CH302 Chapter 7 notes part 3b

CH302 Chapter 7 notes part 3b - Protonation State We can...

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Protonation State We can estimate the fraction of each species present for a given acid or base for a given pH. Here pH is being controlled by the action of another chemical (see Ch. 8) Example 1: Acetic acid at pH’s 2, 4.7, 7, and the methylamonium ion at pH’s 7, 10, 13. Example 2: The methylamonium ion at pH’s 7, 10, 13. Polyprotic Acids The calculation of equilibria for polyprotic acids is done in a stepwise fashion. There is an ionization constant for each step. Consider arsenic acid, H 3 AsO 4 , which has three ionization constants. K a1 = 2.5 x 10 -4 K a2 = 5.6 x 10 -8 K a3 = 3.0 x 10 -13 The first ionization step for arsenic acid is: Second step: [] [ ] [] 4 4 3 4 2 1 - 4 2 4 3 10 5 . 2 AsO H AsO H H Ka AsO H H AsO H + + × = = ←+ [] [ ] [] 8 - 1 4 2 2 4 a2 - 2 4 - 1 4 2 10 6 . 5 AsO H HAsO H K HAsO H AsO H + + × = = ←+ Third step: [] [ ] [] 13 - 2 4 3 4 a3 - 3 4 - 2 4 10 0 . 3 HAsO AsO H K AsO H HAsO + + × = = ←+ For weak polyprotic acids, K a1 is always bigger than K a2 , etc.
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This note was uploaded on 03/26/2012 for the course CHEM 302 taught by Professor Mccord during the Fall '10 term at University of Texas.

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CH302 Chapter 7 notes part 3b - Protonation State We can...

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