Acid-base-Lecture 7(05)

1

2

Acids release H+ in solutionBases release OH- in solution.Examples: HCl H+ + Cl-

NaOH Na+ + OH-

Arrhenius Definition

Acids Acids & & BasesBases

3

• Arrhenius definition of acids and bases has limitations.

• It only applies to aqueous solutions.

• Bronsted and Lowry proposed a definition based on acid base reactions transferring H+ ion from one substance to another.

4

• H+ ions are bare protons

• H+ will react strongly with the nonbonding pair of electrons in a water molecule to form: Hydronium ion (H3O+).

5

Often times use of the H+ or H3O+ are used interchangeably to represent acid solutions

• HCl H+ + Cl-

• HCl(g) + H2O(l) H3O+(aq) + Cl-

(aq)

6

Acids are proton donors

Bases are proton acceptors.Example:

HBr + H2O H3O+ + Br-

The Bronsted Definition

A B

7

• An acid and a base always work together to transfer a proton.

• Bronsted acids must have an H+ to transfer.

• Bronsted base must have a nonbonding pair of electrons that can bind with H+.

8

Substances may act as an acid or a base

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

(aq)

NH3(aq)+H2O(l)NH4+

(aq)+ OH-(aq)

In the 1st reaction above water accepts a proton (Bronsted base).

Rx 2, it donates a proton (Bronsted acid)

Amphoteric substances

9

• Acid base equilibrium: – both the forward and reverse

reactions involve proton transfer.

HX + H2O X- + H3O+

Forward: HX is acid/ H2O is the base

Reverse: H3O+ acid/ X- is base

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HX + H2O X- + H3O

• An acid and a base that differ only in the presence or absence of H+ are called a Conjugate acid-base Pair.

• Every acid has a conjugate base.

• Every base has a conjugate acid.– HX is the conjugate acid of X-

– H3O is the conjugate acid of H2O

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Example:Identify Bronsted acid & base; also The conjugate acid and base.

NH3 + H2O NH4+ + OH-

AB CA CB

NH3 acts as a Bronsted base by accepting a proton.

Water acts as a Bronsted acid by donating a proton.

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Example:

Write the formula of the conjugate base of H2SO4 in aqueous solution.

HSO4-

H2SO4 + H2O HSO4- + H3O+

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Example : Consider the reaction

HSO4- + HCO3

- SO4-2 + H2CO3

Identify the acids and bases for the forward and reverse reactions.Identify the conjugate acid-base pairs

A B CB CA

14

Bases are electron donorsAcids are electron acceptors

Lewis Acids and Bases

Example: H2O + NH3 OH- + NH4

+

Is really: H2O + :NH3 OH- + H:NH3

+

Electron pair donor(NH3)

electron pair acceptor(H+)

15

Identify the Lewis acid and Lewis baseIn each of the following reactions:

a) SnCl4(s)+ 2Cl-(aq) SnCl6

2-(aq)

b) Hg2+(aq)+ 4CN- Hg(CN)4

2-(aq)

a)SnCl4 accepts 2 pair of e- from Cl- (Lewis acid) /Cl- is Lewis base.

b) Hg2+ accepts 4 pair of e- (Lewis Acid)

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Lewis system defines acids and bases without a solvent.

Bronsted system defines acids and bases in solvents other than water.

Arrhenius defines acids and bases in water.

Summary of Definitions

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Acid and Base StrengthAcid and Base Strength

• Strong acids completely transfer their protons to water (completely completely ionized).ionized).

• Weak acids are those that only partially dissociate in aqueous solution.

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Know the following STRONG bases:

All hydroxides of the alkali & alkaline earth metals (except Be)

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Strong acids and bases dissociate 100%

Example: NaOH Na+ + OH-

Na+Na+

Na+

OH-

OH-

OH-

1 M 1 M 1 M

20

Example: HCl H+ + Cl-

H+

H+

H+

Cl-

Cl-

Cl-

1 M 1 M 1 M

21

HF

HF

HF

HF

Weak acids and bases dissociate X%

Example: HF H+ + F-

H+

F-

1 M ? ?.0x .0x

X%

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% ionization = amount ionized x 100 initial con’c

Ionization %100

conc Initial

H

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Weak makes strong &

strong makes weak

The stronger the acid, the weaker its conjugate base.

