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(1)

Chapter 14

(2)

Chapter 14

Table of Contents

14.1The Nature of Acids and Bases 14.2Acid Strength

14.3The pH Scale

14.4Calculating the pH of Strong Acid Solutions

14.5 Calculating the pH of Weak Acid Solutions

14.6 Bases

14.7 Polyprotic Acids

14.8 Acid–Base Properties of Salts

14.9 The Effect of Structure on Acid–Base Properties 14.10 Acid–Base Properties of Oxides

14.11 The Lewis Acid–Base Model

(3)

Section 14.1

The Nature of Acids and Bases

Models of Acids and Bases

• Arrhenius: Acids produce H+ ions in solution,

bases produce OH- ions.

• Brønsted–Lowry: Acids are proton (H+) donors,

bases are proton acceptors:

HCl + H2O  Cl- + H

3O+

(4)

Section 14.1

The Nature of Acids and Bases

Brønsted–Lowry Reaction

acid base conjugate conjugate

acid base

(5)

Section 14.1

The Nature of Acids and Bases

Brønsted–Lowry Reaction

acid base conjugate conjugate

acid base

(6)

Section 14.2

Atomic Masses Acid Strength

• Strong acid:

Ionization equilibrium lies far to the right.

Yields a weak conjugate base.

(H

2

O is a stronger base than A

-

.)

Stronger the acid, the weaker its conjugate

base.

• Weak acid:

(7)

Section 14.2

(8)

Section 14.2

Atomic Masses Acid Strength

(9)

Section 14.2

Atomic Masses Acid Strength

Acid dissociation constant, K

a

or [H

+

]

[A

-

]

[HA]

 High Ka = strong acid

 Write a dissociation equation for C5H6NH3+.

 C5H6NH3+ + H

2O  C5H6NH2 + H3O+

 

3

H O

A

(10)

Section 14.2

Atomic Masses Acid Strength

Acid dissociation constant, K

a

 H2SO4 is a strong, diprotic acid.

 Oxyacids: The more oxygens in the formula, the stronger the acid.

(11)

Section 14.2

Atomic Masses Acid Strength

• Water is amphoteric:

 Behaves either as an acid or as a base:  2 H2O(l)  H3O+

(aq) + OH-(aq)

 Or: H2O(l)  H+

(aq) + OH-(aq)

Kw = [H+][OH–] = 1.0 × 10–14 at 25°C.

(12)

Section 14.2

Atomic Masses Acid Strength

• [H

+

] = [OH

] = 1.0 × 10

–7

M;

neutral

solution

• [H

+

] > [OH

];

acidic

solution

(13)

Section 14.2

Atomic Masses Acid Strength

(14)

Section 14.2

Atomic Masses Acid Strength

(15)

Section 14.2

Atomic Masses Acid Strength

Review

If the equilibrium lies to the right, the value for

Ka is __________. large (or >1)

If the equilibrium lies to the left, the value for Ka

is ___________.

(16)

Section 14.2

Atomic Masses Acid Strength

Concept Check

HA(aq) + H2O(l) H3O+(aq) + A–(aq)

If water is a better base than A–, do products

or reactants dominate at equilibrium?

Does this mean HA is a strong or weak acid?

(17)

Section 14.2

Atomic Masses Acid Strength

Concept Check

Consider a 1.0 M solution of HCl.

Order the following from strongest to weakest

base and explain:

H2O(l)

(18)

Section 14.2

Atomic Masses Acid Strength

Let’s Think About It…

• How good is Cl–(aq) as a base?

• Is A–(aq) a good base?

