﻿10 Mr. W. Sutherland on the Molecular 



limiting volume will be too small and will give the value of 

 t too small. According to the principles of "The Laws of 

 Molecular Force " (Phil. Mag. [5 J xxxv.) the limiting volume 

 of a gramme-molecule of HCi is found to be 27. As the 

 value of B for CI is 19, that for H is 8. This is double the 

 limiting volume of a gramme-atom of H in the CH 2 of organic 

 compounds. But the refraction equivalent of H in the 

 halogen acids is 3*5, while that in organic compounds is only 

 1*3. The limiting volume of a gramme of HOI is thus 0*741. 

 Taking B for Br as 26 and for I as 36, the limiting volume 

 of a gramme of HBr is 0*42, and of HI is 0*343. Hence for 

 the coefficient of p 4 (l— p±) in the limiting contraction for 

 HC1, HBr, and HI we have (0*741-0*5), or 0*241, and 

 0*120 and 0*073. For t these give 0*74 for HOI, 0*82 for 

 HBr, and 0*79 for HI. Using the values of t for the 

 halogens, these give for t for H the values 2*13, 2*22, and 

 2*34. These are necessarily too small, and are not incon- 

 sistent with the value 2*66 obtained from HNO s , which will 

 be adopted as the correct apparent one. To find t/(A,KB£)£ 

 for H for comparison with the average 0*222 for positive 

 ions we need K. Now the refraction equivalent (N ™ 1)B 

 for H in the halogen acids is 3*5, and B is 8, so N = 1*44 

 and K = W = 2*07. With r = 2*66 we get the product 

 desired, 0*19. This is not seriously different from the value 

 for other positive ions, but we shall see soon that this 

 approximate agreement is only accidental. When we 

 evaluate t/(\KB*)* for the OH ion with r = 0*29, X = 174, 

 K = 2-07, and B = 10, we get 0'007 in place of the 0*050 

 characteristic of monovalent negative ions. In this respect 

 OH is strongly exceptional. 



It will now be shown that all these seemingly exceptional 

 results for H and OH become quite regular, if it is assumed 

 that these ions ionize H 2 into H and OH. The simplest 

 way to demonstrate this is to calculate first what the regular 

 ionic velocities of H and OH ought to be by the equation 

 XB*K/V = 280. These are 67*5 for H and '62*7 for OH, 

 which are much nearer to the values for ordinary ions than 

 the experimental 318 and 174. The sum of the velocities of 

 the ions of H 2 is 130*2, so that if H ionizes (318-67*5)/130*2 

 or 1*92 times H 2 0, the apparently high ionic velocity 

 assigned to H would be accounted for as being nearly three 

 times the velocity of H with nearly twice that of OH. In 

 the same way OH ionizes (174-62*7)/130*2 or 0*86 times 

 H 2 0, its apparent ionic velocity being nearly twice the true 

 amount with that of H added on. Sow we can readily test 

 these deductions by applying them to the values of r. Again, 