The stronger the base, the weaker its conjugate acid.

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• Substance with negligible acidity (CH4) contain hydrogen but show no acidic behavior in water.

• The conjugate bases are strong bases reacting with water completely and forming OH- ions.

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• When the acid is a stronger acid than H3O+, the equilibrium lies to the right.

HCl + H2OH3O+ + Cl-

When H3O+ is the stronger acid, the equilibrium lies to the left.

HC2H3O2 + H2OH3O+ + C2H3O2-

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In all Acid-base reactions, the equilibrium favors the transfer of the proton from the stronger acid

to the stronger base.

• Which direction is favored:a. PO4

3-(aq) + H2O(l) HPO4

2-(aq)+ OH-

(aq)

b. NH4+

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

a. Left b. right

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Example :

The strengths of the following acids increase in the order HCN < HF < HNO3. Arrange the conjugate bases of these acids in order of increasing base strength.

CN- > F- > NO3-

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• Water can act as a Bronsted acid(donate proton) or base (accept proton) depending on the environment.

• One water can donate a proton to another water molecule, this is the autoionization of water.

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Autoionization is a rapid equilibrium process.

• The equilibrium constant expression:

22

3

OH

OHOHKc

H2O + H2O H3O+ + OH-

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• For pure water [H2O] is a constant. We rearrange the equation with two constants on the left and define a new constant:

• Kw (ion product constant of H2O)

Kw=[H3O+][OH-]=[H+][OH-]

Kw= 1.0 x 10-14 at 25ºC

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• When [H+]=[OH-] the solution is neutral.

• When the concentration of one increases the other decreases, the product of the 2 equals 1.0 x 10-14 at 25ºC.

• [H+]>[OH-]the solution is acidic

• [H+]<[OH-]the solution is basic

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Arrhenius:

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

(aq)

Kw = [H+][OH-]

The Amphoteric Property of Water

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Bronsted: H2O(l) + H2O(l) H3O+

(aq) + OH-(aq)

Kw = [H3O+][OH-]

The Amphoteric Property of Water

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Kw = [H3O+][OH-] = 1.0 x 10-14

The Ion Product Constant of Water

At equilibrium [H+] = [OH-] = 1.0x10-7 M.

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Common way to express [H+]The negative logarithm of the

hydrogen ion concentration in mol/L (M).

pH = -log[H+]

pH

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• pH decrease as [H+] increases.

• pH of a neutral solution– pH= -log (1.0x10-7) = -(-7.0)= 7.0

• What is the pH of a solution with [OH-]=3.0 x 10-4?

OH

KH w 11

4-

14

103.310 x 3.0

1.0x10

x

10.48)log(3.3x10pH 11

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pOH = - log [OH-]

pH + pOH =-log Kw= 14

The negative log of a quantity is the “p” of that quantity.

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Example

In a NaOH solution [OH-] is 2.9 x 10-4M. Calculate the pH.

OHlogpOH

54.310 x 2.9-logpOH -4

10.46 3.54-14 pOH-14pH

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[H+] means that H+ were released into the solution making it acidic.

[OH-] means that OH- were released into the solution making it basic.

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Example

The concentration of OH- ion in a certain household ammonia cleaning solution is 0.0025 M. Calculate the concentration of the H+ ions.

M4.0x10

0.0025

10 x 1.0

OH

KH 12

-14w

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Example: The H+ ion concentration in a certain solution is 5.0 x 10-5 M. What is the OH- ion concentration?

Kw = [H3O+][OH-]

[OH-] = Kw

[H+] = 1.0 x 10-14

5.0 x 10-5M

= 2.0 x 10-10 M

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• pH Meters• Acid-base indicators

–These have an acid form and a base form that differ in color.

