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Percent Dissociation

Percent Dissociation. Chapter 16 part V. Percent Dissociation. This is the method to determine just how much of a weak acid or weak base dissociates or ionizes. % dissociation=([dissociated]/[initial amount])X100 Previously we found in 1.00 HF that [H+]=2.7 X10 -2 What is % dissociate?

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Percent Dissociation

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  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. • % dissociation=([dissociated]/[initial amount])X100 • Previously we found in 1.00 HF • that [H+]=2.7 X10-2 • What is % dissociate? • ((2.7 X10-2)/1.00)X100=2.7%

  3. Percent Dissociation • 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. 0.10M Acetic Acid

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

  5. Answer • Acetic Acid Ka= 1.8 X 10-5 • H2O Kw = 1.0 X 10-14 • Acetic Acid wins! • HC2H3O2 H+ + C2H3O2- • I 1.00 0 0 • C -X +X +X • E 1.00-X X X • Ka= 1.8X10-5 = [H+][C2H3O2-] =X2/(1.00-X) • [HC2H3O2]

  6. Answer • X2/(1.00-X) ≈ X2/(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 X2. • X2 =1.8 X 10-5 • X = 4.2 X 10-3 • Valid? • (0.0042/1.00)X100 = 0.42 % 5% rule • % dissociation ([H+]/[HC2H3O2])X100=0.42%

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

  8. Answer • HC2H3O2 H+ + C2H3O2- • I 0.10 0 0 • C -X +X +X • E 0.10-X X X • Ka= 1.8X10-5 = [H+][C2H3O2-] =X2/(0.10-X) • [HC2H3O2]

  9. Answer • X2/(0.10-X) ≈ X2/(0.10)= 1.8 X 10-5 • X = 1.3 X 10-3 • Still Valid and % Dissociation • (1.3 X 10-3/0.10)X100= 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.

  10. Ka from % dissociation

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