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George Mason UniversityGeneral Chemistry 212

Chapter 18 Acid-Base Equilibria

Acknowledgements

Course Text: Chemistry: the Molecular Nature of Matter andChange, 6th ed, 2011, Martin S. Silberberg, McGraw-Hill

The Chemistry 211/212 General Chemistry courses taught at George

Mason are intended for those students enrolled in a science

/engineering oriented curricula, with particular emphasis on

chemistry, biochemistry, and biology The material on these slides is

taken primarily from the course text but the instructor has modified,condensed, or otherwise reorganized selected material.

Additional material from other sources may also be included.

Interpretation of course material to clarify concepts and solutions to

problems is the sole responsibility of this instructor.7/16/2014

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Chapter 18 – Acid-Base Equilibria Acids & Bases in Water

Arrhenius Acid-Base Definition

Bronsted-Lowry Acid-Base Definition

Lewis Acid-Base Definition

Acid Dissociation Constant

Relative Strengths of Acids & Bases Autoionization of Water and the pH Scale

Autoionization and Kw

The pH Scale

Proton Transfer and the Bronsted-Lowry Acid-BaseDefinition

The Conjugate Acid-Base Pair

Met Direction of Acid-Base Reactions7/16/2014 2

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Chapter 18 – Acid-Base Equilibria Solving Problems Involving Weak-Acid Equilibria

Finding Ka Given Concentrations

Finding Concentrations Given Ka

Extent of Acid Dissociation

Polyprotic Acids

Weak Bases and Their Relation to Weak Acids

Ammonia & The Amines

Anions of Weak Acids

The Relation Between Ka & Kb

Molecular Properties and Acid Strength Nonmetal Hydrides

Oxoacids

Acidity of Hydrated Metal Ions

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Chapter 18 – Acid-Base Equilibria Acid-Base Properties of Salt Solutions

Salts That Yield Neutral Solutions

Salts That Yield Acidic Solutions

Salts That Yield Basic Solutions

Salts of Weakly Acidic Cations and Weakly Basic Anions

Salts of Amphoteric Anions

Generalizing the Bronsted-Lowry Concept: The LevelingEffect

Electron-Pair Donation and the Lewis Acid-Base Definition

Molecules as Lewis Acids

Metal Cations as Lewis Acids

Overview of Acid-Base Definitions

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Acid & Bases Antoine Lavoisier was one of the first chemists

to try to explain what makes a substance acidic

In 1777, he proposed that oxygen was anessential element in acids

The actual cause of acidity and basicity was

ultimately explained in terms of the effect thesecompounds have in water by Svante Arrheniusin 1884

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Acids & Bases

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Several concepts of acid-base theory:

The Arrhenius concept

The Bronsted-Lowry concept

The Lewis concept

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Acids & Bases

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According to the Arrhenius concept of acids

and bases An acid is a substance that, whendissolved in water, increases theconcentration of Hydrogen ion, H+(aq)

The H+(aq) ion, called the Hydrogen ion isactually chemically bonded to water, that is,H3O

+, called the Hydronium ion

A base is a substance that whendissolved in water increases theconcentration of the Hydroxide ion,

OH

-

(aq)

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Acids & Bases

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More About the Hydronium Ion

The Hydronium Ion (H3O+) actually exists as a hydrogen

bonded H9O4+ cluster. The formation of Hydronium ions is

a complex process in aqueous solution

The Hydronium ion has trigonal

pyramidal geometry because all three H-

O-H dihedral bond angles are identical

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Acids & Bases

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An Arrhenius acid is any substance which increases the concentration of Hydrogen ions (H+) in solution.

Acids generally have a sour taste

HBr(aq) H+(aq) + Br-(aq)

An Arrhenius base is any substance which increases the concentration of Hydroxide ions (OH-) in solution.

Bases generally have a bitter taste KOH(s) K +(aq) + OH-(aq)

Hydrogen and Hydroxide ions are important in aqueoussolutions because - Water reacts to form both ions

H2O(l) H+(aq) + OH-(aq)

HBr and KOH are examples of a strong acid and astrong base ; strong acids & bases dissociatecompletely

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Acids & Bases Monoprotic acids are those acids that are able to donate

one proton per molecule during the process of

dissociation (sometimes called ionization) as shown below(symbolized by HA):

HA(aq) + H2O(l) ⇌ H3O+(aq) + A −(aq)

Common examples of monoprotic acids in mineral acids

include Hydrochloric Acid (HCl) and Nitric Acid (HNO3) Polyprotic acids are able to donate more than one proton

per acid molecule.

Diprotic acids have two potential protons to donate

H2 A(aq) + H2O(l) ⇌ H3O+(aq) + HA −(aq)

Triprotic acids have three potential protons to donate

H3 A(aq) + H2O(l) ⇌ H3O+(aq) + H2 A −(aq)

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Acids & Bases

Oxoacids

Oxoacids with one oxygen have the structure: HOYHOCl HOBr HOI

Oxoacids with multiple oxygen have the structure:

(OH)m YOn

HOClO3 (HClO4) HOClO2 (HOCLO3)

HOClO (HCLO2) HOCL (HClO)

The more oxygen atoms, the greater the acidity; thusHOCLO3 is a stronger acid than HOCl

Acidity refers to the relative acid strength, i.e. the degreeto which the acid dissociates to form H+ (H3O

+) & OH-

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Bronsted-Lowry Acids & Bases Brønsted-Lowry Concept of Acids and Bases

In the Brønsted-Lowry concept:

An Acid is a species that donates protons

A Base is a species that accepts protons

Acids and bases can be ions as well asmolecular substances

Acid-base reactions are not restricted toaqueous solution

Some species can act as either acids or bases(Amphoteric species) depending on what theother reactant is

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Bronsted-Lowry Acids & Bases Amphoteric Species