–Litmus• Red indicate a pH of less than 5• Blue indicates pH above 8

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Ionization of Water

Occasionally, in water, a H+ is transferred between H2O molecules

. . . . . . . .H:O: + :O:H H:O:H + + :O:H-

. . . . . . . . H H H

water molecules hydronium hydroxide ion (+) ion (-)

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Pure Water is Neutral

Pure water contains small, but equal amounts of ions: H3O+ and OH-

H2O + H2O H3O+ + OH-

hydronium hydroxide

ion ion

1 x 10-7 M 1 x 10-7 M

H3O+ OH-

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Ion Product of Water Kw

[ ] = Molar concentration

Kw = [ H3O+ ] [ OH- ]

= [ 1 x 10-7 ][ 1 x 10-7 ]

= 1 x 10-14

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Acids

Increase H+

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

(aq)

More [H3O+] than water > 1 x 10-7M

As H3O+ increases, OH- decreases

[H3O+] > [OH-]H3O+

OH-

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Bases

Increase the hydroxide ions (OH-)

H2O

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

More [OH-] than water, [OH-] > 1 x 10-7M

When OH- increases, H3O+ decreases

[OH] > [H3O+]

H3O+OH-

48

Using Kw

The [OH- ] of a solution is 1.0 x 10- 3 M. What is the

[H3O+]?

Kw = [H3O+ ] [OH- ] = 1.0 x 10-14

[H3O+] = 1.0 x 10-14

[OH-]

[H3O+] = 1.0 x 10-14 = 1.0 x 10-11 M

1.0 x 10- 3

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Learning Check pH-1

The [H3O+] of lemon juice is 1.0 x 10-3 M.

What is the [OH-] of the solution?

1) 1.0 x 103 M

2) 1.0 x 10-11 M

3) 1.0 x 1011 M

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Solution pH-1

The [H3O+] of lemon juice is 1.0 x 10- 3 M.

What is the [OH-]?

[OH- ] = 1.0 x 10 -14 = 1.0 x 10-11 M 1.0 x 10 - 3

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Using the Calculator

1.0 x 10 -14

4.0 x 10-5

Enter 1.0 EE +/- 14 4.0 EE +/- 5

= 2.5 x 10 -10

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Learning Check pH-2

The [OH-] of a solution is 5 x 10 -5 M. What is the [H3O+ ] of the solution?

1) 2 x 10- 5 M

2) 1 x 1010 M

3) 2 x 10-10 M

53

Solution pH-2

The [OH-] of a water solution is 5 x 10-5 M.

What is the [H3O+] in the solution?

[ H3O+] = 1.0 x 10 -14

5 x 10- 5

On some calculators:

1.0 EE +/- 14 5 EE +/- 5 = 2 x 10 -10 M

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Learning Check pH-3

A.The [OH-] when [H3O+ ] of 1 x 10- 4 M

1) 1 x 10-6 M

2) 1 x 10-8 M

3) 1 x 10-10 M

B.The [H3O+] when [OH- ] of 5 x 10-9 M

1) 1 x 10- 6 M

2) 2 x 10- 6 M

3) 2 x 10-7 M

55

Solution pH-3

Kw = [H3O+ ][OH-] = 1.0 x 10 14

A. (3) [OH- ] = 1.0 x 10 -14 = 1.0 x 10 -10

1.0 x 10- 4

B. (2) [H3O+] = 1.0 x 10 -14 = 2 x 10 - 6

5 x 10- 9

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pH

Indicates the acidity [H3O+] of the solution

pH = - log [H3O+]

From the French pouvoir hydrogene

(“hydrogen power” or power of

hydrogen)

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In the expression for [H3O+]

1 x 10-exponent

the exponent = pH

[H3O+] = 1 x 10-pH M

pH

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pH Range

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Neutral

[H+]>[OH-] [H+] = [OH-] [OH-]>[H+]

Acidic Basic

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Some [H3O+] and pH

[H3O+] pH

1 x 10-5 M 5

1 x 10-9 M 9

1 x 10-11 M 11

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pH of Some Common Acids

gastric juice 1.0

lemon juice 2.3

vinegar 2.8

orange juice 3.5

coffee 5.0

milk 6.6

61

pH of Some Common Bases

blood 7.4

tears 7.4

seawater 8.4

milk of magnesia 10.6

household ammonia 11.0

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Learning Check pH-4

A. The [H3O+] of tomato juice is 1 x 10-4 M.

What is the pH of the solution?

