1 acids and bases copyright © the mcgraw-hill companies, inc. permission required for reproduction...

1

Acids and Bases

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2

Acids

Have a sour taste. Vinegar owes its taste to acetic acid. Citrusfruits contain citric acid.

React with certain metals to produce hydrogen gas.

React with carbonates and bicarbonates to produce carbon dioxide gas

Have a bitter taste.

Feel slippery. Many soaps contain bases.

Bases

3

A Brønsted acid is a proton donorA Brønsted base is a proton acceptor

acidbase acid base

acidconjugate

basebase conjugate

acid

4

O

H

H + O

H

H O

H

H H OH-+[ ] +

Acid-Base Properties of Water

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

H2O + H2O H3O+ + OH-

acid conjugate

base

base conjugate

acid

autoionization of water

5

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

The Ion Product of Water

Kc =[H+][OH-]

[H2O][H2O] = constant

Kc[H2O] = Kw = [H+][OH-]

The ion-product constant (Kw) is the product of the molar concentrations of H+ and OH- ions at a particular temperature.

At 250CKw = [H+][OH-] = 1.0 x 10-14

[H+] = [OH-]

[H+] > [OH-]

[H+] < [OH-]

Solution Is

neutral

acidic

basic

6

What is the concentration of OH- ions in a HCl solution whose hydrogen ion concentration is 1.3 M?

Kw = [H+][OH-] = 1.0 x 10-14

[H+] = 1.3 M

[OH-] =Kw

[H+]

1 x 10-14

1.3= = 7.7 x 10-15 M

7

pH – A Measure of Acidity

pH = -log [H+]

[H+] = [OH-]

[H+] > [OH-]

[H+] < [OH-]

Solution Is

neutral

acidic

basic

[H+] = 1 x 10-7

[H+] > 1 x 10-7

[H+] < 1 x 10-7

pH = 7

pH < 7

pH > 7

At 250C

pH [H+]

8

pOH = -log [OH-]

[H+][OH-] = Kw = 1.0 x 10-14

-log [H+] – log [OH-] = 14.00

pH + pOH = 14.00

Other important relationships

pH Meter

9

The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater?

pH = -log [H+]

[H+] = 10-pH = 10-4.82 = 1.5 x 10-5 M

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

pH + pOH = 14.00

pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60

pH = 14.00 – pOH = 14.00 – 6.60 = 7.40

10

Strong Electrolyte – 100% dissociation

NaCl (s) Na+ (aq) + Cl- (aq)H2O

Weak Electrolyte – not completely dissociated

CH3COOH CH3COO- (aq) + H+ (aq)

Strong Acids are strong electrolytes

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

HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)

HClO4 (aq) + H2O (l) H3O+ (aq) + ClO4- (aq)

H2SO4 (aq) + H2O (l) H3O+ (aq) + HSO4- (aq)

11

HF (aq) + H2O (l) H3O+ (aq) + F- (aq)

Weak Acids are weak electrolytes


HSO4- (aq) + H2O (l) H3O+ (aq) + SO4

2- (aq)

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

Strong Bases are strong electrolytes

NaOH (s) Na+ (aq) + OH- (aq)H2O

KOH (s) K+ (aq) + OH- (aq)H2O

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)H2O

12

F- (aq) + H2O (l) OH- (aq) + HF (aq)

Weak Bases are weak electrolytes

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

Conjugate acid-base pairs:

• H3O+ is the strongest acid that can exist in aqueous solution.

• The OH- ion is the strongest base that can exist in aqeous solution.

13

Strong Acid (HCl) Weak Acid (HF)

14

What is the pH of a 2 x 10-3 M HNO3 solution?

HNO3 is a strong acid – 100% dissociation.


pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7

Start

End

0.002 M

0.002 M 0.002 M0.0 M

0.0 M 0.0 M

What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?

Ba(OH)2 is a strong base – 100% dissociation.

