Dissociating Powers of Free and of Combined Water. 163 



In calculating the percentage dissociations by the conductivity 

 method 



Here /*i = molecular conductivity of the solution of potassium chloride, 

 and // 2 = that of the solution of calcium chloride. Further, ^^ con- 

 ductivity at infinite dilution for the potassium chloride and jU 2w = same 

 for calcium chloride. 



From the method of Kohlrausch for calculating conductivity, 



KI and K 2 are cell constants. a\, bi, a 2 , and > 2 are readings on the bridge 

 for the corresponding resistances. 



Substituting the values of (3) in (2) we obtain 



K\ di V* Ko er 2 v 



a=r- -and/3=;- (4) 



&i u>i MI M ^2 u\ ju 2oo 



Substituting these values in (1) 



For potassium chloride m = l and for calcium chloride n = 2. If the 

 same cell be used then Ki = K 2 . And 



GI_ 2q 2 



At 25 /*! = 137 1 and /^ = 246.5. 

 Therefore, 



a i _ 2 q a , . 



1376i ^ 246.56 2 w 2 

 Whence, 



0.8996 X A- = -A- 



Now, by measuring a and 6 of the solution of potassium chloride for 

 the resistance Wi, the left-hand side of the equation becomes a constant. 

 A concentrated solution of calcium chloride is now taken in different 



portions and diluted in different amounts until 7^- becomes equal to 



6 2 w? 2 



the value for the left-hand side. 



The conductivity of the potassium chloride solution was found to be 

 103.6. The specific conductivity of the calcium chloride solution was 

 found to be 93. 12. The calcium chloride solution upon analysis proved 

 to be 0.6951 molar. 



Carnegie Inst. Wash. Pub. No. 170, p. 20 (1912). 



