1. Percent Dissociation Chapter 16 part V

2. Percent Dissociation This is the method to determine just how much of a weak acid or weak base dissociates or ionizes. This is the method to determine just how much of a weak acid or weak base dissociates or ionizes. % dissociation=([dissociated]/[initial amount])X100 % dissociation=([dissociated]/[initial amount])X100 Previously we found in 1.00 HF Previously we found in 1.00 HF that [H+]=2.7 X10 -2 that [H+]=2.7 X10 -2 What is % dissociate? What is % dissociate? ((2.7 X10 -2 )/1.00)X100=2.7% ((2.7 X10 -2 )/1.00)X100=2.7%

3. Percent Dissociation Example: Example: Find the % Dissociate of Acetic Acid (Ka =1.8 X 10 -5 ) in the following two examples. Find the % Dissociate of Acetic Acid (Ka =1.8 X 10 -5 ) in the following two examples. A. 1.00 M Acetic Acid A. 1.00 M Acetic Acid B. 0.10M Acetic Acid B. 0.10M Acetic Acid

4. First: Major species Equations Equations Ka Expression Ka Expression I C E

5. Answer Acetic Acid Ka= 1.8 X 10 -5 Acetic Acid Ka= 1.8 X 10 -5 H 2 O Kw = 1.0 X 10 -14 H 2 O Kw = 1.0 X 10 -14 Acetic Acid wins! Acetic Acid wins! HC 2 H 3 O 2  H+ + C 2 H 3 O 2 - HC 2 H 3 O 2  H+ + C 2 H 3 O 2 - I1.0000 I1.0000 C-X +X +X C-X +X +X E 1.00-X X X E 1.00-X X X Ka= 1.8X10 -5 = [H + ][C 2 H 3 O 2 - ] =X 2 /(1.00-X) Ka= 1.8X10 -5 = [H + ][C 2 H 3 O 2 - ] =X 2 /(1.00-X) [HC 2 H 3 O 2 ] [HC 2 H 3 O 2 ]

6. Answer X 2 /(1.00-X) ≈ X 2 /(1.00)= 1.8 X 10 -5 X 2 /(1.00-X) ≈ X 2 /(1.00)= 1.8 X 10 -5 Why? X is insignificant relative to the value 1, Why? X is insignificant relative to the value 1, But it IS significant relative to Zero, so we keep the X 2. But it IS significant relative to Zero, so we keep the X 2. X 2 = 1.8 X 10 -5 X 2 = 1.8 X 10 -5 X = 4.2 X 10-3 X = 4.2 X 10-3 Valid? Valid? (0.0042/1.00)X100 = 0.42 % 5% rule (0.0042/1.00)X100 = 0.42 % 5% rule % dissociation ([H+]/[HC 2 H 3 O 2 ])X100=0.42% % dissociation ([H+]/[HC 2 H 3 O 2 ])X100=0.42%

7. Part B Major species Major species Equations Equations Ka Expression Ka Expression I C E

8. Answer HC 2 H 3 O 2  H+ + C 2 H 3 O 2 - HC 2 H 3 O 2  H+ + C 2 H 3 O 2 - I0.1000 I0.1000 C-X +X +X C-X +X +X E 0.10-X X X E 0.10-X X X Ka= 1.8X10 -5 = [H + ][C 2 H 3 O 2 - ] =X 2 /(0.10-X) Ka= 1.8X10 -5 = [H + ][C 2 H 3 O 2 - ] =X 2 /(0.10-X) [HC 2 H 3 O 2 ] [HC 2 H 3 O 2 ]

9. Answer X 2 /(0.10-X) ≈ X 2 /(0.10)= 1.8 X 10 -5 X 2 /(0.10-X) ≈ X 2 /(0.10)= 1.8 X 10 -5 X = 1.3 X 10 -3 X = 1.3 X 10 -3 Still Valid and % Dissociation Still Valid and % Dissociation (1.3 X 10 -3 /0.10)X100= 1.3% (1.3 X 10 -3 /0.10)X100= 1.3% The duh Factor The duh Factor As the initial []’s of a weak acid is smaller, the % dissociation gets larger. As the initial []’s of a weak acid is smaller, the % dissociation gets larger. We can also find the Ka from % dissociation. We can also find the Ka from % dissociation.

10. Ka from % dissociation