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16-6 Polyprotic Acids

16-6 Polyprotic Acids. Phosphoric acid: A triprotic acid. K a = 7.1  10 -3. H 3 PO 4 + H 2 O H 3 O + + H 2 PO 4 -. K a = 6.3  10 -8. H 2 PO 4 - + H 2 O H 3 O + + HPO 4 2-. K a = 4.2  10 -13. HPO 4 2- + H 2 O H 3 O + + PO 4 3-. Phosphoric Acid.

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16-6 Polyprotic Acids

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  1. 16-6 Polyprotic Acids Phosphoric acid: A triprotic acid. Ka = 7.110-3 H3PO4 + H2O H3O+ + H2PO4- Ka = 6.310-8 H2PO4- + H2O H3O+ + HPO42- Ka = 4.210-13 HPO42- + H2O H3O+ + PO43- General Chemistry: Chapter 16

  2. Phosphoric Acid • Ka1 >> Ka2 • All H3O+ is formed in the first ionization step. • H2PO4- essentially does not ionize further. • Assume [H2PO4-] = [H3O+]. • [HPO42-]  Ka2regardless of solution molarity. General Chemistry: Chapter 16

  3. General Chemistry: Chapter 16

  4. EXAMPLE 16-9 Calculating Ion Concentrations in a Polyprotic Acid Solution. For a 3.0 M H3PO4 solution, calculate: (a) [H3O+]; (b) [H2PO4-]; (c) [HPO42-] (d) [PO43-] H3PO4 + H2O H2PO4- + H3O+ Initial conc. 3.0 M 0 0 Changes -x M +x M +x M Equilibrium (3.0-x) M x M x M Concentration General Chemistry: Chapter 16

  5. EXAMPLE 16-9 H3PO4 + H2O H2PO4- + H3O+ [H3O+] [H2PO4-] x · x = 7.110-3 Ka= = [H3PO4] (3.0 – x) Assume that x << 3.0 x2 = (3.0)(7.110-3) x = 0.14 M [H2PO4-] = [H3O+] = 0.14 M General Chemistry: Chapter 16

  6. EXAMPLE 16-9 [H3O+] [HPO42-] y · (0.14 + y) = 6.310-8 Ka= = [H2PO4-] (0.14 - y) H2PO4- + H2O HPO42- + H3O+ Initial conc. 0.14 M 0 0.14 M Changes -y M +y M +y M Equilibrium (0.14 - y) M y M (0.14 +y) M Concentration y << 0.14 M y = [HPO42-] = 6.310-8 General Chemistry: Chapter 16

  7. EXAMPLE 16-9 HPO4- + H2O PO43- + H3O+ [H3O+] [HPO42-] (0.14)[PO43-] = 4.210-13 M Ka= = 6.310-8 [H2PO4-] [PO43-] = 1.910-19 M General Chemistry: Chapter 16

  8. Sulfuric Acid Sulfuric acid: A diprotic acid. H2SO4 + H2O H3O+ + HSO4- Ka = very large HSO4- + H2O H3O+ + SO42- Ka = 1.96 General Chemistry: Chapter 16

  9. [NH3] [H3O+] = ? Ka= [NH4+] 1.010-14 KW [NH3][H3O+] [OH-] = 5.610-10 Ka= = = 1.810-5 Kb [NH4+] [OH-] 16-7 Ions as Acids and Bases CH3CO2- + H2OCH3CO2H + OH- base acid NH4+ + H2O NH3 + H3O+ base acid KaKb = Kw General Chemistry: Chapter 16

  10. Hydrolysis • Water (hydro) causing cleavage (lysis) of a bond. Na+ + H2O → Na+ + H2O No reaction No reaction Cl- + H2O → Cl- + H2O NH4+ + H2O → NH3 + H3O+ Hydrolysis General Chemistry: Chapter 16

  11. 16-8 Molecular Structure and Acid-Base Behavior • Why is CH3CO2H a stronger acid than CH3CH2OH? • There is a relationship between molecular structure and acid strength. • Bond dissociation energies are measured in the gas phase and not in solution. General Chemistry: Chapter 16

  12. Strengths of Oxoacids • Factors promoting electron withdrawal from the OH bond to the oxygen atom: • High electronegativity (EN) of the central atom. • A large number of terminal O atoms in the molecule. H-O-Cl H-O-Br ENCl = 3.0 ENBr= 2.8 Ka = 2.910-8 Ka = 2.110-9 General Chemistry: Chapter 16

  13. - ·· - ·· O O ·· ·· ·· ·· 2+ + S O H O H S O H O H ·· ·· ·· ·· ·· - ·· ·· ·· ·· ·· ·· O ·· ·· ·· ·· ·· ·· ·· O O ·· ·· ·· ·· S O H O H S O H O H ·· ·· ·· ·· ·· O Strengths of Oxoacids Ka 103 Ka =1.310-2 General Chemistry: Chapter 16

  14. ·· ·· Strengths of Organic Acids O H H H ·· ·· C C C H O H C H O H ·· ·· H H H acetic acidethanol Ka = 1.810-5 Ka =1.310-16 General Chemistry: Chapter 16

  15. H H H H H H ·· ·· ·· ·· ·· ·· ·· ·· C C C C C C O H H H H H H C - O ·· Structural Effects H O Ka = 1.810-5 Ka = 1.310-5 C C H - O H ·· H C H H General Chemistry: Chapter 16

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