Aqueous Equilibria: Acids and Basesdfard.weebly.com/uploads/1/0/5/3/10533150/ch14.pdf · Chapter 14 •Aqueous Equilibria: Acids and Bases. Slide 2 Acid–Base Concepts 01 Arrhenius

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Chapter 14Chapter 14

•Aqueous Equilibria: Acids and Bases

•Aqueous Equilibria: Acids and Bases

Slide 2

Acid–Base Concepts 01Acid–Base Concepts 01

Arrhenius Acid: A substance which dissociates to form hydrogen ions (H+) in solution.

HA(aq) → H+(aq) + A–(aq)

Arrhenius Base: A substance that dissociates in, or reacts with water to form hydroxide ions (OH–).

MOH(aq) → M+(aq) + OH–(aq)

Slide 3


• Brønsted–Lowry Acid: Substance that can donate H+

• Brønsted–Lowry Base: Substance that can accept H+

• Chemical species whose formulas differ only by one proton are said to be conjugate acid–base pairs.

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Slide 5

Hydrated Protons and HydroniumIonsHydrated Protons and HydroniumIons

H1+(aq) + A1-(aq)HA(aq)

[H(H2O)n]1+

For our purposes, H1+ is equivalent to H3O1+.

n = 4 H9O41+

n = 1 H3O1+

n = 2 H5O21+

n = 3 H7O31+

Due to high reactivity of the hydrogen ion, it is actually hydrated by one or more water molecules.

Slide 6

Acid–Base Concepts (1-Acid Dissociation)Acid–Base Concepts (1-Acid Dissociation)

Slide 7

Lewis Acid–Base ConceptsLewis Acid–Base Concepts

Slide 8

• A Lewis Acid is an electron-pair acceptor. These are generally cations and neutral molecules with vacant valence orbitals, such as Al3+, Cu2+, H+, BF3.

• A Lewis Base is an electron-pair donor. These are generally anions and neutral molecules with available pairs of electrons, such as H2O, NH3, O2–.

• The bond formed is called a coordinate bond.


Slide 9


- +

Slide 10

Lewis Acids and BasesLewis Acids and Bases

Lewis Base: An electron-pair donor.

Lewis Acid: An electron-pair acceptor.

Slide 11

Lewis Acids and BasesLewis Acids and Bases

Lewis Base: An electron-pair donor.

Lewis Acid: An electron-pair acceptor.

Slide 12


• Write balanced equations for the dissociation of each of the following Brønsted–Lowry acids.(a) H2SO4 (b) HSO4

– (c) H3O+

• Identify the Lewis acid and Lewis base in each of the following reactions:

(a) SnCl4(s) + 2 Cl–(aq) æ SnCl62–(aq)(b) Hg2+(aq) + 4 CN–(aq) æ Hg(CN)4

2–(aq)(c) Co3+(aq) + 6 NH3(aq) æ Co(NH3)6

3+(aq)

Slide 13

Dissociation of Water 01Dissociation of Water 01

• Water can act as an acid or as a base.

H2O(l) æ H+(aq) + OH–(aq)

• This is called the autoionization of water.

H2O(l) + H2O(l) æ H3O+(aq) + OH–(aq)

Slide 14


• This equilibrium gives us the ion product constant for water.

Kw = Kc = [H+][OH–] = 1.0 x 10–14

• If we know either [H+] or [OH–] then we can determine the other quantity.

Slide 15


• The concentration of OH– ions in a certain household

ammonia cleaning solution is 0.0025 M. Calculate the

concentration of H+ ions.

• Calculate the concentration of OH– ions in a HCl

solution whose hydrogen ion concentration is 1.3 M.

Slide 16

pH – A Measure of Acidity 01pH – A Measure of Acidity 01

• The pH of a solution is the negative logarithm of the hydrogen ion concentration (in mol/L).

pH = –log [H+], [H+] = 10-pH

pH + pOH = 14

Acidic solutions: [H+] > 1.0 x 10–7 M, pH < 7.00Basic solutions: [H+] < 1.0 x 10–7 M, pH > 7.00Neutral solutions: [H+] = 1.0 x 10–7 M, pH = 7.00

Slide 17


• Nitric acid (HNO3) is used in the production of fertilizer, dyes, drugs, and explosives. Calculate the pH of a HNO3 solution having a hydrogen ion concentration of 0.76 M.

• The pH of a certain orange juice is 3.33. Calculate the H+ ion concentration.

