274 



HYDROGEN ION CONCENTRATION 



This yields as a balance B for the boundary layer: 



B = (m - mi) K+ + (ai - a) A" + (n - m) H+ + (bi - b) OH" (II) 



or totalled in terms of molecules: 



B = (m - m,) KA + (bi - b) H2O + [(ai - a ) - (m - mi)] A-\ 



+ [(n - m) - (b, - b )] H+j ^^^^^ 



Since according to the original definition m + n + a + b = mi + ni 

 + ai + bi, consequently, 



(ai - a) - (m — mO = (n - nO - (bi - b) (I) 



Hence 



B = (m - mi) KA + (bi - b) H2O + [(n - nO - (bi - b)] HA (IV) 



TABLE 41 



Free solution 



pos. pole 



+ m K + 

 + nH + 



— a A~ 



- bOH- 





C 

 O 



Diaphragm 



mi K+ - 

 niH+ - 

 a A- + 

 bi OH- + 



neg. pole 



HA is the free acid, KA is the salt. Therefore, not only did a 

 change in the concentration occur, but also free acid appears or 

 disappears, according as to whether the sign of the factor before HA 

 is positive or negative; this depends on the values of n, ni, b and bi. 



Equation II may also be stated in molecular terms as follows : 



B = (ai - a) KA + (n - nO H2O + [(m - mi) - (ai - a )] K 



(V) 



+ [(b, - b ) - (n - ni)] OH 

 It follows from (1) 



(m — mi) — (ai — a) = (bi — b) — (n — ni) 

 therefore 



B = (ai - a) KA + (n - Hi) H2O + [(bi - b) - (n - nO] KOH (VI) 



and from this we can obtain the increase or decrease in the amount of 

 free base KOH. 