1) - 4 2) 4 3) 8

B. The [OH-] of an ammonia solution is

1 x 10-3 M. What is the pH of the solution?

1) 3 2) 11 3) -11

63

Solution pH-4

A. pH = - log [ 1 x 10-4] = -(- 4) = 4

B. [H3O+] = 1 x 10-11

pH = - log [ 1 x 10- 11] = -(- 11) = 11

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Learning Check pH-5

The pH of a soap is 10. What is the [H3O+] of the soap solution?

1) 1 x 10 - 4 M

2) 1 x 1010 M

3) 1 x 10 - 10 M

65

Solution pH-5

The pH of a soap is 10. What is the [H3O+] of

the soap solution?

[H3O+] = 1 x 10-pH M

= 1 x 10-10 M

66

pH on the Calculator

[H3O+] is 4.5 x 10-6 M

pH = 4.5 x EXP(or EE) 6+/- LOG +/-

= 5.35

67

Learning Check pH-6

A soap solution has a [H3O+] = 2 x 10-8 M. What is the pH of the

solution?

1) 8

2) 7.7

3) 6

68

Learning Check pH-7

Identify each solution as

1. acidic 2. basic 3. neutral

A. _____ HCl with a pH = 1.5

B. _____ Pancreatic fluid [H+] = 1 x 10-8 M

C. _____ Sprite soft drink pH = 3.0

D. _____ pH = 7.0

E. _____ [OH- ] = 3 x 10-10 M

F. _____ [H+ ] = 5 x 10-12

69

Solution pH-7

Identify each solution as

1. acidic 2. basic 3. neutral

A. _1__ HCl with a pH = 1.5

B. _2__ Pancreatic fluid [H+] = 1 x 10-8 M

C. _1__ Sprite soft drink pH = 3.0

D. _3__ pH = 7.0

E. _1__ [OH-] = 3 x 10-10 M

F. _2__ [H+] = 5 x 10-12

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• Methyl violet

• Thymol Blue

• Methyl orange

• Methyl red

• Bromthymol blue

• Phenophthalein

• Alizarin yellow R

71

• Both are strong electrolytes

– Completely ionize in aqueous solution.

• Strong acids

– 6 of the 7 are monoprotic (HX)• Exist as only H ion and x ions

HI(aq) H+(aq) + I-

(aq)

2M HI = [H+]= [I-] = 2M

72

Determining the pH of a Strong AcidDetermining the pH of a Strong Acid

pH = - log[H+]

For a monoprotic strong acid[A] = [H+]

73

All hydroxides of the alkali & alkaline earth metals (except Be).

Strong basic solutions can be formed when substances react with water to form OH-.• Ionic metal oxides (Na2O,CaO)• Each mole of O-2 forms 2 moles OH-

• Na2O(s) + H2O(l)2Na+(aq)+2OH-

(aq)

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• Ionic hydrides and nitrides also react with water in this way.

• Because the anions O2-,H-, and N3-, are all stronger bases than OH-(the conjugate base of H2O) they are able to remove a proton from water.

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• Partially ionize in aqueous solution. HX(aq) + H2O(l) H3O+

(aq)+ X+(aq)

HX(aq) H+(aq)+ X-

(aq)

• Equilibrium constant expression: HX

XHKa

76

• Ka is the acid-dissociation constant or The Acid Ionization Constant

• Subscript a indicates this is the equilibrium constant of an acid.

• The magnitude of Ka indicates the tendency of Hydrogen to ionize.

