1. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 1 of 52 16-6 Polyprotic Acids H 3 PO 4 + H 2 O H 3 O + + H 2 PO 4 - H 2 PO 4 - + H 2 O H 3 O + + HPO 4 2- HPO 4 2- + H 2 O H 3 O + + PO 4 3- Phosphoric acid: A triprotic acid. K a = 7.1  10 -3 K a = 6.3  10 -8 K a = 4.2  10 -13

2. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 2 of 52 Phosphoric Acid  K a1 >> K a2 ◦All H 3 O + is formed in the first ionization step.  H 2 PO 4 - essentially does not ionize further. ◦Assume [H 2 PO 4 - ] = [H 3 O + ].  [HPO 4 2- ]  K a2 regardless of solution molarity.

3. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 3 of 52

4. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 4 of 52 Calculating Ion Concentrations in a Polyprotic Acid Solution. For a 3.0 M H 3 PO 4 solution, calculate: (a) [H 3 O + ]; (b) [H 2 PO 4 - ]; (c) [HPO 4 2- ] (d) [PO 4 3- ] H 3 PO 4 + H 2 O H 2 PO 4 - + H 3 O + Initial conc.3.0 M00 Changes-x M+x M+x M Equilibrium(3.0-x) M x Mx M Concentration EXAMPLE 16-9

5. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 5 of 52 H 3 PO 4 + H 2 O H 2 PO 4 - + H 3 O + [H 3 O + ] [H 2 PO 4 - ] [H 3 PO 4 ] Ka=Ka= x · x (3.0 – x) = Assume that x << 3.0 = 7.1  10 -3 x 2 = (3.0)(7.1  10 -3 ) x = 0.14 M [H 2 PO 4 - ] = [H 3 O + ] = 0.14 M EXAMPLE 16-9

6. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 6 of 52 H 2 PO 4 - + H 2 O HPO 4 2- + H 3 O + [H 3 O + ] [HPO 4 2- ] [H 2 PO 4 - ] Ka=Ka= y · (0.14 + y) (0.14 - y) = = 6.3  10 -8 Initial conc.0.14 M00.14 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 = [HPO 4 2- ] = 6.3  10 -8 EXAMPLE 16-9

7. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 7 of 52 HPO 4 - + H 2 O PO 4 3- + H 3 O + [H 3 O + ] [HPO 4 2- ] [H 2 PO 4 - ] Ka=Ka= (0.14)[PO 4 3- ] 6.3  10 -8 = = 4.2  10 -13 M [PO 4 3- ] = 1.9  10 -19 M EXAMPLE 16-9

8. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 8 of 52 Sulfuric Acid Sulfuric acid: A diprotic acid. H 2 SO 4 + H 2 O H 3 O + + HSO 4 - HSO 4 - + H 2 O H 3 O + + SO 4 2- K a = very large K a = 1.96

9. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 9 of 52 16-7 Ions as Acids and Bases NH 4 + + H 2 O NH 3 + H 3 O + base acid CH 3 CO 2 - + H 2 O CH 3 CO 2 H + OH - base acid [NH 3 ] [H 3 O + ] [OH - ] Ka=Ka= [NH 4 + ] [OH - ] [NH 3 ] [H 3 O + ] Ka=Ka= [NH 4 + ] = ? = KWKW KbKb = 1.0  10 -14 1.8  10 -5 = 5.6  10 -10 K a K b = K w

10. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 10 of 52 Hydrolysis  Water (hydro) causing cleavage (lysis) of a bond. Na + + H 2 O → Na + + H 2 O NH 4 + + H 2 O → NH 3 + H 3 O + Cl - + H 2 O → Cl - + H 2 O No reaction Hydrolysis

11. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 11 of 52 16-8 Molecular Structure and Acid-Base Behavior  Why is CH 3 CO 2 H a stronger acid than CH 3 CH 2 OH?  There is a relationship between molecular structure and acid strength.  Bond dissociation energies are measured in the gas phase and not in solution.

12. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 12 of 52 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-ClH-O-Br EN Cl = 3.0EN Br = 2.8 K a = 2.9  10 -8 K a = 2.1  10 -9

13. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 13 of 52 Strengths of Oxoacids S O O O O HH ·· - 2+ ·· - S O O O HH - + S O O O O HH S O O O HH K a  10 3 K a =1.3  10 -2

14. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 14 of 52 Strengths of Organic Acids C O C O HH ·· H H O C HH H H C H H K a = 1.8  10 -5 K a =1.3  10 -16 acetic acidethanol

15. Prentice-Hall © 2007 General Chemistry: Chapter 16 Slide 15 of 52 Structural Effects C H H C O C O H - ·· H H C H H H C O O - C H H C H H C H H C H H C H H K a = 1.8  10 -5 K a = 1.3  10 -5