Lecture 35. Polyprotic

Lecture 35. Polyprotic - Chapter 11: Polyprotic Acids and...

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Unformatted text preview: Chapter 11: Polyprotic Acids and Bases Equilibrium Constants - Diprotics H2A HAA2- + H2O HA- + H2O H+ + HAH+ + A2HA- + OHH2A + OHKa1 x Kb2 = Kw Ka1 Ka2 Kb1 Kb2 Daniel C. Harris, Exploring Chemical Analysis, 2nd ed., W.H. Daniel Freeman and Company, New York, 2001. Freeman Ka2 x Kb1 = Kw Fumaric Acid Fumaric α Fractions of a Polyprotic Acid? Fractions => What were those α fractions? => Fraction of Dissociation [A- ] Fraction Dissociated = [A- ] + [HA] Daniel C. Harris, Exploring Chemical Analysis, 2nd ed., W.H. Daniel Freeman and Company, New York, 2001. Freeman α Fractions of Diprotic Acids Fractions 1.0 1.0 0.8 0.8 0.6 pKa1 = pKa2 0.6 0.4 0.4 0.2 pKa1 + 3 = pKa2 0 2 4 6 8 10 12 14 0.2 0.0 0 2 4 6 8 10 12 14 0.0 1.0 1.0 0.8 0.8 0.6 0.6 0.4 0.2 pKa1 + 1 = pKa2 0 2 4 6 8 10 12 14 0.4 pKa1 + 4 = pKa2 0 2 4 6 8 10 12 14 0.2 0.0 0.0 1.0 1.0 0.8 0.8 0.6 0.4 pKa1 + 2 = pKa2 0 2 4 6 8 10 12 14 0.6 0.4 pKa1 + 5 = pKa2 0 2 4 6 8 10 12 14 0.2 0.2 0.0 0.0 α Fractions of Diprotic Acids Fractions 1.0 0.8 0.6 0.4 pKa1 = pKa2 ; α 0 = α1 = α 2 0.2 0.0 0 2 4 6 8 10 12 14 α 1.0 H2A HA- A2- 0.8 0.6 0.4 pH = pKa1 pH = pKa2 0.2 0.0 0 2 4 6 8 10 12 14 Finding pH in Diprotic Systems Daniel C. Harris, Exploring Chemical Analysis, 2nd ed., W.H. Daniel Freeman and Company, New York, 2001. Freeman Finding pH in Diprotic Systems Finding [H + ]2 α0 = = [H 2 A] c tot [H + ]2 + K1 [H + ] + K1 K 2 a1 = a2 = K1 [H + ] [H ] + K1 [H ] + K1 K 2 K1 K 2 [H ] + K1 [H ] + K1 K 2 +2 + +2 + Charge Balance for the solution Charge of a diprotic acid: of [H ] = [HA ] + 2 [ A ] + [OH ] + Ð Ð2 Ð = [HAÐ] c tot = [ AÐ2] c tot => => [H ] = + K1 [H + ] [H ] + K1 [H ] + K1 K 2 +2 + c tot + 2 K1 K 2 [H ] + K1 [H ] + K1 K 2 +2 + c tot + K w [H + ] Correct solutions can be found by solving the above equations for [H+] . => Complicated enough for diprotic acids (4th order equation), this quickly => becomes very complicated for tri- and tetraprotic acids … becomes Fumaric Acid Fumaric Finding pH in Diprotic Systems Daniel C. Harris, Exploring Chemical Analysis, 2nd ed., W.H. Daniel Freeman and Company, New York, 2001. Freeman Finding pH in Diprotic Systems Finding [H + ]2 α0 = = [H 2 A] c tot [H + ]2 + K1 [H + ] + K1 K 2 a1 = a2 = K1 [H + ] [H ] + K1 [H ] + K1 K 2 K1 K 2 [H ] + K1 [H ] + K1 K 2 +2 + +2 + Charge Balance for the solution Charge of a diprotic acid: of [H ] = [HA ] + 2 [ A ] + [OH ] + Ð Ð2 Ð = [HAÐ] c tot = [ AÐ2] c tot => => [H ] = + K1 [H + ] [H ] + K1 [H ] + K1 K 2 +2 + c tot + 2 K1 K 2 [H ] + K1 [H ] + K1 K 2 +2 + c tot + K w [H + ] Correct solutions can be found by solving the above equations for [H+] . => Complicated enough for diprotic acids (4th order equation), this quickly becomes very complicated for tri- and tetraprotic acids … becomes Finding pH in Diprotic Systems Finding Out[12]= : H ®- K1 4 + i K12 1 . j ctot K1 + j - K1 K2 + H- ctot K1 + K1 K2 - Kw + Kw+ I 21K3 I H- ctot K1 + K1 K2 - Kw 2 - 12 K1 K2 Kw- 3 K1 H- 2 ctot K1 K2 - K1 Kw M K L L LM j 2 4 3 k 1 27 H- 2 ctot K1 K2 - K1 Kw 2 + , I - 4 I H- ctot K1 + K1 K2 - Kw 2 - 12 K1 K2 Kw- 3 K1 H- 2 ctot K1 K2 - K1 Kw M + I 2 H- ctot K1 + K1 K2 - Kw 3 - 27 K13 K2 Kw+ L L L3 L 72 K1 K2 H- ctot K1 + K1 K2 - Kw Kw- 9 K1 H- ctot K1 + K1 K2 - Kw H- 2 ctot K1 K2 - K1 Kw + 27 H- 2 ctot K1 K2 - K1 Kw 2M M L L L L 2M 1 1K3 J 3 I 2 H- ctot K1 + K1 K2 - Kw 3 - 27 K13 K2 Kw+ 72 K1 K2 H- ctot K1 + K1 K2 - Kw Kw- 9 K1 H- ctot K1 + K1 K2 - Kw H- 2 ctot K1 K2 - K1 Kw + L L L L N+ 3 ´ 21K3 I 2 H- ctot K1 + K1 K2 - Kw 3 - 27 K13 K2 Kw+ 72 K1 K2 H- ctot K1 + K1 K2 - Kw Kw- 9 K1 H- ctot K1 + K1 K2 - Kw H- 2 ctot K1 K2 - K1 Kw + L L L L 27 H- 2 ctot K1 K2 - K1 Kw 2 + , I - 4 I H- ctot K1 + K1 K2 - Kw 2 - 12 K1 K2 Kw- 3 K1 H- 2 ctot K1 K2 - K1 Kw M + I 2 H- ctot K1 + K1 K2 - Kw 3 - 27 K13 K2 Kw+ L L L3 L 72 K1 K2 H- ctot K1 + K1 K2 - Kw Kw- 9 K1 H- ctot K1 + K1 K2 - Kw H- 2 ctot K1 K2 - K1 Kw + 27 H- 2 ctot K1 K2 - K1 Kw 2M M L L L L 2M i K12 1 j . j ctot K1 + - K1 K2 + Kw+ Hctot K1 - K1 K2 + Kw - I 21K3 I H- ctot K1 + K1 K2 - Kw 2 - 12 K1 K2 Kw- 3 K1 H- 2 ctot K1 K2 - K1 Kw M K L L LM j j 2 2 3 k 1 J 3 I 2 H- ctot K1 + K1 K2 - Kw 3 - 27 K13 K2 Kw+ 72 K1 K2 H- ctot K1 + K1 K2 - Kw Kw- 9 K1 H- ctot K1 + K1 K2 - Kw H- 2 ctot K1 K2 - K1 Kw + L L L L 27 H- 2 ctot K1 K2 - K1 Kw 2 + , I - 4 I H- ctot K1 + K1 K2 - Kw 2 - 12 K1 K2 Kw- 3 K1 H- 2 ctot K1 K2 - K1 Kw M + I 2 H- ctot K1 + K1 K2 - Kw 3 - 27 K13 K2 Kw+ L L L3 L 72 K1 K2 H- ctot K1 + K1 K2 - Kw Kw- 9 K1 H- ctot K1 + K1 K2 - Kw H- 2 ctot K1 K2 - K1 Kw + 27 H- 2 ctot K1 K2 - K1 Kw 2M M L L L L 2M 1 3 ´ 21K3 1K3 1K3 y z z+ z { N- I 2 H- ctot K1 + K1 K2 - Kw 3 - 27 K13 K2 Kw+ 72 K1 K2 H- ctot K1 + K1 K2 - Kw Kw- 9 K1 H- ctot K1 + K1 K2 - Kw H- 2 ctot K1 K2 - K1 Kw + L L L L 27 H- 2 ctot K1 K2 - K1 Kw 2 + , I - 4 I H- ctot K1 + K1 K2 - Kw 2 - 12 K1 K2 Kw- 3 K1 H- 2 ctot K1 K2 - K1 Kw M + I 2 H- ctot K1 + K1 K2 - Kw 3 L L L3 L 27 K13 K2 Kw+ 72 K1 K2 H- ctot K1 + K1 K2 - Kw Kw- 9 K1 H- ctot K1 + K1 K2 - Kw H- 2 ctot K1 K2 - K1 Kw + 27 H- 2 ctot K1 K2 - K1 Kw 2M M L L L L 2M i j I - K13 + 4 K1 H- ctot K1 + K1 K2 - Kw - 8 H- 2 ctot K1 K2 - K1 Kw MK j 4 . L L j j k 2 i j ctot K1 + K1 - K1 K2 + 1 H- ctot K1 + K1 K2 - Kw + Kw+ j L j 4 3 k 1K3 + I 21K3 I H- ctot K1 + K1 K2 - Kw 2 - 12 K1 K2 Kw- 3 K1 H- 2 ctot K1 K2 - K1 Kw M K L LM J 3 I 2 H- ctot K1 + K1 K2 - Kw 3 - 27 K13 K2 Kw+ 72 K1 K2 H- ctot K1 + K1 K2 - Kw Kw- 9 K1 H- ctot K1 + K1 K2 - Kw H- 2 ctot K1 K2 - K1 Kw + L L L L 27 H- 2 ctot K1 K2 - K1 Kw 2 + , I - 4 I H- ctot K1 + K1 K2 - Kw 2 - 12 K1 K2 Kw- 3 K1 H- 2 ctot K1 K2 - K1 Kw M + I 2 H- ctot K1 + K1 K2 - Kw 3 - 27 K13 K2 Kw+ L L L3 L 72 K1 K2 H- ctot K1 + K1 K2 - Kw Kw- 9 K1 H- ctot K1 + K1 K2 - Kw H- 2 ctot K1 K2 - K1 Kw + 27 H- 2 ctot K1 K2 - K1 Kw 2M M L L L L 2M 1K3 N+ 1 3 ´ 21K3 I 2 H- ctot K1 + K1 K2 - Kw 3 - 27 K13 K2 Kw+ 72 K1 K2 H- ctot K1 + K1 K2 - Kw Kw- 9 K1 H- ctot K1 + K1 K2 - Kw H- 2 ctot K1 K2 - K1 Kw + L L L L 27 H- 2 ctot K1 K2 - K1 Kw 2 + , I - 4 I H- ctot K1 + K1 K2 - Kw 2 - 12 K1 K2 Kw- 3 K1 H- 2 ctot K1 K2 - K1 Kw M + I 2 H- ctot K1 + K1 K2 - Kw 3 - 27 K13 K2 Kw+ L L L3 L 72 K1 K2 H- ctot K1 + K1 K2 - Kw Kw- 9 K1 H- ctot K1 + K1 K2 - Kw H- 2 ctot K1 K2 - K1 Kw + 27 H- 2 ctot K1 K2 - K1 Kw 2M M L L L L 2M 1K3 yyy zzz zz zzz> zzz {{{ Finding pH in Diprotic Systems Finding pH of diprotic acid solutions; no base added 7 pH 6 5 pK1 = 4.0 pK2 = 6.0 4 pK1 = pK2 = 6.0 3 pK1 = pK2 = 4.0 -8 -6 -4 -2 Log [conc] Finding pH in Diprotic Systems: Simplified Approach • Treat Acidic Form as a monoprotic acid. Treat (works well if pK1 < pK2 + 2) pK • Treat Basic Form as a monoprotic base. Treat Daniel C. Harris, Exploring Chemical Analysis, 2nd ed., W.H. Daniel Freeman and Company, New York, 2001. Freeman The Intermediate Form (Amphiprotic) Remember autoprotolysis: H2O H+ + OH- K a1K w + K a1K a2 [HA - ] [H+ ] = K a1 + [HA - ] [H+ ] ≈ K a1K a2 pK a1 + pK a2 pH ≈ 2 Daniel C. Harris, Exploring Chemical Analysis, 2nd ed., W.H. Daniel Freeman and Company, New York, 2001. Freeman When is this true? Summary for Diprotics (Simplified Approach) Summary • Treat the fully protonated species (e.g. H2A) as a Treat A) monoprotic weak acid. monoprotic • Treat the fully deprotonated species (e.g. A2-) as a Treat as monoprotic weak base. monoprotic • Treat the amphiprotic species (e.g. HA-) as an Treat as intermediate ( pH ≈ ½(pKa1 + pKa2) ). Titration of a Diprotic Acid Titration Lowest: pKa1 = 4.00 pKa2 = 6.00 Middle: pKa1 = 4.00 pKa2 = 8.00 Top: pKa1 = 4.00 pKa2 = 10.00 Daniel C. Harris, Exploring Chemical Analysis, 2nd ed., W.H. Freeman and Company, New York, 2001. Freeman Triprotics Triprotics Example: Citric Acid Example: Arginine Daniel C. Harris, Exploring Chemical Analysis, 4th Daniel ed., W.H. Freeman and Company, New York, 2001. ed., Triprotics Triprotics Daniel C. Harris, Exploring Chemical Analysis, 2nd ed., W.H. Freeman and Company, New York, 2001. Freeman • Treat H3A as a monoprotic weak acid. Treat • Treat H2A- as an intermediate Treat (pH ≈ ½(pKa1 + pKa2)). (pH • Treat HA2- as an intermediate Treat (pH ≈ ½(pKa2 + pKa3)). (pH • Treat A3- as a monoprotic weak base. Treat Triprotics: Simplified Approach Triprotics: Daniel C. Harris, Exploring Chemical Analysis, 2nd ed., W.H. Freeman and Company, New York, 2001. Freeman ...
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