PRS Question from Wed
 What must you do to increase the value of K
for the following exothermic reaction?
2 H2(g) + O2(g)  2 H2O(g)
1. Increase the total pressure
2. Decrease the total pressure
3. Increase the temperature
[H2 O]2
[H2 ] 2 x [O2 ]
Thus anything that shifts
equilibrium to the RIGHT
will increase K
4. Decrease the temperature
5. Two of these
Week 9
K=
CHEM 1310 - Sections L and M
1
PRS Question from Wed
 What must you do to increase the value of K
for the following exothermic reaction?
2 H2(g) + O2(g)  2 H2O(g)
PV = nRT
As P ↑, V↓
Stress is relieved by shifting equilibrium to direction
where there are fewer molecules…
2 H2(g) + O2(g)  2 H2O(g)
LEFT → RIGHT
3 mol
Week 9
2 mol
CHEM 1310 - Sections L and M
2
PRS Question from Wed
 What must you do to increase the value of K
for the following exothermic reaction?
2 H2(g) + O2(g)  2 H2O(g)
PV = nRT
Consequently, lowering P…
“Stress” of lower pressure (i.e. greater volume)
causes shift in equilibrium to side with more molecules!
2 H2(g) + O2(g)  2 H2O(g)
3 mol
Week 9
RIGHT → LEFT
2 mol
CHEM 1310 - Sections L and M
3
1
PRS Question from Wed
 What must you do to increase the value of K
for the following exothermic reaction?
2 Hs(g) + O2(g)  H2O(g)
What impact would temperature changes have?
Consider the analogy with H2O…
Increasing T shifts equilibrium
LEFT to RIGHT
Ice → Water
Exothermic
Decreasing T shifts equilibrium
RIGHT to LEFT
Week 9
CHEM 1310 - Sections L and M
4
PRS Question from Wed
 What must you do to increase the value of K
for the following exothermic reaction?
2 Hs(g) + O2(g)  H2O(g)
1. Increase the total pressure
2. Decrease the total pressure
Both of these result
in more H2O(g)
3. Increase the temperature
4. Decrease the temperature
K=
5. Two of these
Week 9
[H2 O]2
[H2 ] 2 x [O2 ]
CHEM 1310 - Sections L and M
5
Ch 7: Acids and Bases
 Nature of Acids and Bases
 Acid Strength
 The pH Scale
 Calculating pH
 Involving Strong Acid Solutions
 Involving Weak Acid Solutions
Week 9
CHEM 1310 - Sections L and M
6
2
Classification of Acids & Bases
 Arrhenius Definition
 Acids donate H+ or increase [H+ ] in solution
 Bases donate OH- or increase [OH- ] in solution
 Bronsted-Lowry Definition
 Acids donate H+
 Bases accept H+
 Lewis Definition (later…)
Week 9
CHEM 1310 - Sections L and M
7
Examples of Arrhenius Acids
Arrhenius acids can be any of the following:
Binary acids: HCl or H2 S
Polyprotic acids: H2 SO4 or H 3PO4
Oxoacids: HNO3 or H2 SO4 or H 3 PO4
In each case there is an H+ group to donate!
Week 9
CHEM 1310 - Sections L and M
8
More Arrhenius Acids
Some Arrhenius acids do not contain H at all!
Why are they considered acids?
SO3
+ H2O
H 2SO4
A non-metal oxide (or acid anhydride)
Because they increase the concentration
of H+ in solution!
Week 9
CHEM 1310 - Sections L and M
9
3
Examples of Arrhenius Bases
NaOH is an Arrhenius base.
Some Arrhenius bases do not contain OH at all!
Why are they considered bases?
NH3
NH4+ + OH-
+ H2O
Because they increase the concentration
of OH- in solution!
