Chem 1011 – Intersession 2011
Class #26
06-Jun-11
1
Class 26: Autoionization of H2O / Find
pH and [H3O+]
•
Sec 15.5 – Autoionization of Water
▫
▫
•
The pH Scale: A Way to Quantify Acidity or
Basicity
pOH and other p Scales
Sec 15.6 – Finding the [H3O+] and pH of
Strong and Weak Acid Solutions
▫
▫
▫
Strong Acids
Weak Acids
Percent Ionization of a Weak Acid
2
Autoionization of H2O
•
Self-ionization
▫
•
an acid-base reaction in which one molecule acts as
an acid and donates a proton to a molecule of the
same kind acting as a base
Water undergoes self-ionization to produce a
hydronium ion and a hydroxide ion.
H2O(l) + H2O(l) ⇌ H3O+(aq) + OH–(aq)
▫
Note that this equilibrium lies strongly to the left
(that is, much more H2O is present than H3O+ and
OH– at any time)
3
Autoionization of H2O
Autoionization of H2O / Find pH and [H3O+]
1
Chem 1011 – Intersession 2011
Class #26
06-Jun-11
4
Autoionization of H2O
•
We can now write the equilibrium expression for the
self-ionization of water, Kw, (also known as the ion
constant product of water).
▫
Recall that for equilibrium expressions, species in the
liquid or solid phase are omitted.
Kw = [H3O+][OH–] = 1.0 x 10-14
•
(@ 25oC)
This relationship is useful because it tells us that in
every aqueous solution
▫
▫
▫
both H3O+ and OH– are present
in pure water [H3O+] = [OH–] = 1.0 x 10-7 (neutral)
the product of the ion concentrations is constant
5
Autoionization of H2O
•
As a result, we can classify an aqueous
solution as acidic, basic or neutral depending
on the relative concentrations of the ions.
•
That is:
▫
▫
▫
If [H3O+] > [OH–], the solution is acidic
If [H3O+] < [OH–], the solution is basic
If [H3O+] = [OH–], the solution is neutral
6
Autoionization of H2O
•
We can also express these relationships in terms
of molar concentrations of [H3O+] and [OH-]:
•
In a neutral aqueous solution at 25oC:
▫
•
[H3O+] = [OH–] = 1.0 x 10-7 mol/L.
In an acidic aqueous solution:
▫
•
[H3O+] > 1.0 x 10-7 mol/L and [OH–] < 1.0 x 10-7 mol/L.
In a basic aqueous solution:
▫
[H3O+] < 1.0 x 10-7 mol/L and [OH–] > 1.0 x 10-7 mol/L.
Autoionization of H2O / Find pH and [H3O+]
2
Chem 1011 – Intersession 2011
Class #26
06-Jun-11
7
Problem
•
Calculate [H3O+] at 25oC for the following
solutions and determine if they are neutral,
acidic or basic:
(a) [OH–] = 1.5 x 10–2 M
(b) [OH–] = 1.5 x 10–10 M
8
The pH Scale – A Way to Quantify
Acidity
•
▫
pH – ―potential of hydrogen ion‖
this is the shorthand notation to represent the [H3O+] in
solution.
pH = –log[H3O+]
•
The pH scale ranges in values from 0 to 14.
•
The [OH–] can be expressed similarly as:
pOH = –log[OH–]
•
Another useful relationship is:
pH + pOH = 14.00
ONLY @ 25oC
9
Working Backwards
• If you start with the pH (or pOH) and want to find
[H3O+] (or [OH–]):


