1. Slide 1 of 52 16-3 The Self-Ionization of Water and the pH Scale

2. Slide 2 of 52 Ion Product of Water Kc=Kc= [H 2 O][H 2 O] [H 3 O + ][OH - ] H 2 O + H 2 O H 3 O + + OH - baseacid conjugate acid conjugate base K W = K c [H 2 O][H 2 O] = = 1.0  10 -14 [H 3 O + ][OH - ]

3. Slide 3 of 52 pH and pOH  The potential of the hydrogen ion was defined in 1909 as the negative of the logarithm of [H + ]. pH = -log[H 3 O + ]pOH = -log[OH - ] -logK W = -log[H 3 O + ]-log[OH - ]= -log(1.0  10 -14 ) K W = [H 3 O + ][OH - ]= 1.0  10 -14 pK W = pH + pOH= -(-14) pK W = pH + pOH = 14

4. Slide 4 of 52 pH and pOH Scales

5. Slide 5 of 52 16-4 Strong Acids and Bases

6. Slide 6 of 52 16-5 Weak Acids and Bases Lactic AcidGlycine General Chemistry: Chapter 16Prentice-Hall © 2007 Acetic Acid

7. Slide 7 of 52 Acetic Acid Weak Acids Ka=Ka= = 1.8  10 -5 [CH 3 CO 2 H] [CH 3 CO 2 - ][H 3 O + ] pK a = -log(1.8  10 -5 ) = 4.74 General Chemistry: Chapter 16Prentice-Hall © 2007

8. Slide 8 of 52 Weak Bases Kb=Kb= = 4.3  10 -4 [CH 3 NH 2 ] [CH 3 NH 3 + ][HO - ] pK b = -log(4.2  10 -4 ) = 3.37

9. Slide 9 of 52 Table 16.3 Ionization Constants of Weak Acids and Bases

10. Slide 10 of 52 Determining a Value of K A from the pH of a Solution of a Weak Acid. Butyric acid, HC 4 H 7 O 2 (or CH 3 CH 2 CH 2 CO 2 H) is used to make compounds employed in artificial flavorings and syrups. A 0.250 M aqueous solution of HC 4 H 7 O 2 is found to have a pH of 2.72. Determine K A for butyric acid. HC 4 H 7 O 2 + H 2 O C 4 H 7 O 2 + H 3 O + K a = ? EXAMPLE 16-5

11. Slide 11 of 52 HC 4 H 7 O 2 + H 2 O C 4 H 7 O 2 + H 3 O + Initial conc.0.250 M00 Changes-x M+x M+x M Equilibrium (0.250-x) M x Mx M Concentration EXAMPLE 16-5 Solution: For HC 4 H 7 O 2 K A is likely to be much larger than K W. Therefore assume self-ionization of water is unimportant.

12. Slide 12 of 52 Log[H 3 O + ] = -pH = -2.72 HC 4 H 7 O 2 + H 2 O C 4 H 7 O 2 + H 3 O + [H 3 O + ] = 10 -2.72 = 1.9  10 -3 = x [H 3 O + ] [C 4 H 7 O 2 - ] [HC 4 H 7 O 2 ] Ka=Ka= 1.9  10 -3 · 1.9  10 -3 (0.250 – 1.9  10 -3 ) = K a = 1.5  10 -5 Check assumption: K a >> K W. EXAMPLE 16-5

13. Slide 13 of 52 Percent Ionization HA + H 2 O H 3 O + + A - Degree of ionization = [H 3 O + ] from HA [HA] originally Percent ionization = [H 3 O + ] from HA [HA] originally  100%

14. Slide 14 of 52 Percent Ionization K a = [H 3 O + ][A - ] [HA] K a = n H3O+H3O+ A-A- n HA n 1 V