77

Ka = [H+][X-] [HX]

HX H+ + X-

The larger the Ka, the stronger the acid.

The Acid Ionization Constant (Ka)

78

Determining the pH of a weak Acid

pH = - log[H+]

For a weak acid.0x[A] = [H+]

79

Calculations with Ka & pH

• A given pH of a solution will always represents an equilibrium condition.

• Proton transfer reactions are usually very rapid.

80

Using Kato calculate pH• Calculate the pH of 0.30M acetic

acid at 25ºC if the Ka= 1.8 x 10-5.

5

232

232 108.1

xOHHC

OHCHKa

Step 2

Step 1 HC2H3O2 H+ + C2H3O2-

81

HC2H3O2 H+ + C2H3O2-

Initial 0.30M 0.0 0.0

-x +x +x

Equil. (0.30-x) x x

Step 3

82

Step 4

x

xx

OHHC

OHCHKa

30.0108.1

25

232

232

We are able to make the assumption that 0.30-x is equal to 0.30.

83

Solving for x

30.0108.1

25 x

x 62 104.5 xx

3103.2 xx 3103.2 xxH

64.2103.2log 3 xpH

84

What is the % ionization of the solution in the previous problem?

%77.01000.30M

0.0023M Ionization % x

85

• The Ka for niacin is 1.5 x10-5. What is the pH of a 0.010M solution of niacin (HC6H4O2N)?

• For structure of niacin see organic text book.

• Step 1

HC6H4O2N H+ + C6H4O2N-

86


• Step 2

5

246

246 105.1

xNOHHC

NOHCHKa

87


Initial 0.010M 0.0 0.0

-x +x +x

Equil. (0.010-x) x x

88


x

xx

NOHHC

NOHCHKa

010.0105.1

25

246

246

010.0105.1

25 x

x Solve for x

72 105.1 xxX=3.9 x10-4

41.3109.3log 4 xpH

89

• More than 1 ionizable H atom.• These H atoms ionize in

separate steps;• The acid dissociation constants

for each H atom are designated Ka1, Ka2, and so on.

90

H3PO4 H+ + H2PO4- (Ka1)

H2PO4- H+ + HPO4

2- (Ka2)

• It is always easier to remove the first proton.– Thus Ka1 is always larger than Ka2

– Ka2 is larger than Ka3

91

Ionizations Occur in Steps

H2SO3 H+ + HSO3-

HSO3- H+ + SO3

-2

Ka1 = 1.3 x 10-2

Ka2 = 6.3 x 10-8

SO3-2

H2SO3

H2SO3H2SO3

H2SO3H+

H+

HSO3- HSO3

-

92

Calculate the concentrations of all species at equilibrium in a 0.10M solution of oxalic acid.

• Oxalic acid Ka1= 5.9 x 10-2

Ka2= 6.4 x 10-5

H2C2O4 H+ + HC2O4-

93

H2C2O4 H+ + HC2O4-

Initial 0.10 0.0 0.0

-x +x +x

Equil. (0.10-x) x x

94

H2C2O4 H++ HC2O4-

2

422

421 109.5

10.0

)(

xx

xx

OCH

OHCHKa

10.0

109.52

2

x

xx

x2 + 5.9 x 10-2x –5.9 x 10-3 = 0

x= 0.053M

95

• When the equilibrium for the first stage of ionization is reached the concentrations are:

[H+] = 0.053M[C2HO4

-] = 0.053M

[C2H2O4] = (0.10-0.053)=0.047MNext consider 2nd stage of ionization

96

HC2O4- H++ C2O4

2-

Initial 0.053 0.053 0

Change -y +y +y

Equil. 0.053-y 0.053+y y

97

HC2O4- H++ C2O4

2-

5

42

242

1 104.6053.0

)(053.0

x

y

yy

OHC

OCHKa

My

x 5-5 10 x 6.4 yor 053.0

)(053.0104.6

98

• At equilibrium:[C2H2O4] = 0.047

[C2HO4-] = (0.053-6.4x10-5)=0.053

[H+]= (0.053+6.4x10-5) = 0.053[C2O4

2-]= 6.4 x 10-5

[OH-]= Kw/[H+]=1.9 x 10-13

99

• In water many substances act as weak bases.– Weak base + H2O Conj. Acid + OH-

• Remember that ammonia (NH3) is a weak base.– NH3(aq)+ H2O(l) NH4

+ + OH-(aq)

100

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

(aq)

• The equilibrium constant expression:

OHNH

OHNHK

23

4

3

42 NH

OHNHOHKKb

•Because water concentration is constant, it is incorporated into the creating Kb.

