a supplementary note.ppt

文档介绍：It is beyond the scope of this class to understand the following.
If only one equilibration is in effect, the pH of the solution is
determined by one Henderson-Hasselbalch equation.
pH=pK1+log[A-]/[HA] (1)
If two equilibrations are in effect, the pH is determined by two Henderson-Hasselbalch equations (eq. (2) & (3)).
H2A+ --> H+ + HA --> 2H+ + A- first pK1, second pK2
pH=pK1+log[HA]/[H2A+] (2)
pH=pK2+log[A-]/[HA] (3)
The sum of eq. (2) and (3) gives,
2pH=pK1+pK2+log[A-]/[H2A+] (4)
Without proving it mathematically, which is beyond the scope of this class, it is simply stated that [A-]=[H2A+] when the overall reaction proceeds to the same extent in both directions, which is also the point when the concentration of the intermediate, [HA] is maximized. Because log(1)=0, pH=(pK1+pK2)/2 which the pI for HA.
When more equilibrations are in