Ch 11: Titrations in Diprotic Systems

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

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 = 10-2.328 = 4.7 x 10-3 Ka2 = 10-9.744 = 1.8 x 10-10 So assume the pH depends only on H2L+ and ignore the contribution of H+ from HL.

Calculate the pH of 0.050M H2L+

2. The basic form L- Ka1 = 10-2.328 = 4.7 x 10-3 Ka2 = 10-9.744 = 1.8 x 10-10 Strengths of conjugate bases: for L- Kb1 = Kw/Ka2 = 1.01 x 10-14/1.8 x 10-10 = 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.

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

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

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

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.