• The OH– ion concentration of a blood sample is 2.5 x 10–7 M. What is the pH of the blood?

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Strength of Acids and Bases 03Strength of Acids and Bases 03

HClO4

HIHBrHClH2SO4

HNO3

H3O+

HSO4–

HSO4–

HFHNO2

HCOOHNH4

+

HCNH2ONH3

ClO4–

I–

Br –

Cl –

HSO4 –

NO3 –

H2OSO4

2–

SO42–

F –

NO2 –

HCOO –

NH3

CN –

OH –

NH2 –

ACID CONJ. BASE ACID CONJ. BASE

Incr

easi

ng A

cid

Stre

ngth

Incr

easi

ngA

cid

Stre

ngth

Slide 21


• Stronger acid + stronger base →

weaker acid + weaker base

• Predict the direction of the following:

HNO2(aq) + CN–(aq) æ HCN(aq) + NO2–(aq)

HF(aq) + NH3(aq) æ F–(aq) + NH4+(aq)

Slide 22

Acid Ionization Constants 01Acid Ionization Constants 01

• Acid Ionization Constant: the equilibrium constant for the ionization of an acid.

HA(aq) + H2O(l) æ H3O+(aq) + A–(aq)

• Or simply: HA(aq) æ H+(aq) + A–(aq)

[HA]]][A[H −+

=aK

Slide 23

Conjugate Base Ionization ConstConjugate Base Ionization Const

[HA] [OH−][A-]

Kb =

A- + H2O(l) HA(aq) + OH−(aq)

Ka × Kb = Kw

[HA] [OH−][A-]

Kb =Ka × [HA]]][A[H −+

× = Kw

Slide 24


7.1 x 10 –4

4.5 x 10 –4

3.0 x 10 –4

1.7 x 10 –4

8.0 x 10 –5

6.5 x 10 –5

1.8 x 10 –5

4.9 x 10 –10

1.3 x 10 –10

HFHNO2

C9H8O4 (aspirin)HCO2H (formic)C6H8O6 (ascorbic)C6H5CO2H (benzoic)CH3CO2H (acetic)HCNC6H5OH (phenol)

F–

NO2 –

C9H7O4 –

HCO2 –

C6H7O6 –

C6H5CO2 –

CH3CO2 –

CN –

C6H5O –

ACID Ka CONJ. BASE Kb

1.4 x 10 –11

2.2 x 10 –11

3.3 x 10 –11

5.9 x 10 –11

1.3 x 10 –10

1.5 x 10 –10

5.6 x 10 –10

2.0 x 10 –5

7.7 x 10 –5

Slide 25


(a) Arrange the three acids in order of increasing value of Ka.(b) Which acid, if any, is a strong acid?(c) Which solution has the highest pH, and which has the

lowest?

Slide 26

HA æ H+ + A

(M): 0.50 0.00 0.00(M): –x +x +x

Equilib (M): 0.50 –x x x

Acid Ionization ConstantsDetermine the pH of 0.50 M HA

solution at 25°C. Ka = 7.1 x 10–4 05

Acid Ionization ConstantsDetermine the pH of 0.50 M HA

solution at 25°C. Ka = 7.1 x 10–4 05

• Initial Change Equilibrium Table:.

InitialChange

(aq) (aq)-(aq)

Slide 27

What is the pH of a 0.5 M HF solution (at 250C)?

HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-]

[HF] = 7.1 x 10-4

HF (aq) H+ (aq) + F- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.50 0.00

-x +x

0.50 - x

0.00

+x

x x

Ka =x2

0.50 - x = 7.1 x 10-4

Ka ≈x2

0.50 = 7.1 x 10-4

0.50 – x ≈ 0.50100•Ka < Co ?100 x 7.1 x 10-4

= 0.071 < 0.5x2 = 3.55 x 10-4 x = 0.019 M

[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72[HF] = 0.50 – x = 0.48 M

Slide 28


• pH of a Weak Acid (Cont’d):

1. Substitute equilibrium concentrations into equilibrium expression.

2. If 100•Ka < Co then (C0 – x) approximates to (C0).

3. The equation can now be solved for x and pH.

4. If 100•Ka is not significantly smaller than Co the quadratic equation must be used to solve for x and pH.

Slide 29


• The Quadratic Equation:• The expression must first be rearranged to:

• The values are substituted into the quadratic and solved for a positive solution to x and pH.

aacbbx

242 −±−

=

02 =++ cbxax

Slide 30


• Percent Dissociation: A measure of the strength of an acid.