101

Kb (base dissociation constant)

• It is the equilibrium concentration in which the base reacts with water to form the conjugate acid and OH-.

• A lone pair of electrons is needed to bond with the H+

.

102

The Base Ionization ConstantKb

Kb = [OH-][BH+] [B]

B + H2O OH- + BH+

Indicates strength of the base(as Kb goes up more ionization)

103

What is the pH of a 0.400M solution of ammonia?Kb= 1.8 x 10-5

H2O(l) + NH3(aq) NH4+

(aq)+OH-(aq)

Initial 0.400 0.00 0.00

Change -x +x +x

Equil. 0.400-x x x

104

H2O(l) + NH3(aq) NH4+

(aq)+OH-(aq

5

3

4 108.1

xNH

OHNHKb

552

108.1400.0

or 108.1400.0

2

xx

xx

x

equilibrumat 107.2 3MxxOH

pOH= -log(2.7x10-3) =2.57

105

Calculate the pH of a 0.050M HNO2 (nitrous acid) solution.

Ka= 4.5 x 10-4

HNO2(aq) H+(aq) + NO2

-(aq)

Initial 0.050 0.00 0.00Change -x +x +xEquil. (0.050-x) x x

106

4

2

2 105.4

xHNO

NOHKa


-(aq)

42

105.4050.0

xx

x

0x103.210x5.4 -542 xx

107

)1(2

)103.2)(1(4)105.4( 105.4 524-

4

XXx

[H+]X= 4.6 x 10-3 or –5.0 x 10-3

pH=-log(4.6x10-3)=2.34


-(aq)

0x103.210x5.4 -542 xx

108

• Divided into 2 categories1. Neutral substances that have a

nonbonding pair of electrons. Most of these bases contain N,

these are called amines.

109

2. Second category is composed of anions of weak acids.

NaClO in aqueous solution: Na+ + ClO-

In an acid base reaction, the Na+ is a spectator ion.

The ClO- is the conjugate base of a weak acid. (ClO- acts as a weak base)

ClO-(aq)+ H2O(l) HClO(aq) + OH-

(aq)

110

Consider NH4+&NH3 (conjugate

acid-base pair)

NH4+

(aq) NH3(aq)+ H+(aq)

NH3(aq)+ H2O(l) NH4+

(aq)+OH-(aq)

More attention with Ka & Kb

3

4

4

3 NH

OHNHK

NH

HNHK ba

111

When 2 reactions are added together to give a 3rd, the equilibrium constant for the 3rd is equal to the product of the 2 reactions.

NH4+



(aq)+OH-(aq)

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

(aq)

112

Reaction 1 + Reaction 2 = Rx 3K1 x K2 = K3

NH4+



(aq)+OH-(aq)

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

(aq)

3

4

4

3ba K x K

NH

OHNH

NH

HNH

113

• Ka x Kb= Kw

• As Ka gets larger, Kb gets smaller.• It is possible to calculate Kb for

any weak base if the Ka for the conjugate acid is known.

wKOHH ba K x K

pKa + pKb = pKw = 14.00

114

KaKb = Kw

HC2H3O2 H+ + C2H3O2- Ka

Ka = [H+][C2H3O2-]

[HC2H3O2]

C2H3O2- + H2O HC2H3O2 + OH- Kb

Kb = [HC2H3O2][OH-] [C2H3O2

-]

H2O H+ + OH- KaKb = Kw

115

Ka Kb = [H+][C2H3O2-]

[HC2H3O2]X [HC2H3O2][OH-] [C2H3O2

-]

H2O H+ + OH- KaKb = Kw

= [H+][OH-]Kw

116

Salt solutions are strong electrolytes and ionize in water

Acid-base properties of salt solutions are a result of the anions and cations formed.