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)

Start

End

0.018 M

0.018 M 0.036 M0.0 M

0.0 M 0.0 M

pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6

15

HA (aq) + H2O (l) H3O+ (aq) + A- (aq)

Weak Acids (HA) and Acid Ionization Constants

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

Ka =[H+][A-][HA]

Ka is the acid ionization constant

Ka

weak acidstrength

17

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 0.50Ka << 1

x2 = 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

18

percent ionization = Ionized acid concentration at equilibrium

Initial concentration of acidx 100%

For a monoprotic acid HA

Percent ionization = [H+]

[HA]0

x 100% [HA]0 = initial concentration

19

Molecular Structure and Acid Strength

H X H+ + X-

The stronger the bond

The weaker the acid

HF << HCl < HBr < HI

acidityincreases

21

Arrhenius acid is a substance that produces H+ (H3O+) in water

A Brønsted acid is a proton donor

A Lewis acid is a substance that can accept a pair of electrons

A Lewis base is a substance that can donate a pair of electrons

Definition of An Acid

H+ H O H••••

+ OH-••••••

acid base

N H••

H

H

H+ +

acid base

N H

H

H

H+

22

Lewis Acids and Bases

N H••

H

H

acid base

F B

F

F

+ F B

F

F

N H

H

H

No protons donated or accepted!

24

Chemical Kinetics


25

Chemical Kinetics

Thermodynamics – does a reaction take place?

Kinetics – how fast does a reaction proceed?

Reaction rate is the change in the concentration of a reactant or a product with time (M/s).

A B

rate = -[A]t

rate = [B]t

[A] = change in concentration of A over time period t

[B] = change in concentration of B over time period t

Because [A] decreases with time, [A] is negative.

26

A B

rate = -[A]t

rate = [B]t

27

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

time

393 nmlight

Detector

[Br2] Absorption

red-brown

t1< t2 < t3

28

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

average rate = -[Br2]t

= -[Br2]final – [Br2]initial

tfinal - tinitial

slope oftangent

slope oftangent slope of

tangent

instantaneous rate = rate for specific instance in time

29

2H2O2 (aq) 2H2O (l) + O2 (g)

PV = nRT

P = RT = [O2]RTnV

[O2] = PRT1

rate = [O2]t RT

1 Pt=

measure P over time

31

Reaction Rates and Stoichiometry

2A B

Two moles of A disappear for each mole of B that is formed.

rate = [B]t

rate = -[A]t

12

aA + bB cC + dD

rate = -[A]t

1a

= -[B]t

1b

=[C]t

1c

=[D]t

1d

32

Write the rate expression for the following reaction:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

rate = -[CH4]

t= -

[O2]t

12

=[H2O]

t12

=[CO2]

t

33

The Rate Law

The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.

aA + bB cC + dD

Rate = k [A]x[B]y

Reaction is xth order in A

Reaction is yth order in B

Reaction is (x + y)th order overall

34

F2 (g) + 2ClO2 (g) 2FClO2 (g)

rate = k [F2]x[ClO2]y

Double [F2] with [ClO2] constant

Rate doubles

x = 1

Quadruple [ClO2] with [F2] constant

Rate quadruples

y = 1

rate = k [F2][ClO2]

35

F2 (g) + 2ClO2 (g) 2FClO2 (g)

rate = k [F2][ClO2]

Rate Laws

• Rate laws are always determined experimentally.

• Reaction order is always defined in terms of reactant (not product) concentrations.

• The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.

1

36

Determine the rate law and calculate the rate constant for the following reaction from the following data:S2O8

2- (aq) + 3I- (aq) 2SO42- (aq) + I3

- (aq)

Experiment [S2O82-] [I-]

Initial Rate (M/s)

1 0.08 0.034 2.2 x 10-4

2 0.08 0.017 1.1 x 10-4

3 0.16 0.017 2.2 x 10-4

rate = k [S2O82-]x[I-]y

Double [I-], rate doubles (experiment 1 & 2)

y = 1

Double [S2O82-], rate doubles (experiment 2 & 3)

x = 1

k = rate

[S2O82-][I-]

=2.2 x 10-4 M/s

(0.08 M)(0.034 M)= 0.08/M•s

rate = k [S2O82-][I-]

37

Summary of the Kinetics of Zero-Order, First-Orderand Second-Order Reactions

Order Rate LawConcentration-Time

Equation Half-Life

0

1

2

rate = k

rate = k [A]

rate = k [A]2

ln[A] = ln[A]0 - kt

1[A]

=1

[A]0

+ kt

[A] = [A]0 - kt

t½ln 2k

=

t½ =[A]0

2k

t½ =1

k[A]0

38

Exothermic Reaction Endothermic Reaction

The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction.

