What are Acids and Bases?AP Chemistry Unit 14: Guided Notes Part 1

Unit 14: Acids and BasesPart 1:

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

Part 2: 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

14.1 The Nature of Acids and Bases

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

Conjugate acid-base pair

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HA(aq)+H2O( l )⇔H3O+(aq)+A

−(aq)

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. is the conjugate base. The result of the donated proton.

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

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

**Note

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 acidB- acetic acidC- the ammonium ionD- the anilinium ion (C6H5NH3

+)E- the hydrated aluminum (III) ion [Al(H2O)6]3+

14.2 Acid Strength

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HA(aq)+H2O( l)⇒H 3O+(aq )+A

−(aq )

Ka=[ H3O

+ ] [ A− ][HA ]

=[H+ ][ A− ][HA ]

Ka=[ H3O

+ ] [ A− ][HA ]

=[H+ ][ A− ][HA ]

Acid StrengthThe 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. o

A strong acid produces a weak conjugate base.o

A weak acid has an equilibrium that lies to the left. o

A weak acid produces a strong conjugate base.o

Acid strength summary

Acid strength and conjugate base strength are indirectly related.

Common Strong Acids1- HCl2- HNO3

3- HClO4

4- H2SO4

Common weak oxyacids

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1- H3PO4

2- HNO2

3- HOCl

Organic acidsOrganic 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:

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-.

Water as an Acid and a Base

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

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

Using the ion-product constantIt 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-] : [H+] > [OH-] : [H+] < [OH-] :

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+

Practice problem #4 At 60°C, the value of Kw is 1 x 10-13.

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H2O(l)+H2O( l)⇔H3O+(aq)+OH

−(aq )

Kw=[H 3O+ ][OH− ]=[ H+ ] [OH− ]

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.

14.3 The pH Scale

pHThe pH scale represents solution acidity with a log based

scale.

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

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

o 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.

pOHThe pOH scale is similar to the pH scale

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

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pH=−log [H+ ]

pOH=−log [OH− ]

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-

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.

14.4 Calculating the pH of strong acid solutions

Solving acid-base problemsStrategies: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.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.

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Practice problem #7, part 1 Calculate the pH of 0.10 M HNO3.major species:

concentration of species contributing:

Practice problem #7, part 2 Calculate the pH of 1.0 x 10-10 M HCl. major species:

concentration of species contributing:

14.5 Calculating the Ph of weak acid solutions

Calculating the pH of a weak acidWe 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

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Example: Calculating the pH of a weak acidWhat is the pH of a 1.00 M solution of HF, Ka= 7.2x10-4

Major Species:

Concentration of species contributing:

Initial Change Equilibrium

H+

F-

HF

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

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Ka=7 .2x 10−4=[H+ ][F− ]

[HF ]

Ka=7 .2x 10−4=

[ x ][ x ][1 .0−x ]

1 .0−x≃1 .0

x[HA ]o

x100%

2 .7 x10−2

1 .0Mx 100%=2.7%

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). Caculate the pH of a 0.100M aqueous solution of hypochlorous acid.

Major species:

Narrow down contributing species and its concentration. Write balanced equation

Write equilibrium expression.

ICE

Initial Change Equilibrium

H+

OCl-

HOCl

Use approximations when able:

Check validity

pH:

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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.

Major species:

Narrow down contributing species and its concentration. Write balanced equation

Write equilibrium expression.

ICE

Initial Change Equilibrium

H+

NO2-

HNO2

Use approximations when able:

Check validity.

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But wait….what about CN-? 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

Percent dissociationThe percent dissociation is:

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

Initial Change Equilibrium

H+

C2H3O2-

HC2H3O2

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

b. 0.100 M HC2H3O2

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Ka=6 .2x 10−10=[H+ ][CN− ]

[HCN ]

6 .2x 10−10=[4 .5 x10−2 ][CN− ][1 .00 ]

% dissociation= [ amount dissociated ][ initial concentration ]

x 100%

Initial Change Equilibrium

H+

C2H3O2-

HC2H3O2

Percent dissociationThe previous example is a perfect example of the percent dissociation relationship.

Practice Problem #11 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.

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