221-chapter-10

221-chapter-10 - Polyprotic Acid-Base Equilibria...

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Unformatted text preview: Polyprotic Acid-Base Equilibria Introduction 1.) Polyprotic systems Acid or bases that can donate or accept more than one proton Proteins are a common example of a polyprotic system- why activity of proteins are pH dependent Polymer of amino acids- Some amino acids have acidic or basic substituents Introduction 1.) Polyprotic systems Amino acids- Carboxyl group is stronger acid of ammonium group- R is different group for each amino acid Amino acids are zwitterion molecule with both positive and negative charge- At low pH , both ammonium and carboxy group are protonated- At high pH , neither group is protonated- Stabilized by interaction with solvent Polyprotic Acid-Base Equilibria (acidic) (basic) Overall charge is still neutral Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 2.) Multiple Equilibriums Illustration with amino acid leucine (HL) Equilibrium reactions low pH high pH Carboxyl group Loses H + ammonium group Loses H + Diprotic acid: 1 1 a K K 2 2 a K K Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 2.) Multiple Equilibriums Equilibrium reactions Diprotic base: 1 b K 2 b K Relationship between K a and K b w 2 b 1 a K K K w 1 b 2 a K K K pKa of carboxy and ammonium group vary depending on substituents Largest variations Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.) General Process to Determine pH Three components to the process Acid Form [H 2 L + ] Basic Form [L- ] Intermediate Form [HL] low pH high pH Carboxyl group Loses H + ammonium group Loses H + Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.) General Process to Determine pH Acid Form (H 2 L + ) Illustration with amino acid leucine H 2 L + is a weak acid and HL is a very weak acid K 1 =4.70x10-3 K 2 =1.80x10-10 2 1 K K Assume H 2 L + behaves as a monoprotic acid 1 a K K = Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.) General Process to Determine pH 0.050 M leucine hydrochloride + H + H 2 L + HL H + 0.0500 - x x x K 1 =4.70x10-3 ] H [ ] HL [ M 10 x 32 . 1 x x F x L H H HL 10 7 . 4 K 2 2 2 3 a +- + +- = = = - = = = ] [ ] ][ [ 88 . 1 ) M 10 23 . 1 log( ] H log[ pH 2 = - =- =- + M 10 68 . 3 x F ] L H [ 2 2- + =- = Determine [H + ] from K a : Determine pH from [H + ]: Determine [H 2 L + ]: Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.) General Process to Determine pH Acid Form (H 2 L + ) What is the concentration of L- in the solution? [L ] is very small, but non-zero. Calculate from K ] [ ] [ ] [ ] [ ] ][ [ +-- + = = H HL K L HL L H K 2 a 2 a ) ( 2 a 10- 2- 2- 10- K 10 80 . 1 ) 10 32 . 1 ( ) 10 32 . 1 ( ) 10 80 . 1 ( L = = =- ] [ Approximation [H + ] [HL], reduces K a2 equation to [L- ]=K a2 ] HL [ 10 32 . 1 10 80 . 1 ] L [ 2 10 = << =--- Validates assumption Polyprotic Acid-Base Equilibria Diprotic Acids and Bases 3.) General Process to Determine pH For most diprotic acids, K 1 >> K...
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This note was uploaded on 05/09/2010 for the course CHEM 221 taught by Professor Dr.robertpowers during the Fall '07 term at San Diego.

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221-chapter-10 - Polyprotic Acid-Base Equilibria...

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