A Question from Test

1
A Question from Test 2 2004

• Using the following information
• 2SO2(g) O2(g) ? 2SO3(g) K 2.30 x 104
• SO2Cl2(g) ? SO2(g) Cl2(g) K 2.9 x 10-2
• CO(g) Cl2(g) ? COCl2(g) K 21.9
• (a) What is the equilibrium constant for the
  following reaction
• 2SO2Cl2(g) O2(g) ? 2SO3(g) 2Cl2(g)
•  
•  
•  

2
A Question from Test 2 2004

• Using the following information
• 2SO2(g) O2(g) ? 2SO3(g) K 2.30 x 104
• SO2Cl2(g) ? SO2(g) Cl2(g) K 2.9 x 10-2
• CO(g) Cl2(g) ? COCl2(g) K 21.9
• (a) What is the equilibrium constant for the
  following reaction
• 2SO2Cl2(g) O2(g) ? 2SO3(g) 2Cl2(g)
•  
•  
•  

• K P(SO2Cl2)2 P(O2)/P(SO3)2P(Cl2)2
• K P(SO3)P(Cl2)/P(SO2Cl2)2 P(O2)
• K P(SO3)2P(Cl2)2/P(SO2Cl2)2 P(O2)

3
A Question from Test 2 2004

• Using the following information
• 2SO2(g) O2(g) ? 2SO3(g) K 2.30 x 104
• SO2Cl2(g) ? SO2(g) Cl2(g) K 2.9 x 10-2
• CO(g) Cl2(g) ? COCl2(g) K 21.9
• (a) What is the equilibrium constant for the
  following reaction
• 2SO2Cl2(g) O2(g) ? 2SO3(g) 2Cl2(g)
•  
•  
•  

• K P(SO3)2P(Cl2)2/P(SO2Cl2)2 P(O2)
• What is the numerical value of this K.
• This reaction is reaction 1 2xreaction 2.
• The equilibrium constant should be K(reaction 1)
  x K(reaction 2)2
• Therefore, K 2.30 x 104 x (2.9 x 10-2)2
  19.343

4
Chapter 17 Acids and Bases
5
Brønsted-Lowry Theory of Acids and Bases
The most general theory for common aqueous
acids and bases is the BRØNSTED - LOWRY theory

• An acid is a proton donor.
• A base is a proton acceptor.

conjugate acid
conjugate base
base
acid
NH3 H2O NH4 OH-
NH4 OH- NH3 H2O
base
acid
6
Acid-Base Theories
The Brønsted definition means NH3 is a BASE and
hydrogen fluoride is an ACID
NH3 (aq) HF NH4(aq) F-(aq) base acid
acid base
7
Conjugate Pairs
Conjugate acid/base pairs are related through
gain or loss of a proton HA H2O ? H3O A- A-
H2O ? HA OH-
NH3 (aq) H2O NH4(aq) OH-(aq) base
acid
acid base
In this example, NH3 is a base and NH4 is its
conjugate acid ... Water is an acid and OH- is
the conjugate base
EVERY acid has a conjugate base and vice-versa!
8
An extreme case
Acid Base conj. Acid
conj. base
9
Base Ionization Constant
base
acid
Conjugate acid
NH3 H2O NH4 OH-
Conjugate base
This constant Kb follows the rules for
equilibrium constants, it has the subscript b
to indicate that it is the dissociation of an
base.
10
Acid Ionization Constant
base
acid
Conj. acid
Conj. base
CH3CO2H H2O CH3CO2- H3O
This constant Ka follows the rules for
equilibrium constants, it has the subscript a
to indicate that it is the dissociation of an
acid.
11
Question
In the equilibrium system described by HPO42-
H2O ? H2PO4- OH- Brønsted-Lowry theory would
designate
12
Strong Acids
We generally divide acids and bases up into two
groups, STRONG and WEAK STRONG ACIDS are
considered to be 100 dissociated in
water HNO3(aq) H2O(liq) ? H3O(aq) NO3-(aq)
13
Weak Acids
Weak acids are much less than 100 ionized in
water. A prime example of a weak acid is acetic
acid. It is primarily un-dissociated in water
(99.58 un-dissociated) HOAc(aq) H2O(liq)
OAc-(aq) H3O(aq) This is acetic acids
structure (un-dissociated)
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Strong vs. Weak Acids
HA H A- (note H3O and H are considered to
be the exact same thing) Strong Acid Weak
Acid Ka Ka is large Ka is small Dissociation
far to the right far to the left Equilibrium
H H HAo H ltlt HAo Strength of A-
much weaker A- much strongerconjugate base
base than water base than water
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Strong Bases
Strong Base 100 dissociated in water NaOH(aq)
? Na (aq) OH - (aq)
Other common strong bases include KOH and Ca(OH)2
CaO (lime) H2O ? Ca(OH)2 (slaked lime) (gives
2 equivalents of OH-)
16
Weak Bases
Weak base less than 100 ionized in water One of
the best known weak bases is ammonia
NH3 (aq) H2O NH4(aq) OH-(aq)
base acid
acid base
ammonia ammonium
17
Conjugate Pairs
Strong Acid - Weak Base
A strong acid is 100 dissociated. Therefore, a
STRONG ACID, a good H donor, has a WEAK
CONJUGATE BASE, a poor H acceptor.
HNO3 (aq) H2O(liq) H3O
(aq) NO3 - (aq)
Strong a. Weak b.
weak a. Weak b.
Notice that every A-B reactions has two acids and
two bases!
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Conjugate Pairs
We know from experiment that HNO3 is a strong
acid 1. It is a stronger acid than H3O 2. H2O
is a stronger base than NO3 -
19
Conjugate Pairs
Acetic acid is only 0.42 ionized when HOAc
1.0 M It is a weak acid
HOAc (aq) H2O(liq) H3O (aq)
OAc- (aq) weak base acid
strong acid base
Because H3O is small, this must mean 1. H3O
is a stronger acid than HOAc 2. OAc- is a
stronger base than H2O
20
The Self-Ionization of Water and the pH Scale
H2O can function as both an ACID and a BASE. It
is an amphiprotic material.
In pure water this is called SELF-IONIZATION
21
Ion Product of Water
Conj. acid
Conj. base
base
acid
H2O H2O H3O OH-
In a neutral solution H3O OH- and so
H3O OH- 1.00 x 10 -7 M
(note SELF-IONIZATION can happen in other PROTIC
solvents, like ammonia)
22
pH and pOH

