1/19/2015 1 George Mason University General Chemistry 212 Chapter 18 Acid-Base Equilibria...

1/19/2015 1

George Mason UniversityGeneral Chemistry 212

Chapter 18Acid-Base Equilibria

Acknowledgements

Course Text:Course Text: Chemistry: the Molecular Nature of Matter and Chemistry: the Molecular Nature of Matter and Change, 7Change, 7thth edition, 2011, McGraw-Hill edition, 2011, McGraw-Hill

Martin S. Silberberg & Patricia AmateisMartin S. Silberberg & Patricia Amateis

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.

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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-Base Definition The Conjugate Acid-Base Pair Met Direction of Acid-Base Reactions

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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 Leveling Effect

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 an

essential element in acids The actual cause of acidity and basicity was

ultimately explained in terms of the effect these compounds have in water by Svante Arrhenius in 1884

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Acids & Bases Several concepts of acid-base theory:

The Arrhenius concept

The Bronsted-Lowry concept

The Lewis concept

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Acids & Bases According to the Arrhenius concept of acids

and bases

An acid is a substance that, when dissolved in water, increases the concentration of Hydrogen ion, H+(aq)

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

A base is a substance that when dissolved in water increases the concentration of the Hydroxide ion, OH-(aq)

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Acids & Bases 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 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

aqueous solutions because - Water reacts to form both ions

H2O(l) H+(aq) + OH-(aq) HBr and KOH are examples of a strong acid and a

strong base; strong acids & bases dissociate completely

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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

H2A(aq) + H2O(l) ⇌ H3O+(aq) + HA−(aq)

Triprotic acids have three potential protons to donate

H3A(aq) + H2O(l) ⇌ H3O+(aq) + H2A−(aq)

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Acids & Bases Oxoacids

Oxoacids with one oxygen have the structure: HOY

HOCl HOBr HOI

Oxoacids with multiple oxygen have the structure:

(OH)mYOn

HOClO3 (HClO4) HOClO2 (HOCLO3)

HOClO (HCLO2) HOCL (HClO)

The more oxygen atoms, the greater the acidity; thus HOCLO3 is a stronger acid than HOCl

Acidity refers to the relative acid strength, i.e. the degree to 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 as molecular substances

Acid-base reactions are not restricted to aqueous solution

Some species can act as either acids or bases (Amphoteric species) depending on what the other 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 to a

base

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

+ -3 2 4N H (a q ) + H O (l) N H (a q ) + O H (a q )

H+

+2 3H F (a q ) + H O (l) (a q ) + H O (a q )F

H+

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

Proton transfer as the essential feature of a Brønsted-Lowry acid-base reaction

AcidH+ donor

BaseH+ acceptor

BaseH+ acceptor

AcidH+ donor

Lone pair binds H+

Lone pair binds H+

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

HNO2(aq) + H2O(l) H3Oaq) + 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

BronstedBase

H3O+ is conjugate

acid of base H2ONO2

- is conjugatebase of acid HNO2

BronstedAcid

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

base acidacid base

NH4+ is conjugate

acid of base NH3

OH- is conjugatebase of acid H2O

BronstedBase

BronstedAcid

Donates Proton

AcceptsProton

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H2S + NH3 NH4+

+ HS- acid base baseacid

HS- is conjugateBase of acid H2S

NH4+ is conjugate

acid of base NH3

BronstedBase

BronstedAcid

Donates Proton

AcceptsProton

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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 they form 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 or base by reacting with it to form water ionization products

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Practice Problem Identify the conjugate acid-base pairs in the

following: - -2 - 2-

2 4 3 3 4H 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

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Lewis Acids & Bases Lewis Acids & Bases

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

The Lewis acid-base concept expands the acid class

Such reactions involve a “sharing” of electron pairs between 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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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 covalent bond (shared electron pair)

+ A + : B A - B

Acid BaseElectrophile Nucleophile Adduct

F

B

F F

H

N

H H

+

F

B

F F

H

N

H H

An acid is an electron-pair

acceptor

A base is an electron-pair

donor

Adduct

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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 an electron pair

The donated proton of a Bronsted-Lowry acid acts as a Lewis acid by accepting an electron pair donated by the base: The Lewis acids are Yellow and the Lewis bases are

