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Titrating Polyfunctional Acids and Bases 9202071http:\\asadipour.kmu.ac.ir 40 slides.

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Presentation on theme: "Titrating Polyfunctional Acids and Bases 9202071http:\\asadipour.kmu.ac.ir 40 slides."— Presentation transcript:

1 Titrating Polyfunctional Acids and Bases http:\\asadipour.kmu.ac.ir 40 slides

2 1. Treating Complex Acid-Base Systems Complex systems are defined as solutions made up of: (1) An acid or base that has two or more acidic protons or basic functional groups H 3 PO 4 Ca(OH) 2 (2) Two acids or bases of different strengths HCl + CH 3 COOH NaOH + CH 3 COO http:\\asadipour.kmu.ac.ir 40 slides

3 )3) An amphiprotic substance that is capable of acting as both acid and base HCO H 2 O  CO H 3 O + HCO H 2 O  H 2 CO 3 + OH - NH 3 + CH 2 COO - + H 2 O  NH 2 CH 2 COO - + H 3 O + NH 3 + CH 2 COO - + H 2 O  NH 3 + CH 2 COOH + OH http:\\asadipour.kmu.ac.ir 40 slides

4 http:\\asadipour.kmu.ac.ir 40 slides K b3 K b2 K b1 K a1 ×K b3 =K w K a2 ×K b2 =K w K a3 ×K b1 =K w K a1 =1×10 -2 >K a2 =1×10 -7 > K a3 =1× K total =K a1 ×K a2 × K a3 =1×10 -21

5 pH of H 3 PO 4 1.Calculate the pH of 0.100M H 3 PO 4 solution http:\\asadipour.kmu.ac.ir 40 slides H + is not negligible

6 pH of HA - pH of HA - solution HA -  A 2- + H + HA -  H 2 A + OH http:\\asadipour.kmu.ac.ir 40 slides K a2 K b2 K a1

7 pH of HA - Calculate the pH of 0.100M NaHCO 3 solution. K a2 ×C HA - =1× ×1.00>>K w …….K w is negligible http:\\asadipour.kmu.ac.ir 40 slides K a1 =1×10 -6 >K a2 =1×10 -10

8 pH of HA - Calculate the pH of M NaH 2 PO 4 solution http:\\asadipour.kmu.ac.ir 40 slides K a1 =1×10 -2 >K a2 =1×10 -7 > K a3 =1× K a2 ×C HA - =1×10 -7 ×0.01>>K w …….K w is negligible

9 pH of HA - Calculate the pH of 1.00×10 -3 M Na 2 HPO 4 solution. K a2 ×C HA - =1× ×0.001=1× K w isnot negligible http:\\asadipour.kmu.ac.ir 40 slides K a1 =1×10 -2 > K a2 =1×10 -7 > K a3 =1×10 -12

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25 Sulfuric acid is unusual in that one of its protons behaves as a strong acid in water and the other as a weak acid (K a2 = 1.02 X ). Let us consider how the hydronium ion concentration of sulfuric acid solutions is computed using a M solution as an example. H 2 SO 4 →H + +HSO4 -  SO H + We will first assume that the dissociation of HSO 4 is negligible because of the large excess of H 3 0+ resulting from the complete dissociation of H 2 SO 4. Therefore, This result shows that [SO 4 - ] is not small relative to [HSO 4 ], and a more rigorous so­lution is required. From stoichiometric considerations, it is necessary that Mixture of weak and strong acids http:\\asadipour.kmu.ac.ir 40 slides [SO 4 ] = [H + ] [H + ] = [SO 4 2- ] C H 2 SO 4, = = [HS0 4 - ] + [SO 4 2- ] [HSO 4 - ] = [H 3 O + ] [H + ] ≈ [HSO 4 ] ≈ M

26 Sulfuric acid is unusual in that one of its protons behaves as a strong acid in water and the other as a weak acid (K a2 = 1.02 X ). Let us consider how the hydronium ion concentration of sulfuric acid solutions is computed using a M solution as an example. H 2 SO 4 →H + +HSO4 -  SO H + We will first assume that the dissociation of HSO 4 is negligible because of the large excess of H 3 0+ resulting from the complete dissociation of H 2 SO 4. Therefore, Mixture of weak and strong acids http:\\asadipour.kmu.ac.ir 40 slides [H + ] = [SO 4 2- ] [HSO 4 - ] = [SO 4 2- ]