• Because the concentrations of H3O and OH- are so
  small we use the following definitions (1909)

pH -logH3O
pOH -logOH-
KW H3OOH- 1.0x10-14 Using these
definitions we can derive the relationship
-logKW -logH3O-logOH- -log(1.0?10-14)
pKW pH pOH -(-14)
pKW pH pOH 14
23
pH and pOH Scales
24
You add 0.0010 mol of NaOH to 1.0 L of pure
water. Calculate H3O and OH- (assume
complete dissociation of the NaOH).
Solution OH- NaOH 2 H2O(liq) H3O
(aq) OH- (aq) initial 0 0.0010 change x
x equilib x 0.0010 x
Kw (x) (0.0010 x) 10-14 Because x ltlt 0.0010
M, assume, OH- 0.0010 M
H3O Kw / 0.0010 1.0 x 10-11 M and OH-
0.0010M
pH -logH3O -log(1.0 x 10-11 ) 11 pOH
-logOH- -log(0.0010) 3
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Strong Acids and Bases
HCl CH3CO2H
Thymol Blue Indicator pH lt 1.2 lt pH lt 2.8 lt pH
26
Weak Acids and Bases
Acetic Acid
HC2H3O2 or CH3CO2H
27
Weak Acids
See Table 17.3 for Ionization Constants of Weak
Acids/Bases
28
Approaching Acid-Base Equilibrium Calculations

• List major species in solution.
• Choose species that can produce/consume H and
  write reactions.
• Write equilibrium expression for dominant
  equilibrium. (usually we can neglect Kw)
• List initial concentrations in dominant
  equilibrium.
• Define change at equilibrium (as x).
• Write equilibrium concentrations in terms of x.
• Substitute equilibrium concentrations into
  equilibrium expression.
• Solve for x the easy way.
• Verify assumption.
• Calculate H and pH.

29
An Acid/Base Question! DO NOT PANICK!!
A solution with the total volume 250.0 mL is
prepared by diluting 20.0 mL of glacial acetic
acid with water (pure acetic acid, d 1.05
g/mL). Calculate H and pH. (Ka 1.8 x 10-5 )
1. Determine concentration 20 mL x 1.05 (g/mL) /
60.05 (g/mol) 0.350 mol CH3CO2H 0.350 mol /
0.250 L 1.40 M acetic acid solution in water
CH3CO2H (aq) CH3CO2- (aq)
H(aq)
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Acetic acid 1.40 M, Ka 1.80 x 10-5. Calc. H
and pH.
CH3CO2H CH3CO2- H Initial
1.40 -- -- Change -x x x Equilibrium
1.40-x x x
31
Acetic acid 1.40 M, Ka 1.80 x 10-5. Calc. H
and pH.
CH3CO2H CH3CO2- H Initial
1.40 -- -- Change -0.00501 0.00501 0.00501
Equilibrium 1.40 0.00501 0.00501
H 0.00501 M pH -log H 2.30
Sometimes the assumption does not work. You will
then need to rely on solving the quadratic
equation.