117

Salt - an ionic compound formed by the reaction of an acid and a base

NaCl

NaOH HCl

strongbase

strongacid

118

NaNO3

NaOH HNO3

strongbase

strongacid

119

NaC2H3O2

NaOH HC2H3O2

strongbase

weakacid

Remember! Weak strong conjugate

120

Hydrolysis

• Many ions react with water to form H+ or OH-

• This reaction is hydrolysis.• The anions from weak acids react

with water to form OH-.• Conjugate base is strong.

C2H3O2-+H2O(l) HC2H3O2 + OH-

121

• The anion of a strong acid (NO3-)

forms a weak conjugate base.–It does not react with water and

does not affect pH.

• Anions with an ionizable proton (HPO4

2-) are amphoteric.–Reaction with water will be

determined by relative Ka and Kb.

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Will a solution of the salt Na2HPO4 be acid or basic?

• HPO42-

(aq) H+(aq) + PO4

3-(aq)

–Ka=4.2 x 10-13

• HPO42-

(aq) H2PO4-(aq) +OH-

–Calculate Kb from the Ka of the conjugate acid (H2PO4

-)

123

• Ka x Kb = Kw

Ka (HPO42-)=4.2 x 10-13

Ka (H2PO4-)= 6.2x 10-8

Kb(HPO42-) x Ka(H2PO4

-)=1.0x10-14

Kb (HPO42-) = 1.6 x 10-7

HPO42-

(aq) H2PO4-(aq) +OH-

Solution is basic

124

• Cations of alkali metal and alkaline earth metals do not hydrolyze in water.

• They do not influence pH.• The pH of a salt solution can be

qualitatively predicted by considering the cation and anion composing the salt.

125

Consider the relative strengths of the acids and bases that the salt is

derived from.

1. Salts derived from strong acid and strong base. NaCl and NaNO3

Neither the cation or anion hydrolyses.

No change in pH (pH=7)

126

NaNO3

NaOH HNO3

strongbase

strongacid

Strong weak conjugates

NO HYDROLYSIS

127

2. Salts of strong base and weak acid. NaClO and Ba(C2H3O2)

Anion hydrolyses/ Cation does not

Basic (pH above 7)

128

NaC2H3O2

NaOH HC2H3O2

strongbase

weakacid

Weak strong conjugate

C2H3O2- + H2O HC2H3O2 + OH-

NaC2H3O2 is a basic salt

129

3. Salt of a weak base and strong acid. NH4Cl and Al(NO3)3

The cation hydrolyses and the anion does not.

Acid (pH less than 7)

130

NH4Cl

NH3 HCl

weakbase

strongacid

Weak strong conjugate

NH4+ + H2O NH3 + H3O+

NH4Cl is an acidic salt

131

4. Salt of a weak base and weak acid. NH4C2H3O2 and FeCO3

Both cation and anion exhibit hydrolyses.

pH depends on the extent of each hydrolysis.

Relative Ka and Kb of ions.

132

Salts formed by weak acid + weak base can be either acidic OR basic

Kb < Ka acidic

Kb > Kabasic

Kb Kanearly neutral

133

Example 1: Write net ionic equations to show which of the following ions hydrolyze in aqueous solution:a. NO3

-

NO3- + H2O NO RXN

134

b. NO2-

NO2- + H2O HNO2 + OH-

BASIC

135

c. NH4+

NH4+ + H2O NH3 + H3O+

ACIDIC

136

Example 2: Predict whether the following aqueous solutions will be acidic, basic, or neutral:

KINH4IKC2H3O2

(answers: neutral, acid, base)

137

138

• When substances are placed in water some act as acids, some as bases, and others are neutral.