A species that can act as either an acid or base is:

Amphoteric Water is an important amphoteric species in the acid-

base properties of aqueous solutions

Water can react as an acid by donating a proton toa base

Water can also react as a base by accepting aproton from an acid

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+ -3 2 4NH (aq) + H O(l) NH (aq) + OH (aq)

H+

+2 3HF(aq) + H O(l) (aq) + H O (aq)F

H+

d d &

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Bronsted-Lowry Acids & Bases

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AcidH+ donor

BaseH+ acceptor

Lone pairbinds H+

Base

H+ acceptor

Acid

H+ donor

Brønsted-Lowry Concept of Acids and Bases

Proton transfer as the essential feature of a

Brønsted-Lowry acid-base reaction

Lone pair

binds H+

A id & B

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Acids & Bases

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Bronsted-Lowery Concept – Conjugate Pairs

In Bronsted theory, acid and base reactants form:

Conjugate Pairs

The acid (HA) donates a proton to water leaving the

conjugate Base (A-)

The base (H2O) accepts a proton to form the conjugate

acid (H3O+)

HA(aq) + H2O(l) ⇌ A−(aq) + H3O+(aq)

Acid Base Conjugate Conjugate

Base Acid

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HNO2(aq) + H2O(l) H3O+ aq) + NO2-(aq)

Conjugate acid-base pairs are shown connected

Every acid has a conjugate base

Every base has a conjugate acid The conjugate base has one fewer H and one more

“minus” charge than the acid

The conjugate acid has one more H and one fewer

minus charge than the base

acid acidbase base

Bronsted-Lowery Acid-Base Conjugate Pairs

BronstedBase

H3O+ is conjugate

acid of base H2O

NO2- is conjugate

base of acid HNO2

Bronsted Acid

B t d L A id & B

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Bronsted-Lowry Acids & Bases Bronsted-Lowery Acid-Base Conjugate Pairs

NH3(aq) + H2O(l) NH4+ aq) + OH-(aq)

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base acidacid base

NH4+ is conjugate

acid of base NH3

OH- is conjugate

base of acid H2O

Bronsted

Base

Bronsted

Acid

Donates

Proton

Accepts

Proton

B t d L A id & B

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H2S + NH3 NH4+

+ HS-

acid base baseacid

HS- is conjugateBase of acid H2S

NH4+ is conjugate

acid of base NH3

Bronsted-Lowery Acid-Base Conjugate Pairs

Bronsted

Base

Bronsted

Acid

Donates

Proton

Accepts

Proton

B t d L A id & B

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Bronsted-Lowry Acids & Bases Bronsted_Lowry Concept – The Leveling Effect

All Bronsted_Lowry acids yield H3O+ ions (Cation)

All Bronsted_Lowry bases yield OH- ions (Anion)

All strong acids are equally strong because all of them form

the strongest acid possible - H3O+

Similarly, all strong bases are equally strong because theyform the strongest base possible - OH-

Strong acids and bases dissociate completely yielding H3O+

and OH-

Any acid stronger than H3O+ simply donates a proton to H2O

Any base stronger than OH- simply accepts proton from H2O

Water exerts a leveling effect on any strong acid orbase by reacting with it to form water ionization

products 7/16/2014 19

P ti P bl

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Practice Problem Identify the conjugate acid-base pairs in the following:

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- -2 - 2-

2 4 3 3 4

H PO (aq) + CO (aq) HCO (aq) + HPO (aq)

- + 2-2 4 4

2- + -

3 3

H PO has one more H than HPO and

CO has one fewer H than HCO

- -

2 4 3

2- 2-

4 3

H PO & HCO are acids

HPO & CO are bases

- 2-

2 4 4H PO / HPO is a conjugate acid -base pair

- 2-

3 3HCO / CO is a conjugate acid -base pair

Acid Conjugate base

Base Conjugate Acid

L i A id & B

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Lewis Acids & Bases Lewis Acids & Bases

Some acid base reactions don’t fit the Bronsted-Lowryor Arrhenius classifications

The Lewis acid-base concepts expands the acid class

Such reactions involve a “sharing” of electron pairsbetween atoms or ions

Lewis Acid – An electron deficient species (Electrophile)

that accepts an electron pair

Lewis Base – An electron rich species (Nucleophile)

that donates an electron pair

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L i A id & B

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Lewis Acids & Bases Lewis Acids & Bases

The product of any Lewis acid-base reaction is called an

“Adduct” , a single species that contains a new covalentbond (shared electron pair)

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+ A + : B A - B

Acid Base

Electrophile Nucleophile Adduct

F

B

F F

H

N

H H

+

F

B

FF

H

N

HH

An acid is an

electron-pair

acceptor

A base is an

electron-pair

donor

Adduct

L i A id & B

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Lewis Acids & Bases Species that do not contain Hydrogen in their formulas, such

as CO2 and Cu++ function as Lewis acids by accepting anelectron pair

The donated proton of a Bronsted-Lowry acid acts as a Lewisacid by accepting an electron pair donated by the base:

The Lewis acids are Yellow and the Lewis bases are Blue infollowing reaction

BF3 + :NH3 BF3NH3 Fe3+ + 6 H2O Fe(H2O)6

3+

HF + H2O F- + H3O+

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+ -

L i A id & B

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Lewis Acids & Bases Solubility Effects

A Lewis acid-base reaction between nonpolar Diethyl

Ether and normally insoluble Aluminum Chloride

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The solubility of Aluminum Chloride in Dimethyl Etherresults from the Oxygen acting as a base by donatingan electron pair to the Aluminum acting as an acidforming a water-soluble polar covalent bond

+

-

Practice Problem

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Practice ProblemThe following shows ball-and-stick models of the reactantsin a Lewis acid-base reaction.