pH  log H3O
 pH  log H3O



10pH  10logH3O 
10 pH  H3O

• Similarly; 10

 pOH

 OH

Autoionization of H2O / Find pH and [H3O+]
3
Chem 1011 – Intersession 2011
Class #26
06-Jun-11
10
Sig Figs and Logs
• When using pH, pay close attention to the sig fig
rules for logs:
11
The pH Scale
•
A solution is acidic if pH < 7
▫
•
pH < 7 indicates [H3O+] > 10–7 M
A solution is neutral if pH = 7
▫
•
pH = 7 indicates [H3O+] = 10–7 M
A solution is basic if pH >7
▫
pH > 7 indicates [H3O+] < 10–7 M
12
Problems
•
What is the pH of a solution that is 0.0025 mol/L
HCl?
•
Students found a sample of yogurt to have a pH of
2.85. What are [H3O+] and [OH–] in the sample?
Autoionization of H2O / Find pH and [H3O+]
4
Chem 1011 – Intersession 2011
Class #26
06-Jun-11
13
Another “p” Scale
•
Another common p-scale is pKa (or pKb)
pKa = –log(Ka)
•
pKb = –log(Kb)
The pKa of a weak acid also allows us to quantify its
strength.
▫
•
The smaller the pKa, the stronger the acid.
The pKb of a weak base also allows us to quantify its
strength.
▫
The smaller the pKb, the stronger the base.
14
Finding [H3O+] and pH in Weak Acids
•
There are two sources of H3O+ in an aqueous solution
of a strong acid — the acid and the water.
•
There are two sources of OH− in an aqueous solution of
a strong base — the base and the water.
•
For a strong acid or base, the contribution of the water
to the total [H3O+] or [OH−] is negligible.
•
There are also two sources of H3O+ in an aqueous
solution of a weak acid — the acid and the water.
•
However, finding [H3O+] is complicated by the fact
that the acid undergoes only partial ionization.
15
Finding [H3O+] and pH in Weak Acids
• Calculating [H3O+] for a weak acid requires solving an
equilibrium problem for the reaction that defines the
acidity of the acid → ICE Table!
▫ Write the reaction for the acid with water
 HA(aq) + H2O(l) ⇌ A–(aq) + H3O+(aq)
▫ Build an ICE Table for the reaction, and enter the initial
concentrations
 We assume [H3O+] from the autoionization of water is
negligible, ie, [H3O+] ≈ 0
Autoionization of H2O / Find pH and [H3O+]
5
Chem 1011 – Intersession 2011
Class #26
06-Jun-11
16
Example
• Find the pH of 0.200 M HNO2(aq) solution at 25°C
▫ Ka of HNO2 = 4.6 x 10–4
• Write the reaction for the acid in water:
▫ HNO2(aq) + H2O(l) ⇌ NO2(aq) + H3O+(aq)
• Make an ICE Table with all initial concentrations:
▫ Assume [H3O+] ≈ 0
HNO2(aq)
[I]
+
H2O(l)
⇌
0.200 M
NO2(aq)
+
H3O+(aq)
0
≈0
[C]
[E]
17
Example
• Represent changes in terms of x
HNO2(aq)
+
H2O(l)
⇌
NO2(aq)
+
H3O+(aq)
[I]
0.200 M
0
≈0
[C]
–x
+x
+x
[E]
0.200 M – x
x
x
• Substitute into the equilibrium constant expression and
solve for x
NO H O 

Ka 
2

3
HNO2 
Substitute
4.6 10- 4 
x x 
0.200  x 
18
Example
x x 
0.200  x 
4.6 10-4 0.200  x  x2
4.6 10- 4 
x 2  4.6 104 x  9.2 105  0
• Quadratic equation...
 b  b 2  4ac
2a
2
 4.6 10 4  4.6 10 4   41 9.2 10 5 
x
21
x
Autoionization of H2O / Find pH and [H3O+]
6
Chem 1011 – Intersession 2011
Class #26
06-Jun-11
19
Example
 4.6 104  2.12 107  3.68 104
2
 4.6 104  3.682 104
x
2
 4.6 104  0.0192
 4.6 104  0.0192
x
x
2
2
x
x  0.00983 M
x  0.00936 M
20
Example
• In order to find pH, we need to know [H3O+] at
equilibrium
• From the ICE Table, [H3O+] = x = 0.00936 M


pH   logH O 
x  H3O  0.00936 M

3
pH   log0.00936
pH  2.03
21
Problem
•
Find the pH and pOH of a 0.250 M HF
solution.
▫
•
▫
Ka = 3.5 x 10-4
Problem: Find the pH and pOH of a 0.010 M
HNO2 solution.
Ka = 4.6 x 10-4
Autoionization of H2O / Find pH and [H3O+]
7
Chem 1011 – Intersession 2011
Class #26
06-Jun-11
22
Percent Ionization
• Percent ionization
▫ the percentage of acid molecules that ionize when dissolved
in water. It is another way to measure the strength of an
acid
▫ The higher the percent ionization, the stronger the
acid.
Percent Ionization 
molarity of ionized acid
 100%
initial molarity of acid
• since [ionized acid]equil = [H3O+]equil
23
Problem
• Find the percent ionization of a 0.250 M
HC2H3O2 solution. (Ka = 1.8 x 10-5)
24
Don’t Forget!
• NO LAB this afternoon – Labs resume on
Wednesday
• Everyone is to go to the first tutorial session
today at 2:00pm
• Midterm Review Session on Thursday, June 9th
from 12:00pm – 2:00pm (C-3033)
• Midterm #2 on Friday, June 10th at 9:00am in
C-4002
Autoionization of H2O / Find pH and [H3O+]
8
Chem 1011 – Intersession 2011
Class #26
06-Jun-11
25
Wednesday
•
Sec 15.7 – Base Solutions
▫
▫
•
Strong and Weak Bases
Finding the [OH–] and pH of a Basic Solution
Sec 15.8 – The Acid-Base Properties of Ions and Salts
▫
▫
▫
•
Anions as Weak Bases
Cations as Weak Acids
Classifying Salt Solutions
Sec 15.10 – Lewis Acids and Bases
▫
▫
Molecules that Act as Lewis Acids
Cations that Act as Lewis Acids
Autoionization of H2O / Find pH and [H3O+]
9