《生物化学》课程PPT教学课件（英文版）a supplementary note

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=pKI+log[A/[HAI If two equilibrations are in effect, the ph is determined by two Henderson-Hasselbalch equations(eq(2)&3)) HA H++Ha-->2H+A first pKI, second pK2 pH=pKI+logIHA/H2A 2 pH=pK2+logIA/[HAI (3) The sum of eq()and 3)gives, 2pH=pKI+pK2+logan,a( Without proving it mathematically, which is beyond the scope of this class, it is simply stated that [a]=lh,a 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=(pKI+pk2)/2 which the pI for HA

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 pKI pK2, pK3,... etc are contributing at the same time to the proton concentration HI the pl can not be calculated by the sum average. In general, pKI, 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

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)