Write the complete equation for the reaction, including theproduct

Identify each reactant as a Lewis acid or Lewis base

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AlCl3 + :NH3 AlCl3:NH3

Lewis acid Lewis base Adduct

Acceptse- pair

Donatese- pair

Acids & Bases (Summary)

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Acids & Bases (Summary)

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Arrhenius acid concept

acid = proton (H+) producer; base = OH- producer

Formation of Water (H2O) from H+ & OH-

Bronsted-Lowry concept

Stronger Acid (HA) transfers proton to a stronger base (H2O)

to form weaker acid H3O+) and weaker base (A-)

Stronger base (NH3) accepts proton from stronger acid (H

2O)

Lewis Concept

Donation and acceptance of an electron pair to form

a covalent bond in an adduct

+

2 3

-HA(aq) + H O(l) H O (aq) + A (aq)

+ -HCl(aq) H (aq) + Cl (aq)

+ -

3 2 4NH (aq) + H O(l) NH (aq) + OH (aq)

+ -KOH(s) K (aq) + OH (aq)

Acids & Bases

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Acids & Bases Self Ionization of Water

Pure water undergoes auto-ionization to produceHydronium and Hydroxide ions

2 H2O(l) H3O+(aq) + OH-(aq)

The extent of this process is described by an auto-ionization (ion-product) constant, K w

The concentrations of Hydronium and Hydroxide ions inany aqueous solution must obey the auto-ionizationequilibrium

Classification of solutions using K w:

Acidic: [H3O+] > [OH-]

Basic: [H3O+] < [OH-]

Neutral: [H3O+

] = [OH-

]7/16/2014 27

+ - -14w 3K = [H O ][OH ] = 1.0 10

Acids & Bases

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Acids & Bases Because of the auto-ionization constant of water, the

amounts of Hydronium and Hydroxide ions are alwaysrelated

For example, pure water at 25 oC,

The Definition of pH The amount of Hydronium ion (H3O

+) in solution isoften described by the pH of the solution

The Definition of pOH

Hydroxide (OH-) can be expressed as pOH

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+3pH = - log[H O ]

+ - -14w 3K = [H O ][OH ] = 1.0 10

+ -pH

3[H O ] = 1 10

+ - -73[H O) ] = [OH ] = 1.0 10

-10pOH = - log [OH ]

- -pOH[OH ] = 1 x 10

Acids & Bases

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Acids & Bases

pH scale ranges from 0 to 14

The pH of pure water = -log(1.00 X 10-7) = 7.00

pH = 0.00 - 6.999+ = acidic [H3O+] > [OH-]

pH = 7.00 = neutral [H3O+] = [OH-]

pH = 7.00+ - 14.00 = basic [H3O+] < [OH-]

Acidic solutions have a lower pH (higher [H3O+] and

a higher pOH (lower [OH-] than basic solutions

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Acids & Bases

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Acids & Bases Relations among pH, pOH, and pK w

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owpK = pH + pOH = 14 (at 25 C)

+ - -14w 3K = [H O ][OH ] = 1.0 10

+ - -14w 3-log(K ) = - log([H O ][OH ]) = - log(1 10 )

+ - -14w 3-log(K ) = - log[H O ] - log[OH ] = - log(1 10 )

Equilibrium & Acid Base Conjugates

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Equilibrium & Acid-Base Conjugates

Equilibrium Constants (K) for acids & bases can also beexpressed as pK

Reaction of an Acid (HA) with water as a base

Reaction of a Base (A -) with water as an acid

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a 10 apK = - log [K ]

+ -2 3HA(aq) + H O(l) H O (aq) + A

+ -3

a

[H O ][A ]K =

[HA]

-

b -

[HA][OH ]K =

[A ]

- -2A (aq) + H O(l) HA(aq) + OH (aq)

]b 10 bpK = - log [K

Acid Base Conjugate Acid Conjugate Base

Base Acid Conjugate Acid Conjugate Base


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Equilibrium & Acid-Base Conjugates Relationship between:

K a and HA

K b and A -

Setup two dissociation reactions:

The sum of the two dissociation reactions is theautoionization of water

The overall equilibrium constant is the product of theindividual equilibrium constants

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+ -

2 3HA + H O H O + A

- -

2A + H O HA + OH+ -

2 32H O H O + OH (autoionization)

a b w K × K = K

+ - -

3

-

[H O ][A ] [HA][OH ] ×

[HA] [A ]+ -

3= [H O ][OH ]


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Equilibrium & Acid-Base Conjugates

From this relationship, K a of the acid in a conjugate paircan be computed from K b and vice versa

Reference tables typically have K a & K b values formolecular species, but not for ions such as F- or

CH3NH3

+

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w a bK = K × K

w a b-log(K ) = - log([K ][K ])

w a b-log(K ) = - log[K ] - log[K ]

o

w a bpK = pK + pK = 14 (at 25 C)

owpK = pH + pOH = 14 (at 25 C)


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Equilibrium & Acid-Base Conjugates Acid strength [H3O

+] increases with increasing value of K a

Base strength [OH-] increases with increasing value of k b

A low pK corresponds to a high value of K

A reaction that reaches equilibrium with mostly productspresent (proceeds to far right) has a “low pK” (high K)

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+ -3

a

[H O ][A ]K =

[HA]

-

b -

[HA][OH ]K =

[A ]

a 10 apK = - log [K ]

]b 10 bpK = - log [K

Practice Problem

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Practice Problem Ex. Find K b value for F-

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-4

aK (HF) = 6.8 10 (K values of compounds from table)

a b w

-K (HF) K (F ) = K

-14

-11w

b -4

a

- K 1.0 10

K (F ) = = = 1.5 10K (HF) 6.8 10

2

+ -

3HF + H O H O + F

Acids & Bases

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Acids & Bases

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The Relationship Between K a and pK a

