1. 4.8-4.9: Ka, Kb and the Conjugate Pair
Chemistry 12

2. 4.8 – 4.9
Keq =
Ksp =
Kw =
Ka =
Kb =
equilibrium constant solubility product eqb exp for ionization of water acid ionization constant base ionization constant

3. Concept 1
Using Kw to calculate unknown [H3O+] and [OH-] from a strong acid or a strong base.

4. Eg. ) At 250C an HCl solution has a concentration of 0. 0100 M
Eg.) At 250C an HCl solution has a concentration of M. What is the [OH-]?
HCl + H20 ↔ H3O+ + Cl-
[H3O+] = M
[H3O+][OH-] = Kw
M [OH-] = 1.00 x 10-14
[OH-] = 1.00 x M

5. Concept 2: Equilibrium equations and Ka and Kb Expressions
What is the Ka expression for oxalic acid? (Hint: use the table) H2O(l) + H2C2O4(aq) ↔ H3O+(aq) + HC2O4-(aq) Ka = [H3O+][HC2O4-] [H2C2O4]

6. What is the equilibrium equation for HC2O4- acting as a base and the corresponding Kb expression?
HC2O4-(aq) + H2O(l) ↔H2C2O4(aq) +OH-(aq)
Kb = [H2C2O4][OH-]
[HC2O4-]

7. Concept 3: Finding Ka and calculating Kb
What is Ka value for the following?
(simply use the table)
H2C2O4 Ka = 5.9 x 10-2
HC2O4- Ka = 6.4 x 10 -5

8. …but solving for Kb is not that easy.

9. What is the Kb value of HPO42-
What is the Kb value of HPO42-? … get out your B-L Table and follow along.

10. We must use the Ka value of H2PO4-
to determine Kb of HPO42-
HPO42- is acting as a base
HPO42- is acting as an acid
Ka value of HPO42- so we cannot use this to calculate Kb

11. Calculation
Kb of HPO42- = Kw/ Ka(H2PO4-) = 10-14/6.2 x 10-8 = 1.6 x10-7
You should notice that this is the Ka value of the conjugate acid for HPO42-

12. Now you try
Kb for HC2O4-? Find HC2O4- on right side of table Ka = 5.9 x Kw = Ka x Kb 1.0 x = 5.9 x x Kb Kb = 1.7 x 10-13
Actually Ka of H2C2O4

13. Concept 4: Explaining the connection between Kw, Ka and Kb.
Why do we use the formula Kw = Ka x Kb? H2O(l) + HC2O4-(aq) ↔ H3O+(aq) + C2O42-(aq) Acid dissociation Base dissociation
This can be explained using HC2O4- but other weak acids could also be used

14. 4.8 – 4.9
To get Kb we must figure out the base dissociation equation of C2O42-.

15. 4.8 – 4.9
H2O(l) + HC2O4-(aq) ↔ H3O+(aq) + C2O42-(aq) Invert H3O+(aq) + C2O42-(aq) ↔ H2O(l) + HC2O4-(aq) + OH- (aq) ↔ OH-(aq) 2 H2O(l) H2O(l) + C2O42-(aq) ↔ OH-(aq) + HC2O4-(aq)

16. 4.8 – 4.9
H2O(l) + C2O42-(aq) ↔ OH-(aq) + HC2O4-(aq) H2O(l) + HC2O4-(aq) ↔ H3O+(aq) + C2O42-(aq) Kb = [HC2O4-][OH-] [C2O42-] Ka = [H3O+][C2O42-] [HC2O4-]

17. 4.8 – 4.9  if Then Ka x Kb [HC2O4-][OH-] x [H3O+][C2O42-]
Kb = [HC2O4-][OH-] Ka = [H3O+][C2O42-]
[C2O42-] [HC2O4-]
Then Ka x Kb
[HC2O4-][OH-] x [H3O+][C2O42-]
[C2O42-] [HC2O4-]
 Ka x Kb = [OH-][H3O+] = Kw

18. Learning Check
Can you calculate [H3O+] and [OH-] of a strong acid or base using Kw?
Can you write equilibrium equations for weak acids and bases?
Can you write out Ka and Kb expressions?
Can you find Ka on the B-L table?
Can you calculate Kb using the B-L table?

19. Read & Highlight Self Notes p. 3 - 5 Hebden # 31-37