Conductivity of Organic Acids in Ethyl Alcohol. 133 



98Q23 

 reciprocal of this ratio, or '-^~ - - Similarly, the normality at 35 



1^28 



N 25 ., , 1.01114 



would be -I-I-IA an d the volume would be 



RESULTS. 



In the table of conductivity results (table 75), V m is the volume at 

 which the solutions were made up ; V c is the corrected volume. The 

 corrections were applied in the manner just described, both for expan- 

 sion or contraction of the alcohol and for change in the concentration 

 of the acid due to the formation of ester. Molecular conductivity, 

 /* was calculated in the usual manner. 



A word of explanation should be added in regard to the method of 

 calculating temperature coefficients of conductivity. The formulae gen- 

 erally employed in the determination of these coefficients are T = 



rr\ 



, ~ and A = where fj. t and fj. t > represent the molecular conductivi- 

 t t fj.t 



ties of the same solution at t and t', respectively (t'>t); T is the tem- 

 perature coefficient expressed in conductivity units, and A is the per- 

 centage temperature coefficient. But these formulae as such are not 

 applicable to alcoholic solutions of organic acids, because the true 

 volume (V c ) of a given solution is different at different temperatures 

 (due to the causes already described). For example, in the case of 

 benzoic acid (see table 75), the N/8 solution was found to have a 

 volume of 7.945 at 15, of 8.05 at 25, and of 8.15 at 35. Before the 

 conductivities of these three solutions can be compared they must be 

 reduced to values corresponding to the same volume. For the sake of 

 simplicity V c at 25 is taken as the standard of reference. The reduc- 

 tion of ju, at 15 to V c at 25 is made in the following way: The specific 



conductivity at 15 (^y^ 92 = 0.000364) is multiplied by the differ- 



ence in volume at 15 and 25 (8.05-7.945 = 0.105). The product 

 (0.000038) is added to the molecular conductivity at 15 (0.002892), 

 giving 0.002930, which represents the molecular conductivity at 15 

 of a solution of volume 8.05. This value (0.002930) and the value of 

 ju, at 25 (0.004073) can then be substituted for ^ and /v in the above 

 equations, and T and A are found to be 0.0001143 and 3.90 respectively. 



/O 005444 \ 



Similarly, the specific conductivity at 35 (-' . - : = 0. 000668 j is 



multiplied by the difference in volume at 25 and 35 (8.15-8.05 = 0.10). 

 The product (0.000067) is subtracted from the molecular conductivity 

 at 35 (0.005444), giving 0.005377, which represents the molecular 

 conductivity at 35 of a solution of volume 8.05. This value (0.005377) 

 and the value of JU B at 25 (0.004073) can then be substituted for /v 

 and n t in the formulae, and T and A are determined to be 0.0001304 and 

 3.20 respectively. 