The bases from strongest to weakest are:

A–, H

(19)

Section 14.2

Atomic Masses Acid Strength

Concept Check

Consider a solution of NaA where A– is the

anion from weak acid HA:

A–(aq) + H

2O(l) HA(aq) + OH–(aq)

base acid conjugate conjugate

(20)

Section 14.2

Atomic Masses Acid Strength

Concept Check

Consider a solution of NaA where A– is the

anion from weak acid HA:

A–(aq) + H

2O(l) HA(aq) + OH–

(aq)

base acid conjugate conjugate

acid base

b) Is the value for Kb greater than or less than

(21)

Section 14.2

Atomic Masses Acid Strength

Concept Check

Consider a solution of NaA where A– is the anion from

weak acid HA:

A–(aq) + H

2O(l) HA(aq) + OH–(aq)

base acid conjugate conjugate

acid base

(22)

Section 14.2

Atomic Masses Acid Strength

Concept Check

Acetic acid (HC2H3O2) and HCN are both weak acids. Acetic acid is a stronger acid than HCN.

Arrange these bases from weakest to strongest

and explain your answer:

H2O Cl– CN– C

(23)

Section 14.2

Atomic Masses Acid Strength

Let’s Think About It…

• H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

acid base conjugate conjugate

acid base

• At 25C, Kw = 1.0 x 10–14

The bases from weakest to strongest are:

(24)

Section 14.2

Atomic Masses Acid Strength

Concept Check

Discuss whether the value of K for the reaction: HCN(aq) + F–(aq) CN–(aq) + HF(aq)

is <1 >1 =1

(Ka for HCN is 6.2×10–10; K

a for HF is 7.2×10–4.)

(25)

Section 14.2

Atomic Masses Acid Strength

Concept Check

Calculate the value for K for the reaction:

HCN(aq) + F–(aq) CN–(aq) + HF(aq)

(Ka for HCN is 6.2×10–10; K

a for HF is 7.2×10–4.)

(26)

Section 14.3

The Mole

The pH Scale

• pH = –log[H+] (Also: pOH = –

log[OH-] )

• An easy way to represent acidity. • pH decreases as [H+] increases.

• Significant figures: The number of places after the decimal in the log is equal to the number of sig figs in the original number.

(27)

Section 14.3

The Mole

The pH Scale

pH Range

• pH = 7; neutral

• pH > 7; basic

 Higher pH, more basic. • pH < 7; acidic

(28)

Section 14.3

The Mole

The pH Scale

The pH scale and pH values of some

(29)

Section 14.3

The Mole

The pH Scale

Exercise

Calculate the pH for each of the following solutions.

a) 1.0 × 10–4 M H+

pH = 4.00

b) 0.040 M OH–

(30)

Section 14.3

The Mole

The pH Scale

Exercise

The pH of a solution is 5.85. What is the [H+]

for this solution?

(31)

Section 14.3

The Mole

The pH Scale

pH and pOH

• Recall:

Kw = [H+][OH–]

–log Kw = –log[H+] – log[OH–]

(32)

Section 14.3

The Mole

The pH Scale

Exercise

Calculate the pOH for each of the following solutions.

a) 1.0 × 10–4 M H+

pOH = 10.00

b) 0.040 M OH–

(33)

Section 14.3

The Mole

The pH Scale

Exercise

The pH of a solution is 5.85. What is the [OH–] for this solution?

(34)

Section 14.4

Calculating the pH of Strong Acid Solutions

Thinking about acid–base problems

• What are the major species in solution?

• What is the dominant reaction that will take place?

 Is it an equilibrium reaction or a reaction that will go essentially to completion?

 React all major species until you are left with an equilibrium reaction.

(35)

Section 14.4

Calculating the pH of Strong Acid Solutions

Concept check

Consider an aqueous solution of 2.0 x 10–3 M

HCl.

What are the major species in solution?

H+, Cl–, H 2O

What is the pH?

(36)

Section 14.4

Calculating the pH of Strong Acid Solutions

Calculate the pH of a 1.5 x 10–11 M solution of

HCl.

pH = 7.00

(37)

Section 14.4

Calculating the pH of Strong Acid Solutions

Calculate the pH of a 1.5 x 10–2 M solution of

HNO3.

