MEMBRANE POTENTIALS 133 



is bivalent, e.g., S0 4 , in the case of gelatin sulphate. Let x be the 

 concentration of hydrogen ions in the outside solution, y 



the hydrogen ion concentration in the inside solution; then | 



is the concentration of the SO 4 ions in the outside, and | the 



concentration of the SO 4 ions of the free H 2 SO 4 in the inside 

 (gelatin) solution. The concentration of SO 4 ions in combination 



with gelatin becomes ~- Then the following two dissociation 

 equations must hold : 



x 2 ^ = K I H 2 SO 4 J undissociated (outside) 



y 2 (! + !)= K[H 2 SO 4 ] undissociated (inside) 



Since the undissociated H 2 SO 4 must be distributed equally on 

 both sides of the membrane, x 3 = y 2 (y + 2); x = [y 2 (y -f 2)]*. 



x 

 The value which interests us is -, i.e., the ratio of the hydrogen ion 



concentrations. 



Substituting \y z (y + 2)]* for x in - we get 



The P.D. is, therefore, in the case of gelatin sulphate, 

 P.D. = y log (l + p millr 



millivolts 



while in the case of gelatin chloride it was 



CO f -V 



P.D. = 7f loglH--J millivolts 

 z \ y/ 



Hence, the P.D. of gelatin sulphate solutions should be only 

 two-thirds of the value of a gelatin chloride solution of the 

 same pH and the same concentration of originally isoelectric 

 gelatin. 



This theoretical deduction is actually fulfilled, as the Tables 

 XI, XII, and XIII show. Thus in Table XII we find that for 



