1. ACIDS AND BASES:
Ionization of Water

2. OUTCOME QUESTION(S):
C ACID/BASE EQUILIBRIUM
Describe the relationship between the hydronium and hydroxide ion concentrations in water
Include: the ion product of water, Kw
Write the equilibrium expression (Ka or Kb) and solve problems involving pH, pOH, and percent dissociation.
Vocabulary & Concepts 

3. Remember: water is amphoteric
When water particles collide - ions form
H2O(l) + H2O(l) H3O+(aq) + OH¯(aq)
Remember: water is amphoteric
Water also dissociates into ions - self-ionization
H2O(l) H+(aq) + OH¯(aq)
Equilibrium is reached between water and its ions:
KW =
[H+][OH¯]

4. KW = [H+][OH¯] = 1.00 x 10-14 Ion product constant for , Kw
The Kw is very small as equilibrium favours the reverse reaction – ions quickly reform
Since ions are at a 1:1 ratio, and water is neutral:
[H+]e = [OH-]e = 1.00 x 10-7 M
Like all equilibrium constants, the value of Kw
varies with temperature

5. Like any reversible reaction, Le Chatelier's applies:
Acids and bases are typically solutions made in water
Acids and bases contribute H+ or OH- to the solution
Acids and bases can be considered a stress to water’s equilibrium
H2O H+ +
OH-
KW =
[H+]
[OH-]
= 1.00 x 10-14
Adding base: ↑[OH-] – shift left – Kw is constant
Adding acid: ↑[H+] – shift left – Kw is constant
This means that H+ and OH- are BOTH present in any solution – whether “acidic” or “basic”

6. H2O H+ + OH- Kw = [H+] [OH-] 1.0 x 10-14 = [OH-] [2.0 ]
How does the addition of 2.0 M of H+ ions affect water’s equilibrium and the [OH-]?
H2O H+ +
OH-
Equation 6:
Of course [OH-] is small ~ H+ stress caused shift left
Kw =
[H+]
[OH-]
Tiny amount contributed by self-ionization of water (can be ignored)
1.0 x =
[OH-]
[2.0 ]
+ 1.0 x 10-7]
[OH-] = 5.0 x M

7. pH/pOH Formulae Equation 5: Equation 6: pH [H+] pH = - log [H+]
[H+][OH-] = 1.0x10-14
Equation 6:
Equation 3:
pOH = - log [OH-]
pOH
[OH-]
Equation 4:
[OH-] = 10-pOH

8. n V M = HCl (g) H+(aq) + Cl¯(aq) [HCl]i = [H+]e = 0.50 M 2.5 mol =
If 2.5 moles of hydrochloric acid is dissolved in 5.0 L of water, what is the [hydroxide ions]?
HCl (g) H+(aq) + Cl¯(aq)
Strong acid…
1. What is the acid/base contributing? n V M =
5.0 L =
0.50 mol/L
2.5 mol
[HCl]i = [H+]e = 0.50 M

9. H2O H+ + OH- Kw = [H+] [OH-] 1.0 x 10-14 = [OH-] [0.50 ]
2. How does the contribution affect water’s equilibrium?
H2O H+ +
OH-
[H+]e = 0.50M
Equation 6:
Kw =
[H+]
[OH-]
1.0 x =
[OH-]
[0.50 ]
+ 1.0 x 10-7]
[OH-] = 2.0 x M

10. = NaOH (s) Na+(aq) + OH¯(aq) [NaOH]i = [OH-]e = 0.010 M 0.40 g 0.010 M
0.40 g of NaOH is dissolved in water to a volume of 1.0 L. What is the hydronium ion concentration?
NaOH (s) Na+(aq) + OH¯(aq)
1. What is the acid/base contributing?
Strong base…
NaOH = 40.0 g/mol
0.40 g
1 mol =
0.010 M
1 L
40.0 g
[NaOH]i = [OH-]e = M

11. Of course [H+] is small ~ OH- stress caused shift left
2. How does the contribution affect water’s equilibrium?
H2O H+ +
OH-
[OH-]e = 0.010M
Equation 6:
Kw =
[H+]
[OH-]
Of course [H+] is small ~ OH- stress caused shift left
1.0 x =
[H+]
[0.010 ]
+ 1.0 x 10-7]
[H+] = 1.0 x M

12. Find the [OH-] if the pH of a solution is 7.50.
Questions can be solved multiple ways… pH
pOH
[OH-] pH
[OH-]
[H+]
pOH = pH
[H+] = 10-pH
pOH = – 7.50
= 6.50
[H+] =
= 3.2 x 10-8
[OH-] = 10-pOH
[OH-] = 1.0x10-14
[H+] =
= 3.16 x 10-7 M
[OH-] = 1.0 x 10-14
3.2 x 10-8
= 3.16 x 10-7 M
[OH-] and [H+] should be close ~ the solution is nearly neutral

13. CAN YOU / HAVE YOU?
C ACID/BASE EQUILIBRIUM
Describe the relationship between the hydronium and hydroxide ion concentrations in water
Include: the ion product of water, Kw
Write the equilibrium expression (Ka or Kb) and solve problems involving pH, pOH, and percent dissociation.
Vocabulary & Concepts