W. J. V. OSTERHOUT 



419 



For reasons given in a former paper,^ we assume that this reaction 

 occurs at the surface of the cell and that CaCl2 accumulates in the 

 surface to a greater degree than NaCl. The increase in concentration 

 of CaCU in the surface is supposed to be ten times as great as the 

 corresponding increase of NaCl, so that the proportions in the surface 

 are those given in Table III. For example, when the proportions in 

 the solution are 97.56 NaCl + 2.44 CaCl2, the proportion of NaCl 

 to CaCl2 in the surface is as 97.56 to 24.40, which is equivalent^ to 

 80 NaCl + 20 CaClg. 



TABLE III. 



Amount of NaiXCa. 



We calculate the amount of Na4XCa by the usual formula: 



(CNa^XCa) (CzCh)^ 



K = 



(CNaCl)* (CcaCh) (CxZ^) 



but since 2CNa4X ca = Czcb we may write 



(CNa4XCa) (2 CNa^XCa)^ 



K = 



(CNaCl)* (CcaCla) (Cxz^) 



^ Osterhout, W. J., V., Proc. Am. Phil. Soc, 1916, Iv, 533. 



^ As explained in a former paper,^ it is assumed that the reaction takes place 

 in a surface which is saturated with respect to NaCl and CaCl2, so that while 

 one of these may be displaced by the other (in case their relative proportions in 

 the solution are altered) the total concentration does not change; for convenience 

 this concentration is taken as 100 and the sum of NaCl + CaCl2 is therefore always 

 equal to 100. 



