119_6_Polyprotic_Acids_06F

119_6_Polyprotic_Acids_06F - 1 VI Polyprotic Acids Harris...

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Unformatted text preview: 1 VI. Polyprotic Acids Harris, Chap 10 2 Acid/Base Equilibria Weak Polyprotic Acids Acid K A1 K A2 Maleic 1.23*10-2 4.66*10-7 cis-HOOCCH=CHCOOH Oxalic 5.60*10-2 5.42*10-5 HOOCCOOH o-Phthalic 1.12*10-2 3.90*10-6 C 6 H 4 (COOH) 2 ( protonated) Glycine 4.47*10-3 1.67*10-10 CH 2 (NH 3 + )COOH (protonated) Leucine 4.69*10-3 1.79*10-10 NH 3 COOH 3 Acid/Base Equilibria: Weak Diprotic Acids Neutral Acid : 1. H 2 A + H 2 O ⇔ H 3 O + + HA – K A1 = [H 3 O + ][HA- ]/[HA] 2. HA- + H 2 O ⇔ H 3 O + + A 2- K A2 = [H 3 O + ][A 2- ]/[HA- ] Amino Acid : 1. + H 3 NXCOOH + H 2 O ⇔ H 3 O + + + H 3 NXCOO-– AAH + AA (Zwitterion) K A1 = [H 3 O + ][AA]/AAH + ] 2. + H 3 NXCOO – + H 2 O ⇔ H 3 O + + H 2 NXCOO – AA (AA-H)- K A2 = [H 3 O + ][(AA-H) – ]/[AA] 4 Acid/Base Equilibria: Weak Triprotic Acids K A1 K A2 K A3 H 3 PO 4 7.1*10-3 6.3*10-6 7.1*10-13 Aspartic acid 1.0*10-2 1.3*10-4 1.0*10-10 HOOCCH 2 CH(NH 3 + )COOH = protonated R Lysine 9.1*10-3 8.3*10-10 2.0*10-11 + H 3 N(CH 2 ) 4 CH(NH 3 + )COOH = diprotonated K 5 Acid/Base Equilibria: Weak Triprotic Acids Neutral Acid H 3 A + H 2 O ⇔ H 3 O + + H 2 A- K A1 = [H 3 O + ][H 2 A- ]/[H 3 A] H 2 A- + H 2 O ⇔ H 3 O + + HA 2- K A2 = [H 3 O + ][HA 2- ]/[H 2 A- ] HA 2- + H 2 O ⇔ H 3 O + + A 3- K A3 = [H 3 O + ][A 3- ]/[HA 2- ] 6 Acid/Base Equilibria: Weak Triprotic Acids Aspartic Acid: HOOCCH 2 CH(NH 3 + )COOH + H 2 O ⇔ H 3 O + + HOOCCH 2 CH(NH 3 + )COO- AAH + AA(Zwitterion) K A1 = [H 3 O + ][AA]/[AAH + ] HOOCCH 2 CH(NH 3 + )COO- + H 2 O ⇔ H 3 O + + - OOCCH 2 CH(NH 3 + )COO- AA(Zwitterion) (AA-H)- K A2 = [H 3 O + ][(AA-H)- ]/[AA]- OOCCH 2 CH(NH 3 + )COO- ⇔ H 3 O + + - OOCCH 2 CH(NH 2 )COO- (AA-H)- (AA-2H) 2- K A3 = [H 3 O + ][(AA-2H) 2- ]/[(AA-H)- ] 7 Acid/Base Equilibria: Weak Triprotic Acids Lysine: + H 3 N(CH 2 ) 4 CH(NH 3 + )COOH ⇔ + H 3 N(CH 2 ) 4 CH(NH 3 + )COO- + H 3 O + AAH 2 2+ AAH + K A1 = [H 3 O + ][AAH + ]/[AAH 2 2+ ] + H 3 N(CH 2 ) 4 CH(NH 3 + )COO- ⇔ + H 3 N(CH 2 ) 4 CH(NH 2 )COO- + H 3 O + AAH + AA(Zwitterion) K A2 = [H 3 O + ][AA]/[AAH + ] + H 3 N(CH 2 ) 4 CH(NH 2 )COO- ⇔ H 2 N(CH 2 ) 4 CH(NH 2 )COO- + H 3 O + AA(Zwitterion) (AA-H)- K A3 = [H 3 O + ][(AA-H)- ]/[AA] 8 Acid/Base Equilibria: Weak Diprotic Acids Solutions of weak acids only : If K A2 and K A1 are << 1 and K A2 << K A1 , the first step will occur only to a small extent and the second step will occur to an even smaller (and negligible) extent. Therefore, [H 3 O + ] is formed primarily by the first ionization step and we assume it is formed only by the first ionization step to simplify the calculations. Therefore, nothing new is involved in the analysis compared with weak monoprotic acids. 9 Acid/Base Equilibria: Weak Diprotic Acids Calculate the pH of 0.109 F maleic acid, H 2 Ma....
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This note was uploaded on 04/13/2008 for the course CHEM 119H taught by Professor Munson during the Fall '07 term at University of Delaware.

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119_6_Polyprotic_Acids_06F - 1 VI Polyprotic Acids Harris...

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