( 141 ) 



In each of (lie saturation points three saturation surfaces and there- 

 fore also three saturation lines meet each other; such a point there- 

 fore represents a solution saturated with three solid substances. 

 From this it follows that the solution represented: 

 by h is saturated with Cu S0 4 . 5 H,0, Li, S0 4 . H,0 and D Cu 

 ., I „ „ ,, D L i, Li, S0 4 . H,0 and D Cll 



„ w ,, „ „ D L i, (NH 4 ), S0 4 and D Cn . 



This shows that each of these solutions is saturated with Cu S0 4 . 

 (NH 4 ), S0 4 . 6 H,0. 



With the aid of this figure we may readily draw some conclusions. 

 Let us therefore observe the external lines, for instance dk. The 

 point d represents a ternary solution saturated at 30° with Cu S0 4 . 

 5 H,0 and Li, S0 4 . H,0. To this solution we add (NHj, S0 4 ; the 

 solution will now alter its composition until at last a third solid 

 phase appears. What is this phase? (NHJ, S0 4 forms a double salt 

 with copper as well as with lithium sulphate and the question now 

 arises which of these two will appear first. The experiment shows 

 that Cu S0 4 . (NH 4 ),SO . 6 H,0 is formed. If we start from the 

 ternary solution h which is saturated at 30° with Cu S0 4 . 5 H,0 

 and Cu S0 4 . (NH 4 ), S0 4 . 6 H,0 and if Li, S0 4 . H,0 is added the 

 solution undergoes the changes represented by points of the line hk 

 until finally the third solid phase occurs in k in this case Li, S0 4 . H,0. 



If we start from the ternary solution ƒ saturated at 30° with 

 (NH 4 ), S0 4 and Li, S0 4 . (NH 4 ), S0 4 and if we add Cu S0 4 . 5 H,0 

 and represent the solution by in Cu S0 4 . (NH 4 ), S0 4 . 6 H,0 is formed 

 as the third solid phase; if we start from the ternary solution g 

 which is saturated with (NH 4 ), S0 4 and Cu S0 4 . (NH 4 ), SO, . 6 H,0 

 and add Li, S0 4 . H,0, Li, S0 4 . (NH 4 ), S0 4 will form in m as the 

 third phase. 



If we start from the ternary solution e which is saturated with 

 Li, S0 4 . H,0 and Li, S0 4 . (NH 4 ), S0 4 and add Cu S0 4 . 5 H,0 the 

 solution traverses the branch el; in I however a new solid phase is 

 formed, namely, Cu S0 4 . (NH 4 ), S0 4 . 6 H,0. 



Suppose a plane is passed through the points W, Cu and Dlï 

 of the tetrahedron ; the points of this plane represent solutions with 

 a constant proportion of the components Li, S0 4 and (NH 4 ), S0 4 ; this 

 ratio is the same as that in which they occur in the double salt. 

 This plane intersects the saturation surface leq/m of this double 

 salt, so that this is not only soluble without decomposition in water 

 but also in solutions of copper sulphate of a definite concentration. 



