Ap chem unit 14 presentation part 1

AP CHEMIS

TRY

UNIT 1

4

WH

AT

AR

E A

CI D

S A

ND

BA

SE

S?

UNIT 14: ACIDS AND BASES

• The Nature of Acids and Bases• Acid Strength• The pH Scale• Calculating the pH of Strong Acid Solutions• Calculating the pH of Weak Acid Solutions• Bases• Polyprotic Acids• Acid-Base Properties of Salts• The Effect of Structure on Acid-Base Properties• Acid-Base Properties of Oxides• The Lewis Acid-Base Model• Strategy for Solving Acid-Base Problems

THE

NATURE

OF ACID

S AND

BASES

14.1

ARRHENIUS CONCEPT

The first person to recognize the essential nature of acids and bases was Svante Arrhenius. He stated that acids produce hydrogen ions in aqueous solutions, while bases produce hydroxide ion.

• Major step in quantifying acid-base chemistry.

• Concept is very limited because it applies to aqueous solutions only and it only allows for only one kind of base – the hydroxide ion.

BRONSTED-LOWRY MODEL

A more general definition of acids and bases was suggested by the Danish chemist, Bronsted, and the English chemist Lowry. An acid is defined as a proton (H+) donor, and a base is a proton acceptor.

• includes reactions of gases and not just aqueous reactions.

WATER AS A BASE

Water can act as a base because the water molecule has two unshared electrons pairs that allow a covalent bond with a proton (H+).

• When water accepts a proton, H3O+, hydronium is formed.

• HA is the acid (proton donor) and H2O is the base (proton acceptor).

CONJUGATE ACID-BASE PAIR

A conjugate acid-base pair consists of two substances that are related to each other by the donation and acceptance of a proton.

• H3O+ is the conjugate acid. The result of the accepted proton.

• A- is the conjugate base. The result of the donated proton.

• These two components are considered conjugate acid base pairs.


H2O and A-, the base and conjugate base are both competing for a proton.

• If H2O is a stronger base, the reaction will lie to the right because water will accept the proton before A- will.


H2O and A-, the base and conjugate base are both competing for a proton.

• If A- is a stronger base, the reaction will lie to the left because A- will accept the proton before H2O will.

ACID DISSOCIATION CONSTANT

The equilibrium expression for an acid-base reaction gives Ka, the acid dissociation constant:

**Note that H2O is not in the expression and H+ is commonly substituted for hydronium.

ACID DISSOCIATION CONSTANT

Strong acids have large dissociation constants.

Weak acids have small dissociation constants.

PRACTICE PROBLEM #1

Write the simple dissociation (ionization) reaction (omitting water) for each of the following acids.

a. hydrochloric acid

b. acetic acid

c. the ammonium ion

d. the anilinium ion (C6H5NH3+)

e. the hydrated aluminum (III) ion [Al(H2O)6]3+

ACID S

TREN

GTH

14.2

ACID STRENGTH

The strength of an acid is defined by the equilibrium position of its dissociation reaction.

• A strong acid has an equilibrium that lies to the right. •Most HA has dissociated.• A strong acid produces a weak conjugate

base.•Water is the stronger base

ACID STRENGTH

The strength of an acid is defined by the equilibrium position of its dissociation reaction.

• A weak acid has an equilibrium that lies to the left. •Most HA is still present.• A weak acid produces a strong conjugate

base.•Water is the weaker base

ACID STRENGTH SUMMARY

Acid strength and conjugate base strength are indirectly related.

COMMON STRONG ACIDS

1. HCl

2.HNO3

3.HClO4

4. H2SO4

Sulfuric acid is a diprotic acid, an acid with two acidic protons. H2SO4 is a strong acid and dissociates completely, but HSO4

- is a weak acid.

COMMON WEAK OXYACIDS

1.H3PO4

2.HNO2

3.HOCl• Most acids are oxyacids, in which

the acidic proton is attached to an oxygen atom.

ORGANIC ACIDS

Organic acids are those with a carbon atom “backbone”, commonly contain the carboxyl group:

Acids of this type are usually weak.

1. acetic acid (HC2H3O2 or CH3COOH)

2. benzoic acid (C6H5COOH)

*note: the rest of the hydrogens are not acidic

COMMON MONOPROTIC ACIDS

PRACTICE PROBLEM #2

Using Table 14.2, arrange the following species according to their strengths as bases: H2O, F-, Cl-, NO2

-, and CN-.

Cl- < H2O < F- < NO2-, < CN-

WATER AS AN ACID AND A BASE

A substance that can behave either as an acid or as a base is called amphoteric.

• Water is the most common amphoteric substance by autoionization.

WATER AS AN ACID AND A BASE

The equilibrium expression for this reaction is the dissociation constant of water, also called the ion-product constant.

