
Acids are substances that:
◦
◦
◦
◦
◦
◦
taste sour.
react w/ bases to form salts and water.
are electrolytes.
turn blue litmus red.
produce H3O+1 ions in aqueous solution.
donate H+1 ions (protons).

Bases are substances that:
◦
◦
◦
◦
◦
◦
taste bitter.
react w/ acids to form salts and water.
are electrolytes.
turn red litmus blue.
produce OH-1 ions in aqueous solutions.
accept H+1 ions (protons).

H2O + H2O  H3O+1 + OH-1

In pure water, [H3O+1] = 1.0 x 10-7 Molar.
◦ Extremely small Keq.
◦ Keq = Kw = 1.0 x 10-14
◦ Only 0.00001% of H2O molecules autoionize.
◦ [OH-1] also equals 1.0 x 10-7 M.
 Less than 2 H3O+1 and OH-1 ions per billion water
molecules.
◦ Since [H3O+1] = [OH-1], water is pH neutral.

Adding an acid or a base upsets this
balance.
O
H
H
-
O
H
H
+
Adding an Acid to Water
O
O
H2O
H2O
-1+1
HH3AO
2O
H2O
H2O
HA
HA
HA
HA
HA
H2O
H2O
H2O
H2O
H2O
H3O+1
H2O
H2O
H2O
-1+1
HH3AO
2O
H2O
H2O
-1+1
HH3AO
2O
-1+1
HH3AO
2O
H2O
H2O
H2O
H2O
H2O
H2O
OH
HA2-1O-1
H2O
H2O
H2O
H2O
Adding a Base to Water
O
O
H2O
H2O
-1
OH
HHA
2O
H2O
H2O
A-1
A-1
A-1
A-1
A-1
H2O
H2O
H2O
H2O
H2O
OH-1
H2O
H2O
H2O
-1
OH
HHA
2O
H2O
H2O
-1
OH
HHA
2O
-1
OH
HHA
2O
H2O
H2O
H2O
H2O
H2O
H2O
+1
HHHA
3O
2O
H2O
H2O
H2O
H2O
1x10-1 M
1x10-1 M
1x10-2 M
1x10-2 M
1x10-3 M
1x10-3 M
1x10-4 M
1x10-4 M
1x10-5 M
1x10-5 M
1x10-6 M
1x10-6 M
1x10-7 M
1x10-7 M
1x10-8 M
1x10-8 M
1x10-9 M
1x10-9 M
1x10-10 M
1x10-10 M
1x10-11 M
1x10-11 M
1x10-12 M
1x10-12 M
1x10-13 M
1x10-13 M
Neutral
Solution
[OH-1]
[H3O+1]
As [H3O+1] Increases, [OH-1] Decreases
Acid added
to neutral
solution
Base added
to neutral
solution

In any aqueous solution:
◦ [H3O+1] [OH-1] = 1x10-14
◦ As [H3O+1] goes up, [OH-1] must decrease.
◦ As [OH-1] goes up, [H3O+1] must decrease.
 In other words, adding an acid to water causes the solution
to become more acidic and less basic.
 Adding a base to water causes the solution to become less
acidic and more basic.

If [H3O+1] = 1x10-3 M, what is [OH-1]?
◦
◦
◦
◦

[H3O+1][OH-1] = 1x10-14
(1x10-3 M)[OH-1] = 1x10-14
[OH-1] = (1x10-14) / (1x10-3)
[OH-1] = 1x10-11 M
If [OH-1] = 1x10-8 M, what is [H3O+1]?
◦
◦
◦
◦
[H3O+1][OH-1] = 1x10-14
[H3O+1](1x10-8 M) = 1x10-14
[H3O+1] = (1x10-14) / (1x10-8 M)
[H3O+1] = 1x10-6 M

Acidity
◦ How much H3O+1 is dissolved in a sol’n.
 Remember, acids increase [H3O+1] in solutions.
◦ Acidity = pH.

pH = power of Hydrogen
◦ negative logarithmic (powers of ten) scale.

pH = -log10[H3O+1]
◦ If [H3O+1] = 1x10-1 M,
 pH = -log(1x10-1 M) = 1
◦ If [H3O+1] = 1x10-2 M,
 pH = -log(1x10-2 M) = 2
◦ If [H3O+1] = 1x10-3 M,
 pH = -log(1x10-3 M) = 3

The logarithm of a number is the power to
which you would have to raise a base to equal
that number.
◦ Unless otherwise indicated, assume the base is 10.

log(100) = 2

log(1000) = 3

log(0.001) = -3

log(0.000 001) = -6
◦ because 102 = 100
◦ because 103 = 1000
◦ because 10-3 = 0.001
◦ because 10-6 = 0.000 001

The [H3O+1] and [OH-1] of an aqueous solution can
vary by a very large degree.
◦ [H3O+1] = 1 M for a very acidic soln
◦ [H3O+1] = 1x10-7 M for a neutral soln
◦ [H3O+1] = 1x10-14 M for a very basic soln

1 M is ten million times greater than 1x10-7 M.
◦ If you tried to plot both concentrations on the same graph,
1x10-7 M would barely register above zero.
◦ If 1x10-7 M was 1 mm above the 0 mark, the axis would
have to be ten kilometers (six miles) tall to show 1 M.

