1. 8.2 – Equilibrium of Weak Acids and Bases
Acids and bases dissociate in aqueous solutions to form ions that interact with water.
pH of a solution is determined by the equilibrium of the reaction between water and the ions of the acid or base.

2. Dissociation of Water
Pure water contains a few ions from the dissociation of water.
2H20(l) ↔ H3O+(aq) OH-(aq)
At 25°C ~ 2 molecules/1 billion dissociate
Due to the 1:1 ratio of H3O+ to OH- the [H3O+] = [OH-]
At 25°C, [H3O+] = [OH-] = 1.0x10-7 mol/L

3. The Ion Product Constant for Water
The equilibrium constant for water
Kc = [H3O+] [OH-]
[H2O]2
The equilibrium value of [H3O+] [OH-] at 25°C is called the ion constant for water (Kw).
Kw = [H3O+] [OH-]
= 1.0 x 10-7 mol/L x 1.0 x 10-7 mol/L
= 1.0 x (units are dropped)

4. Strong Acids and Bases
With strong acids and bases, [H3O+] and [OH-] from water is negligible compared to that from the acid or base, so the dissociation of water is ignored.
Consider 0.1 M HCl,
All of the HCl dissociates, forming [H3O+] = 0.1 mol/L.
2H20(l) ↔ H3O+(aq) OH-(aq)
This forces the equation for dissociation of water to the left (←) (LeChâtelier’s)
Therefore ,[H3O+] from water is, < 1.0 x 10-7 mol/L
So it can be ignored

5. [H3O+] and [OH-] at 25°C Acidic Solutions Neutral Solutions
[H3O+] > 1.0 x 10-7 mol/L
[OH-] < 1.0 x 10-7 mol/L
Neutral Solutions
[OH-] = [H3O+] = 1.0 x 10-7 mol/L
Basic Solutions
[H3O+] < 1.0 x 10-7 mol/L
[OH-] > 1.0 x 10-7 mol/L

6. Determining [H3O+] and [OH-]
Find [H3O+] and [OH-] in 0.16 M Ba(OH)2
Ba(OH)2 is a strong base, therefore it dissociates completely.
Therefore use [Ba(OH)2] to find [OH-]
H2O
Ba(OH)2  Ba2+ + 2OH-
0.16 mol/L Ba(OH)2 x 2 mol OH = 0.32 mol/L OH-
1 mol Ba(OH)2

7. Determining [H3O+] and [OH-]
Use Kw = [H3O+] [OH-] = 1.0 x to find [H3O+]
[H3O+][OH-] = 1.0 x 10-14
[H3O+] = 1.0 x 10-14
[OH-]
[H3O+] = 1.0 x = 3.1 X mol/L
0.32 mol/L
Therefore the [H3O+] = 3.1 x mol/L, and the
[OH-] = 0.32 mol/L

8. Practice Finding [OH-] and [H3O+] for strong acids and bases
Practice problems on Pg. 537 # 4, 5
Pg. 540 # 10

9. Calculating pH and pOH This should be review from SCH3U
pH = -log[H3O-]
pOH = -log[OH-]
Kw = [H3O+] [OH-] = 1.0 x at 25°C
Therefore, pH + pOH = 14

10. Problems involving pH and pOH
A liquid shampoo has a hydroxide ion concentration of 6.8 x 10-5 mol/L at 25°C
Is the shampoo acidic, basic or neutral
[OH-] = 6.8 x 10-5 mol/L > 1.0 x 10-7 mol/L
Therefore, the shampoo is basic
Calculate the hydronium ion concentration.
[H3O+] = 1.0 x = 1.5 x mol/L
6.8 x 10-5

11. Problems involving pH and pOH
A liquid shampoo has a hydroxide ion concentration of 6.8 x 10-5 mol/L at 25°C
What is the pH and the pOH of the shampoo?
pH = -log[H3O+] = -log[1.5 x 10-10] = 9.83
pOH = -log[OH-] = -log[6.8 x 10-5] = 4.17
Note: With pH and pOH values. The numbers to the left of the decimal do not count as sig. digits.

