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

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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