Chapter 8: Acids and Bases

Chapter 8: Acids and Bases 1

Acids and Bases Acid: any substance which when dissolved in

H2O makes [H3O+] (hydronium)

HNO3 + H2O H3O+ + NO3

- acid conjugate base

Base: any substance which when dissolved in

H2O increases [OH-], hydroxide ion NaOH Na+ + OH-

base conjugate acid

Both examples above are strong acid/base

– indicates 100% ionization

2

ACID NOMENCLATURE

Chapter 4: Suppliment- Naming Acids

Naming Acids: Simple acids (Binary)

Rule: if anion ends with “—ide” acid is named “hydro_________ic acid”

Anion Anion name Acid formula Acid name

Cl-

F-

S2-

CN-

Hydrobromic acid

Hydroiodic acid

Chapter 4: Suppliment- Naming Acids

Naming Acids: Ternary (Polyatomic Ion containing- If you haven’t already done so, you

MUST memorize the polyatomic ions)

• If the anion ends with “_ate” acid is named “________ic acid” (note the absence of hydro)

Anion Anion name Acid formula Acid name

NO3-

Sulfate

Phosphoric acid

Chapter 4: Suppliment- Naming Acids

Naming Acids: Ternary (Polyatomic Ion containing- If you haven’t already done so, you MUST

memorize the polyatomic ions)

If the anion ends with “___ite”, the acid is name “_________ous acid”

NO2-

Sulfurous

acid

Anion Anion name Acid formula Acid name

Chapter 4: Suppliment- Naming Acids

Bronsted-Lowry Acids/Bases

• Acid: a species which donates a proton.

• Base: a species which accepts a proton.

– Thus, acid base rxns involve the transfer of a proton. CH3COOH + H2O H3O+ + CH3COO-

acid

7

base base acid

NH3 + H2O NH4+ + OH-

base acid acid base

Concept of Equilibrium

• A + B C + D

• A + B C + D

• A + B C + D

• A + B C + D

Chapter 8: Acids and Bases 8

]][[

]][[

BA

DCK

K = VERY LARGE # Reaction is almost 100% complete

K =1 equal amounts of reactant and product

K <1 More reactant (equilibrium lies to the LEFT)

K >1 More product (equilibrium lies to the RIGHT)

Meaning of K

Strength of Acids and Bases

• Strong Acid – Ionizes completely in aqueous solution

• Strong Base – Ionizes completely in aqueous solution

9

The Strong Acids/Bases

• Strong acids

–HClO4

–H2SO4

–HNO3

–HCl

–HBr

–HI

• Strong bases

–LiOH

–NaOH

–KOH

–Ca(OH)2

–Sr(OH)2

–Ba(OH)2

10 If it does not appear on this list it is WEAK!

Strength of Acids and Bases

• Weak Acid – Partially ionizes in aqueous solution

• Weak Base – Partially ionizes in aqueous solution

11

Conjugate Pairs

o Bronsted – Lowry define:

o Acid – Proton donor

o Base – Proton acceptor

o Acid – Base Reaction – Proton transfer reaction

o Compounds involved in the proton transfer are a conjugate acid – base pair

12

Conjugate Pairs

13

Weak Acids

• Monoprotic acid: Acid that gives up one proton

• Diprotic acid: Acid that gives up to two protons

• Triprotic acid: Acid that gives up to three protons

14

Weak Acids

• Amphoteric: a species which can act as an acid or base by either gaining or losing protons (amphiprotic) or OH- groups.

– Most important example is water

– Example 8.1

• Remember – Bronsted–Lowry acids can give up a hydrogen

– Not all hydrogens can be given up*

15

Introduction to Acid-Base Equilibrium

• Strong Acids have equilibrium, just far to the right

– Example of HCl in water

• Weak acids have an equilibrium more to the left

– Only a few acetic acid molecules react with water to give acetate and hydronium ions

16

Weak Acids/Bases: Equilibria

19

Acid HA + H2O H3O+ + A-

Base BOH B+ + OH-

acid base acid base

base acid base

K = [H3O

+] [A-]

[HA] [H2O]

Ka = [H3O

+] [A-]

[HA]

But H2O is a pure liquid so

[H2O] = 1

Ka - the acid dissociation constant.

The size of Ka tells the strength of

the acid.

The Ionization Constant

• Ka for acetic acid is 1.8 x 10-5

• We manipulate the number to make it easier to use by taking the negative logarithm of Ka

• So acid strength is represented as –log Ka which we call the pKa

• pKa = -log Ka

• The pKa for acetic acid is 4.75, a nicer number to deal with

• Example 8.3 and 8.4

20

Examples

• Ka for benzoic acid is 6.5 X 10-5. What is the pKa for this acid

• Which is the stronger acid:

– Benzoic acid with a Ka of 6.5 X 10-5 or hydrocyanic acid with a Ka of 4.9 X 10-5

Chapter 8: Acids and Bases 21

Calculator

Acid-Base Reactions

• Several important reactions of acids and bases:

1. They react with each other – neutralization

2. Reaction with metals – Strong acids react with active metals to produce H2 gas and salt

23

Acid-Base Reactions

3. Reaction with Ammonia and Amines – Acids stronger than NH4

+ will react with NH3 to form a salt:

HCl (aq) + NH3 (aq) NH4+ (aq) + Cl- (aq)

(very important in biological systems)

24

Acidic and Basic Properties of Pure Water

• Acids make H3O+ and bases produce OH- ions

• What goes on in pure water with no acid or base?

