1. Ch 11: Titrations in Diprotic Systems
Biological Applications - Amino Acids (Sec. 11-1)
low pH high pH
R = (CH3)2CHCH2 -

2. Finding the pH in Diprotic Systems (Sec. 11-2)
1. The acidic form H2L+
The strength of H2L+ as an acid is much, much greater than HL -
Ka1 = = 4.7 x 10-3
Ka2 = = 1.8 x 10-10
So assume the pH depends only on H2L+ and ignore the contribution of H+ from HL.

3. Calculate the pH of 0.050M H2L+

4. 2. The basic form L-
Ka1 = = 4.7 x Ka2 = = 1.8 x 10-10
Strengths of conjugate bases:
for L- Kb1 = Kw/Ka2 = 1.01 x 10-14/1.8 x = 5.61 x 10-5
for HL Kb2 = Kw/Ka1 = 1.01 x 10-14/4.7 x 10-3 = 2.1 x 10-12
Since the second conj. base HL is so weak, we'll assume all the OH- comes from the L- form.

5. Example: Calculate the pH of a 0.050M solution of sodium leucinate

6. The Intermediate Form The pH of a Zwitterion Solution - Leucine (HL form)

9. assume:
KwKa1 << Ka1Ka2CHL
Ka1 << CHL
[H+]2 = Ka1 Ka2
-log [H+]2 = - log Ka1 - log Ka2
2 pH = pKa1 + pKa2
pH of a solution of a diprotic zwitterion

10. Example: pH of the Intermediate Form of a Diprotic Acid
Potassium hydrogen phthalate, KHP, is a salt of the intermediate form of phthalic acid. Calculate the pH pf 0.10M KHP and 0.010M KHP.