Slide 1 Chapter 14 Aqueous Equilibria: Acids and Bases.

Chapter 14Chapter 14

•Aqueous Equilibria: Acids and Bases

•Aqueous Equilibria: Acids and Bases

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)


• 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.

Strong vs. Weak acids 03Strong vs. Weak acids 03

Hydrated Protons and Hydronium IonsHydrated Protons and Hydronium Ions

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.

Acid–Base Concepts Acid–Base Concepts

Lewis Acid–Base ConceptsLewis Acid–Base Concepts

• 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.



- +

Lewis Acids and BasesLewis Acids and Bases

Lewis Base: An electron-pair donor.

Lewis Acid: An electron-pair acceptor.


• 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)42–(aq)

(c) Co3+(aq) + 6 NH3(aq) æ Co(NH3)63+(aq)

Dissociation of Water 01Dissociation of Water 01

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

H2O(l) æ H+(aq) + OH–(aq)

• Kc = [H+][OH–]

• This is called the autoionization of water.

H2O(l) + H2O(l) æ H3O+(aq) + OH–(aq)


• 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.


• 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.

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 < 7.00

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

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


• 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?


Color of Tea: Polyphenols, Thearubigins

Color of Red Cabbage: Anthocyanin

Strength of Acids and Bases03Strength of Acids and Bases03

HClO4

HI

HBr

HCl

H2SO4

HNO3

H3O+

HSO4–

HSO4–

HF

HNO2

HCOOH

NH4+

HCN

H2O

NH3

ClO4–

I–

Br –

Cl –

HSO4 –

NO3 –

H2O

SO42–

SO42–

F –

NO2 –

HCOO –

NH3

CN –

OH –

NH2 –

ACID CONJ. BASE ACID CONJ. BASE

Incr

easi

ng A

cid

Str

engt

h

Incr

easi

ng A

cid

Str

engt

h


• 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)

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

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


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

HF

HNO2

C9H8O4 (aspirin)

HCO2H (formic)

C6H8O6 (ascorbic)

C6H5CO2H (benzoic)

CH3CO2H (acetic)

HCN

C6H5OH (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


(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?

(42/2) = 8 12/5= 0.2 Very LargeK =

HA æ H+ + A

(M): 0.50 0.00 0.00 (M): –x +x +x

Equilib (M): 0.50 –x x x

Acid Ionization Constants Determine the pH of 0.50 M HA solution at 25°C. Ka = 7.1 x 10–4 05

Acid Ionization Constants Determine 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)

What is the pH of a 0.50 M Citric acid solution (at 250C)?

HA (aq) H+ (aq) + A- (aq) Ka =[H+][A-][HA]

= 7.1 x 10-4

HA (aq) H+ (aq) + A- (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+] = [A-] = 0.019 M pH = -log [H+] = 1.72

[HA] = 0.50 – x = 0.48 M


• 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.


• 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.

aacbb

x2

42

02 cbxax


• 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 %

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

Percent dissociation of a weak acid decreases as its concentration increases

Percent dissociation of a weak acid decreases as its concentration increases

• Concentration Dependence:

Weak Bases: Base Ionization Constants 01Weak 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+].

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.(caffeine)

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

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

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)

[NH3], M [NH4+], M [OH−], M

Initially 0.15 0 0

At Equilibrium 0.15 - x x x


(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


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

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

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

H2SO4

HSO4–

C2H2O4

C2HO4–

H2SO3

HSO3–

H2CO3

HCO3–

H2SHS–

H3PO4

H2PO4–

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

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 < HI299 kJ/mol for HI567 kJ/mol for HF


• 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.


• 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.


• 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.


• 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.

Polar Covalent Bonds 02Polar Covalent Bonds 02

Pauling ElectronegativitiesPauling Electronegativities

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

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


• Oxoacids of Chlorine:


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

a) HClO, HBrO, and HIO.

b) HNO3 and HNO2.

c) H3PO3 and H3PO4.

Acid-Base Properties of SaltsAcid-Base Properties of Salts

Strong basesStrong bases

• Strong bases:• The following metals make strong hydroxy base

• Alkali metal cations of group 1A • Alkaline earth metal cations of group 2A

except for Be

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.

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


Salts 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).

Salts That Contain Cation from a Weak Acid 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 10-4

5.6 x 10-11

1.0 x 10-14

Kb for CO32- =

Ka for HCO31-

Kw

=

= 5.6 x 10-10

1.8 x 10-5

1.0 x 10-14

Ka for NH41+ =

Kb for NH3

Kw

=

Basic, Ka < Kb

Hydrated Cation of Al3+


• Metal Ion Hydrolysis:


• 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

Slide 1 Chapter 14 Aqueous Equilibria: Acids and Bases.

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Transcript of Slide 1 Chapter 14 Aqueous Equilibria: Acids and Bases.