• Small amounts of ions are present!

25

Amphoteric Water

26

Kw, the ion product of water

Soooo...

H2O + H2O H3O+ + OH-

K = [H3O+] [OH-]

[H2O] [H2O] Kw = [H3O+] [OH-]

Auto-ionization of water

Auto-Ionization of Water

• Kw = [H3O+][OH-]

• Kw = 1.0 x 10-14

• At equilibrium the concentrations are equal, so…

• Example 8.5

The [OH-] of an aqueous solution is 1.0 X 10-4 M. What is it’s [H3O+]? 27

pH

• We generally represent the hydronium ion concentration as:

• pH = - log [H3O+]

Rules – A solution is acidic if the pH is less than 7.0

– A solution is basic if its pH is greater than 7.0

– A solution is neutral if the pH is equal to 7.0

28

Acid/Base Calculations

29

[H+] = (1 x 10-14)/[OH-]

[OH-] = (1 x 10-14)/[H+]

If given [H+] or [OH-], then use these to solve for the other.

[H+] x [OH-] = 1 x 10-14

pH = - log[H+]

Once you have [H+] you can use this for pH.

OR given pH you can solve for [H+]. [H+] = 10

-pH

Example

Chapter 8: Acids and Bases 32

The [H3O+] of a certain liquid detergent is 1.4 X 10-9 M. What is its pH? Is this solution acidic basic or neutral? The pH of tomato juice is 4.1. What is its [H3O+]? Is this solution acidic, basic, or neutral?

pOH and pH Relationship

• Same treatment for the concentration of OH-

• pOH = - log[OH-]

• Kw = 1 x 10-14 = [H+][OH-]

• 14 = pH + pOH

• If we know the pH, we can calculate the pOH

• Example 8.7 – The [OH-] of a strongly basic solution is 1.0 X 10-2 M.

What are the pOH and pH of this solution?

33

Example

• The [OH-] of a strongly basic solution is 1.0 X 10-2. What is the pOH and pH of this solution? Is it acidic/basic/ or neutral?

Chapter 8: Acids and Bases 35

Titration: Stoichiometry with Solutions (8.8-8.9)

36 Chapter 8: Acids and Bases

Titrations 1. Acid of unknown concentration in the flask, known volume!

2. A known concentration of base and known volume is added by

buret

3. When the indicator changes from colorless to pink the acid has

been completely neutralized the base

37

Titrations Requirements 1. Know the balanced chemical equation, must know the

molar ratio to calculate the unknown concentration

2. Reaction must be fast and go to completion

3. When reactants combine completely you have to visualize

the completion, indicator and equivalence point

4. Measure CAREFULLY and ACCURATELY to obtain good

data

Problem 8.8

38

Use Molarity to convert between Moles and Liters of solution

Mole A Liters of

solution A __Moles A* = 1 Liter of Solution

Example: 3.0 M HCl = solution HCl L 1

HCl moles 0.3

Molarity (M)

39 Chapter 8: Acids and Bases

Stoichiometry

40

Mole A Mass A

(grams)

# of particles*

of A

*Atom (element) * Molecules (molecular) *Formula Units (ionic Compounds

6.022 X 1023 particles*A=1mole A __g A** = 1 mole A

**Use molar mass as a conversion PERIODIC TABLE!!!!

Mole B Mass B

(grams)

# of particles*

of B

6.022 X 1023 particles*B=1mole B __g B** = 1 mole B

_mo

l A**

*

= _

_m

ol B

***

Coefficients in balanced equation represent the MOLE RATIO

Chapter 8: Acids and Bases

Stoichiometry

41

Mole A Mass A

(grams) __g A** = 1 mole A

Mole B Mass B

(grams) __g B** = 1 mole B

_mo

l A**

*

= _

_m

ol B

***

Coefficients in balanced equation represent the MOLE RATIO

__Moles A* = 1 Liter of Solution

__Moles B* = 1 Liter of Solution

Molarity (M)

Molarity (M)

Liters of Solution

containing A

Liters of Solution

containing A

Chapter 8: Acids and Bases

Example 8.8

Chapter 8: Acids and Bases 42

Following are data for the titration of 0.108 M H2SO4 with a solution of NaOH of unknown concentration. What is the concentration of the NaOH solution?

Volume of 0.108 M H2SO4

Volume of NaOH

Trial 1 25.0 mL 33.48 mL

Trial 2 25.0 mL 33.46 mL

Trial 3 25.0 mL 33.50 mL

Titration Vocabulary

• Titration: Technique for determining concentration of an unknown reaction

• End point

– Indicator turns color

– Slightly Past the Equivalence point

• Equivalence point

– Acid and Base have been added in STOICHIOMETRIC proportion

43 Chapter 8: Acids and Bases

HC2H3O2 + NaOH H2O + NaC2H3O2