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
◦ Extremely small Keq.◦ Keq = Kw = 1.0 x 10-14
◦ Only 0.00001% of H2O molecules autoionize. In pure water, [H3O+1] = 1.0 x 10-7 Molar.
◦ [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.
HO
H
HO
H
- +
Adding an Acid to Water
H2OH2OH2O H2O H2OH2OH2OH2O H2OH2OH2O
H2OH2OH3O+1 H2O H2OH2OH2OH2O H2OH2OH2O
H2OH2OH2O H2O H2OOH-1H2OH2O H2OH2OH2O
H2OH2OH2O H2O H2OH2OH2OH2O H2OH2OH2O
HA HA HA HA HA
A-1H3O+1
H3O+1A-1
H3O+1A-1
H2O
H3O+1A-1
H2OA-1
H3O+1
H2O
H2O
H2O
H2OH2O
H2O
OH-1
OH-1
OH-1
Adding a Base to Water
H2O H2O H2OH2OH2OH2O H2OH2OH2O
H2OOH-1 H2O H2OH2OH2O H2OH2OH2O
H2OH2OH2O H2OH2OH2O H2OH2OH2O
H2OH2OH2O H2O H2OH2OH2OH2O H2OH2OH2O
A-1 A-1 A-1 A-1
H2O
A-1
OH-1HA
HA
HA
HA
HA
As [H3O+1] Increases, [OH-1] Decreases
1x10-13 M
1x10-12 M
1x10-11 M
1x10-10 M
1x10-9 M
1x10-8 M
1x10-7 M
1x10-6 M
1x10-5 M
1x10-4 M
1x10-3 M
1x10-2 M
1x10-1 M
1x10-13 M
1x10-12 M
1x10-11 M
1x10-10 M
1x10-9 M
1x10-8 M
1x10-7 M
1x10-6 M
1x10-5 M
1x10-4 M
1x10-3 M
1x10-2 M
1x10-1 M
[H3O
+1]
[OH
-1]
Neutral Solution
Acid added to neutral solutionBase 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◦ because 102 = 100
log(1000) = 3◦ because 103 = 1000
log(0.001) = -3◦ because 10-3 = 0.001
log(0.000 001) = -6◦ 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 vs [H3O+1]
0
2
4
68
10
12
14
16
0.00E+00
1.00E-03
2.00E-03
3.00E-03
4.00E-03
5.00E-03
6.00E-03
7.00E-03
8.00E-03
9.00E-03
1.00E-02
[H3O+1]
pH
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.
pH vs [H3O+1]
0
2
4
68
10
12
14
16
1.00E-14
1.00E-13
1.00E-12
1.00E-11
1.00E-10
1.00E-09
1.00E-08
1.00E-07
1.00E-06
1.00E-05
1.00E-04
1.00E-03
1.00E-02
[H3O+1]
pH
In this graph the x-axis is logarithmic. It allows a much greater range of data to be displayed in a readable format.
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 is 10 times more acidic. 3 is 100 times more acidic. 2 is 1000 times more acidic. 1 is 10,000 times more acidic. 0 is 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 13 14…15
Acidic 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 10-13 10-14
A lot of H3O+ Not a lot of H3O
+
Acidic Basic
pOH ScaleThe 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 13 14…15
Basic 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 10-13 10-14
A lot of OH- Not a lot of OH-
Basic Acidic
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-14
1.00E-12
1.00E-10
1.00E-08
1.00E-06
1.00E-04
1.00E-02
1.00E+00
1 2 3 4 5 6 7 8 9 10 11 12 13
pH
Mo
lar [H3O+1]
[OH-1]
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
04/11/23 31
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
#H+a Ma Va = #OH-
b Mb Vb
• #H+ in acid formula
• M= Molarity
• V= Volume used to neutralize
• #OH- in base formula
ExampleYou 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)
• 6 M