NAME: ___________________________________

UNIT #11: Acids and Bases

pH and pOH

Neutralization Reactions

Oxidation and Reduction

1. SELF-IONIZATION OF WATER

a) Water molecules collide, causing a very small number to ionize in a reversible reaction:

H2O(l) + H2O(l) ↔ H3O+(aq) + OH

-(aq)

water molecules hydronium ion hydroxide ion

b) The hydronium ion (H3O+) consists of a water molecule attached to a hydrogen ion

(H+) by a covalent bond; thus, H3O

+ and H

+ can be used interchangeably in a chemical

equation to represent a hydrogen ion in aqueous solution.

Simplified equation for self-ionization of water:

H2O(l) ↔ H+(aq) + OH

-(aq)

c) Water is considered neutral since it produces equal numbers of H+ and OH

- ions.

Hydrogen ion concentration, shown as [H+], for water = 1.0 x 10

-7M

Hydroxide ion concentration, shown as [OH-], for water = 1.0 x 10

-7M

Self-ionization produces a tiny number of ions but explains how pure water can behave

as a very weak electrolyte.

d) Water is the usual solvent for acids and bases; the dissociation of acids or bases in an

aqueous solution increases either the [H+] or [OH

-], resulting in an aqueous solution

that is no longer neutral.

2. DEFINITION OF ACIDS AND BASES

a) Arrhenius Model of Acids and Bases

1. Acid: A substance that contains hydrogen and ionizes to produce hydrogen ions in

aqueous solutions.

Ex. HCl(g) → H+(aq) + Cl

-(aq)

2. Base: A substance that contains a hydroxide group and dissociates to produce

hydroxide ions in aqueous solution.

Ex. NaOH(s) → Na+(aq) + OH

-(aq)

3. The Arrhenius model is limited because it does not describe all bases such as NH3

(ammonia) and NaHCO3 (baking soda/sodium bicarbonate). These substances

dissociate to form OH- in an aqueous solution but do not have the hydroxide

group in their formula and therefore, would not meet the definition of an

Arrhenius base.

NH3(aq) + H2O(l) ↔ NH4+ + OH

-

NaHCO3(s) + H2O(l) ↔ H2CO3 (aq) + Na+(aq) + OH

-(aq)

sodium bicarbonate carbonic acid

b) Bronsted-Lowry Model of Acids and Bases

1. Acid: A substance that is a hydrogen ion (H+) donor; also referred to as a

proton donor since H+ consists of 1 proton only.

2. Base: A substance that is a hydrogen ion (H+) or proton acceptor.

3. Using the example cited above:

NH3(aq) + H2O(l) ↔ NH4+ + OH

-

Bronsted-Lowry acid: H2O since it donates a H+ to NH3

Bronsted-Lowry base: NH3 since it accepts a H+ from H2O

NaHCO3(s) + H2O(l) ↔ H2CO3 (aq) + Na+(aq) + OH

-(aq)

Bronsted-Lowry acid: H2O since it donates a H+ to NaHCO3, with Na

+ released

Bronsted-Lowry base: NaHCO3 since it accepts a H+ from H2O, with Na

+ released

3. CHARACTERISTICS OF ACIDS AND BASES

a) Acids

1. pH between 0 and 6.9; the lower the pH value, the stronger the acid.

2. React with certain metals (aluminum, magnesium and zinc) to produce H2 gas.

3. Acids are corrosive.

4. Acids taste sour.

Ex. Carbonic and phosphoric acids give carbonated beverages their sharp taste.

Citric and ascorbic acids give lemons and grapefruit their tart taste.

Acetic acid makes vinegar taste sour.

5. Causes blue litmus paper to turn pink.

6. Conducts electricity.

b) Bases

1. pH between 7.1 and 14; the higher the pH value, the stronger the base.

2. Taste bitter.

Ex. Soap has a bitter taste.

3. Feel slippery

Ex. Soap has a slippery texture.

4. Causes red litmus paper to turn blue.

5. Conducts electricity.

c) Water: has a neutral pH of 7.0 but can act as a very weak acid or base due to the

tiny degree of self-ionization.

