Lecture_overheads_-_Ch09

Lecture_overheads_-_Ch09 - CHEM 321 Quantitative Analysis...

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CHEM 321 – Quantitative Analysis Ch. 9 – Polyprotic Acid Base Equilibria Amino Acid Chemistry 9-1 Diprotic Acids and Bases Calculating pH of H 2 A Calculating pH of A 2- Calculating pH of HA - 9-2 Diprotic Acid Buffers 9-3 Triprotic Acids and Bases 9-4 What is the Principal Species? 9-5 Fractional Composition
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Up until now, we’ve been dealing with monoprotic systems in which there are only two forms – an acid and its conjugate base We’ll now introduce polyprotic systems in which the equibria and their solutions are more complicated Polyprotic donates >1 proton (general description) Diprotic donates 2 protons (H 2 CO 3 ) Triprotic donates 3 protons (H 3 PO 4 ) Amphiprotic can donate and accept a proton (H 2 O, NaHCO 3 ) (can act as both acid and base) Amino acids common form of polyprotic acid/base (H 2 N-R-COOH) Regardless of whether we’re discussing a polyprotic acid/base or amino acid, we’ll see that it can exist in three or more forms and that the species present and their relative concentrations are markedly dependent on pH
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Amino Acid Chemistry Amino Acid Chemistry Amino group can act as base At physiological pH’s, amino acids present as zwitterions (contain both + and – charge, yet electrically neutral) All amino acids are diprotic carboxylic acid group can act as acid
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9-1 Diprotic Acids and Bases Diprotic acids ex. H 2 S, H 2 CO 3 , H 2 P (related to KHP) Diprotic bases ex. Na 2 S, Na 2 CO 3 , K 2 P H 2 A ↔ H + + HA - K a 1 HA - H + + A 2 - K a 2 [ ][ ] [ ] [ ] [ ] [ ] [ ][ ] - + - - - + = = = OH H HA OH A H A H HA H 2 2 2 1 w b a K K K [ ][ ] [ ] [ ][ ] [ ] [ ][ ] - + - - - - - + = = = OH H A OH HA HA A H 2 2 1 2 w b a K K K A 2 - + H 2 O HA - + OH - K b 1 HA - + H 2 O H 2 A + OH - K b 2 Can still prove K a K b = K w , but must use proper subscripts K b ’s not given, but can compute from appropriate K a value Diprotic amino acids (ALL) ex. H 2 L + (aq) (low pH), L·HCl (s) Fully deprotonated amino acids (ALL) ex. L - (aq) (high pH), NaL (s) H 2 L + H + + HL K a 1 HL H + + L - K a 2 L - + H 2 O HL + OH - K b 1 HL + H 2 O H 2 L + + OH - K b 2 2 1 a w b K K K = 1 2 a w b K K K =
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Amino Acids with Aliphatic R-Groups
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Given solutions of the following formalities, we’ll now compute the pH and relative amounts of all 3 forms of the diprotic amino acid leucine 1. 0.0500 F H 2 L + (or H 2 A) 2. 0.0500 F HL (or HA - ) 3. 0.0500 F L - (or A 2 - )
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Will solve this problem using systematic approach 1. Write chemical reactions and associated K expressions
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Lecture_overheads_-_Ch09 - CHEM 321 Quantitative Analysis...

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