A + B AB C + D++

39

Reaction Mechanisms

The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions.

The sequence of elementary steps that leads to product formation is the reaction mechanism.

2NO (g) + O2 (g) 2NO2 (g)

N2O2 is detected during the reaction!

Elementary step: NO + NO N2O2

Elementary step: N2O2 + O2 2NO2

Overall reaction: 2NO + O2 2NO2

+

40

2NO (g) + O2 (g) 2NO2 (g)

Mechanism:

41

Elementary step: NO + NO N2O2

Elementary step: N2O2 + O2 2NO2

Overall reaction: 2NO + O2 2NO2

+

Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation.

An intermediate is always formed in an early elementary step and consumed in a later elementary step.

The molecularity of a reaction is the number of molecules reacting in an elementary step.

• Unimolecular reaction – elementary step with 1 molecule

• Bimolecular reaction – elementary step with 2 molecules

• Termolecular reaction – elementary step with 3 molecules

42

Unimolecular reaction A products rate = k [A]

Bimolecular reaction A + B products rate = k [A][B]

Bimolecular reaction A + A products rate = k [A]2

Rate Laws and Elementary Steps

Writing plausible reaction mechanisms:

• The sum of the elementary steps must give the overall balanced equation for the reaction.

• The rate-determining step should predict the same rate law that is determined experimentally.

The rate-determining step is the slowest step in the sequence of steps leading to product formation.

43

Sequence of Steps in Studying a Reaction Mechanism

44

The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:

Step 1: NO2 + NO2 NO + NO3

Step 2: NO3 + CO NO2 + CO2

What is the equation for the overall reaction?

NO2+ CO NO + CO2

What is the intermediate?

NO3

What can you say about the relative rates of steps 1 and 2?

rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2

45

A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.

Ea k

ratecatalyzed > rateuncatalyzed

Ea < Ea′

Uncatalyzed Catalyzed

)/( RTEaeAk

46

In heterogeneous catalysis, the reactants and the catalysts are in different phases.

In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid.

• Haber synthesis of ammonia

• Ostwald process for the production of nitric acid

• Catalytic converters

• Acid catalysis

• Base catalysis

47

Catalytic Converters

CO + Unburned Hydrocarbons + O2 CO2 + H2Ocatalytic

converter

2NO + 2NO2 2N2 + 3O2

catalyticconverter

48

Enzyme Catalysis

49

Binding of Glucose to Hexokinase

50

rate = [P]t

rate = k [ES]

Enzyme Kinetics

51

All of the following may be true concerning catalysts and the reaction which they catalyze EXCEPT

a. catalysts are not used up by the reactionb. catalysts lower the activation energyc. catalysts increase the rate of the reverse reactiond. catalysts shift the reaction equlibrium to the right

52

As the temperature is increased in an exothermic gaseous reaction, all of the following increase EXCEPT

a. reaction rateb. rate constantc. activation energyd. all of the above

53

Which of the following changes to a reaction will always increase rate constant for that reaction?

a. decreaseing the temperature b. increasing the temperature

c. increasing the concentration of the reactants

d. increasing the concentration of the catalysts

54

When a radioactive isotope undergoes nuclear decay, the concentration of the isotope decreases exponentially with constant half live. It can be determined from this that radioactive decay is a

a. zeroth order reactionb. first order reactionc. second order reactiond. third order reactin

56

Chemical Equilibrium


57

Equilibrium is a state in which there are no observable changes as time goes by.

