LAWS OF ELECTROLYTIC DISSOCIATION 75 



the five above equations. In carrjdng out the calculation for [H+], 

 we get from (I), (II) and (III) 



and from (II, III, V): 



k 



a 



H+ OH- 

 By substituting this value of U in (VI) we obtain; 



ka kb 



* \H+ OH-/ 



H+ OH- H+ - OH- 



in which we can further substitute kw/[H+] for [0H~]. This gives 

 an equation of the fourth order for [H+] as given below, where x 

 stands for [H+j 



X* + x3 ^^ + A^ + x^ ^ (ka - kb) - X ^ (kaA + kj 



- 1^ . ka • kw = iVII) 

 kb 



The general solution of this equation is difficult. It can have only- 

 one real positive root, because it contains only one sign change, 

 irrespective of the sign of the coefficient of x- (Descartes' law). 

 The practical solution rests upon the permissibility of disregarding 

 the first (x*) or the last member of the equation, depending upon 

 the numerical relationships of the individual constants. Further- 

 more equation (VII) may be factored in the following way (this 

 may be tested by simply multiplying out) : 



(x2-k„jx(x2 + x^ + x.A + ^jkw-xXAXkw(^-l j=0 



or 



Q X R - S = 0> 



By letting S = 0, i.e., by making ka = kb, equation (VIII) is 

 satisfied when either Q or R = 0. When R = 0, no real positive 