27 Curves for the titration of strong acid / weak acid mixture with M NaOH. Each titration is on ml of a solution that is M in HCl and M in HA http:\\asadipour.kmu.ac.ir 40 slides

28 Curves for the titration of ml of polyprotic acid with M NaOH solution. A) M H 3 PO 4, B) M oxalic acid, C) M H 2 SO http:\\asadipour.kmu.ac.ir 40 slides K a1 =5.6 × and K a2 = 5.4 x 10-5 K a1 =1×10 -2 >K a2 =1×10 -7 > K a3 =1× K a2 = 1.02 × 10 -2

29 Titration of ml of M H 2 A with M NaOH. For H 2 A, K a1 = 1.00 × 10 –3 and K a2 = 1.00 × 10 –7. Titration curves for polyfunctional acids http:\\asadipour.kmu.ac.ir 40 slides

30 Titration of ml of M maleic acid with M NaOH http:\\asadipour.kmu.ac.ir 40 slides HOOC-C=C-COOH pKa1=1.89,pKa2=6.23

31 Fractional composition diagram for fumaric acid (trans-butenedioic acid). Fractional composition diagram for maleic acid (Cis-butenedioic acid) http:\\asadipour.kmu.ac.ir 40 slides Z-HOOC-C=C-COOH E-HOOC-C=C-COOH pK a1 =3.05,pK a2 =4.49 pK a1 =1.89,pK a2 =6.23

32 amino acids alanine The amine group behaves as a base, while the carboxyl group acts as an acid. Aspartic acid http:\\asadipour.kmu.ac.ir 40 slides

33 1-Determining the pK values for amino acids Amino acids contain both an acidic and a basic group http:\\asadipour.kmu.ac.ir 40 slides NH 2 -CH 2 -COOH  + NH3-CH2-COO - Zwitterion formation + NH 3 -CH 2 -COO - + H 2 O  NH 2 -CH 2 -COO - + H 3 O + + NH 3 -CH 2 -COO - + H 2 O  + NH 3 -CH 2 -COOH + OH - Ka×Kb= ??!!!!

34 2-Determining the pK values for amino acids Amino acids contain both an acidic and a basic group http:\\asadipour.kmu.ac.ir 40 slides NH 2 -CH 2 -COOH  + NH3-CH2-COO - Zwitterion formation + NH 3 -CH 2 -COOH  + NH 3 -CH 2 -COO -  NH 2 -CH 2 -COO - KbKb KaKa K a1 =5×10 -3 K a2 =2×10 -10

35 Curves for the titration of 20.00ml of M alanine with A) M NaOH B) M HCl http:\\asadipour.kmu.ac.ir 40 slides A B

36 The zwitterion of an amino acid, containing as it does a positive and a negative charge, has no tendency to migrate in an electric field, whereas the singly charged anionic and cationic species are attracted to electrodes of opposite charge. NH 2 -CH 2 -COO - + NH 3 -CH 2 -COOH No net migration of the amino acid occurs in an electric field when the pH of the solvent is such that [anionic] = [cationic], which is pH dependent. The pH at which no net migration occurs is called the isoelectric point; this point is an important physical constant for characterizing amino acids. The isoelectric point is readily related to the ionization constants for the species. Thus, for glycine, Iso electric point: The pH at which the average charge of the polyprotic acid is zero http:\\asadipour.kmu.ac.ir 40 slides + NH 3 -CH 2 -COO -

37 1-Determining iso electric point for amino acids http:\\asadipour.kmu.ac.ir 40 slides + NH3-CH2-COO - Zwitterion formation

38 2-Determining iso electric point for amino acids http:\\asadipour.kmu.ac.ir 40 slides + NH 3 -CH 2 -COOH  + NH 3 -CH 2 -COO -  NH 2 -CH 2 -COO - pK a1 =2.35 K a2 =9.87

39 Method1=method2 Ka=Ka2,,,,,,,,,,,,Kb=Kb2

40 For simple amino acids, K a and K b are generally so small that their quantitative determination by neutralization titrations is impos­sible. Amino acids that contain more than one carboxyl or amine group can sometimes be determined. If the K a values are different enough (10 4 or more), stepwise end points can be obtained just like other polyfunctional acids or bases as long as the K a values Formol titration http:\\asadipour.kmu.ac.ir 40 slides + NH 3 -CH 2 -COO - + OH -  Product + NH 3 -CH 2 -COO - + HCOH  CH2=NCH2COOH


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