderable evolution of nitric oxide on treatment with ferrous 
sulphate and sulphuric acid. 
The quantitative experiment was commenced merely to obtain a 
rough idea of the increased rate of decomposition, and was not 
carried out in a very strict manner, the titrations being performed 
* Writing the three equations in comparable terms we have : — 
(1) 24(NH 4 ) 2 S 2 0 8 + 24H 2 0 = 24(NH 4 ) 2 S0 4 + 24H 2 S0 4 + 120 2 . 
(2) 24(NH 4 ) 2 S 2 0 8 - 16(NH 4 ) 2 S0 4 + 32H 2 S0 4 + 8N 2 . 
(3) 24(NH 4 ) 2 S 2 0 8 + 1 8H 2 0 — 21(NH 4 ) 2 S0 4 + 27H 2 S0 4 + 6 HN 0 3 . 
These give respectively 48H*, 64H’, and 60H*, for the same quantity of persul- 
phate, corresponding to the ratio:— 1, 1*33, 1*25. 