Blue in following reaction BF3 + :NH3 BF3NH3

Fe3+ + 6 H2O Fe(H2O)63+

HF + H2O F- + H3O+

+ -

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Lewis Acids & Bases Solubility Effects

A Lewis acid-base reaction between nonpolar Diethyl Ether and normally insoluble Aluminum Chloride

The solubility of Aluminum Chloride in Dimethyl Ether results from the Oxygen acting as a base by donating an electron pair to the Aluminum acting as an acid forming a water-soluble polar covalent bond

+

-

-

+

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

Write the complete equation for the reaction, including the product

Identify each reactant as a Lewis acid or Lewis base

AlCl3 + :NH3 AlCl3:NH3

Lewis acid Lewis base Adduct

Acceptse- pair

Donatese- pair

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Acids & Bases (Summary) 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 (H2O)

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)

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Acids & Bases Self Ionization of Water Pure water undergoes auto-ionization to produce

Hydronium and Hydroxide ions 2 H2O(l) H3O+(aq) + OH-(aq)

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

The concentrations of Hydronium and Hydroxide ions in any aqueous solution must obey the auto-ionization equilibrium

Classification of solutions using Kw:Acidic: [H3O+] > [OH-]Basic: [H3O+] < [OH-]Neutral: [H3O+] = [OH-]

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

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

the amounts of Hydronium and Hydroxide ions are always related For example, pure water at 25 oC,

The Definition of pH

The amount of Hydronium ion (H3O+) in solution

is often described by the pH of the solution

The Definition of pOH Hydroxide (OH-) can be expressed as pOH

+3pH = - log[H O ]

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

+ -pH3[H O ] = 1 10

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

-10pOH = - log [OH ]

- -pOH[OH ] = 1 10

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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 Relations among pH, pOH, and pKw

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 )

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Equilibrium & Acid-Base Conjugates Equilibrium Constants (K) for acids & bases can also

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

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

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:

Ka and HA Kb and A-

Setup two dissociation reactions:

The sum of the two dissociation reactions is the autoionization of water

The overall equilibrium constant is the product of the individual equilibrium constants

+ -

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, Ka of the acid in a conjugate

pair can be computed from Kb and vice versa

Reference tables typically have Ka & Kb values for molecular species, but not for ions such as F- or CH3NH3

+

w a bK = K × K

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

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

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


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Equilibrium & Acid-Base Conjugates Acid strength [H3O+] increases with increasing value of

Ka

Base strength [OH-] increases with increasing value of kb

A low pK corresponds to a high value of K

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

+ -3

a[H O ][A ]

K = [HA]

-

b -

[HA][OH ]K =

[A ]

a 10 apK = - log [K ]

]b 10 bpK = - log [K

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Practice Problem Ex. Find Kb value for F-

-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 10K (F ) = = = 1.5 10

K (HF) 6.8 10

2

+ -3

HF + H O H O + F

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Acids & Bases The Relationship Between Ka and pKa

Acid Name (Formula) Ka at 25oC pKa

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

Acid

S

tren

gth

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Acids & Bases

The pH Scale

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

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

In the case of HCl, the dissociation lies very far to the right because the conjugate base (Cl-) is extremely weak, 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

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Acids & Bases

Activity Series

Strengths ofacids and bases

Strength determined by

K, not concentration

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Practice Problem A strong monoprotic acid is dissolved in a beaker of

water. Which of the following pictures best represents the acid solution

Ans: B

A B C

A solution of a strong acid contains equal amounts of H3O+ and the acid anion, such as a “Cl-” The monoprotic acid is completely dissociated

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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 - F bond and the very high electronegativity of the Fluoride ion

When you take into account the Enthalpy, Entropy, and Free Energy properties (Chapter 19), the numbers clearly show a distinct difference between HF and the other Hydrohalic acids

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Hydrohalic Acids – Weak vs Strong These properties can be summarized as

follows: Very strong Hydrogen Bonding between

the un-ionized Hydrogen Fluoride (HF) molecules and water molecules

The energy required to break the H-F bond, is much greater than for the bonds of the

other Hydrohalic Acid bonds A large decrease in Entropy when the

Hydrogen Fluoride molecules react with water

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Hydrohalic Acids – Weak vs Strong This is particularly noticeable with

