Kw of Water
Chapter 16 part III
Kw of Water
•From the auto-ionization
expression of water we
get the equilibrium
constant expression.
•2 H2O  H3O+ + OH•Kw= [H3O+ ][OH-]
•Why not H2O?
•Kw= [H+ ][OH-]
Kw is the ion-product
 Kw
is aka dissociation constant of water
 It has been found experimentally that at 25
°C,
 [H+ ]=[OH-] and they both equal 1X10-7M
 Since Kw= [H+ ][OH-] = [1X10-7M]2
 Then Kw = 1X 10-14.
What does this mean?
 This
means that any aqueous solution at
25 ° C, no matter what it (the water)
contains, the product of [H+] and [OH-]
MUST always equal 1.0 X 10-14.
 There are 3 possibilities:
 A neutral solution where [H+]=[OH-].
 An acid solution where [H+]>[OH-].
 A basic solution where [H+]<[OH-].
In all 3 of the situations:
= [H+][OH-] = 1.0 X 10-14.
 So in any given aqueous situation, one
may calculate the [H+] or [OH-] as
required for any solution at 25°C.
 State if Acidic, Basic or Neutral.
 A. 1.0X10-5 M OH- 1.0 X10-9 M H+
 B. 1.0X10-7 M OH- 1.0 X10-7 M H+
 C. 1.0X10-15 M OH- 10.0 M H+
 Kw
Answers
 A.
Basic
 B. Neutral
 C. Acid
How to Solve for the [ions]
 Kw
= [H+][OH-]
 1.0 X 10-14= [H+][OH-]
 [H+] = [OH-]/(1.0 X 10-14)
Kw varies with Temperature
if the Kw at 60° C is 1.0 X 10-13
 Then, using Le Châtlier’s principle predict
whether the ionization of water is endo or
exothermic.
• 2 H2O  H3O+ + OH• Kw increases
• from 1.0 X 10-14 to 1.0 X 10-13
• Therefore the system adjusted to form
more product in the presence of heat.
• This indicates the reaction is endothermic
 So
Example
 Calculate
the [OH-] and [H+] at 60° C in a
neutral solution.
 Neutral means what about [H+] to [OH-]?
 Answer: [H+] = [OH-]
 So: Kw= [H+] x [OH-] = 1.0 X 10-13
 [H+] = [OH-] = (1.0 X 10-13 )½
 =3.0 X 10-7 M
pH scale is an easy way to
represent acidity.
 pH
= -log[H+]
 At a neutral solution at 25 °C
 [H+] = [OH-] = (1.0 X 10-14)½ = 1.0 X 10-7
 What is the pH of this?
 pH = -log[H+]
 = -log(1.0 X 10-7)
 Take out you calculator and what do you
get?
 7.00
Sig Figs in Log problems
 The
number of sig figs in an original
number equals the number of decimal
places in the pH.
 Example:
 If sample is Kw= [H+] = 1.0 X 10-7
 How many sig fig?
2
 This pH is …
7.00 2 decimal places for the two sig figs.
pH vs. pOH
 If
pH is = -log [H+]
 Then pOH = -log [OH-]
 And pK = -log K
 Note that pH changes by 1 for every
power of 10 in the change of
concentration.
Examples
 Calculate
the pH and pOH of each
 1.0 X 10-3 M OH pOH = 3.00 pH = 11.00
 H+ = Kw/[OH-] = (1.0 X 10-14)/(1.0 X 10-3)
 = 1.0 x 10-11
 1.0 M H+
 pH = 0.00
[OH-]=Kw/[H+]= 1.0x10-14/1
 pOH = 14
Remember
 Kw
=[H+][OH-]
 And therefore,
 -log K = -log [H+] + -log [OH-]
 log K = log[H+] + log[OH-]
 pKw = pH = pOH
 At 25 °C pKw = 14.00 (1.0 X 10-14)
 Thus pH = pOH = 14 at 25 °C
Summary of General Strategies
 Think
Chemistry: focus on solution and
the components. It is usually easy to
identify one reaction that is not important.
 Be systematic: Acid-base problems
require step by step approach.
 Be Flexible: Although all acid-base
problems are similar, important differences
do occur. Do not force a given problem
into matching a problem you have solved
before.
Summary of General Strategies
 Be
patient: the complete solution to a
complicated problem cannot be seen
immediately in all its detail. Pick the
problem apart into workable steps.
 Understand & Think: don’t just memorize.