Prentice-Hall © 2007General Chemistry: Chapter 16Slide 1 of 52

16-6 Polyprotic Acids

H3PO4 + H2O H3O+ + H2PO4-

H2PO4- + H2O H3O+ + HPO4

2-

HPO42- + H2O H3O+ + PO4

3-

Phosphoric acid:

A triprotic acid.

Ka = 7.110-3

Ka = 6.310-8

Ka = 4.210-13

Prentice-Hall © 2007General Chemistry: Chapter 16Slide 2 of 52

Phosphoric Acid

Ka1 >> Ka2

◦ All H3O+ is formed in the first ionization step.

H2PO4- essentially does not ionize further.

◦ Assume [H2PO4-] = [H3O+].

[HPO42-] Ka2

regardless of solution molarity.

Prentice-Hall © 2007General Chemistry: Chapter 16Slide 3 of 52

Prentice-Hall © 2007General Chemistry: Chapter 16Slide 4 of 52

Calculating Ion Concentrations in a Polyprotic Acid Solution. For a 3.0 M H3PO4 solution, calculate:

(a) [H3O+]; (b) [H2PO4-]; (c) [HPO4

2-] (d) [PO43-]

H3PO4 + H2O H2PO4- + H3O+

Initial conc. 3.0 M 0 0

Changes -x M +x M +x M

Equilibrium (3.0-x) M x M x MConcentration

EXAMPLE 16-9

Prentice-Hall © 2007General Chemistry: Chapter 16Slide 5 of 52

H3PO4 + H2O H2PO4- + H3O+

[H3O+] [H2PO4-]

[H3PO4] Ka=

x · x

(3.0 – x)=

Assume that x << 3.0

= 7.110-3

x2 = (3.0)(7.110-3) x = 0.14 M

[H2PO4-] = [H3O+] = 0.14 M

EXAMPLE 16-9

Prentice-Hall © 2007General Chemistry: Chapter 16Slide 6 of 52

H2PO4- + H2O HPO4

2- + H3O+

[H3O+] [HPO42-]

[H2PO4-]

Ka=y · (0.14 + y)

(0.14 - y)= = 6.310-8

Initial conc. 0.14 M 0 0.14 M

Changes -y M +y M +y M

Equilibrium (0.14 - y) M y M (0.14 +y) MConcentration

y << 0.14 M y = [HPO42-] = 6.310-8

EXAMPLE 16-9

Prentice-Hall © 2007General Chemistry: Chapter 16Slide 7 of 52

HPO4- + H2O PO4

3- + H3O+

[H3O+] [HPO42-]

[H2PO4-]

Ka=(0.14)[PO4

3-]

6.310-8 = = 4.210-13 M

[PO43-] = 1.910-19 M

EXAMPLE 16-9

Prentice-Hall © 2007General Chemistry: Chapter 16Slide 8 of 52

Sulfuric Acid

Sulfuric acid:

A diprotic acid.

H2SO4 + H2O H3O+ + HSO4-

HSO4- + H2O H3O+ + SO4

2-

Ka = very large

Ka = 1.96

Prentice-Hall © 2007General Chemistry: Chapter 16Slide 9 of 52

16-7 Ions as Acids and Bases

NH4+ + H2O NH3 + H3O+

baseacid

CH3CO2- + H2O CH3CO2H + OH-

base acid

[NH3] [H3O+] [OH-] Ka= [NH4

+] [OH-]

[NH3] [H3O+] Ka= [NH4

+] = ?

=KW

Kb

= 1.010-14

1.810-5= 5.610-10

Ka Kb = Kw

Prentice-Hall © 2007General Chemistry: Chapter 16Slide 10 of 52

Hydrolysis

Water (hydro) causing cleavage (lysis) of a bond.

Na+ + H2O → Na+ + H2O

NH4+ + H2O → NH3 + H3O+

Cl- + H2O → Cl- + H2O

No reaction

No reaction

Hydrolysis

Prentice-Hall © 2007General Chemistry: Chapter 16Slide 11 of 52

16-8 Molecular Structure and Acid-Base Behavior

Why is CH3CO2H a stronger acid than CH3CH2OH?

There is a relationship between molecular structure and acid strength.

Bond dissociation energies are measured in the gas phase and not in solution.

Prentice-Hall © 2007General Chemistry: Chapter 16Slide 12 of 52

Strengths of Oxoacids

Factors promoting electron withdrawal from the OH bond to the oxygen atom: High electronegativity (EN) of the central atom. A large number of terminal O atoms in the molecule.

H-O-Cl H-O-Br

ENCl = 3.0 ENBr= 2.8

Ka = 2.910-8 Ka = 2.110-9

Prentice-Hall © 2007General Chemistry: Chapter 16Slide 13 of 52

Strengths of Oxoacids

S OO

O

O

H H····

····

-

2+

··

··

·· ···· ··

-

S OO

O

H H····

····

-

+

··

··

·· ··

S OO

O

O

H H····

····

·· ···· ··

S OO

O

H H····

···· ··

·· ··

Ka 103 Ka =1.310-2

Prentice-Hall © 2007General Chemistry: Chapter 16Slide 14 of 52

Strengths of Organic Acids

C OC

O

H H····

·· ··H

H

OCH H····

H

H

C

H

H

Ka = 1.810-5 Ka =1.310-16

acetic acid ethanol

Prentice-Hall © 2007General Chemistry: Chapter 16Slide 15 of 52

Structural Effects

C

H

H

C

O

C

O

H

-··

····

H

H

····

CH

H

H

C

O

O

-··

····

····

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

Ka = 1.810-5

Ka = 1.310-5