16-6 polyprotic acids
DESCRIPTION
16-6 Polyprotic Acids. Phosphoric acid: A triprotic acid. K a = 7.1 10 -3. H 3 PO 4 + H 2 O H 3 O + + H 2 PO 4 -. K a = 6.3 10 -8. H 2 PO 4 - + H 2 O H 3 O + + HPO 4 2-. K a = 4.2 10 -13. HPO 4 2- + H 2 O H 3 O + + PO 4 3-. Phosphoric Acid. - PowerPoint PPT PresentationTRANSCRIPT
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