• Structure of the compound plays a role in the behaviors.

139

Acid StrengthAcid strength of binary acids depends on:

1) Polarity of the bond (P = acidity) H X

2) Strength of the bond Strong bonds are more difficult to

break than weak bonds.(strength= acidity)

This explains why the very polar HF is a weak acid.

3) Stability of conjugate base (X-)

140

Binary versus Oxyacids

Example: HCl HClO3

141

• The strength of the HX bond is usually the determining factor of acid strength when the X is from the same group in the periodic table.– Strength of the bond decreases as the

element (X) increases in size – acidity with radius

–HI > HBr > HCl

142

• Bond strength changes much less across a period.

• Bond polarity becomes the determining factor for binary acids in the same row.

• Acidity increases with increasing electronegativity across a row.– Acidity increase left to right.

143

Which is stronger?HCl or HFHCl or HBrHCl or H2SCH4, NH3 ,H2O ,or HF

144

Contain 1 or more O-H bonds

One or more OH groups or additional oxygen atoms are bonded to the central atom.

145

• Acid strength of oxyacids acids depends on the central atom

1) If the oxyacid has the same number of O atoms and the same number of OH groups; strength increases with increasing electronegativity of the central atom.

2) Oxyacids with same central atomAcid strength increases as the number of

O atoms attached to central atom increases.

146

Example:

Which is stronger?HClO4 or HBrO4

HClO4 or HClO3

HClO4

HClO4

147

• Contain carboxyl groups–COOH

• Strength of carboxylic acids increase as the number of electronegative atoms in the acid increases.

148

• Carboxylic acids are the largest group of organic acids.

• Acetic acid is a carboxylic acid–HC2H3O2 (CH3COOH)

• Which is the stronger acid–CF3COOH or HC2H3O2

CF3COOH

149

Predict the relative strengths of the oxoacids in each of the following groups:

HClO, HBrO, and HIO

HClO > HBrO > HIO

HNO3 and HNO2HNO3>HNO2

150

Acid Base Reactions

Case 1: strong acid + strong baseNaOH + HCl NaCl + HOH

Na+ + OH- + H+ + Cl- Na+ + Cl- + H2O

Net: H+ + OH- H2OEquimolar solution will be NEUTRAL

Write the net ionic equation and decide whether solution is acidic, basis, or neutral

151

Case 2: weak acid + strong base

HC2H3O2 + NaOH NaC2H3O2 + HOH

HC2H3O2

Net: HC2H3O2 + OH- C2H3O2-+ H2O

solution will be BASIC

+ Na+ + OH- Na+ + C2H3O2

- + H2O

152

HC2H3O2 + OH- C2H3O2-+ H2O

C2H3O2-+ H2O HC2H3O2 + OH-

WEAK ACID STRONG CONJUGATE BASE

153

+ NH3

Case 3: strong acid + weak base

HCl + NH3 NH4Cl

H+ + Cl-

Net: H+ + NH3 NH4+

solution will be ACIDIC

NH4+ + Cl-

154

H+ + NH3 NH4+

NH4+ + H2O NH3 + H3O+

WEAK BASE STRONG CONJUGATE ACID

155

Case 4: weak acid + weak base

?Both form strong conjugates. It depends on the the reaction

156

Bases are electron donorsAcids are electron acceptors

Lewis Acids and Bases

Example: H2O + NH3 OH- + NH4

+

Is really: H2O + :NH3 OH- + H:NH3

+

Electron pair donor(NH3)

electron pair acceptor(H+)

157

Hydrolysis of Metal Ions• The Lewis concept helps explain why

some metal ions display acidic properties in solution.

• Positively charged metal ions attract unshared electrons pairs in water molecules.– This attraction is what causes hydrolysis,

dissolving of salts.

158

• Water molecules bound to the metal ion are more acidic than those in the solution.

• This effect usually has increasing acidity (Ka hydrolysis) with increasing charge of the metal ion and decreasing radius of the ion.

159

•end

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Acid-base-Lecture 7(05)