(38)

Section 14.4

Calculating the pH of Strong Acid Solutions

Let’s Think About It…

• When HNO3 is added to water, a reaction takes place immediately:

(39)

Section 14.4

Calculating the pH of Strong Acid Solutions

Let’s Think About It…

• Why is this reaction not likely?

NO3–(aq) + H

(40)

Section 14.4

Calculating the pH of Strong Acid Solutions

Let’s Think About It…

• What reaction controls the pH?

• H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

• In aqueous solutions, this reaction is always taking place.

• But is water the major contributor of H+ (H

3O+)?

(41)

Section 14.4

Calculating the pH of Strong Acid Solutions

End of lesson

(42)

Section 14.4

Calculating the pH of Strong Acid Solutions

14.5-6 pH of a weak acid solution

Objective

:

Students will be able to calculate the

pH and the concentrations of all

(43)

Section 14.5

Calculating the pH of Weak Acid Solutions

Solving Weak Acid Equilibrium Problems

1. List the major species in the solution.

2. Choose the species that can produce H+, and

write balanced equations for the reactions producing H+.

3. Using the values of the equilibrium constants for the reactions you have written, decide which

equilibrium will dominate in producing H+.

(44)

Section 14.5

Calculating the pH of Weak Acid Solutions

Solving Weak Acid Equilibrium Problems

5. List the initial concentrations of the species participating in the dominant equilibrium.

6. Define the change needed to achieve equilibrium; that is, define x.

7. Write the equilibrium concentrations in terms of x.

(45)

Section 14.5

Calculating the pH of Weak Acid Solutions

Solving Weak Acid Equilibrium Problems

9. Solve for x the “easy” way, that is, by

assuming that [HA]0x about equals [HA]0. 10.Use the 5% rule to verify whether the

(46)

Section 14.5

Calculating the pH of Weak Acid Solutions

The easy way to find pH of a weak acid…

Ex. 1 Find the [H3O+], [HCO

3-], [H2CO3], and pH of a

0.20 M solution of H2CO3 (p. A23: Ka = 4.3 x 10-7).

Note: H2CO3 gives off more H+ than H

2O does… So:

H2CO3 + H2O H3O+ + HCO 3

-Initial 0.20 M ~ 0 ~ 0

Change –x +x +x

(47)

Section 14.5

Calculating the pH of Weak Acid Solutions

Ka = [H3O+] [HCO 3-]

[H2CO3]

4.3 x 10–7 = x2

H2CO3 + H2O H3O+ + HCO 3

-Initial 0.20 M ~ 0 ~ 0

Change –x +x +x

Equilibrium 0.20–x x x

4.3 x 10–7 = x2

0.20

x = √(4.3 x 10–7)(0.20)

Very small (because Ka is so small) so we ignore x in this case.

(48)

Section 14.5

Calculating the pH of Weak Acid Solutions

[HCO

3-

] = [H

3

O

+

] = 2.9 x 10

-4

M

[H

2

CO

3

] = 0.20 - 2.9 x 10

-4

M ≈ 0.20 M

(49)

Section 14.5

Calculating the pH of Weak Acid Solutions

Concept Check

Consider a 0.80

M

aqueous solution of the

weak acid HCN (

K

a

= 6.2 x 10

–10

).

(50)

Section 14.5

Calculating the pH of Weak Acid Solutions

Let’s Think About It…

(51)

Section 14.5

Calculating the pH of Weak Acid Solutions

Consider this:

HCN(aq) + H2O(l) H3O+(aq) + CN–(aq)

Ka = 6.2 x 10-10

H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

Kw = 1.0 x 10-14

(52)

Section 14.5

Calculating the pH of Weak Acid Solutions

Exercise

(53)

Section 14.5

Calculating the pH of Weak Acid Solutions

Let’s think about it…

• What are the major species in solution?

HF, H

2

O

(54)

Section 14.5

Calculating the pH of Weak Acid Solutions

Let’s think about it…

• What are the possibilities for the dominant reaction?

