CH302 Chapter 7 notes part 3b answers

# CH302 Chapter 7 notes part 3b answers - Example: Acetic...

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3 Example: Acetic Acid: protonation state Using pH of 7 we would get: 3 7 5 - COOH CH 10 x 525 . 5 10 10 x 1.8 1 1 f 3 = + = 994 . 0 1 10 x 1.8 10 1 f 5 - 7 COO CH - 3 = + = Now its mostly (99.4%) the acetate ion. . Example: Methylammonium ion: protonation state at pH = 7? [] 11 3 3 2 3 a 2 3 3 3 10 28 . 2 NH CH NH CH H K NH CH H NH CH + + + + × = = + Scenario: we dont know the conc. but we have been told pH. We are interested in the fraction of each form of methylamine present: [ ] 2 3 3 3 3 3 NH CH NH CH NH CH NH CH f 3 3 + = + + + [ ] 2 3 3 3 2 3 NH CH NH CH NH CH NH CH f 2 3 + = + Example: Methylammonium ion: protonation state at pH = 7? Math trick: divide top and bottom of each of these by the number on the top, THEN the single fraction can be expressed in terms of a K a and a [H+], which we do know! + + + = + = + H K 1 1 NH CH NH CH 1 1 f a 3 3 2 3 NH CH 3 3 1 K H 1 1 NH CH NH CH 1 f a 2 3 3 3 NH CH 2 3 + + = + = + Example: Methylammonium ion: protonation state at pH = 7? Finally feed in the K a (2.28 x 10 -11 ) and the [H + ] = 10 -7 9997 . 0 10 10 x 2.28 1 1 f 7 11 - NH CH 3 3 = + = + 4 11 - 7 NH CH 10 x 28 . 2 1 10 x 2.28 10 1 f 2 3 = + = This says at pH = 7, that 99.97% of the methylamine would be as the methylamonium ion, and 0.023% would be as methylamine.

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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 at Austin.

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CH302 Chapter 7 notes part 3b answers - Example: Acetic...

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