1. ACIDS AND BASES
Dissociation Constants

2. Write the equilibrium expression (Ka or Kb) from a balanced chemical equation.
Use Ka or Kb to solve problems for pH, percent dissociation and concentration.
Additional KEY Terms

3. Larger Ka : stronger acid : more product : more H+
HA(aq) H+(aq) A-(aq)
Strong Acid
HA(aq) + H2O(l) H3O+(aq) A-(aq)
Weak Acid
Ka - acid dissociation constant
Larger Ka : stronger acid : more product : more H+

4. Larger Kb : stronger base : more product : more OH-
BOH (aq) B+(aq) OH-(aq)
Strong Base
B (aq) + H2O(l) BH+(aq) OH-(aq)
Weak Base
Kb - base dissociation constant
Larger Kb : stronger base : more product : more OH-

5. Type III – all initial and one equilibrium concentration
Initially a 0.10 M solution of acetic acid, it reaches equilibrium with a [H3O+] = 1.3 x 10-3 M. What is the acid dissociation constant, Ka?
CH3COOH(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq) I
C x x x 10-3
Ka = 1.7 x 10-5
Ignore the units for K.
E x 10-3
1.3 x 10-3
Type III – all initial and one equilibrium concentration

6. Type IV– all initial and NO equilibrium
HA is a weak acid with a Ka of 7.3 x What are the equilibrium concentrations if the initial [HA] is 0.50 M?
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
[I]
[C] -x +x +x
[E] 0.5-x x x
Ka = [H3O][A-]
[HA]
Type IV– all initial and NO equilibrium

7. √ √ 7.3 x 10-8 = [x][x] 0.50 - x (7.3 x 10-8)(0.50) = x2
*Ka is small - assume that x is
negligible compared to 0.50
- x
(7.3 x 10-8)(0.50) = x2 √ √
3.65 x = x2
1.9 x 10-4 = x
[H3O+] = [A-] = 1.9 x 10-4 M
[HA] = x
= x 10-4
= M
*Ka is small – OK to ignore it
0.50 M

8. Ka = [H3O+][HS-] [H2S] H2S (aq) + H2O (l) H3O+(aq) + HS-(aq)
Calculate the pH of a 0.10 mol/L hydrogen sulfide solution. (Ka=1.0 x 10-7)
H2S (aq) + H2O (l)
H3O+(aq) + HS-(aq)
[I]
[C] -x +x +x
[E] x x x
Ka = [H3O+][HS-]
[H2S]

9. √ √ 1.0 x 10-7 = [x][x] 0.10 - x (1.0 x 10-7)(0.10) = x2
*Ka is small - x is negligible
- x
(1.0 x 10-7)(0.10) = x2 √ √
1.0 x = x2
1.0 x = x
[H3O+] = [HS-] = 1.0 x 10-4 M
pH = - log [H3O+] = - log(1.0 x 10-4)
pH = 4.00

10. Each acid/base has K associated with it
polyprotic acids lose their hydrogen one at a time - each ionization reaction has separate Ka
Sulfuric acid H2SO4
H2SO4(aq) H+(aq) + HSO4-(aq)
Ka1
HSO4- (aq) H+(aq) + SO4-2(aq)
Ka2

11. Percent Dissociation Ka / Kb represent the degree of dissociation
(how much product has formed)
Another way to describe dissociation is by percent dissociation

12. CH2O2H (aq) + H2O(l) H3O+(aq) + CH2O2¯(aq)
Calculate the percent dissociation of a solution of formic acid (CH2OOH) if the hydronium ion concentration is
0.100 M
4.21 x 10-3 M
CH2O2H (aq) + H2O(l) H3O+(aq) + CH2O2¯(aq)

13. Use the %diss formula to find [OH-]
Calculate the Kb of hydrogen phosphate ion (HPO42-) if 0.25 M solution of hydrogen phosphate dissociates 0.080%.
HPO42- + H2O H2PO4- + OH-
Use the %diss formula to find [OH-]

14. HPO42- + H2O H2PO4- + OH- Kb = [H2PO4-][OH-] [HPO42-]
[I]
[C] x x x
[E] x x
[OH-] = [H2PO4-] = x 10-4 M
Kb = [H2PO4-][OH-]
[HPO42-]
Kb= [2.0 x 10-4][2.0 x 10-4]
0.25
Kb = 1.6 x 10-7

15. The smaller the Ka or Kb, the weaker the acid / base
percent dissociation describes the amount of
acid/base dissociated

16. CAN YOU / HAVE YOU?
Write the equilibrium expression (Ka or Kb) from a balanced chemical equation.
Use Ka or Kb to solve problems for pH, percent dissociation and concentration.
Additional KEY Terms