4. CALCULATING pH and pOH FROM SIMPLE CONCENTRATIONS

a) pH and pOH : Since [H+] and [OH

-] concentrations are usually very small numbers

expressed in scientific notation, scientists have adopted a shorthand method to

express these concentrations in an aqueous solution. This is known as the pH scale

and ranges from 0 to 14.

b) pH = -log[H+] and pOH = -log[OH

-]

Additionally, pH + pOH = 14

c) For concentrations of H+ and OH

- that are 1.0 x 10

n, calculating the pH or pOH of the

solution is relatively easy. The log (n) is the number to which 10 is raised.

1. [H+] = 1.0 x 10

-4 the log is -4

pH = -log[1.0 x 10-4

] or –log[10-4

]

pH = -(-4) = 4 this represents an acidic solution with a pH of 4

Since pH + pOH = 14, then the pOH is 10

2. [OH-] = 1.0 x 10

-2 the log is -2

pOH = -log[1.0 x 10-2

] or –log[10-2

]

pOH = -(-2) = 2

Since pH + pOH = 14, then the pH is 12 this represents a basic solution with a

pH of 12

3. For concentrations with the coefficient 1.0, the pH and pOH will be a simple whole

number between 0 and 14.

d) Whether a solution is acidic, basic or neutral is based only on pH, a measurement of

the number of H+ ions in solution.

1. Acidic solutions have more H+ ions than OH

- ions and thus, have pH’s below 7

2. Basic solutions have more OH- ions than H

+ ions and thus, have pH’s above 7.

3. Neutral solutions have equal numbers of H+ ions and OH

- ions and thus, have a pH

of 7.

5. CALCULATING pH and pOH FROM MORE COMPLEX CONCENTRATIONS (HONORS ONLY)

a) Calculating pH and pOH from concentrations in which the coefficient of the [H+] and

[OH-] is not 1.0 is more difficult and requires a calculator with a LOG function since the

calculation must include both the coefficient and the power to which 10 is raised.

b) [H+] = 3.8 x 10

-10

In your calculator, enter the -log of 3.8 x 10-10

to calculate the pH.

In a TI-83 or equivalent, punch in (-), followed by the LOG button and a left parenthesis

“(“ will appear; enter 3.8, followed by the 2nd

function button, followed by the button

beneath EE, followed by (-), followed by 10, followed by the right parenthesis “)” and

press ENTER.

ANSWER: pH = 9.4 and pOH = 4.6 (Remember, pH + pOH = 14) A basic solution

c) [OH] = 7.9 x 10-14

In your calculator, enter the -log of 7.9 x 10-14

to calculate the pOH as instructed

above.

ANSWER: pOH = 13.1 and pH = 0.9 An acidic solution

d) For concentrations with a coefficient other than 1.0, the pH and pOH will be a more

precise number usually carried out to 1 decimal place.

6. CALCULATING CONCENTRATIONS FROM pH and pOH

a) The [H+] and [OH

-] values can be calculated from the pH or pOH using the reverse

process. This is relatively simple for pH or pOH values that are whole numbers, such as

1, 2, 3, etc.

Ex. pH = 6.0

Since pH = -log[H+], then the [H

+] must equal 1.0 x 10

-6

If pH = 6.0, then pOH = 8; therefore, [OH-] must equal 1.0 x 10

-8

b) HONORS ONLY

1. For pH and pOH values that are not whole numbers, the calculation is more

difficult. This requires a calculator with a LOG and ANTILOG function.

2. If pH = -log[H+], then multiplying both sides of the equation by -1 gives:

-pH = log [H+]

To calculate [H+] using this equation, you must take the antilog of both sides of the

equation:

antilog (-pH) = [H+]; likewise, antilog (-pOH) = [OH

-]

3. pH = 3.6 and pOH = 10.4

antilog -3.6 = [H+] and antilog -10.4 = [OH

-]

In your calculator, enter the antilog of -3.6 to calculate the [H+].

In a TI-83 or equivalent, punch the 2nd

button, followed by the LOG button and the

left parenthesis “(“ will appear; punch in (-), followed by 3.6, followed by the right

parenthesis “)” and press ENTER.

ANSWER: [H+] = 2.5 x 10

-4

In your calculator, enter the antilog of -10.4 to calculate the [OH-] using the same

method.

ANSWER: [OH-] = 3.98 x 10

-11

7. NEUTRALIZATION REACTIONS

a) A neutralization reaction is a reaction in which an acid and a base react in aqueous

solution to produce a salt and water.

