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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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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 = 10 -2.328 = 4.7 x 10 -3 K a2 = 10 -9.744 = 1.8 x 10 -10 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 +

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Calculate the pH of 0.050M H 2 L +

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2. The basic form L - K a1 = 10 -2.328 = 4.7 x 10 -3 K a2 = 10 -9.744 = 1.8 x 10 -10 Strengths of conjugate bases: for L - K b1 = K w /K a2 = 1.01 x 10 -14 /1.8 x 10 -10 = 5.61 x 10 -5 for HLK b2 = K w /K a1 = 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.

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

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

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[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

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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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