Acid Name (Formula) K a at 25oC pK a

Hydrogen Sulfate ion (HSO4-) 1.0x10-2

Nitrous Acid (HNO2)

Acetic Acid (CH3COOH)

Hypobromous Acid (HBrO)

Phenol (C6H5OH)

7.1x10-4

1.8x10-5

2.3x10-9

1.0x10-10

1.99

3.15

4.74

8.64

10.00

A c i d

S t r e n g t h

Acids & Bases

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Acids & Bases

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The pH Scale

Acids & Bases

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Acids & Bases Strong Acids & Bases

Strong acids and bases dissociate completely in

aqueous solutionsStrong acids: HCl, HBr, HI, HNO3, H2SO4, HClO4

Strong bases: all Group I & II Hydroxides

NaOH, KOH, Ca(OH)2, Mg(OH)2

The term strong has nothing to do with theconcentration of the acid or base, but rather theextent of dissociation

In the case of HCl, the dissociation lies very far to the

right because the conjugate base (Cl-) is extremelyweak, much weaker than water

HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)

H2O is a much stronger base than Cl- and “out

competes it” for available protons 7/16/2014 38

Acids & Bases

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Acids & Bases

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Activity Series

Strengths of

acids and bases

Strengthdetermined by K,not concentration

Practice Problem

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Practice Problem

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A strong monoprotic acid is dissolved in a beaker ofwater. Which of the following pictures best

represents the acid solution.

Ans: B

A B C

A solution of astrong acid containsequal amounts ofH3O

+ and the acid

anion, such as a “Cl-”The monoproticacid is completelydissociated

Hydrohalic Acids – Weak vs Strong

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Hydrohalic Acids Weak vs Strong The reason Hydrofluoric acid is weak relative to

the other Hydrohallic Acids (I, Br, Cl) is quite

complicated The material being somewhat beyond the scope

of the discussions we have in the chem 211 &chem 212 courses

It starts out with the short length of the H - Fbond and the very high electronegativity of theFluoride ion

When you take into account the Enthalpy,Entropy, and Free Energy properties (Chapter19), the numbers clearly show a distinctdifference between HF and the other Hydrohalic

acids7/16/2014 41


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Hydrohalic Acids Weak vs Strong These properties can be summarized as follows:

Very strong Hydrogen Bonding between theun-ionized Hydrogen Fluoride (HF) moleculesand water molecules

This energy required to break the H-F bond,

is much greater then for the bonds of theother Hydrohalic Acid bonds

A large decrease in Entropy when theHydrogen Fluoride molecules react with

water

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Hydrohalic Acids Weak vs Strong This is particularly noticeable with hydrogen

fluoride because the attractiveness of the

very small fluoride ions produced imposes alot of order on the surrounding watermolecules, and also on nearby hydroxoniumions

The effect falls as the halide ions get bigger

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Hydrohalic Acids Weak vs Strong Note the values below. Hydrofluoric acid clearly

stands out

The Ka values clearly suggest a weak acid

H TS G Ka

(kJ mol-1) (kJ mol-1) (kJ mol-1) (mol dm-3)

HF -13 -29 +16 1.6 x 10-3

HCl -59 -13 -46 1.2 x 10+8

HBr -63 -4 -59 2.2 x 10

+10

HI -57 +4 -61 5.0 x 10+10

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Practice Problem

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Practice ProblemWhat are the concentrations of Hydronium Ions (H3O

+)

and Hydroxide (OH-) ions in 1.2 M HBr?

HBr + H2O H3O+ + Br-

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- -15[OH ] = 8.3 x 10 M

[HBr] = 1.2 mol / L

+3[H O ] = 1.2 mol / L

+ - -14w 3K = [H O ][OH ] = 1 x 10

-14- w

+3

K 1 x 10 mol / L[OH ] = =1.2 mol / L[H O ]

HBr is a strong acid

and is completelydissociated

Autoionization of Water

Practice Problem

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Practice ProblemWhat are the concentrations of Hydronium Ions (H3O

+)

and Hydroxide (OH-) ions in 0.085 M Ca(OH)2?

Ca(OH)2 Ca+2 + 2OH-

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+ -14

3[H O ] = 5.8 x 10 mol / L

2[Ca(OH)] = 0.085 mol / L

-[OH ] = 2 0.085 = 0.17 mol / L

+ - -14w 3K = [H O ][OH ] = 1 10

-14+ w

3 -

K 1 10[H O ] = =

0.17 mol / L[OH ]

Ca(OH)2 is a strongbase and is completelydissociated

Autoionization of Water

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Practice Problem

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Practice ProblemWhat is the pOH of the following solution:?

[H3O+] = 9.3 x 10-4 M

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+ -w 3K = [H O ][OH ]

-14- w

+ -4

3

K 1.0 10[OH ] = = = 1.0753 M

[H O ] 9.3 10

- -1110 10pOH = -log [OH ] = -log (1.075268 10 )

-1110 10pOH = -(log 1.075268 + log 10 )

pOH = -(0.031516 - 11.0) = 11.0 - 0.031516

pOH = 11

Practice Problem

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Practice ProblemWhat is the pH of a solution containing 0.00256 M HCl?

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+ -2 3HCl + H O H O + Cl

+ -3Thus : [HCl] = [H O ] = [Cl ] = 0.00256M

Hydrochloric Acid (HCl) is a strong acid

It dissociates completely in aqueous solution

+3pH = - log[H O ]

pH = - log[0.00256]

pH = - (-2.59) = 2.59

Practice Problem

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Practice ProblemWhat is the pH of a solution containing 0.00813 M Ca(OH)2?

