12 WORK OF C. M. STINE. 



considerably, in many cases, from that of the single solutions assumed to have the 

 same normality. This would involve changes in friction in the movement of the 

 ions. Also, the amount of water combined as hydrate, i. e., the dimensions of the 

 atmosphere about the ion, continually changed from one normality to another. 

 This involved a varying degree of freedom of movement of the ions. Again, the 

 variation in the number of particles present, ions and molecules, must involve 

 continual variation in the amount of friction between the ions. As a comparatively 

 short, and at the same time sufficiently accurate method, we have adopted the 

 following in apportioning the diminution in conductivity to the constituent salts: 

 Since we have dealt in every case with mixtures in which the two electrolytes have 

 a common ion, we have based the method on the number of this common ion yielded 

 by the constituent salts in separate solutions at the same normality. By the same 

 normality we mean that if equal volumes of two solutions are mixed, and any dimi- 

 nution in volume which may take place upon mixing is compensated for by the addi- 

 tion of pure solvent, then the normality of each solution in the mixture may be 

 assumed to be exactly half of that of the constituent solutions. If suppression of 

 the ionization is dependent upon the common ion, then it will be inversely as the 

 number of this common ion furnished by each of the salts in a mixture. This number 

 was obtained thus : The figure representing the normality of the individual solutions, 

 multiplied by the value of a for the solution at the normality in question, would give 

 the gram-atomic weight of the common ion. Suppose the case to be one where a 

 solution of potassium chloride has been added to an equal volume of a solution 



of calcium chloride. It being assumed that calcium chloride dissociates into 



++ - - 



Ca, and CI, CI, the figure representing the gram-atomic weight of the common ion 



would have to be doubled for a ternary electrolyte, such as calcium chloride. Since 

 two chlorine ions are required to re-form a molecule of calcium chloride, the figure 

 expressing the number of chlorine ions driven back, in the case of calcium chloride, 

 must be divided by two, in order to get the relative number of molecules of calcium 

 chloride re-formed. 



Let us take the values of {j.*, as expressing the molecular conductivity of the com- 

 pletely dissociated molecules of the various salts. If the figures for the relative 

 numbers of molecules of the two salts re-formed be multiplied by the proper value of 

 /Xoo, we shall then have the relative losses in conductivity. From this the absolute 

 losses are readily obtained. An example may suffice to make this clear: Take the 

 case of the mixture of equal volumes of a solution of 1.8 N calcium chloride and of 

 1.7 N potassium chloride. The resulting solution, any diminution of volume having 

 been compensated for, will consist of 0.9 N calcium chloride and 0.85 N potassium 

 chloride. The dissociation of a 0.9 N solution of the specimen of calcium chloride 

 employed was found, by conductivity measurements, to be 53.2 per cent; of 0.85 N 

 potassium chloride, 8.32 per cent. Doubling the value of a in the case of calcium 

 chloride and multiplying by the normality, we get 0.9576. Similarly, for potassium 

 chloride, 0.70G8. These represent the relative numbers of chlorine ions which 

 would be furnished by each salt in the mixture if no suppression of the ionization 

 took place upon mixing. That is, if driving back occurs, according to the law of 