Logarithms allow us to compare numbers that are
widely different by thinking of them as powers of
ten.
pH
pH vs [H3O+1]
This graph shows pH as a function of hydrogen
ion concentration. It isn’t a very useful graph
because it is hard to get accurate information for
[H3O+1] below 1x10-3 M.
1.00E-02
9.00E-03
8.00E-03
7.00E-03
[H3O+1]
6.00E-03
5.00E-03
4.00E-03
3.00E-03
2.00E-03
1.00E-03
0.00E+00
16
14
12
10
8
6
4
2
0
pH
pH vs [H3O+1]
In this graph the x-axis is
logarithmic. It allows a much
greater range of data to be
displayed in a readable format.
1.00E-02
1.00E-03
1.00E-04
1.00E-05
[H3O+1]
1.00E-06
1.00E-07
1.00E-08
1.00E-09
1.00E-10
1.00E-11
1.00E-12
1.00E-13
1.00E-14
16
14
12
10
8
6
4
2
0


Each unit decrease in pH is a 10-fold increase
in acidity.
Imagine a soln with a pH of 5.
◦ A soln with a pH of:





4
3
2
1
0
is
is
is
is
is
10 times more acidic.
100 times more acidic.
1000 times more acidic.
10,000 times more acidic.
100,000 times more acidic.
Where does the pH scale come
from?
pH scale
-1…0
1
2
3
4
5
6
7
8
9
10
11
12
Acidic
13
14…15
Basic
[H3O+]
1 0.1 0.01 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-12
A lot of H3O+
Acidic
10-13 10-14
Not a lot of H3O+
Basic
pOH Scale
The pOH scale indicates the hydroxide ion
concentration, [OH-] or molarity, of a solution. (In
other words how many OH- ions are in the solution.
If there are a lot we assume it is a base, if there are
very few it is an acid.)
Two chemists meet for the first time at a
symposium. One is American, one is British. The
British chemist asks the American chemist, "So
what do you do for research?" The American
responds, "Oh, I work with aerosols." The
British chemist responds, "Yes, sometimes my
colleagues get on my nerves also."
pOH scale
-1…0
1
2
3
4
5
6
7
8
9
10
11
12
Basic
13
14…15
Acidic
[OH-]
1 0.1 0.01 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-12
A lot of OHBasic
10-13 10-14
Not a lot of OHAcidic
Example:
1. Lemon juice (citric acid) pH = 2, pOH = 12
2. Pure water pH = 7, pOH = 7
3. Milk of magnesia pH = 10, pOH = 4
The last words of a chemist:
1. And now for the taste test.
2. I wonder if this is hot?
3. And now a little bit from this...
4. And now shake it a bit.
Concentrations of Hydronium and Hydroxide Ions
1.00E+00
1.00E-02
Molar
1.00E-04
1.00E-06
[H3O+1]
1.00E-08
[OH-1]
1.00E-10
1.00E-12
1.00E-14
1
2
3
4
5
6
7
pH
8
9
10
11
12
13
4. Calculations Involving pH, pOH, [H3O+],
and [OH-] of strong Acids and Bases
1st: determine which ion will be produced,
either OH or H3O+ (Acids produce H3O+
and bases produce OH-). 2nd: use formula to
determine pH or pOH. 3rd: check to see if
answer is reasonable.
pH = -log [H3O+]
pOH = -log [OH-]
pOH + pH = 14

What are the pH values of the following
solutions?
◦ 1x10-1 M H3O+1
 pH = -log(1x10-1 M) = 1
◦ 1x10-3 M H3O+1
 pH = -log(1x10-3 M) = 3
◦ 1x10-5 M H3O+1
 pH = -log(1x10-5 M) = 5
◦ 1x10-1 M OH-1
 [H3O+1] = (1x10-14) / (1x10-1 M) = 1x10-13 M
 pH = -log(1x10-13 M) = 13




Given pH, you can calculate [H3O+1] and [OH-1].
[H3O+1] = 10-pH
[H3O+1] [OH-1] = 1x10-14
If pH = 2,
◦ [H3O+1] = 1x10-2 M
◦ [OH-1] = 1x10-12 M
Titrations
• Determining the pH of an unknown
solution using the pH of a known solution
• Titrations take a very long time and you
have to have excellent lab technique
• You add small amounts of known solution
until a pre-determined endpoint is reached
3/21/2016
31
#H+a Ma Va = #OH-b Mb Vb
•
•
•
•
#H+ in acid formula
M= Molarity
V= Volume used to neutralize
#OH- in base formula
Example
You have 50 drops CH3COOH & it takes 5
drops 5M NaOH reach the endpoint.
What is the molarity of the acetic acid?
• (1 H+)(Ma)(50 drops)=(1 OH-)(5 M)(5 drops)
• Ma = 0.5 M
Example 2
• What is the molarity of sulfuric acid if it
takes 12 mL of H2SO4 to neutralize 30 mL of
5 M NaOH.
• (2 H+)(Ma)(12 drops)=(1 OH-)(5 M)(30
drops)
• 6M