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2Calcium Hydroxide (Ca(OH) ) is a strong base

It dissociates completely in aqueous solution

+2 -

2Ca(OH) Ca + 2OH

+ -w 3K = [H O ][OH ]

-14+ -13

3 -

Kw 1.0 x 10[H O ] = = = 6.15006 x 10 M

0.01626[OH ]

+2 -2Thus : [Ca(OH) ] = [Ca ] = 2[OH ] = 0.00813 M

-[OH ] = 2× 0.00813 = 0.01626 M

+3pH = - log[H O ]

-13pH = - log[6.15006 x 10 ]

pH = 12.1

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Weak Acid Equilibria Problems

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Weak Acid Equilibria Problems Construct a “Reaction” table

Make assumptions that simplify calculations

Usually, “x” is very small relative to initialconcentration of HA

Assume [HA]init ≈ [HA]eq & Apply 5% rule

If “x” < 5% [HA]init ; assumption is valid

Substitute values into K a expression, solve for “x”

The Notation System

Brackets – [HA]; [HA]diss; [H3O+]

indicate “Molar” concentrations

Bracket with “no” subscript refers to molar

concentration of species at equilibrium7/16/2014

Weak Acid Equilibria Assumptions

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Weak Acid Equilibria Assumptions Assumptions

The [H3O+] from the “autoionization” of water is

“Negligible”

It is much smaller that the [H3O+] from the

dissociation of HA

[H3O+] = [H3O

+]from HA + [H3O+]from H2O ≈ [H3O

+]from HA

A weak acid has a small K a

Therefore, the change in acid concentration can beneglected in the calculation of the equilibriumconcentration of the acid

[HA] = [HA]init - [HA]dissoc ≈ [HA]init7/16/2014 54

+ -2 3HA(aq) + H O(l) H O (aq) + A

Weak Acid Equilibria Assumptions

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Weak Acid Equilibria Assumptions Assumption Criteria - 2 approaches

7/16/2014 55

initinit

[HA]if < 400, the assumption [HA] [HA] is not justified

Kc Neglecting x introduces an error > 5%

initinit

[HA]if > 400, the assumption [HA] [Ha] is justified

Kc Neglecting x introduces an error < 5%

For [HA]diss relative to [HA]init

diss init initif [HA] < 5% [HA] , the assumption ([HA] [HA] ) is valid

+ -2 3HA(aq) + H O(l) H O (aq) + A

For Kc relative to [HA]init (or Kp relative to P A(init))

+ -3H O A

K =HA

Practice Problem

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Practice ProblemWhat is the K a of a 0.12 M solution of Phenylacetic acid (HPAc)

with a pH of 2.62

HPAc(aq) + H2O(l) [H3O+

](aq) + PAc-

(aq)[H3O

+]from HPAc + [H3O+]from H2O ≈ [H3O

+]from HPAc

pH = - log[H3O+]

[H3O+] = 10-pH = 10-2.62 = 2.4 x 10-3 M

7/16/2014 56

Concentration HPAc(aq) + H2O(l) H3O+ + PAc-

Initial Ci 0.12 - 0 0

- X - + X +X

Equilibrium C f 0.12 - X - X X

Final Cf 0.12 2.4x10-3 2.4x10-3

[H3O+] = x = 2.4 x 10-3 M = [PAc]

HPAc is weak acid HPAc = 0.12 M – x ≈ 0.12 M+ -3 -3

3a

-[H O ][PAc ] (2.4 x 10 )(2.4 x 10 )K =

[HPAc] 0.12

-5aK = 4.8 x 10

Practice Problem

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Practice ProblemWhat is the [H3O

+] of a 0.10 M solution of Propanoic Acid(HPr)? K a = 1.3 x 10-5

7/16/2014 57

+-53

a

-[H O ][Pr ]K = = 1.3 x 10

[HPr]

HPr is a weak acid = [HPr]init = 0.10 MLet x = [HPr]diss = [H3O

+]from HPr = [Pr -]from HPr

Concentration HPr(aq) + H2O(l) H3O+ + Pr -


- X - + X +XEquilibrium C f 0.10- X - X X

Final Cf 0.10 X X

Assumption: K a is very small and x is small compared to[HPr]

init

HPr = 0.10 M – x ≈ 0.10 M Con’t

+ -2 3HPr(aq) + H O H O (aq) + Pr (Aq)

Practice Problem (con’t)

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Practice Problem (con t) Substituting into the K a expression and solving for x

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-3 + -3 dissx = 1.1 x 10 M = [H O ] = [Pr ] = [HPr]

+

-53a

-[H O ][Pr ] (x)(x)

K = = 1.3 10[HPr] 0.10

-5x = 0.10 1.3 10

Checking the Assumption: [HBr]diss < 5% [HBr]init

1.1% < 5% Assumption Valid

-3diss

dissinit

[HPr] 1.1 10 M

For [HPr] : = x 100 = 1.1%[HPr] 0.10M

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Weak Acid Equilibria

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Weak Acid Equilibria Polyprotic Acids

Acids with more than one ionizable proton arepolyprotic acids.

One proton at a time dissociates from the acidmolecule

Each dissociation step has a different K aH3PO4(aq) + H2O(l) H2PO4

- + H3O+(aq)

H2PO4-(aq) + H2O(l) HPO4

-2 + H3O+(aq)

HPO4-2(aq) + H2O(l) PO4

-3 + H3O+(aq)

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- +-32 4 3

a13 4

[H PO ][H O ]K = = 7.2 x 10

[H PO ]

-2 +-8

4 3a2 -2 4

[HPO ][H O ]K = = 6.3 x 10[H PO ]

-3 +-134 3

a3 -4

[PO ][H O ]K = = 4.2 x 10

[HPO ]


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7/16/2014 61

Successive dissociation equilibrium constants (Ka)for polyprotic acids have significantly lower values

Ka1 >> Ka2 >> Ka3

Practice Problem

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act ce ob e Calculate the following for 0.05 M Ascorbic Acid (H2 Asc):