• Stronger acids have higher percent dissociation.• Percent dissociation of a weak acid decreases as its concentration increases.

100[HA]

][HonDissociati % ×=+

Slide 31

Percent dissociation of a weak acid decreases as its concentration increases

Percent dissociation of a weak acid decreases as its concentration increases

• Concentration Dependence:

Slide 32

Weak Bases:Base Ionization Constants 01

Weak Bases:Base Ionization Constants 01

• Base Ionization Constant:The equilibrium constant for the ionization of a base. • The ionization of weak bases is treated in the same way as the ionization of weak acids.

B(aq) + H2O(l) æ BH+(aq) + OH–(aq)• Calculations follow the same procedure as used for a weak acid but [OH–] is calculated, not [H+].

Slide 33

Base Ionization Constants 02Base Ionization Constants 02

5.6 x 10 –4

4.4 x 10 –4

4.1 x 10 –4

1.8 x 10 –5

1.7 x 10 –9

3.8 x 10 –10

1.5 x 10 –14

C2H5NH2 (ethylamine)CH3NH2 (methylamine)C8H10N4O2 (caffeine)NH3 (ammonia)C5H5N (pyridine)C6H5NH2 (aniline)NH2CONH2 (urea)

C2H5NH3+

CH3NH3+

C8H11N4O2+

NH4+

C5H6N+

C6H5NH3+

NH2CONH3+

BASE Kb CONJ. ACID Ka

1.8 x 10 –11

2.3 x 10 –11

2.4 x 10 –11

5.6 x 10 –10

5.9 x 10 –6

2.6 x 10 –5

0.67

Note that the positive charge sits on the nitrogen.

http://en.wikipedia.org/wiki/File:Caffeine.svg

Slide 34

Base Ionization Constants 03Base Ionization Constants 03

• Product of Ka and Kb: multiplying out the expressions for Ka and Kb equals Kw.

Ka × Kb = Kw

Slide 35

pH of Basic SolutionspH of Basic Solutions

What is the pH of a 0.15 M solution of NH3?

[NH4+] [OH−]

[NH3]Kb = = 1.8 × 10−5

NH3(aq) + H2O(l) NH4+(aq) + OH−(aq)

Slide 36


Tabulate the data.

[NH3], M [NH4+], M [OH−], M

Initially 0.15 0 0

At Equilibrium 0.15 - x x x

Slide 37


(1.8 × 10−5) (0.15) = x2

2.7 × 10−6 = x2

1.6 × 10−3 = x2

(x)2

(0.15 - x )1.8 × 10−5 =

100 x Kb < C0 ?1.8 × 10−3< 0.150.15 –x = 0.15

Slide 38


Therefore,X = [OH−] = 1.6 × 10−3 MpOH = −log (1.6 × 10−3)pOH = 2.80pH = 14.00 − 2.80pH = 11.20

Slide 39

Diprotic & Polyprotic Acids 01Diprotic & Polyprotic Acids 01

• Diprotic and polyprotic acids yield more than one hydrogen ion per molecule.

• One proton is lost at a time. Conjugate base of first step is acid of second step.

• Ionization constants decrease as protons are removed.

H2SO4

H3PO4

Slide 40

Diprotic & Polyprotic Acids 02Diprotic & Polyprotic Acids 02

Very Large1.3 x 10 –2

6.5 x 10 –2

6.1 x 10 –5

1.3 x 10 –2

6.3 x 10 –8

4.2 x 10 –7

4.8 x 10 –11

9.5 x 10 –8

1 x 10 –19

7.5 x 10 –3

6.2 x 10 –8

4.8 x 10 –13

H2SO4HSO4

–

C2H2O4C2HO4

–

H2SO3HSO3

–

H2CO3HCO3

–

H2SHS–

H3PO4H2PO4

–

HPO42–

ACID Ka CONJ. BASE Kb

HSO4 –

SO4 2–

C2HO4–

C2O42–

HSO3 –

SO3 2–

HCO3–

CO3 2–

HS–

S 2–

H2PO4–

HPO42–

PO43–

Very Small7.7 x 10 –13

1.5 x 10 –13

1.6 x 10 –10

7.7 x 10 –13

1.6 x 10 –7

2.4 x 10 –8

2.1 x 10 –4

1.1 x 10 –7

1 x 10 –5

1.3 x 10 –12

1.6 x 10 –7

2.1 x 10 –2

Slide 41

Molecular Structure and Acid Strength 01Molecular Structure and Acid Strength 01

• The strength of an acid depends on its tendency to ionize.

