SOLUTIONS. MACGREGOR. 



71 



(2.) As equation (2; may be written : 



+ 



we may proceed as follows : 



Plot a new curve D 

 (Fig. 2) with the same 

 abscissae as A and B, 

 but with ordinates 

 equal to the sum of 

 the ordinates of A and 

 N 2 /N L times the or- 

 dinates of B. Draw 

 K L parallel to the 

 ionic-concentration 

 axis and at a distance 

 1/N 1 from it, and let 

 it cut D in L. Draw 



L Q parallel to the dilution axis and cutting A and B in 



P and M respectively. P and M are the two points required. 



For they have the same abscissa Q, and their ordinates, P Q 



and M Q, are such that 



Then 



NX N/ 



a, = OQ.PQ,anda 2 = OQ.MQ. 



(3) Plot a new curve 

 E (Fig 3), having the 

 same abscissae as A and 

 B, but with ordinates 

 equal to Na/Nj times the 

 ordinates of B. Draw 

 R S parallel to the axis 

 of ionic concentrations 

 and at a distance from 

 it of l/(2 N t ). Find, by 

 inspection, the line T Y 

 parallel to the axis of 