[H2 Asc], [HAsc-], [Asc2-], pH

K a1 = 1.0 x 10-5 K a2 = 5 x 10-12

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- +2 2 3H Asc(aq) + H O(l) HAsc (aq) + H O (aq)

- 2- +2 3HAsc (aq) + H O(l) Asc (aq) + H O (aq)

-] +-53

a12

[HAsc ][H O ]K = = 1.0 10

[H Asc]

2-] +-123

a2-

[Asc ][H O ]K = = 5.0 10

[HAsc ]

+ + -a1 a2 3 2 3K >> K [H O ] from H Asc > > [H O ] from HAsc

a1 2

2 init 2 eq

Because K is small, the amount of H Asc that dissociates can be neglected

relative to initial conc : [H Asc] [H Asc] = 0.05 M Con’t


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( )

7/16/2014 63

Concentration H2Asc(aq) + H2O(l) H3O+ + HAsc-


- X - + X +X


Final C f 0.05 7.1x10-4

7.1x10-4

2

2

+ +a2 a1 3 from HAsc 3 from H Asc

+ +

3 from H Asc 3

Assumptions :

Because K << K , [H O ] [H O ]

[H O ] [H O ]

a1 2 init 2 2 init

2

Because K is small, [H Asc] - x = [H Asc] [H Asc] Thus,

[H Asc] = 0.050M - x 0.050 M

+ - 2

-53

a1 2

[H O ][HAsc ] xK = = 1 x 10

[H Asc] 0.050

- + -43x = [HAsc ] [H O ] 7.1 10 M

+ -43pH = - log[H O ] = - log(7.1 10 ) = 3.15

Con’t


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( ) Calculate [Asc2-]

7/16/2014 64

2-] +-123

a2 -

[Asc ][H O ]

K = = 5.0 x 10[HAsc ]

-2-] a2

+3

(K )[HAsc ][Asc ] =

[H O ]

-12 -42-] -12

-4

(5 x 10 )(7.1 x 10 )[Asc ] = = 5 10 M

7.1 x 10

Weak Bases & Relation to Weak Acids

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Base – Any species that accepts a proton (Bronsted)

In water the base accepts a proton from water acting as

an acid (water is amphoteric) and leaves behind a OH- ion

7/16/2014 65

+ -

c2

[BH ][OH ]K =

[B][H O]

The water term is treated as a constant in aqueousreactions and is incorporated into the value of K c

+ -

c 2 b

[BH ][OH ]K [H O] = K =

[B]

b bpK = -logK

Base strength increases with increasing K b (pK b deceases)

+ -2B(aq) + H O(l) BH (aq) + OH (aq)

Weak Base Equilibria

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q Ammonia is the simplest Nitrogen-containing compound

that acts as a weak base in water

Ammonium Hydroxide (NH4OH) in aqueous solutionconsists largely of unprotonated NH3 molecules, as itssmall K b indicates:

7/16/2014 66

+ -3 2 4NH (aq) + H O(l) NH (aq) + OH (aq)

+ - + -4 4

b

3 2 3

NH OH NH OHK = =

NH H O NH

-5bK = 1.76×10


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qIn a 1.0 M NH3 solution:

The NH3 is about 99.58% undissociated

If one or more of the Hydrogen atoms in ammonia isreplaced by an “organic” group (designated as R), an “Amine” results:

The Nitrogen atom in each compound has a “lone-pair”of unbonded electrons, a Bronsted Base characteristic,i.e., the electron pair(s) can accept protons

7/16/2014 67

- + -34[OH ] = [NH ] = 4.2 x 10 M

Primary

Amine

Secondary

Amine

Tertiary

Amine


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q

7/16/2014 68

Practice Problem

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What is the pH of a 1.5 M solution of Dimethylamine(CH3)2NH? K b = 5.9 x 10-4

7/16/2014 69

+ -3 2 2 3 2 2(CH ) NH(aq) + H O(l) (CH ) NH (aq) + OH (aq)

+ -3 2 2

b3 2

[(CH ) NH ][OH ]K =

[(CH ) NH]

Concentration(CH3)2NH(aq

)

+ H2O(l) (CH3)2NH2(aq) + H3O+


- X - + X +X


-

b wBecause K > > K , the [OH ] from the autoionization of water is negligible,thus,

3 2 init 3 2 reacting 3 2 3 2 init[(CH ) NH] - [(CH ) NH] = [(CH ) NH] [(CH ) NH]

Con’t

- + -from base 3 2 2[OH ] = [(CH ) NH ] = [OH ]

bBecause K is small, assume the amount of Amine reacting is small, thus


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( )Since K b is small: [(CH3)2NH]init [(CH3)2NH]: thus,

7/16/2014 70

1.5 M - x 1.5 M

+ - 2

-43 2 2b

3 2

[(CH ) NH ][OH ] xK = = 5.9 x 10

[(CH ) NH] 1.5

- -2x = [OH ] 3.0 x 10

-2

3.0 x 10 M x 100 = 2.0% (< 5%; assumption is justified)

1.5 M

Checking the assumptions:

3init

-4

b

[B] 1.5= = 2.5 x 10 > 400

K 5.9 x 10

Calculating pH:

+ -13

3pH = - log[H O ] = - log(3.3 10 ) = (13 - 0.52) = 12.48

-14

+ -13w

3 - - 2

K 1.0 10[H O ] = = = 3.3 10 M

[OH ] 3.0 10

Anions of Weak Acids & Bases

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The anions of weak acids are Bronsted-Lowry bases

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- -2F (aq) + H O(l) HF(aq) + OH (aq)

- -2A (aq) + H O(l) HA(aq) + OH (aq)

-F , anion of weak acid HF, acts as weak base

and accepts a proton to form HF

-

b -

[HF][OH ]K =

[F ]

