1. Chapter 14: Aqueous Equilibria: Acids and Bases
5/17/2018
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2. Polyprotic Acids Acids that contains more than one dissociable proton
Dissociate in a stepwise manner
Each dissociation step has its own Ka
Stepwise dissociation constants decreases in the order Ka1 > Ka2 > Ka3
More difficult to remove a positively charge proton from negative ion

3. Polyprotic Acids
Diprotic acid solutions contain a mixture of acids: H2A, HA, H2O
Strongest acid – HA
Principle reaction – dissociation of H2A
All of H3O+ come from the first ionization

4. Chapter 14: Aqueous Equilibria: Acids and Bases
5/17/2018
Polyprotic Acids
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5. Equilibria in Solutions of Weak Bases
Chapter 14: Aqueous Equilibria: Acids and Bases
Equilibria in Solutions of Weak Bases
5/17/2018
BH1+(aq) + OH1-(aq)
B(aq) + H2O(l)
Base
Acid
Acid
Base
[BH1+][OH1-]
[B]
Kb =
Base-Dissociation Constant:
NH41+(aq) + OH1-(aq)
NH3(aq) + H2O(l)
[NH41+][OH1-]
[NH3]
Kb =
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6. Equilibria in Solutions of Weak Bases
Chemistry: McMurry and Fay, 6th Edition
Chapter 14: Aqueous Equilibria: Acids and Bases
5/17/2018 2:59:53 AM
Equilibria in Solutions of Weak Bases
Copyright © 2011 Pearson Prentice Hall, Inc.

7. Equilibria in Solutions of Weak Bases
Chapter 14: Aqueous Equilibria: Acids and Bases
Equilibria in Solutions of Weak Bases
5/17/2018
Calculate the [-OH] and pH of a 0.40 M NH3 solution. At 25 °C, Kb = 1.8 x 10-5.
While the initial hydroxide ion concentration is not 0, it’s insignificant when compared to that which comes from the dissociation of NH3.
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8. Example
Morphine (C17H19NO3), a narcotic used in painkillers, is a weak organic base. If the pH of a 7.0 x 10-4 M solution of morphine is 9.5, what is the value of Kb?

9. Relation Between Ka and Kb
Chapter 14: Aqueous Equilibria: Acids and Bases
Relation Between Ka and Kb
5/17/2018
H3O1+(aq) + NH3(aq)
NH41+(aq) + H2O(l) Ka
NH41+(aq) + OH1-(aq)
NH3(aq) + H2O(l) Kb
H3O1+(aq) + OH1-(aq)
2H2O(l) Kw
[H3O1+][NH3]
[NH41+][OH1-]
Ka x Kb = x
= [H3O1+][OH1-] = Kw
[NH41+]
[NH3]
= (5.6 x 10-10)(1.8 x 10-5) = 1.0 x 10-14
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10. Relation Between Ka and Kb
Chapter 14: Aqueous Equilibria: Acids and Bases
Relation Between Ka and Kb
5/17/2018
Ka x Kb = Kw
conjugate acid-base pair Kb Kw Ka Kw
Ka =
Kb =
pKa + pKb = pKw = 14.00
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11. Example
Find the pH of a M NaCHO2 solution. The salt completely dissociate into Na+(aq) and CHO2-(aq) and Na+ ion has no acid or base properties. Ka (HCHO2)= 1.8 x 10-4

12. Acid-Base Properties of Salts
pH of a salt solution is determined by the acid-base properties of the consistuent cations and anions
In an acid-base reaction, the influence of the stronger partner is predominant
Strong acid + Strong Base  Neutral solution
Strong acid + Weak Base  Basis solution
Weak acid + Strong Base  Acidic solution

13. Acid-Base Properties of Salts
Acidic cation + neutral anion  Acidic salt
Neutral cation + neutral anion  neutral salt
Neutral cation + basic anion  basic salt

14. Acid-Base Properties of Salts
Acidic cation + basic anion  (50 :50 mixture) must compare Ka and Kb
Ka > Kb: The solution will contain an excess of H3O1+ ions (pH < 7).
Ka < Kb: The solution will contain an excess of OH1- ions (pH > 7).
Ka ≈ Kb: The solution will contain approximately equal concentrations of H3O1+ and OH1- ions (pH ≈ 7).

16. Acid-Base Properties of Salts
Chapter 14: Aqueous Equilibria: Acids and Bases
5/17/2018
Acid-Base Properties of Salts
Salts That Yield Acidic Solutions
Hydrated cations of small, highly charged metal ions, such as Al3+.
The cations listed previously are slightly acidic even though we ignore them at this level. Acidity increases as the charge increases and also as the ion size decreases.
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17. Acid-Base Properties of Salts
Chapter 14: Aqueous Equilibria: Acids and Bases
5/17/2018
Acid-Base Properties of Salts
Copyright © 2008 Pearson Prentice Hall, Inc.
Copyright © 2008 Pearson Prentice Hall, Inc.

18. Examples
Calculate Ka for the cation, and Kb for the anion in an aqueous NH4CN solution. Is the solution acidic, basic or neutral? Write the hydrolysis reaction of the salt (Kb for NH3 = 1.8 x 10-5, Ka for HCN = 4.9 x 10-10)
Predict whether 0.35 M NH4Br solution is neutral, basic or acid. Write the hydrolysis equation and calculate the pH

19. Chapter 14: Aqueous Equilibria: Acids and Bases
5/17/2018
Lewis Acids and Bases
Lewis Acid: An electron-pair acceptor.
Include cations and neutral molecule having vacant valence orbitals that can accept a share in a pair of electrons from a Lewis Base
Lewis Base: An electron-pair donor.
All Lewis bases are Bronsted-Lowry bases
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20. Lewis Acids and Bases Chemistry: McMurry and Fay, 6th Edition
Chapter 14: Aqueous Equilibria: Acids and Bases
5/17/2018 2:59:53 AM
Lewis Acids and Bases
Copyright © 2011 Pearson Prentice Hall, Inc.

21. Chapter 14: Aqueous Equilibria: Acids and Bases
5/17/2018
Lewis Acids and Bases
Copyright © 2008 Pearson Prentice Hall, Inc.

22. Chapter 14: Aqueous Equilibria: Acids and Bases
5/17/2018
Lewis Acids and Bases
Copyright © 2008 Pearson Prentice Hall, Inc.

23. Examples
For each of the following reactions, identify the Lewis acid and the Lewis base
CO2(g) OH(aq)  HCO3-(aq)
AlCl3(aq) + Cl-(aq)  AlCl4-(aq)