Fydrogen Fluoride because the attractiveness of the very small Fluoride ions produced imposes a lot of order on the surrounding water molecules, and also on nearby hydroxonium ions

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 as a weak acid The Ka values clearly suggest a weak acid The larger the value of the equilibrium

constant (Ka), the stronger the acid

ΔH TΔS ΔG Ka

(kJ mol-1) (kJ mol-1) (kJ mol-1) (mol/L)

HF -13 -29 +16 6.3 x 10-4

HCl -59 -13 -46 2.00 x 10+6

HBr -63 -4 -59 5.01 x 10+8

HI -57 +4 -61 2.0 x 10+9

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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-

- -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 10

-14- w

+3

K 1 x 10 mol / L[OH ] = =

1.2 mol / L[H O ]

HBr is a strong acid and is completely dissociated

Autoionization of Water

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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-

+ -143[H O ] = 5.8 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 strong base and is completely dissociated

Autoionization of Water

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Practice ProblemWhat is the pH of the following solution:?

[OH-] = 5.6 x 10-10 M

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

-14+ -5w

3 - -10

K 1.0 10[H O ] = = = 1.786 10 M

[OH ] 5.6 10

+ -510 3 10pH = - log [H O ] = - log (1.786 10 )

-510 10pH = -(log 1.786 + log 10 )

pH = - (0.252 - 5) = 5 - 0.252 = 4.75

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Practice ProblemWhat is the pOH of the following solution:?

[H3O+] = 9.3 x 10-4 M

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

-14- w

+ -43

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 = 11.0

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

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

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

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 10[H O ] = = = 6.15006 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 ] = 12.2

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Practice ProblemWhat is [H3O

+] in a solution with a pOH of 5.89?

wpK = pH + pOH = 14.00

wpH = pK - pOH = 14.00 - 5.89 = 8.11

+10 3pH = - log [H O ]

+10 3log [H O ] = -pH

+ -93[H O ] = 7.76 x 10 M

+ (-pH) -8.113[H O ] = 10 = 10

1/19/2015 52

Weak Acid Equilibria Problems Solving Problems involving Weak-Acid Equilibria

Types Given equilibrium concentrations, find Ka

Given Ka and some concentrations, find other concentrations

Approach Determine “Knowns” and “Unknowns” Write the Balanced Equation Set up equilibrium expression –

Define “x” as the unknown change in [HA] Typically, x = [A]diss (conc. of HA that

dissociates)x = [A]diss = [H3O

+] = [A-]

[ ] [ ] .....

[ ] [ ] ....

c d

a a b

C DK

A B

1/19/2015 53

Weak Acid Equilibria Problems Construct a “Reaction” table Make assumptions that simplify calculations

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

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

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

Substitute values into Ka 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 equilibrium

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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 Ka

Therefore, the change in acid concentration can be neglected in the calculation of the equilibrium concentration of the acid

[HA] = [HA]init - [HA]dissoc ≈ [HA]init

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

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Weak Acid Equilibria Assumptions Assumption Criteria - 2 approaches

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 PA(init))

+ -3H O A

K = HA

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Practice ProblemWhat is the Ka 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 MConcentration HPAc(aq) + H2O(l) H3O

+ + PAc-

Initial Ci 0.12 - 0 0

- X - + X +X

Equilibrium Cf 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

-53a

-[H O ][PAc ] (2.4 x 10 )(2.4 x 10 )K = = 4.8×10

[HPAc] 0.12

1/19/2015 57

Practice ProblemWhat is the [H3O

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

+-53

a

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

[HPr]

HPr is a weak acid = [HPr]init = 0.10 M

Let x = [HPr]diss = [H3O+]from HPr = [Pr-]from HPr

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


- X - + X +X

Equilibrium Cf 0.10- X - X X

Final Cf 0.10 X X

Assumption: Ka 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)

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Practice Problem (con’t)Substituting into the Ka expression and solving for x

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

+-53

a

-[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 MFor [HPr] : = x 100 = 1.1%

[HPr] 0.10M

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Weak Acid Equilibria Effect of Concentration on Extent of Acid Dissociation

Case 1: [HPr]init = 0.10 M

Case 2: [HPr]init = 0.01 M

-3diss

diss -1init

[HPr] 1.1 x 10 M% HA = = x 100 = 1.1%

[HPr] 1.0 x 10 M

-4

diss -2

3.6 x 10 M% HA = x 100 = 3.6%

1.0 x 10 M

-4 + -3 dissx = 3.6 x 10 M = [H O ] = [Pr ] = [HPr]

