Transcript Chapter 14 - part 2

Chapter 14
Aqueous Equilibria: Acids and Bases
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
Polyprotic Acids

Diprotic acid solutions contain a mixture of acids: H2A, HA,
H2 O
◦ Strongest acid – HA
 Principle reaction – dissociation of H2A
 All of H3O+ come from the first ionization
H2SO4(aq) + H2O(l)  H3O+(aq) + HSO4-(aq)
HSO4-(aq) + H2O(l)
H3O+(aq) + SO42-
Polyprotic acids
H2CO3(aq) + H2O(l)
Ka1 = 4.3 x 10-7

HCO3-(aq) + H2O(l)
Ka2 = 4.8 x 10-11
H3O+( aq) + HCO3-(aq)
H3O+( aq) + CO32-(aq)
Polyprotic Acids
Equilibria in Solutions of Weak
Bases
B(aq) + H2O(l)
Base
Acid
Base-Dissociation Constant:
BH1+(aq) + OH1-(aq)
Acid
[BH1+][OH1-]
Kb =
NH41+(aq) + OH1-(aq)
NH3(aq) + H2O(l)
[NH41+][OH1-]
Kb =
[NH3]
Base
[B]
Equilibria in Solutions of Weak
Bases
Equilibria in Solutions of Weak
Bases
Calculate the [-OH] and pH of a 0.40 M NH3 solution. At 25 °C,
Kb = 1.8 x 10-5.
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?
Relation Between Ka and Kb
NH41+(aq) + H2O(l)
H3O1+(aq) + NH3(aq)
Ka
NH3(aq) + H2O(l)
NH41+(aq) + OH1-(aq)
Kb
H3O1+(aq) + OH1-(aq)
Kw
2H2O(l)
Ka x Kb =
[H3O1+][NH3]
[NH41+]
x
[NH41+][OH1-]
[NH3]
= (5.6 x 10-10)(1.8 x 10-5) = 1.0 x 10-14
= [H3O1+][OH1-] = Kw
Relation Between Ka and Kb
Ka x Kb = Kw
conjugate acid-base pair
Ka =
Kw
Kb
Kb =
pKa + pKb = pKw = 14.00
Kw
Ka
Example

Find the pH of a 0.100 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
Example

What is the pH of 0.10M sodium nicotinate at 25oC? The Ka
for nicotinic acid is 1.4 x 10-5.
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
Acid-Base Properties of Salts

Acidic cation + neutral anion  Acidic salt
NH4+ + ClNH4Cl

Neutral cation + neutral anion  neutral salt
Na+
ClNaCl

Neutral cation + basic anion  basic salt
Na+
CNNaCN
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).
Acid-Base Properties of Salts
Salts That Yield Acidic Solutions
Hydrated cations of small, highly charged metal ions, such as Al3+.
Acid-Base Properties of Salts
Examples

Classify each of the following salt solution as acidic basic or
neutral. Write a hydrolysis equation for each ion.
◦ KBr
◦ NaNO2
◦ NH4Br
Example
Calculate the pH of a 0.10M solution of sodium fluoride
(NaF) at 25oC.
Ka = 7.1 x 10-4
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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)
Example

Predict whether 0.35M NH4Br solution is acidic, basic or
neutral. Calculate its pH. Kb = 1.8 x10-5
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
Lewis Acids and Bases
Lewis Acids and Bases
Lewis Acids and Bases
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)