Transcript The Ion Product Constant for Water (Kw)

8.2 Strong and Weak Acids
and Bases
Learning Goals …
…determine [H30+] and [OH-] using Kw, pH or pOH
…determine Ka or Kb using Kw
…find pH and pOH of a substance
Water as an Acid and a Base
•Pure water dissociates according to the following reaction:
2 H2O(l)

H3O+(aq) + OH-(aq)
• There is an equal amount of H+ and OH- ions in solution
(neutral, pH = 7)
• at 25°C [H3O+] = [OH-] = 1x10-7 mol/L
• equilibrium constant for the dissociation of water: Kw
Kw = [H3O+][OH-]
= (1x10-7)(1x10-7)
= 1x10-14
* small k, reactants are favoured
(does not go to completion)
@ 25°C
acids: [H3O+] > [OH-] [H3O+] > 1x10-7
[OH-] < 1x10-7
bases: [OH-] > [H3O+] [H3O+] < 1x10-7
[OH-] > 1x10-7
We can use Kw to calculate [H3O+] and [OH-] in
solutions
[H3O+][OH-]= 1x10-14
The Relationship Between Kw, Ka and Kb
HA (aq) + H2O (l)  H3O+ (aq) + A- (aq)
Ka = [H3O+][A-]
[HA]
A- (aq) + H2O (l)  HA (aq) + OH- (aq)
Kb = [HA][OH-]
[A-]
[H3O+][A-]  [HA][OH-]
[HA]
[A-]
= [H3O+][OH-]
= Kw
Ka x Kb =
Kw = Ka x Kb
• A strong acid or base (large Ka or Kb) will have a very weak
conjugate (very small Kb or Ka)
• A weak acid or base (small Ka or Kb) has a weak conjugate
(small Ka or Kb)
Ex) What is the Kb value for F- if the Ka value of HF is 6.6x10-4?
HF(aq) + H2O(l) 
H3O+(aq) + F-(aq)
acid
conjugate base
Kw = Ka Kb
Kb = Kw / Ka
= (1.0x10-14)/(6.6x10-4)
= 1.5x10-11
POLYPROTIC ACIDS
• acids that have more than one proton
• only one proton is transferred at a time
Ex) H2CO3 + H2O
HCO3- + H2O
 H3O+ + HCO3 H3O+ + CO32-
Ka1 = 4.4x10-7
Ka2 = 4.7x10-11
HCO3- is amphoteric. It can act as an acid or a base.
HCO3- + H2O
 H2CO3 + OH-
Kb = ?
Kb = Kw / Ka1
= (1.0x10-14)/(4.4x10-7)
= 2.3x10-8
Since Kb > Ka, HCO3- is more likely to act as a base
pH and pOH
pH = -log [H3O+]
pOH = -log [OH-]
In neutral water,
pH = -log [H3O+] = -(log(1 x 10-7) = 7
pOH = -log [OH-] = -(log(1 x 10-7) = 7
Note: pH + pOH = 14, always, regardless of solution!
Another way to calculate [H3O+] & [OH-] in solution:
[H3O+] = 10-pH
[OH-] = 10-pOH
Ex)
A liquid shampoo has a [OH-] of 6.8x10-5 mol/L
(a) Is the shampoo acid, basic or neutral?
(b) What is [H3O+]?
(c) What is the pH and pOH of the shampoo?
(a) [OH-] = 6.8x10-5
> 1.0x10-7, basic
(b) [H3O+] = Kw / [OH-] = (1.0x10-14)/(6.8x10-5)
= 1.5x10-10 mol/L
(c)
pOH = -log [OH-]
= -log [6.8x10-5]
= 4.17
pH = 14 – 4.17
= 9.83
Self Check
How prepared am I to start my homework? Can I …
…determine [H+] and [OH-] using Kw, pH or pOH
…determine Ka or Kb using Kw
…find pH and pOH of a substance
HOMEWORK
p509 #1, 2, 3ab, 4ab, 5ab, 6ab, 7-9