1. EQUILBRIUM OF Acids and Bases
To play the movies and simulations included, view the presentation in Slide Show Mode.
Chapter 17

2. Water Equilibrium constant for water Kw = [H3O+] [OH-] =
H2O can function as both an ACID and a BASE.
Water
Equilibrium constant for water
Kw = [H3O+] [OH-] =
1.00 x at 25 oC

6. Equilibria Involving Weak Acids and Bases
Aspirin is a good example of a weak acid, Ka = 3.2 x 10-4

7. Weak Acids and Bases
Acid Conjugate Base
acetic, CH3CO2H CH3CO2-, acetate
ammonium, NH NH3, ammonia
bicarbonate, HCO3- CO32-, carbonate
A weak acid (or base) is one that ionizes to a VERY small extent (< 5%).

9. Weak Acids and Bases acetic acid, CH3CO2H (HOAc)
HOAc H2O D H3O OAc-
Acid Conj. base
(K is designated Ka for ACID)
[H3O+] and [OAc-] are SMALL, Ka << 1.

10. Equilibrium Constants for Weak Acids
Weak acid has Ka < 1
Leads to small [H3O+] and a pH of 2 - 7

12. ?

13. Equilibria Involving A Weak Acid
You have 1.00 M HOAc. Calc. the pH. And the equilibrium concs. of EACH
Step 1. ICE table.
[HOAc] [H3O+] [OAc-] I C E

14. Equilibria Involving A Weak Acid
[HOAc] [H3O+] [OAc-] I
C x +x x
E x x x
Note that we neglect [H3O+] from H2O.

15. Equilibria Involving A Weak Acid
Step 2. Write Ka expression
This is a quadratic.
Use quadratic formula or
method of approximations
(see Appendix A).
HOWEVER

16. Equilibria Involving A Weak Acid
Assume x is very small because Ka is so small.
Now we can more easily solve this approximate expression.

17. Equilibria Involving A Weak Acid
Step 3. Solve Ka approximate expression
x = [H3O+] = [OAc-] = [Ka • 1.00]1/2
x = [H3O+] = [OAc-] = 4.2 x 10-3 M
pH = -log (4.2 x 10-3) = 2.37

18. [H3O+] = [Ka • Co]1/2

19. Equilibria Involving A Weak Acid
Calculate the pH of a M solution of formic acid, HCO2H.
HCO2H H2O D HCO H3O+
Ka = 1.8 x 10-4
Approximate solution
[H3O+] = [Ka • Co]1/2 = 4.2 x 10-4 M, pH = 3.4
Exact Solution
[H3O+] = [HCO2-] = 3.4 x 10-4 M
[HCO2H] = x 10-4 = M
pH = 3.5

20. Chapter 17 – Acids and Bases
What is the [H+] of a 1.0 M acetic acid if
Ka = 1.8 x 10-5 for the solution:
A M
B M
C M
D x 10-5 M
[H3O+] = [Ka • Co]1/2
Answer: C

21. Chapter 17 – Acids and Bases
A M solution of HX has a pH of is 4.5 Determine the Ka for the acid, HX.
A x 10-8
B x 10-7
C x 10-8
D x 10-9
Answer: A

22. Weak Bases

23. Equilibrium Constants for Weak Bases
Weak base has Kb < 1
Leads to small [OH-] and a pH of

25. Equilibria Involving A Weak Base
You have M NH3. Calc. the pH.
NH3 + H2O D NH OH-
Kb = 1.8 x 10-5
Step 1; ICE table
[NH3] [NH4+] [OH-] I C E

26. Weak Base I 0.010 0 0 C -x +x +x E 0.010 - x x x Step 1. ICE table
[NH3] [NH4+] [OH-] I
C x x +x
E x x x

27. Assume x is small (100•Kb < Co), so The approximation is valid !
Weak Base
Step 2. Solve the equilibrium expression
Assume x is small (100•Kb < Co), so
x = [OH-] = [NH4+] = 4.2 x 10-4 M
[NH3] = x 10-4 ≈ M
The approximation is valid !

28. Kw = Ka* Kb
Acids
Conjugate
Bases

29. Relation of Ka, Kb, [H3O+] and pH