+-53

a

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

[HPr] 0.10

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

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Weak Acid Equilibria Polyprotic Acids

Acids with more than one ionizable proton are polyprotic acids

One proton at a time dissociates from the acid molecule

Each dissociation step has a different Ka

H3PO4(aq) + H2O(l) H2PO4- + H3O

+(aq)

H2PO4-(aq) + H2O(l) HPO4

-2 + H3O+(aq)

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

-3 + H3O+(aq)

- +-32 4 3

a13 4

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

[H PO ]

-2 +-84 3

a2 -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 ]a1 a2 a3K >> K >> K

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Weak Acid Equilibria

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

Ka1 >> Ka2 >> Ka3

1/19/2015 62

Practice Problem Calculate the following for 0.05 M Ascorbic Acid

(H2Asc):

[H2Asc], [HAsc-], [Asc2-], pH

Ka1 = 1.0 x 10-5 Ka2 = 5 x 10-12 - +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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Practice Problem (con’t) Assumptions

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


- X - + X +X


Final Cf 0.05 7.1x10-4 7.1x10-

4

-

2

2

+ +a2 a1 3 from HAsc 3 from H Asc

+ +3 from H Asc 3

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]

[H Asc] = 0.050M - x 0.050 M

+ - 2

-53a1

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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Practice Problem (con’t) Calculate [Asc2-]

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

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Weak Bases & Relation to Weak Acids

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

+ -

c2

[BH ][OH ]K =

[B][H O]

The water term is treated as a constant in aqueous reactions and is incorporated into the value of Kc

+ -

c 2 b[BH ][OH ]

K [H O] = K = [B]

b bpK = -logK

Base strength increases with increasing Kb (pKb deceases)

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

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Weak Base EquilibriaWeak Base Equilibria Ammonia is the simplest Nitrogen-containing

compound that acts as a weak base in water

Ammonium Hydroxide (NH4OH) in aqueous solution consists largely of unprotonated NH3 molecules, as its small Kb indicates:

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

+ - + -4 4

b3 2 3

NH OH NH OHK = =

NH H O NH

-5bK = 1.76×10

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Weak Base EquilibriaWeak Base Equilibria In a 1.0 M NH3 solution:

The NH3 is about 99.58% undissociated

If one or more of the Hydrogen atoms in ammonia is replaced 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

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

PrimaryAmine

SecondaryAmine

TertiaryAmine

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Weak Base EquilibriaWeak Base Equilibria

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Practice ProblemPractice Problem What is the pH of a 1.5 M solution of Dimethylamine

(CH3)2NH? Kb = 5.9 x 10-4

+ -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]

- + -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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Practice Problem (con’t)Since Kb is small: [(CH3)2NH]init [(CH3)2NH]: thus,

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

-23.0 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 10 > 400 (Assumption is justified

K 5.9 10

Calculating pH:

-14+ -13w

3 - - 2

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

[OH ] 3.0 10

+ -133pH = - [H O ] = - log(3.3 ×10 ) = 12.5

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Anions of Weak Acids & Bases The anions of weak acids are Bronsted-Lowry bases

- -

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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Anions of Weak Acids & Bases 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

+ -

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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Anions of Weak Acids & Bases Basicity of A-(aq)

Consider 1 M solution of NaF Salt dissociates completely to yield stoichiometric

concentrations 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 reaction of F- and H2O

04/21/23 73

+ -

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

- (solution basic)

2

-

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

- -

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

b -

[HF][OH ]K =

[F ]

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Anions of Weak Acids & Bases Summary

The relative concentration of HA & A- determine the acidity or basicity of the solution

In an HA solution, [HA] >> [A-] and

[H3O+]from HA >> [OH-]from H3O+

“Solution is Acidic”

In an A- solution, [A-] >> [HA] and

[OH-]from A- >> [H3O+]from H3O+

“Solution is Basic”

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Practice Problem

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

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

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

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Practice Problem pH of 0.25 M Sodium Acetate (Con’t)