Acid + Base → Salt + Water

b) It is a double replacement reaction where the H+ ion in the acid replaces the metal

cation in the base (forming H2O) and the metal cation of the base replaces the H+ ion in

the acid (forming an ionic salt).

Examples:

HNO3 + KOH → KNO3 + H2O

nitric acid potassium hydroxide potassium nitrate water

2HCl + Mg(OH)2 → MgCl2 + 2H2O

hydrochloric acid magnesium hydroxide magnesium chloride water

H2SO4 + 2NaOH → Na2SO4 + 2H2O

sulfuric acid sodium hydroxide sodium sulfate water

8. MONOPROTIC AND POLYPROTIC ACIDS

a) Acids that deliver 1 H+ to solutions are termed monoprotic.

Ex. HNO3 nitric acid → H+(aq) + NO3

-(aq)

HCl hydrochloric acid → H+(aq) + Cl

-(aq)

b) Acids that deliver more than 1 H+ to solutions are termed polyprotic.

Ex. H2SO4 sulfuric acid → 2H+(aq) + SO4

2-(aq) this is diprotic since it delivers 2H

+

H3PO4 phosphoric acid → 3H+(aq) + PO4

3-(aq) this is triprotic since it delivers 3H

+

9. TITRATIONS

a) A titration is a method for determining the unknown concentration of an acid or base

solution by reacting it with a known volume and concentration of an opposing base or

acid.

b) Neutralization occurs when all H+ ions in solution have bonded with all OH

- ions in

solution to produce water, a compound with a neutral pH of 7.0.

c) Titrations are performed by carefully adding an acid in a burette to a base in a

receiving flask until the pH of the base is neutralized to 7.0. The receiving flask

contains a chemical “indicator” which turns color when a neutral pH in the base is

achieved. The titration is complete at this point, referred to as the endpoint. The

same process can be performed by adding a base to an acid in the receiving flask.

d) Equation for calculating an unknown concentration of acid/base from a titration:

(Macid)(Vacid)( # of H+) = (Mbase)(Vbase)( # of OH

-)

Where Macid = Molarity of acid

Vacid = Volume of acid

Mbase = Molarity of base

Vbase = Volume of base

# of H+ = H

+ ions delivered to solution by acid

# of OH- = OH

- ions delivered to solution by base

e) Examples:

1. If it takes 45 mL of 1.0M NaOH solution to neutralize 57 mL of HCl, what is the

concentration of the HCl?

NOTE: HCl delivers 1 H+

ion and NaOH delivers 1 OH- ion based on their formulas.

(Macid)(Vacid)( # of H+) = (Mbase)(Vbase)( # of OH

-)

(Macid)(57 mL)(1) = (1.0M)(45 mL)(1)

Macid = 0.79M

2. If if takes 67 mL of 0.5M H2SO4 to neutralize 15 mL of Al(OH)3, what is the

concentration of the Al(OH)3?

NOTE: H2SO4 delivers 2 H+

ions and Al(OH)3 delivers 3 OH- ions based on their

formulas.

(Macid)(Vacid)( # of H+) = (Mbase)(Vbase)( # of OH

-)

(0.5M)(67 mL)(2) = (Mbase)(15 mL)(3)

Mbase = 1.49M

10. STRENGTHS OF ACIDS AND BASES

a) Strong acids and bases are strong electrolytes; they completely dissociate in solution,

delivering the maximum number of ions to solution.

Strong Acids: HCl, HBr, HNO3, H2SO4

Strong Bases: Group 1 & 2 metals bound to OH-; NaOH, KOH, RbOH, Ca(OH)2

b) Weak acids and bases are weaker electrolytes; they do not dissociate completely in

solution and deliver smaller numbers of ions to solution.

Weak Base: Fe(OH)2

Weak Acids: H2CO3 (carbonic acid in soda), H2S, HF

11. NAMING ACIDS AND BASES

a) Acids

Chemical formula is the general form HX where:

H- Hydrogen

X- an anion (monoatomic or polyatomic)

b) Three Rules for Naming Acids

1. When the anion X is a monoatomic ion and ends in “ide,” the acid name

begins with the prefix “hydro”. The second part of the name consists of the root

name of the anion followed by the suffix “ic,” followed by acid.

