Percent Dissociation Chapter 16 part V Percent Dissociation

Percent Dissociation Chapter 16 part V

Percent Dissociation n n n This is the method to determine just how much of a weak acid or weak base dissociates or ionizes. % dissociation=([dissociated]/[initial amount])X 100 Previously we found in 1. 00 HF that [H+]=2. 7 X 10 -2 What is % dissociate? ((2. 7 X 10 -2)/1. 00)X 100=2. 7%

Percent Dissociation n n Example: Find the % Dissociate of Acetic Acid (Ka =1. 8 X 10 -5) in the following two examples. A. 1. 00 M Acetic Acid B. 0. 10 M Acetic Acid

First: Major species n n n Equations Ka Expression I C E

Answer n n n n n Acetic Acid Ka= 1. 8 X 10 -5 H 2 O Kw = 1. 0 X 10 -14 Acetic Acid wins! HC 2 H 3 O 2 H+ + C 2 H 3 O 2 I 1. 00 0 0 C -X +X +X E 1. 00 -X X X Ka= 1. 8 X 10 -5 = [H+][C 2 H 3 O 2 -] =X 2/(1. 00 -X) [HC 2 H 3 O 2]

Answer n n n n X 2/(1. 00 -X) ≈ X 2/(1. 00)= 1. 8 X 10 -5 Why? X is insignificant relative to the value 1, But it IS significant relative to Zero, so we keep the X 2 = 1. 8 X 10 -5 X = 4. 2 X 10 -3 Valid? (0. 0042/1. 00)X 100 = 0. 42 % 5% rule % dissociation ([H+]/[HC 2 H 3 O 2])X 100=0. 42%

Part B n n n Major species Equations Ka Expression I C E

Answer n n n HC 2 H 3 O 2 H+ + C 2 H 3 O 2 I 0. 10 0 0 C -X +X +X E 0. 10 -X X X Ka= 1. 8 X 10 -5 = [H+][C 2 H 3 O 2 -] =X 2/(0. 10 -X) [HC 2 H 3 O 2]

Answer n n n n X 2/(0. 10 -X) ≈ X 2/(0. 10)= 1. 8 X 10 -5 X = 1. 3 X 10 -3 Still Valid and % Dissociation (1. 3 X 10 -3/0. 10)X 100= 1. 3% The duh Factor As the initial []’s of a weak acid is smaller, the % dissociation gets larger. We can also find the Ka from % dissociation.

Ka from % dissociation