﻿Constitution of Aqueous Solutions. 7 



unit time will be proportional to x 3 . Moreover, the chance of 

 the ions bringing Jf of one stion and \) of another into such a 

 position as favours their forming a bond may be taken pro- 

 portional to (1/K# 2 ) 2 , the square being taken because there 

 are two independent electrons to be controlled by the electric 

 force at the surface of the ion. In (H 2 0) 3 there are 3 bonds 

 to be considered ; but probably the formation of one causes the 

 other two to form almost automatically, because (H 2 0) 3 forms 

 spontaneously with such ease. Thus we are going to measure 

 the rate of formation of (H 2 0) 3 from sHons by x*/(Ka 2 ) 2 . 

 The rate at which a positive ion breaks up (H 2 0) 3 must be 

 proportional to the rate at which the ion hemmed in amongst 

 (H 2 0) 3 molecules moves so as to get near to a bond in such a 

 way as to break it. The ion is kept on the move by alternate 

 attractions and repulsions proportional to 1/Ka 2 generating 

 velocity proportional to \/Ka 2 , where X is the ionic velocity 

 under standard external electric force. Hence for equilibrium 



4*l(Ka*)*=c\/Kjofi, (11) 



where in dilute solutions c may be treated as a constant, 

 though for concentrated ones it becomes appreciably a function 

 of x. Now, though x has been taken to represent a number 

 of H 2 stions, .r/3 really represents the number of (H 2 0) 3 

 molecules destroyed by the presence of the positive ion, the 

 stions JH 2 0(7 existing mostly as (H 2 0) 2 in process of formation 

 from (H 2 0) 3 , or into (H 2 0) 3 . Hence x is proportional to r, 

 and (11) gives us the result that for a positive ion with ace B J , 



t/(\KB*J* is constant, .... (12) 



In a similar way we c*an make the problem of the action of 

 the negative ions tractable by treating it as a process of 

 building up (H 2 0) 2 in excess of its amount in pure water 

 through the agency of an imaginary supply of stions repre- 

 senting the actual increase of flux and reflux between (H 2 0) 2 

 and (H 2 0) 3 caused by the presence of the negative ion. On 

 account of the 2 in the formula of (H 2 0) 2 we obtain for the 

 negative ion instead of (12) 



r/(XKB*)* is constant (13) 



In the next table these relations (12) and (13) are tested 

 by means of the values of t/(\KB^ and t/(\KB^ ; given 

 in the last row from the data, namely t from Table I., Xfrom 

 Kohlrausch (Wied. Ann. lxvi.), B from Table I. in which 

 v±e = B for monovalent ions and = B/2 for divalent, and K from 

 " The Dielectric Capacity of Atoms " (he. cit.), or from the 

 formula there given. 