-Base (A ) accepts a proton from water to form conjuate acid (HA)


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The acidity of HF vs F-

HF is a weak acid, very little dissociates

Very little H3O+ and F- produced

Water autoionization also contributes H3O

+

and OH

-

,but in much smaller amounts than H3O+ from HF

The Acidity of a solution is influenced mainly by the

H3O+ from HF and OH- from water

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+ -

2 32H O(l) H O (aq) + OH (aq)

+ -

2 3HF(aq) + H O(l) H O (aq) + F (aq)

(solution acidic)2

+ -

3 from HF from H O[H O ] >> [OH ]


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Basicity of A -(aq)

Consider 1 M solution of NaF

Salt dissociates completely to yield stoichiometricconcentrations of F- , assume Na+ is spectator ion

Some of the F- reacts with water to form OH-

Water autoionization also contributes H3O+ and OH-,

but in much smaller amounts than OH- from reactionof F- and H2O

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+ -

2 32H O(l) H O (aq) + OH (aq)

- (solution basic)+

2

-

from F 3 from H O[OH ] >> [H O ]

- -

2

F (aq) + H O(l) HF(aq) + OH (aq)

-

b -

[HF][OH ]K =

[F ]

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Practice Problem

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What is the pH of 0.25 M Sodium Acetate (CH3COONa)?

K a of Acetic Acid (HAc) is 1.8 x 10-5

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Sodium Acetate, a Sodium salt, is water soluble

- -init [Ac ] [Ac ] = 0.25 MInitial concentration of Acetate,

-- -

2 b -

[HAc][OH ]Ac (aq) + H O(l) HAc(aq) + OH (aq) K = [Ac ]

Concentration Ac-(aq) + H2O(l) HAc(aq) + OH-


- X - + X +X


-14-10w

b -5a

K 1.0 x 10K = = = 5.6 x 10

K 1.8 x 10

bBecause K is small, 0.25 M - x = 0.25 M Con’t

Practice Problem

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7/16/2014 76

- 2-10b -

[HAc][OH ] XK = = 5.6 x 100.25[Ac ]

- -5X = [OH ] 1.2 x 10 M

-14+ -10

3 - -5

Kw 1 10[H O ] = = = 8.3 10 M

[OH ] 1.2 10

+ -103pH = - log[H O ] = - log(8.3 10 ) = 9.08

pH of 0.25 M Sodium Acetate (Con’t)

Acid Strength

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g Trends in Acid Strength of Nonmetal Hydrides

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As the electronegativity of thenonmetal (E) bonded to theionizable proton increases(left to right), the acidityincreases

As the length of the E-H bondincreases (top to bottom), thebond strength decreases, sothe acidity increases

The “strong” monoprotichalide acids (HCl, HBr, HI) areequally strong

Note: (HF) is a weak acid)

Acid Strength in Oxoacids

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Oxoacids

An Oxoacid is an acid that contains Oxygen

To be more specific, it is an acid that:

contains oxygen

contains at least one other element (S, N, X)

has at least one Hydrogen atom (acid hydrogen)bound to oxygen

forms an ion by the loss of one or more protons

acidity of an Oxoacid is not related to bond strength

as in nonmetal hydrides

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Acid Strength in Oxoacids

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For oxoacids with the same number of oxygen atomsaround E, acid strength increases with the electronegativityof E

K a(HOCl) = 2.9x10-8 K a(HOBr) = 2.3x10-9 K a(HOI) = 2.3x10-11

For oxoacids with different numbers of oxygen atomsaround a given E, acid strength increases with the number

of oxygen atoms.

K a (HOCl Hypochorous acid) = 2.9x10-8 (Weak)

K a (HOCLO Chlorous acid) = 1.12x10-2

K a (HOCLO2 Chloric acid) ≈ 1 K a (HOCLO3 Perchloric acid) = > 107 (Strong)

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Aqueous solution of certain metal ions are acidic

As the MX species dissolves in water, the ions separate

and become bonded to a specific number of watermolecules – Hydration

If the metal ion, M+, is “small and highly charged (high

charge density), the hydrated metal ion transfers an H+ ion from a bonded water molecule to water (Bronstedacid) forming H3O

+

The bound H2O that releases the proton, becomes a

bound OH

-

ion The released proton forms Hydronium Ion with water

making solution Acidic

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n+1 -3 n 2 2 x 3M(NO ) (s) + xH O(l) M(H O) + nNO (aq)

n+1 (n-1)+ +2 x 2 2 x-1 3M(H O) (aq) + H O(l) M(H O) OH (aq) + H O (aq)


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Al(H2O)5OH2+ Al(H2O)63+

The acidic behavior of the hydrated Al3+ ion.

H2O

H3O+

Electron density

drawn toward Al3+

Nearby Base (H2O) attracts proton from one of theWater molecules in the hydrated Al ion producinga Hydronium ion (H3O

+)

Soln is Acidic

+

OH-

Practice Problem

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Explain with equations and calculations whether an aqueous solution of

each salt is acidic, basic, or neutral.