HF(aq) + H2O(l) H3O+(aq) + F–(aq)

Ka=7.2 x 10-4

H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

Kw=1.0 x 10-14

(55)

Section 14.5

Calculating the pH of Weak Acid Solutions

Solving for pH

Ka = [H3O+] [F–]

[HF]

7.2 x 10–4 = x2

0.20 - x

HF(aq) + H2O H3O+(aq) + F(aq)

Initial 0.20 M ~ 0 ~ 0

Change –x +x +x

Equilibrium 0.20–x x x

This time we cannot

ignore x. Ka is too large… > 1% is ionized.

7.2 x 10-4 (0.20 – x) = x2

1.44 x 10-4 - 7.2 x 10-4 x - x2 = 0

x2 + 7.2 x 10-4 x - 1.44 x 10-4 = 0

(56)

Section 14.5

Calculating the pH of Weak Acid Solutions

Percent dissociation

Ex. 3 Find percent dissociation for a 0.20 M HF solution. (See example 2.)

Note: % dissociation = amount dissoc (M) x 100% initial conc (M)

• % dissociation = amount dissoc (M) x 100% initial conc (M)

(57)

Section 14.5

Calculating the pH of Weak Acid Solutions

(58)

Section 14.5

Calculating the pH of Weak Acid Solutions

Exercise

A solution of 8.00 M formic acid (HCHO2) is 0.47% ionized in water.

Calculate the Ka value for formic acid.

(59)

Section 14.5

Calculating the pH of Weak Acid Solutions

Exercise

Calculate the pH of an 8.00 M solution of formic acid. Use the data from the previous slide.

(60)

Section 14.5

Calculating the pH of Weak Acid Solutions

Concept Check

The value of Ka for a 4.00 M formic acid solution should be:

higher than lower than the same as

the value of Ka of an 8.00 M formic acid solution.

(61)

Section 14.5

Calculating the pH of Weak Acid Solutions

Concept Check

The percent ionization of a 4.00 M formic acid solution should be:

higher than lower than the same as

the percent ionization of an 8.00 M formic acid solution.

(62)

Section 14.5

Calculating the pH of Weak Acid Solutions

Concept Check

The pH of a 4.00 M formic acid solution should be:

higher than lower than the same as

the pH of an 8.00 M formic acid solution.

(63)

Section 14.5

Calculating the pH of Weak Acid Solutions

Exercise

Calculate the percent ionization of a 4.00 M

formic acid solution in water.

(64)

Section 14.5

Calculating the pH of Weak Acid Solutions

Exercise

Calculate the pH of a 4.00 M solution of formic acid.

(65)

Section 14.6

Bases

• Arrhenius: bases produce OH– ions.

• Brønsted–Lowry: bases are proton acceptors. • In a basic solution at 25°C, pH > 7.

• Ionic compounds containing OH- are generally

considered strong bases.

 LiOH, NaOH, KOH, Ca(OH)2

• pOH = –log[OH–]

(66)

Section 14.6

Bases

Concept Check

Calculate the pH of a 1.0 x 10–3 M solution of

sodium hydroxide.

(67)

Section 14.6

Bases

Concept Check

Calculate the pH of a 1.0 x 10–3 M solution of

calcium hydroxide.

(68)

Section 14.6

Bases

• Equilibrium expression for weak bases uses Kb.

CN–(aq) + H

2O(l) HCN(aq) + OH–(aq)

(69)

Section 14.6

Bases

• pH calculations for solutions of weak bases are very similar to those for weak acids.

• Kw = [H+][OH–] = 1.0 × 10–14

• pOH = –log[OH–]

(70)

Section 14.6

Bases

Concept Check

Calculate the pH of a 2.0 M solution of ammonia (NH3). (Kb = 1.8 × 10–5)

NH3 + H2O 

 NH4+ + OH

-Initial 2.0 M ~ 0 ~ 0

Change –x +x +x

(71)

Section 14.6

Bases

Kb = [NH4+] [OH-]

[NH3]

1.8 x 10 = x

1.8 x 10–5 = x2

2.0

x = √(1.8 x 10–5)(2.0)

NH3 + H2O NH4+ + OH

-Initial 2.0 M ~ 0 ~ 0

Change –x +x +x

(72)

Section 14.6

Bases

[OH-] = 6.0 x 10-3 M

= 2.22 pOH = antilog (6.0 x 10-3 M)

(73)

Section 14.6

Bases

End of lesson.