12. Alternative Method for finding [H3O+] and [OH-]
[H3O+] = 10-pH
[OH-] = 10-pOH
Ex. If the pH is 5.20, what is the [H3O+] = 10-pH
[H3O+] = = 6.3 x 10-6 mol/L
Practice Problems
Pg. 546 # 12, 13
Pg. 549 # 17, 18

13. Acid Dissociation Constant
Weak acids do not completely dissociate in water.
For a weak monoprotic acid
HA(aq) H2O(aq) ↔ H3O+(aq) + A-(aq)
The equilibrium expression is
Kc = [H3O+][A-]
[HA][H2O]

14. Acid Dissociation Constant
In dilute solutions the [H2O] is almost constant
The expression can be rearranged so both constants are on the same side.
The rearrangement gives Ka, the acid dissociation constant
[H2O]Kc = Ka = [H3O+][A-]
[HA]

15. Acid Dissociation Constant
If you know [acid] and pH, you can find Ka
A table of Ka values is located on Pg. 803 in your text.

16. Calculations with the Acid Dissociation Constant
The smaller Ka is, the less the acid ionizes in aqueous solution
Solving Equilibrium Problems Involving Acids and Bases
Write the balanced chemical equation
Use the equation to write an ICE table
Let x represent the change in concentration of the substance with the smallest coefficient

17. Solving Equilibrium Problems Involving Acids and Bases
If problem gives [inital] of the acid, compare [HA] with Ka
If [HA]/Ka > 100, the change in the [initial], x, is negligible and can be ignored.
If [HA]/Ka < 100 the change in [initial] may not be negligible. The equilibrium equation will be more complex and may require the solution of a quadratic equation.

18. Percent Dissociation
% Ionization = [molecules that ionize]x100% [Initial] Acid

19. Example Problem
Propanoic Acid (CH3CH2COOH) is a weak monoprotic acid. A 0.10 mol/L solution has a pH of What is Ka? What is the percent dissociation?
Given: Initial [CH3CH2COOH] = 0.10 mol/L
pH = 2.96
Write the balanced equation.
CH3CH2COOH(aq) + H2O(l) ↔ CH3CH2COO-(aq) + H3O+ (aq)

20. Write the equation for Ka Ka = [CH3CH2COO-] [H3O+] = (x) (x)__ = x2___
Prepare an ICE table
Write the equation for Ka
Ka = [CH3CH2COO-] [H3O+] = (x) (x)__ = x2___
[CH3CH2COOH] (0.10 – x) (0.10 – x)
x = [H3O+] at equilibrium = = 1.1x10-3mol/L
Concentration (mol/L)
CH3CH2COOH(aq)
H2O(l)
CH3CH2COO-(aq)
H3O+ (aq)
Initial
0.10 ~0
Change -x +x
Equilibrium x

21. Substitute value for x into the Ka expression.
Ka = _ x2___ = (1.1 x 10-3) = 1.2 x 10-5
(0.10 – x) (0.10 – 1.1 x 10-3)
Percent ionization = 1.1 x 10-3 mol/L x 100
0.10 mol/L
= 1.1%
Therefore, Ka for propanoic acid is 1.2 x 10-5 and the percent ionization is 1.1%

22. Practice Problems
Pg. 556 # 3, 5
Pg. 568 # 7, 8

23. Polyprotic Acids To calculate Ka, To calculate [H3O+] and pH
Divide the problem into stages.
Equilibrium [acid] for the first H+ = initial [acid] for the second H+
To calculate [H3O+] and pH
With the exception of sulfuric acid, all polyprotic acids are weak. The second dissociation is even weaker than the first
Therefore, only the [first dissociation] is used to find [H3O+] and pH

24. Practice Problems
Pg. 578 # 14
Section Review:
Pg. 579 # 3, 4, 6, 13