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Ch 11: Titrations in Diprotic Systems Biological Applications - Amino Acids (Sec. 11-1) low pH high pH R = (CH 3 ) 2 CHCH 2 -

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Presentation on theme: "Ch 11: Titrations in Diprotic Systems Biological Applications - Amino Acids (Sec. 11-1) low pH high pH R = (CH 3 ) 2 CHCH 2 -"— Presentation transcript:

1 Ch 11: Titrations in Diprotic Systems Biological Applications - Amino Acids (Sec. 11-1) low pH high pH R = (CH 3 ) 2 CHCH 2 -

2 Finding the pH in Diprotic Systems (Sec. 11-2) The strength of H 2 L + as an acid is much, much greater than HL - K a1 = = 4.7 x K a2 = = 1.8 x So assume the pH depends only on H 2 L + and ignore the contribution of H + from HL. 1. The acidic form H 2 L +

3 Calculate the pH of 0.050M H 2 L +

4 2. The basic form L - K a1 = = 4.7 x K a2 = = 1.8 x Strengths of conjugate bases: for L - K b1 = K w /K a2 = 1.01 x /1.8 x = 5.61 x for HLK b2 = K w /K a1 = 1.01 x /4.7 x = 2.1 x 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)

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9 [H + ] 2 = K a1 K a2 -log [H + ] 2 = - log K a1 - log K a2 2 pH = pK a1 + pK a2 assume: K w K a1 << K a1 K a2 C HL K a1 << C HL 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.


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