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 effect,that is,when pK1,
pK2,pK3,…etc are contributing at the same time to the
proton concentration [H+],the pI can not be calculated by the
sum average,In general,pK1,pK2,and pK3 should be
separated from each other by at least one unit,that is,the
corresponding equilibrium constants differ by one order of
magnitude (10 times more or 10 times less).