Week 9
CHEM 1310 - Sections L and M
10
Bronsted-Lowry Examples
Arrhenius
Acids donate
Bronsted-Lowry
H+
Acids donate H+
Bases donate OH-
Bases accept H+
B-L Acids: H2S, HCl, H2SO4, HNO3
B-L Bases: NH3 , CN-, HCO3-
Week 9
CHEM 1310 - Sections L and M
11
Conjugate Acid-Base Pairs
 Conjugate Base
 Subtract an H+ from the acid
 Conjugate Acid
 Add H+ to the base
 Examples:
 OH- is the conjugate base of H2 O
 H2O is the conjugate base of H3O + (called the
hydrodium ion, sometimes abbreviated H+ )
 H2O is the conjugate acid of OHWeek 9
CHEM 1310 - Sections L and M
12
4
Conjugate Acid-Base Pairs
CH3COOH donates H+ to H2O
CH3COO- + H3O+
CH3COOH + H2O
Acetic Acid
Base
CH3COOH + NaOH
Acetic Acid
Acid
CH3 COONa + H2O
Conjugate
base
Base
Week 9
Conjugate
base
Acid
CHEM 1310 - Sections L and M
13
Amphoterism and Water
H2O can acts as an acid or a base
CH3COOH + H 2O
Acetic Acid
Base
CH3COOH + NaOH
Acetic Acid
Week 9
CH3COO- + H3O+
Conjugate
base
Acid
CH3 COONa + H 2O
Conjugate
base
Base
Acid
CHEM 1310 - Sections L and M
14
Dissociation Constant, Ka
HA + H2 O
Week 9
H3O+ + A-
Ka =
CHEM 1310 - Sections L and M
[H3 O +] x [A-]
[HA]
15
5
Dissociation Constant, Kw
2 H2O
H3O+ + OH-
Kw = [H3 O +] x [OH-]
From this we can
derive the equation
for pH…
Week 9
CHEM 1310 - Sections L and M
16
The pH Scale
Week 9
CHEM 1310 - Sections L and M
17
Interpreting pH
2 H2O
H3O+ + OH-
Kw = [H3 O +] x [OH-]
pH = -log [H3O+]
pH < 7
Acidic Solution
[H3O +] > [OH-]
pH = 7
Neutral Solution
[H3O +] = [OH-]
pH > 7
Basic Solution
[H3O +] < [OH-]
Let’s consider some calculations with pH…
Week 9
CHEM 1310 - Sections L and M
18
6
Calculations with pH
Calculate the pH of a solution in which the
[OH-] = 1.2 x 10-6M
Kw = [H3 O +] x [OH-]
[H3O+] =
[H3 O +] =
1 x 10-14
1.2 x 10-6
Kw
[OH-]
= 8.3 x 10-9 M
pH = -log [H3O+] = - log (8.3 x 10-9) = 8.1
Week 9
CHEM 1310 - Sections L and M
19
Calculations with pH
Water from a soil sample that is rich in calcium
carbonate has a pH of 8.14. What is the [H+]? Is the
sample acidic or basic?
[H+] = 10-pH = 10-8.14 = 7.2 x 10-9M
pH < 7
Acidic Solution
[H3O +] > [OH-]
pH = 7
Neutral Solution
[H3O +] = [OH-]
pH > 7
Basic Solution
[H3O +] < [OH-]
Since pH > 7, the sample is basic.
Week 9
CHEM 1310 - Sections L and M
20
PRS Question
 Which of the following is the conjugate base
of HPO42-?
1. H3PO4
2. H2PO43. HPO424. PO435. None of these
Week 9
CHEM 1310 - Sections L and M
21
7
PRS Question
 Which of the following is the conjugate base
of HPO42-?
1. H3PO4
2. H2PO4Subtract H+ from the acid
3. HPO424. PO435. None of these
Week 9
CHEM 1310 - Sections L and M
22
PRS Question
 What is the concentration of H3O+ in an
aqueous solution with a pH of 9.25?
1. 2.5 x 10-10 M
2. 5.6 x 10-10 M
3. 2.5 x 10-9 M
4. 9.6 x 10-5 M
5. None of these
Week 9
CHEM 1310 - Sections L and M
23
PRS Question
 What is the concentration of H3O+ in an
aqueous solution with a pH of 9.25?
1. 2.5 x 10-10 M
2. 5.6 x 10-10 M
Antilog (-9.25) = [H3 O +]
3. 2.5 x 10-9 M
4. 9.6 x 10-5 M
5. None of these
Week 9
CHEM 1310 - Sections L and M
24
8