1. CHAPTER 13
Acids
and Bases
13.3 Acid-Base Equilibria

2. HCl(aq) → H+(aq) + Cl–(aq)
Strong:
Complete dissociation
HCl(aq) → H+(aq) + Cl–(aq)
1 M
Acids
1 M
< 0.02 M
H2C6H6O6(aq) → 2H+(aq) + C6H6O6–(aq)
Weak:
Partial dissociation
Strong:
Complete dissociation
Ca(OH)2(aq) → Ca2+(aq) + 2OH–(aq)
1 M
2 M
Bases
Weak:
Partial dissociation
1 M
~ M
NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

3. Incomplete dissociations are examples of chemical equilibria
We will learn how to calculate the pH of a weak acid or a weak base solution
1 M
< 0.02 M
H2C6H6O6(aq) → 2H+(aq) + C6H6O6–(aq)
Weak:
Partial dissociation
Acids
Weak:
Partial production of OH-
1 M
~ M
NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)
Bases

4. Dissociation of water and Kw
H2O(l) H+(aq) + OH–(aq)
For dilute solutions there is an equilibrium between [H+] and [OH–]
ion product constant: in dilute aqueous solutions,
[H+] x [OH–] = 1.0 x 10–14 = Kw at 25oC.

5. A certain solution has a concentration of H+ ions of 0. 0005 M
A certain solution has a concentration of H+ ions of M. What is the concentration of OH– ions? Is the solution acidic or basic?

6. A certain solution has a concentration of H+ ions of 0. 0005 M
A certain solution has a concentration of H+ ions of M. What is the concentration of OH– ions? Is the solution acidic or basic?
Asked: [OH–] and pH of the new solution
Given: [H+] = M
Relationships: [H+] x [OH–] = 1.0 x 10–14
pH = –log[H+]

7. A certain solution has a concentration of H+ ions of 0. 0005 M
A certain solution has a concentration of H+ ions of M. What is the concentration of OH– ions? Is the solution acidic or basic?
Asked: [OH–] and pH of the new solution
Given: [H+] = M
Relationships: [H+] x [OH–] = 1.0 x 10–14
pH = –log[H+]
Solve:

8. A certain solution has a concentration of H+ ions of 0. 0005 M
A certain solution has a concentration of H+ ions of M. What is the concentration of OH– ions? Is the solution acidic or basic?
Asked: [OH–] and pH of the new solution
Given: [H+] = M
Relationships: [H+] x [OH–] = 1.0 x 10–14
pH = –log[H+]
Solve:
Answer: The concentration of OH– ions is 2 x 10–11 M.
The solution has a pH of 3.3, and is a moderately strong acid.

9. Ka is the equilibrium constant for an acid
Complete dissociation
Large Ka (1.3 x 106)
Partial dissociation
Small Ka (1.8 x 10–5)
Ka is the equilibrium constant for an acid

10. The Ka for acetic acid is a small number(1.8 x 10–5)
because
[products] < [reactants]

11. pH of weak acids
Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5).
Write down the balanced net ionic equation

12. No product at the beginning of the reaction
pH of weak acids
Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5).
No product at the beginning of the reaction
Initial molarity
Write down the balanced net ionic equation
Set up the equilibrium relationship

13. pH of weak acids
Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5).
An unknown amount is consumed (–x)
For every x of reactant consumed,
x of product is produced (+x), 1:1 ratio
Write down the balanced net ionic equation
Set up the equilibrium relationship

14. Add the molarities from Initial and Change
pH of weak acids
Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5). +
Add the molarities from Initial and Change
Write down the balanced net ionic equation
Set up the equilibrium relationship

15. Finding x will help us calculate pH
pH of weak acids
Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5).
x = [H+] at equilibrium
Finding x will help us calculate pH
Write down the balanced net ionic equation
Set up the equilibrium relationship
Use the RICE table to solve for [H+]

16. pH of weak acids
Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5).

17. pH of weak acids
Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5).

18. pH of weak acids
Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5).
Write down the balanced net ionic equation
Set up the equilibrium relationship
Use the RICE table to solve for [H+]
Calculate the pH from [H+]

19. Same answer as from solving the quadratic equation
pH of weak acids
Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5).
Same answer as from solving the quadratic equation

20. pH of weak acids
This equation only works when MKa > 100

21. Polyprotic acids H2A HA– + H+ A2– + 2H+
A polyprotic acid can dissociate more than once and donates more than one proton:
H2A HA– + H A2– + 2H+
Carbonic acid is
a polyprotic acid:
pH = –log[H+]

22. Equilibrium constants
A weak acid:
HC2H3O2 H+ + C2H3O2– Ka
NH3 + H2O NH OH– Kb
acetic acid
Equilibrium constants
A weak base:
ammonia

23. pH of weak bases
Find the pH of a 1.0 M ammonia solution (Kb = 1.8 x 10–5).
Write down the balanced net ionic equation

24. pH of weak bases
Find the pH of a 1.0 M ammonia solution (Kb = 1.8 x 10–5).
Write down the balanced net ionic equation
Set up the equilibrium relationship

25. pH of weak bases
Find the pH of a 1.0 M ammonia solution (Kb = 1.8 x 10–5).
Write down the balanced net ionic equation
Set up the equilibrium relationship
Use the RICE table to solve for [OH–]

26. pH of weak bases
Find the pH of a 1.0 M ammonia solution (Kb = 1.8 x 10–5).

27. pH of weak bases
Find the pH of a 1.0 M ammonia solution (Kb = 1.8 x 10–5).
Write down the balanced net ionic equation
Set up the equilibrium relationship
Use the RICE table to solve for [OH–]
Calculate the pH using pH = 14 + log[OH–]

28. Use a RICE table to find the pH of a weak acid or a weak base
Once [H+] or [OH–] has been determined, the pH can be calculated