Chemical equilibrium is achieved when:

• the rates of the forward and reverse reactions are equal and

• the concentrations of the reactants and products remain constant

Physical equilibrium

H2O (l)

Chemical equilibrium

N2O4 (g)

H2O (g)

2NO2 (g)

NO2

58

N2O4 (g) 2NO2 (g)

Start with NO2 Start with N2O4 Start with NO2 & N2O4

equilibrium

equilibrium equilibrium

59

constant

60

N2O4 (g) 2NO2 (g)

= 4.63 x 10-3K = [NO2]2

[N2O4]

aA + bB cC + dD

K = [C]c[D]d

[A]a[B]bLaw of Mass Action

61

K >> 1

K << 1

Lie to the right Favor products

Lie to the left Favor reactants

Equilibrium Will

K = [C]c[D]d

[A]a[B]baA + bB cC + dD

62

Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.

N2O4 (g) 2NO2 (g)

Kc = [NO2]2

[N2O4]Kp =

NO2P 2

N2O4P

aA (g) + bB (g) cC (g) + dD (g)

Kp = Kc(RT)n

n = moles of gaseous products – moles of gaseous reactants

= (c + d) – (a + b)

In most cases

Kc Kp

63

Homogeneous Equilibrium

CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)

Kc =′[CH3COO-][H3O+][CH3COOH][H2O]

[H2O] = constant

Kc = [CH3COO-][H3O+]

[CH3COOH]= Kc [H2O]′

General practice not to include units for the equilibrium constant.

64

The equilibrium constant Kp for the reaction

is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm?2

2NO2 (g) 2NO (g) + O2 (g)

Kp = 2PNO PO

2

PNO2

2

PO2 = Kp

PNO2

2

PNO2

PO2 = 158 x (0.400)2/(0.270)2 = 347 atm

65

Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.

CaCO3 (s) CaO (s) + CO2 (g)

[CaCO3] = constant[CaO] = constant

Kc = [CO2] = Kp = PCO2

The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

[CaO][CO2][CaCO3]

Kc =′

[CaCO3][CaO]

Kc x′

66

If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.

Le Châtelier’s Principle

• Changes in Concentration

N2 (g) + 3H2 (g) 2NH3 (g)

AddNH3

Equilibrium shifts left to offset stress

67


• Changes in Concentration continued

Change Shifts the Equilibrium

Increase concentration of product(s) left

Decrease concentration of product(s) right

Decrease concentration of reactant(s)

Increase concentration of reactant(s) right

left

aA + bB cC + dD

AddAddRemove Remove

68


• Changes in Temperature

Change Exothermic Rx

Increase temperature K decreases

Decrease temperature K increases

Endothermic Rx

K increases

K decreases

colder hotter

N2O4 (g) 2NO2 (g)

69

Catalyst lowers Ea for both forward and reverse reactions.

Catalyst does not change equilibrium constant or shift equilibrium.

• Adding a Catalyst• does not change K• does not shift the position of an equilibrium system• system will reach equilibrium sooner


70

Le Châtelier’s Principle - Summary

Change Shift EquilibriumChange Equilibrium

Constant

Concentration yes no

Pressure yes* no

Volume yes* no

Temperature yes yes

Catalyst no no

*Dependent on relative moles of gaseous reactants and products

71

As the temperature is increased, the equilibrium of gaseous reaction will always:

a. shift to the rightb. shift to the leftc. remain constantd. the answer cannot be determined from the

information given

72

All of the following are true concerning a reaction at equilibrium EXCEPT:

a. the rate of the forward reaction equals the rate of the reverse reaction

b. There is no change in the concentrations of both the products and the reactants

c. The activation energy has reached zerod. All reactants will start to move forward at

equilibrium

73

What is the equilibrium expression for the following reaction

CaCO3(s)CaO(s)+ CO2(g)

a. K=[CO2]b. K=[CaO] [CO2]c. K=[CaO] [CO2]/ [CaO]d. K= [CO2]/ [CaO]

1 acids and bases copyright © the mcgraw-hill companies, inc. permission required for reproduction...

Documents

h aq oh aq h

log h h

hf aq h

f aq h

kwkw h

log oh h oh

aq slide

o aq oh aq strong bases