• At 25°C, the ion-product constant, Kw, = 1.0 x 10-14

• [H+] = [OH-]= 1.0 x 10-7 M

USING THE ION-PRODUCT CONSTANT

It is important to recognize the meaning of Ks. In any aqueous solution at 25°C, no matter what it contains, the product of [H+] and [OH-] must always equal 1.0 x 10-14.

• [H+] = [OH-] : neutral solution• [H+] > [OH-] : acidic solution• [H+] < [OH-] : basic solution

PRACTICE PROBLEM #3

Calculate [H+] or [OH-] as required for each of the following solutions at 25°C, and state whether the solution is neutral, acidic, or basic.

a. 1.0 x 10-5 M OH-

b. 1.0 x 10-7 M OH-

c. 10.0 M H+

a. basic b. neutral c. acidic

PRACTICE PROBLEM #4

At 60°C, the value of Kw is 1 x 10-13.

a.Using Le Chatelier’s principle, predict whether the below is exothermic or endothermic.

b.Calculate [H+] and [OH-] in a neutral solution at 60°C.

Kw increases with temperature – endothermic

[H+] =[OH-] = 3 x 10-7 M

THE

PH S

CALE

14.3

PH

The pH scale represents solution acidity with a log based scale.

• [H+] = 1.0 x 10-7

• pH = 7.00

*The number of decimal places in the log is equal to the number of sig figs in the original number

PH

Because the pH scale is based on a scale of 10, the pH changes by 1 for every power of 10 change in [H+]

• example: a solution of pH = 3 has 10 times the concentration of that a solution of pH 4 and 100 times the concentration of a solution of pH 5.

POH

The pOH scale is similar to the pH scale

• [OH-] = 1.0 x 10-7

• pOH = 7.00

*Because the ion product constant of water, pOH + pH always = 14.

PRACTICE PROBLEM #5

Calculate pH and POH for each of the following solutions at 25°C.

a.1.0 x 10-3 M OH-

b.1.0 M OH-

a. pH=11.00, pOH=3.00 b. pH=0.00, pOH=14.00

PRACTICE PROBLEM #6

The pH of a sample of human blood was measured to be 7.41 at 25°C. Calculate pOH, [H+], and [OH-] for the sample.

pOH=6.59, [H+]= 3.9x10-8, [OH-]= 2.6x10-7M

CALCULA

TING T

HE PH

OF

STRONG A

CID S

OLUTI

ONS

14.4

SOLVING ACID-BASE PROBLEMS

Strategies:

1. Think chemistry: Focus on the solution components and their reactions. Choose the reaction that is most important.

2. Be systematic: Acid-base problems require a step by step approach.

3. Be flexible: Treat each problem as a separate entity. Look for both similarities and the differences to other problems.

SOLVING ACID-BASE PROBLEMS

Strategies:

4. Be patient: The complete solution to a complicated problem cannot be seen immediately in all its detail. Pick the problem apart into its workable steps.

5. Be confident: Do not rely on memorizing solutions to problems. Understand and think, do not memorize.

CALCULATING THE PH OF STRONG ACID SOLUTIONSWhen calculating acid-base equilibria, focus

on the main solution components:• example: 1.0 M HCl contains virtually no

HCl molecules. Because HCl is a strong acid, it is completely dissociated. The main solution components are H+, Cl-, and H2O.

• The first step of solving acid-base problems is the writing of the major species in the solution.

PRACTICE PROBLEM #7, PART 1

Calculate the pH of 0.10 M HNO3.

major species:

H+, NO-3, and H2O

concentration of species contributing:

H+ = 0.10 M • H+ from autoionization of water is

negligible compared to H+ from HNO3.

pH

pH = 1.00

PRACTICE PROBLEM #7 PART 2

Calculate the pH of 1.0 x 10-10 M HCl.major species:

H+, Cl-, and H2O

concentration of species contributing: H+ from HCl is negligible.H+ from autoionization of water is

most important. pH

pH = 7.00

CALCULA

TING T

HE PH

OF

WEAK A

CID S

OLUTI

ONS

14.5

CALCULATING THE PH OF A WEAK ACID

We will walk through a systematic approach to solving pH for a weak acid:

• Look for major species.• Narrow down contributing species and its

concentration. Write balanced equation• Write equilibrium expression.• ICE• Use approximations when able• Check validity


What is the pH of a 1.00 M solution of HF, Ka= 7.2x10-4

Major Species:

HF and H2O because both have small dissociation constants.

Concentration of species contributing:

H+ from HF is far more significant than H+ from H2O

Ka = 7.2x10-4 vs Kw= 1.0x10-14



strategy:

pH [H+] Ka

initial [HF] = 1.0 M Equilibrium [H+] = ?

ICE!



