﻿the Adsorption of Ions. 333 



When an acid is added to the water the neutralizations 

 according to scheme 2 (a) will he completely negligible, but 

 those according to scheme 2(b) will not be so. The total 

 number of neutral molecules of water formed in the surface 

 is still equal to " .r," but a number of them is now being 

 formed according to 2 (/>). Corresponding to the number of 

 neutralizations according to 2 (6), a number of hydrogen 

 ions will remain in the surface layer in excess of the number 

 of hydroxyl ions. The rate at which 2 (/>) proceeds thus 

 determines the free charge on the surface. An equivalent 

 number of anions remain unneutralized in the liquid and 

 form the second mobile sheet of the double layer. 



The free charge on the surface will evidently increase with 

 rise in the concentration of hydrogen ions in the solution. 

 There are, however, two factors opposing this increase in the 

 charge of the surface. 



A. The proportion of hydrogen ions striking on the surface 

 diminishes as the positive charge of the surface increases. 

 Only those ions which have sufficient kinetic energy to over- 

 come the electrical repulsion can reach it. If e be the 

 potential of the double layer in C.Gr.S. units, then the number 

 of collisions of the ions per unit surface per second is pro- 

 portional to 



U H o.C E o.^- E /^ ..... (6) 



where H o denotes the concentration of free hydrogen 

 ions in the liquid, "E" is the electronic charge in C.Gr.S. 

 units, T is the absolute temperature, Uho is the mobility of 

 the hydrogen ions in water, and K = E/No, where " R " is the 

 gas constant and No the Avogadro number. 



B. The other factor that tends to diminish the charge of 

 the surface is the electrical adsorption of the anion of the 

 acid added to the solution. That this plays an important 

 part will be evident from the following examples taken from 

 the observations of Perrin : — 



Rate of Electro- 

 Substance. Electrolyte. endosraotic outflow. 



XUOr, M/1000 HC1 +110 



„ M/1000 citric acid + 5 



,, M/1000 HNO :i (or HOI) +100 



„ M/1000 H 2 S0 4 + IS 



OCl 3 M/1000 IIN0 3 + 85 



„ M/1000 H.^SO, + 21 



,. M/500HO1 + 90 



, M/1000 H, C 2 , + 30 



„ Feebly acid with HC1 + 75 



., Solution of KH, (P0 4 )1 



with approximately the 

 same number of free 

 hydrogen ions as above 