Ex. HCl- hydrochloric acid; H2S- hydrosulfuric acid

2. When the anion X is a polyatomic ion and ends in “-ite,” the acid name is the root

name of the anion followed by the suffix “-ous,” followed by acid.

Ex. HNO2- nitrous acid; H2SO3-sulfurous acid

3. When the anion X is a polyatomic ion and ends in “-ate,” the acid name is the root

name of the anion with the suffix “ic,” followed by acid.

Ex: HNO3- nitric acid; H2CO3- carbonic acid

c) Naming Bases

1. Most bases consist of a metal cation bound to a hydroxide ion (OH-).

2. Base names consist of the cation name followed by the anion name.

Ex. NaOH- sodium hydroxide; Mg(OH)2-magnesium hydroxide

12. OXIDATION AND REDUCTION

a) Oxidation and reduction reactions, also called redox reactions, are reactions in which

electrons are transferred from one atom to another.

b) To remember, use the phrase “LEO the lions says GER.”

c) Oxidation occurs when electrons are lost from a substance.

1. “LEO” Loss of Electrons is Oxidation

Neutral metal atoms lose electrons to become positive cations.

Na0 → Na

+ + 1e

-; Na loses 1 electron and thus, is oxidized.

The oxidation number is the charge on the resulting ion. In the example above,

the sodium ion has an oxidation number of 1+.

Al0 → Al

3+ + 3e

-; Al loses 3 electrons and is oxidized to form an ion with an

oxidation number of 3+.

d) Reduction occurs when electrons are gained by a substance.

1. “GER” Gain of Electrons is Reduction

Neutral non-metal atoms gain electrons to become negative anions.

Cl0 + 1e

- → Cl

-; Cl gains 1 electron and thus, is reduced with an oxidation

number of 1-.

P0 + 3e

- → P

3-; P gains 3 electrons and thus, is reduced to form an ion with an

oxidation number of 3-.

e) The oxidizing agent is the substance that facilitates oxidation by accepting lost

electrons; it is the substance that is reduced.

f) The reducing agent is the substance that facilitates reduction by losing electrons; it is

the substance that is oxidized.

g) Analyzing redox reactions

1. A complete balanced equation is given. To determine which substances are

oxidized and reduced, the reaction should be divided into half reactions.

2. Example:

Complete chemical equation: 2KBr + Cl2 → 2KCl + Br2

Looking at this in terms of ions gives:

2K+Br

- + Cl2 (neutral molecule) → 2K

+Cl

- + Br2 (neutral molecule)

Since the 2 potassium ions in this reaction maintain a charge of 1+, it has not been

oxidized or reduced.

2Br- becomes Br2, which means each of the bromine ions have lost their extra

electron to become Br2.

Cl2 becomes 2Cl-, which means each of the chlorine atoms gained an electron to

become 2Cl-.

Net ionic equation: 2Br - + Cl2 → Br2 + 2Cl

–

Half reaction: 2Br - → Br2 + 2e

– Bromine is oxidized and acts as the reducing

agent.

Half reaction: Cl2 + 2e -

→ 2Cl –

Chlorine is reduced and acts as the oxidizing

agent.

NOTE: The 2 electrons were transferred from bromine to chlorine and thus,

oxidation and reduction must occur simultaneously.

APPENDIX: pH SCALE WITH COMMON SUBSTANCES

Unit11NoteQuizQuestion

Unit 11.2: Acid-Base Theory

1. A

2. A

3. A

4. A

5. A

6. A

7. A

8. A

9. A

10. A

Unit 11.3: Titration

1. a

2. a

3. a

4. a

5. a

6. a

7. a

8. a

9. a

10. a

Unit 11.4: Electro-chemistry

1. A

2. A

3. A

4. a

5. A

6. A

7. A

8. A

9. A

10. Which of the following represents a reaction in which the same reactant undergoes both oxidation and

reduction?

a. H2SeO4(aq) + 2Cl-(aq) + 2H

+(aq) → H2SeO3(aq) + Cl2(g) + H2O(l)

b. S8(s) + 8O2(g) → 8SO2(g)

c. 3Br2(aq) + 6OH-(aq) → 5Br

-(aq) + BrO3

-(aq) + 3H2O(l)

d. Ca2+

(aq) + SO42-

(aq) → CaSO4(s)