1. Fe3+

(HCOO1-

)3 (Iron Formate)Fe(HCOO)3(s) + n H2O(l) → Fe(H2O)n

3+(aq) + 3 HCOO – (aq)

Fe(H2O)n3+(aq) + H2O(l) ⇆ Fe(H2O)n-1OH2+(aq) + H3O

+(aq)

HCOO – (aq) + H2O(l) ⇆ OH – (aq) + HCOOH(aq)

K a (Fe3+) = 6 x 10 –3 K a (HCOOH) = 1.8 X 10-4 (Appendix C)

K b (HCOO –) = Kw / K a (HCOOH) = (1.0 x 10 –14) / (1.8 x 10 –4)

K b (HCOO –) = 5.6 x 10 –11 (unrounded)

K a (Fe3+) > K b (HCOO –) (6 x 10 –3 > 5.6 x 10 –11)

Solution is Acidic

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Practice Problem

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2. KHCO3 (Potassium Bicarbonate)

KHCO3(s) + H2O(l) → K +(aq) + HCO3 – (aq)

HCO3 – (aq) + H2O(l) ⇆ H3O

+(aq) + CO32 – (aq)

HCO3 – (aq) + H2O(l) ⇆ OH – (aq) + H2CO3(aq)

Potassium (Group 1A) is fully hydrated in aqueous solution and does

not participate in an acid/base reactionOnly the Bicarbonate (anion of weak acid) is considered in acidity

determination

K a (HCO3 –) = 4.7 x 10-11 From Appendix C

K b (HCO3 –) = K w / Ka (HCO3

-) = (1.0 x 10 –14) / (4.7 x 10 –11)

K b (HCO3 –) = 2.1 x 10-4

Since K b (HCO3 –) > K a (HCO3

–) (2.1 x 10-4 > 4.7 x 10-11)

Solution is basic

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Practice Problem

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3.0 K 2S (Potassium Sulfide)

K 2S(s) + H2O(l) ⇆ 2 K +(aq) + S2 – (aq)

S2 – (aq) + H2O(l) ⇆ OH – (aq) + HS – (aq) basic

Potassium (Group 1A) is fully hydrated in aqueous solution and does

not participate in an acid/base reaction

Only the Sulfide (S2-) (anion of weak acid) is considered in aciditydetermination

K a (S2-) = 9.0 x 10-8 (H2S from table)

K b (S2-) = K w / K a (S

2-) = (1.0 x 10 –14) / (9.0 x 10 –8)

K b (S2-) = 1.1 x 10-7

Since K b (S2-) > K a (S

2-) Solution is basic

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Acid-Base Properties – Salt Solutions

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Neutral Salt Solutions

Salts consisting of the Anion of a Strong Acid and the

Cation of a Strong Base yield a neutral solutionsbecause the ions do not react with water

Nitrate (NO3-) is a weaker base than water (H2O) ;

reaction goes to completion as the NO3

-

becomes fullyhydrated; does not react with water

Sodium becomes full hydrated; does not react with

water

Na+ & NO3- do not react with water, leaving just the

“autoionization of water, i.e., a neutral solution

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2H O(l) + -NaOH(s) Na (aq) + OH (aq)

2H O + - + -

3 3 3NaNO (s) Na (aq) + NO (aq) + H O (aq) + OH (aq)

- +3 2 3 3HNO (l) + H O(l) NO (aq) + H O (aq)

+ -

2 32H O H O + OH

Acid-Base Properties – Salt Soln

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Salts that produce Acidic Solutions

A salt consisting of the anion of a strong acid and the

cation of a weak base yields an acidic solution The cation acts as a weak acid

The anion does not react with water (previous slide)

In a solution of NH4Cl, the NH4+ ion that forms from

the weak base, NH3, is a weak acid The Chloride ion, the anion from a strong acid does

not react with water

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2H O + -

4 2 4NH Cl(s) + H O NH (aq) + Cl

+ +4 2 3 3NH (aq) + H O NH (aq) + H O (Acidic)

Acid-Base Properties – Salt Soln

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Salts that produce Basic Solutions

A salt consisting of the anion of a weak acid and the

cation of a strong base yields a basic solution The anion acts as a weak base

The cation does not react with water

The anion of the weak acid accepts a proton from

water to yield OH- ion, producing a “Basic” solution

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2H O + -

3 2 3CH COONa(s) + H O Na (aq) + CH COO (aq)

(dissoluton & hydration)

- -3 2 3CH COO + H O(l) CH COOH(aq) + OH (Basic)

(reaction of weak base - Acetate Anion))

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Salts of Weakly Acidic Cations andW kl B i A i

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Weakly Basic Anions

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Ex. NH4CN - Acidic or Basic? (Con’t)

Compare K a of NH4+ & K b of CN-

Recall: Molecular compounds only in tables of K a & K b -14

+ -10w

a 4 -5

b 3

K 1.0 x 10K of NH = = = 5.7 x 10

K of NH 1.76 x 10

-14

- -5w

b -5

a

K 1.0 x 10K of CN = = = 1.6 x 10

K of HCN 6.2 x 10

Magnitude of K b (K b > K a) (1.6 x 10-5 / 5.7 x 10-10 = 3 x 104)

Kb of CN- >> Ka of NH4+ Solution is Basic

Acceptance of proton from H2O by CN- proceeds much further than

the donation of a proton to H2O by NH4+

Salts of Weakly Acidic Cations andW kl B i A i

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Weakly Basic Anions

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Equation Summary

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7/16/2014 93

-

2 3HA + H O H O + A

- -

2A + H O HA + OH

+ -

2 32H O H O + OH

a b w

K × K = K = 14

- -

+ -3

3-

[H O+][A ] [HA][OH ] × = [H O ][OH ]

[HA] [A ]

a 10 apK = -log [K ]

+ -

3a [H O ][A ]K =[HA]

+ -

b

[BH ][OH ]K =

[B]]b bpK = -log[K

w a b-log(K ) = - log([K ][K ])

w a b-log(K ) = - log[K ] - log[K ]

a-pK a[K ] = 1 x 10

b-pK

b[K ] = 1 10

ow a bpK = pK + pK = 14 (at 25 C)

Equation Summary

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initinit

[A]if > 400, the assumption [HA] [Ha] is justified.

Kc Neglecting x introduces an error < 5%

diss init initif [HA] < 5% [HA] , the assumption ([HA] [HA] ) is valid

diss

init

[HA] ]Percent HA dissociated = x 100

[HA]

Documents

chap18web