(74)

Section 14.7

Polyprotic Acids

• Acids that can furnish more than one proton.

• Always dissociate one proton at a time.

• The conjugate base of the first dissociation

equilibrium becomes the acid in the second

step.

• For a typical weak polyprotic acid:

K

a1

>

K

a2

>

K

a3
(75)

Section 14.7

Polyprotic Acids

Exercise

Calculate the pH of a 1.00 M solution of H3PO4, given:

Ka1 = 7.5 x 10-3

Ka2 = 6.2 x 10-8

Ka3 = 4.8 x 10-13

(76)

Section 14.7

Polyprotic Acids

Extremely long, challenging question:

Calculate the equilibrium concentration of PO43- in

a 1.00 M solution of H3PO4.

Ka1 = 7.5 x 10-3

Ka2 = 6.2 x 10-8

Ka3 = 4.8 x 10-13

(77)

Section 14.8

Acid–Base Properties of Salts

Salts

Ionic compounds, when dissolved in

(78)

Section 14.8

Acid–Base Properties of Salts

Salts

The salt of a strong acid and a strong

base gives a neutral solution.

(79)

Section 14.8

Acid–Base Properties of Salts

Salts

A basic solution is formed if the anion

of the salt is the conjugate base of a

weak acid.

NaF, KC

2

H

3

O

2

K

b

=

K

w

/

K

a
(80)

Section 14.8

Acid–Base Properties of Salts

Salts

An acidic solution is formed if the cation of

the salt is the conjugate acid of a weak base.

NH

4

Cl

K

a

=

K

w

/

K

b
(81)

Section 14.8

Acid–Base Properties of Salts

Cation

Anion

Acidic

or basic

Example

from

strong

base

from strong

acid

neutral

NaCl

from

strong

base

from weak

acid

basic

NaF

from weak

base

from strong

acid

acidic

NH

4

Cl

from weak

base

from weak

acid

depends

on

K

a

&

K

b
(82)

Section 14.8

Acid–Base Properties of Salts

Qualitative prediction of pH of salt solutions (from weak parents)

Let’s say a salt “came from” an acid and a base, like NaF “came from” HF and NaOH.

(83)

Section 14.8

Acid–Base Properties of Salts

pH of a salt of a weak acid

Ex. 4 Find the pH of a 0.30 M solution of NaClO.

Note: NaClO is not an acid or base, but it is the salt of a weak acid, HClO. A hydrolysis reaction occurs

when it is placed in water, with the salt dissolving in water and leading to changes in [H+] and [OH-].

Step 1: NaClO(s) Na+

(84)

Section 14.8

Acid–Base Properties of Salts

Kb = [HClO] [OH–]

[ClO-]

Step 2: ClO- + H

2O HClO + OH–

Initial 0.30 M 0 ~ 0

Change - x + x + x

Equilibrium 0.30 - x x x

2.9 x 10–7 = x2

0.30 - x

Use Kb = 1 x 10-14

Ka for HClO

See p. A22.

(85)

Section 14.8

Acid–Base Properties of Salts

2.9 x 10–7 = x2

0.30 - x

x2 = (0.30) (2.9 x 10–7)

x = √(0.30) (2.9 x 10–7)

x = √8.7 x 10–8

x = 2.9 x 10–4

[OH-] = 2.9 x 10–4

(86)

Section 14.8

Acid–Base Properties of Salts

Exercise… depending on time, let’s skip ahead to slide #90.