- 2-10

b -

[HAc][OH ] XK = = 5.6 x 10

0.25[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

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Acid Strength Trends in Acid Strength of Nonmetal Hydrides As the electronegativity of

the nonmetal (E) bonded to the ionizable proton increases (left to right), the acidity increases

As the length of the E-H bond increases (top to bottom), the bond strength decreases, so the acidity increases

The “strong” monoprotic halide acids (HCl, HBr, HI) are equally strong

Note: (HF) is a weak acid)

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Acid Strength in Oxoacids 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 Oxoacids with 1 Oxygen - HOE structure:

The more electronegative is E, the stronger the acid

Acid Strength: HOCl > HOBr > HOI

Oxoacids with multiple Oxygens - (OH)mEOn structure

The more Oxygen atoms, the greater the acidity

HOClO3 (HClO4) > HOClO2 (HCLO3) >

HOClO (HCLO2) > HOCL (HClO)

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Acid Strength in Oxoacids For oxoacids with the same number of oxygen atoms

around E, acid strength increases with the electronegativity of E

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

For oxoacids with different numbers of oxygen atoms around a given E, acid strength increases with the number of oxygen atoms. Ka (HOCl Hypochorous acid) = 2.9x10-8

(Weak) Ka (HOCLO Chlorous acid) = 1.12x10-2

Ka (HOCLO2 Chloric acid) ≈ 1

Ka (HOCLO3 Perchloric acid) = > 107 (Strong)

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Acidity of Hydrated Metal Ions 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 water molecules – 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 (Bronsted acid) 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

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)

1/19/2015 82

Acidity of Hydrated Metal Ions The acidic behavior of the hydrated Al3+ ion.

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

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 isAcidic

+

OH-

1/19/2015 83

Practice Problem

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)n3+(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)

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

Kb (HCOO–) = Kw / Ka (HCOOH) = (1.0 x 10–14) / (1.8 x 10–4)

Kb (HCOO–) = 5.6 x 10–11 (unrounded)

Ka (Fe3+) > Kb (HCOO–) (6 x 10–3 > 5.6 x 10–11)

Solution is Acidic

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Practice Problem2. 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 reaction

Only the Bicarbonate (anion of weak acid) is considered in acidity determination

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

Kb (HCO3–) = Kw / Ka (HCO3

-) = (1.0 x 10–14) / (4.7 x 10–11)

Kb (HCO3–) = 2.1 x 10-4

Since Kb (HCO3–) > Ka (HCO3

–) (2.1 x 10-4 > 4.7 x 10-11)

Solution is basic

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Practice Problem3. K2S (Potassium Sulfide)

K2S(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 acidity determination

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

Kb (S2-) = Kw / Ka (S2-) = (1.0 x 10–14) / (9.0 x 10–8)

Kb (S2-) = 1.1 x 10-7

Since Kb (S2-) > Ka (S2-) Solution is basic

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Acid-Base Properties – Salt Solutions

Neutral Salt Solutions Salts consisting of the Anion of a Strong Acid and

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

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

reaction goes to completion as the NO3- becomes

fully hydrated; 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

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)

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Acid-Base Properties – Salt Soln 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 water2

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

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

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Acid-Base Properties – Salt Soln 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

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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Overall acidity of solution depends on relative acid strength (Ka) or base strength (Kb) of the separated ions

Ex. NH4CN - Acidic or Basic?

Write equations for any reactions that occur between the separated ions and water

+ +

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

- -

2CN (aq) + H O(l) HCN(aq) + OH

1/19/2015 90


Ex. NH4CN - Acidic or Basic? (Con’t)

Compare Ka of NH4+ & Kb of CN-

Recall: Molecular compounds only in tables of Ka & Kb

-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 Kb (Kb > Ka) (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

+

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Behavior of Salts in Water

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Equation Summary

-10pOH = -log [OH ]

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

+ -

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

+3pH = - log[H O ] + -pH

3[H O ] = 1 x 10

- -pOH[OH ] = 1 x 10

-

2 3HA + H O H O + A- -

2A + H O HA + OH+ -

2 32H O H O + OH


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

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

w wpK = - log[K ] w-pKw[K ] = 1 x 10

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Equation Summary -

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-pKa[K ] = 1 x 10

b-pK

b[K ] = 1 10

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Equation Summary

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]

1/19/2015 1 George Mason University General Chemistry 212 Chapter 18 Acid-Base Equilibria...

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