• For general acids of the type H–X:1. The stronger the bond, the weaker the acid.2. The more polar the bond, the stronger the acid.

• For the hydrohalic acids, bond strength plays the key role giving: HF < HCl < HBr < HI

299 kJ/mol for HI567 kJ/mol for HF

Slide 42


• The electrostatic potential maps show all the hydrohalic

acids are polar. The variation in polarity is less

significant than the bond strength which decreases

from 567 kJ/mol for HF to 299 kJ/mol for HI.

Slide 43

(pm)

Slide 44


• For binary acids in the same group, H–A bond strength decreases with increasing size of A, so acidity increases.

• For binary acids in the same row, H–A polarity increases with increasing electronegativity of A, so acidity increases.

Slide 45


• For oxoacids bond polarity is more important. If we consider the main element (Y):

Y–O–H• If Y is an electronegative element, the Y–O bond will pull more electrons, the O–H bond will be more polar and the acid will be stronger.

Slide 46


• For oxoacids with different central atoms that are from the same group of the periodic table and that have the same oxidation number, acid strength increases with increasing electronegativity.

Slide 47

Polar Covalent Bonds 02Polar Covalent Bonds 02

Pauling Pauling ElectronegativitiesElectronegativities

Detailed List of Electronegativity; http://environmentalchemistry.com/yogi/periodic/electronegativity.html

http://environmentalchemistry.com/yogi/periodic/electronegativity.html

Slide 48


• Oxoacids of Chlorine:

Slide 49


• Predict the relative strengths of the following groups of oxoacids:

a) HClO, HBrO, and HIO.

b) HNO3 and HNO2.

c) H3PO3 and H3PO4.

Slide 50

Acid-Base Properties of SaltsAcid-Base Properties of Salts

Slide 51

Acid–Base Properties of Salts 01Acid–Base Properties of Salts 01

• Salts that produce neutral solutions are those formed from strong acids and strong bases.

• Salts that produce basic solutions are those formed from weak acids and strong bases.

• Salts that produce acidic solutions are those formed from strong acids and weak bases.

Slide 52

The pH of an ammonium carbonate solution, (NH4)2CO3, depends on the relative acid strength of the cation and the relative base strength of the anion.

Is it acidic or basic?

Salts That Contain Cation from a Weak Base and anion from a Weak Base

Slide 53

Acid-Base Properties of SaltsAcid-Base Properties of SaltsSalts That Contain Acidic Cations and Basic Anions

HCO31-(aq) + OH1-(aq)CO3

2-(aq) + H2O(l) Kb

H3O1+(aq) + NH3(aq)NH41+(aq) + H2O(l) Ka

(NH4)2CO3:

Three possibilities:• Ka > Kb: The solution will contain an excess of

H3O1+ ions , Acidic solution, (pH < 7).• Ka < Kb: The solution will contain an excess of

OH1- ions, Basic solutions, (pH > 7).• Ka ≈ Kb: The solution will contain approximately

equal concentrations of H3O1+ and OH1- ions (pH ≈ 7).

Slide 54

Salts That Contain Cation from a Weak Base and anion from a Weak Base

HCO31-(aq) + OH1-(aq)CO3

2-(aq) + H2O(l) Kb

H3O1+(aq) + NH3(aq)NH41+(aq) + H2O(l) Ka

(NH4)2CO3:

= 1.8 x 10-45.6 x 10-11

1.0 x 10-14

Kb for CO32- =

Ka for HCO31-

Kw=

= 5.6 x 10-101.8 x 10-5

1.0 x 10-14

Ka for NH41+ =

Kb for NH3

Kw=

Basic, Ka < Kb

Slide 55

Acid-Base Properties of SaltsAcid-Base Properties of Salts

Slide 56

Hydrated Cation of Al3+

Slide 57


• Metal Ion Hydrolysis:

Slide 58


• Calculate the pH of a 0.020 M Al(NO3)3 solution

Ka = 1.4 x 10-5.

• Predict whether the following solutions will be acidic, basic, or nearly neutral:

(a) NH4I (b) CaCl2 (c) KCN (d) Fe(NO3)3

Aqueous Equilibria: Acids and Basesdfard.weebly.com/uploads/1/0/5/3/10533150/ch14.pdf · Chapter 14 •Aqueous Equilibria: Acids and Bases. Slide 2 Acid–Base Concepts 01 Arrhenius

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