ICE:Initial Change Equilibri

umH+

F-

HF



ICE:Initial Change Equilibri

umH+ 0 x xF- 0 x xHF 1.0 -x 1.0-x



• Since Ka for HF is so small, HF will dissociate only slightly, and x is expected to be small. If x is very small compared to concentration (2-3+ exponents), the denominator can be simplified.



The validity of this answer needs to be checked due to the approximation made in the calculations. Typically Ka values are known to an accuracy of ±5%. We will use this same accuracy for answers to be considered valid.



Checking validity:

x=2.7x10-2, [HF] = 1.00 M

x is valid. [H+] = 2.7x10-2, pH=1.57

PRACTICE PROBLEM #8

The hypochlorite ion (OCl-) is a strong oxidizing agent often found in household bleaches and disinfectants. It is also the active ingredient that forms when swimming pool water is treated with chlorine. In addition to its oxidizing abilities, the hypochlorite ion has a relatively high affinity for protons (it is much stronger base than Cl-, for example) and forms the weakly acidic hypochlorous acid (HOCl, Ka=3.5x10-8). Calculate the pH of a 0.100M aqueous solution of hypochlorous acid.

PRACTICE PROBLEM #8

Major species:

HOCl and H2O

Narrow down contributing species and its concentration. Write balanced equation

HOCl Ka=3.5x10-8 significant ≅ 0.100M

H2O Kw=1.0x10-14 negligible

PRACTICE PROBLEM #8

Write equilibrium expression.

ICE Initial Change Equilibrium

H+

OCl-

HOCl

PRACTICE PROBLEM #8

Use approximations when able:

x≅5.9x10-5

PRACTICE PROBLEM #8

Check validity

pH:

[H+] = 5.7x10-8, pH=4.23

PRACTICE PROBLEM #9

Calculate the pH of a solution that contains 1.00M HCN (Ka=6.2 x 10-

10) and 5.00 M HNO2 (Ka=4.0 x 10-

4). Also calculate the concentration of cyanide ion (CN-) in this solution at equilibrium.

PRACTICE PROBLEM #9

Major species:


PRACTICE PROBLEM #9

Write equilibrium expression.

ICE Initial Change Equilibrium

H+

NO2-

HNO2

PRACTICE PROBLEM #9


Check validity.

pH= 1.35 But wait….what about CN-?

PRACTICE PROBLEM #9

Calculate the pH of a solution that contains 1.00M HCN (Ka=6.2 x 10-10) and 5.00 M HNO2 (Ka=4.0 x 10-4). Also calculate the concentration of cyanide ion (CN-) in this solution at equilibrium.

We also want to calculate the CN- concentration from the dissociation of HCN

HCN(a)<-> H+(aq)+ CN-

(aq) Ka=6.2x10-

10

PRACTICE PROBLEM #9

The molarity of the HCN is known and the [H+] is equal to the prominent contribution of HNO2. The source of the H+ ions is not important in equilibrium.

[CN-]=1.4x10-8M

PERCENT DISSOCIATION

The percent dissociation is:

For a given weak acid, the percent dissociation increases as the acid becomes more dilute.

PRACTICE PROBLEM #10

Calculate the percent dissociation of acetic acid (Ka = 1.8 x 10-5) in each of the following solutions.

a.1.00 M HC2H3O2

Major species:

HC2H3O2 and H2O


HC2H3O2 Ka = 1.8 x 10-5, H2O Kw = 1.0 x 10-14

HC2H3O2(aq) <-> H+(aq) + C2H3O2

-(aq)



a.1.00 M HC2H3O2

ICEInitial Change Equilibri

umH+

C2H3O2-

HC2H3O

2



a.1.00 M HC2H3O2


Apply Equation



b. 0.100 M HC2H3O2

Major species:

HC2H3O2 and H2O


HC2H3O2 Ka = 1.8 x 10-5, H2O Kw = 1.0 x 10-14

HC2H3O2(aq) <-> H+(aq) + C2H3O2

-(aq)



b. 0.100 M HC2H3O2

ICEInitial Change Equilibri

umH+

C2H3O2-

HC2H3O

2



b. 0.100 M HC2H3O2


Apply Equation

PERCENT DISSOCIATION

The previous example is a perfect example of the percent dissociation relationship.

• For a given weak acid, the percent dissociation increases as the acid becomes more dilute.


Lactic acid (HC3H5O3) is a waste product that accumulates in muscle tissue during exertion, leading to pain and a feeling of fatigue. In a 0.100 M aqueous solution, lactic acid is 3.7% dissociated. Calculate the value of Ka for this acid.


Major species:

HC3H5O3 and H2O

Narrow down contributing species and its concentration. Write balanced equation HC3H5O3(aq) <-> H+

(aq) + C3H5O3

-(aq)


Percent Dissociation

Substitute in the expression

Ka=1.4x10-4

UNIT 1

4 CONTI

NUED IN

PART

2

Ap chem unit 14 presentation part 1

Education