HC2H3O2 Ka = 1.8 x 10-5

HCN Ka = 6.2 x 10-10

Calculate the Kb values for: C2H3O2− and CN−

(87)

Section 14.8

Acid–Base Properties of Salts

Concept Check

Arrange the following 1.0 M solutions from lowest to highest pH.

HBr NaOH NH4Cl NaCN NH3 HCN NaCl HF

Justify your answer.

(88)

Section 14.8

Acid–Base Properties of Salts

Concept Check

Consider a 0.30 M solution of NaF. The Ka for HF is 7.2 x 10-4.

What are the major species?

(89)

Section 14.8

Acid–Base Properties of Salts

• Why isn’t NaF considered a major species? • What are the possibilities for the dominant

(90)

Section 14.8

Acid–Base Properties of Salts

The possibilities for the dominant reactions are:

1. F–(aq) + H

2O(l) HF(aq) + OH–(aq)

2. H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

3. Na+(aq) + H

2O(l) NaOH + H+(aq)

(91)

Section 14.8

Acid–Base Properties of Salts

Let’s Think About It…

• How do we decide which reaction controls the pH?

F–(aq) + H

2O(l) HF(aq) + OH–(aq)

H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

(92)

Section 14.8

Acid–Base Properties of Salts

Ex. 1

Calculate the pH of a 0.30 M solution of NaC2H3O2.

(93)

Section 14.8

Acid–Base Properties of Salts

(Notice it’s like example 4 yesterday…)

1. NaC

2

H

3

O

2(s)

Na

+(aq)

+ C

2

H

3

O

2-(aq)

2. C

2

H

3

O

2-

+ H

2

O



HC

2

H

3

O

2

+ OH

-3. K

b

= K

w

/K

a

= 1.00 x 10

-14

/1.8 x 10

-5

= 5.6 x 10

-10
(94)

Section 14.8

Acid–Base Properties of Salts

Find the pH and the conc of all species at

equilibrium for a 0.50 M solution of CH3NH3Cl. Kb for CH3NH2 = 4.4 x 10-4

-11

(95)

Section 14.9

The Effect of Structure on Acid–Base Properties

Models of acids and bases

• Two factors for acidity in binary compounds:

(96)

Section 14.9

The Effect of Structure on Acid–Base Properties

(97)

Section 14.9

The Effect of Structure on Acid–Base Properties

Oxyacids

• … contain the group H–O–X.

• For a given series, the acid strength

increases with an increase in the number

of oxygen atoms attached to the central

atom.

• The greater the ability of X to draw

(98)

Section 14.9

The Effect of Structure on Acid–Base Properties

Several series of

oxyacids and

(99)

Section 14.9

The Effect of Structure on Acid–Base Properties

Comparison of electronegativity of X and Ka

(100)

Section 14.9

The Effect of Structure on Acid–Base Properties

End of lesson

(101)

Section 14.10

Acid–Base Properties of Oxides

Oxides

Objectives: Students will learn about

acid-base properties of oxides, and use a

(102)

Section 14.10

Acid–Base Properties of Oxides

Oxides

• Acidic oxides (acid anhydrides):

• The O - X bond is strong and covalent. (X

is a nonmetal.)

SO

2

, NO

2

, CO

2
(103)

Section 14.10

Acid–Base Properties of Oxides

Oxides

• Basic oxides (basic anhydrides):

• O - X bond is ionic. (X is a metal.)

K

2

O, CaO

(104)

Section 14.11

The Lewis Acid–Base Model

Lewis acids and bases (Not on AP Exam)

• Lewis acid: electron pair

acceptor

• Lewis base: electron pair

donor

O

H

H

O

H

H

Al

Al

3+

+ 6

6

(105)

Section 14.11

The Lewis Acid–Base Model

(106)

Section 14.11

The Lewis Acid–Base Model

Review the main steps for solving

acid-base equilibrium problems, p. 666 – 667.

(107)

Section 14.11

The Lewis Acid–Base Model

End of lesson

• See homework on board.

References

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