CHEMICAL
EQUILIBRIUM
ERT 207 ANALYTICAL CHEMISTRY
SEMESTER 1, ACADEMIC SESSION 2015/16
OVERVIEW
2
THE CHEMICAL COMPOSITION OF
AQUEOUS SOLUTIONS
 CHEMICAL EQUILIBRIUM
 THE EQUILIBRIUM STATE
 EQUILIBRIUM-CONSTANT EXPRESSIONS
 TYPES OF EQUILIBRIUM CONSTANTS IN
ANALYTICAL CHEMISTRY
 APPLYING THE ION-PRODUCT
CONSTANT FOR WATER

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OVERVIEW
3
SOLUBILITY-PRODUCT CONSTANTS
 LE CHATELIER’S PRINCIPLE

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THE CHEMICAL COMPOSITION OF
AQUEOUS SOLUTIONS
4
Electrolytes:
 It forms ions when dissolved in water (or certain
other solvents) and thus produce solutions that
conduct electricity.
 Strong electrolytes ionize essentially completely in
a solvent, but weak electrolytes ionize only
partially.
 These characteristics mean that a solution of a
weak electrolyte will only conduct electricity as
well as a solution containing an equal
concentration of a strong electrolytes.

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THE CHEMICAL COMPOSITION OF
AQUEOUS SOLUTIONS
5
A salt is produced in the reaction of an acid with
a base.
 Examples include NaCl, Na2SO4, NaOOCH3
Table 1: Classification of Electrolytes

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THE CHEMICAL COMPOSITION OF
AQUEOUS SOLUTIONS
6
Acids and Bases:
 According to Bronsted-Lowry concept, an acid
donates protons and a base accepts protons.
 An acid donates protons only in the presence of a
proton acceptor (a base). Likewise, a base
accepts protons only in the presence of a proton
donor (an acid).
 A conjugate base is formed when an acid loses a
proton.
 Acetate ion is the conjugate base of acetic acid.
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
THE CHEMICAL COMPOSITION OF
AQUEOUS SOLUTIONS
7
acid1 ↔ base1 + proton
 We refer to acid1 and base1 as a conjugate pair.
 Every base accepts a proton to produce a
conjugate acid.
base2 + proton ↔ acid2
 Neutralization reaction:
acid1 + base2 ↔ base1 + acid2
 Many solvents are proton donors or proton
acceptors & can induce basic or acidic behavior in
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solutes dissolved in them.
THE CHEMICAL COMPOSITION OF
AQUEOUS SOLUTIONS
8

In an aqueous solution of ammonia, water can donate
a proton and acts as an acid with respect to the
solute NH3:
NH 3  H 2O  NH 4  OH 
Base1
Acid 2
Conjugate
Acid1
Conjugate
Base2
Ammonia (base1) reacts with water (acid2), to give
the conjugate base (base2) of the acid water.
 Water acts as a proton acceptor, or base, in an
aqueous solution of nitrous acid.



2
H 2O  HNO2  H 3O  NO
Base1
Acid 2
Conjugate
Acid1
Conjugate
Base2
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THE CHEMICAL COMPOSITION OF
AQUEOUS SOLUTIONS
9
The conjugate base of the acid HNO2 is nitrate ion.
 The conjugate acid of water is the hydrated proton,
H3O+ (hydronium ion).

Figure 1:
Possible
structures for
the
hydronium
ion.
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THE CHEMICAL COMPOSITION OF
AQUEOUS SOLUTIONS
10
Amphiprotic species:
 Species that have both acidic and basic
properties.
 Example: Dihydrogen phopshate

4

H 2 PO  H 3O  H 3 PO4  H 2O
Base1
Acid2

4
Acid1

Base2
2
4
H 2 PO  OH  HPO  H 2O
Acid1
Base2
Base1
Acid2
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THE CHEMICAL COMPOSITION OF
AQUEOUS SOLUTIONS
11
The simple amino acids are an important class of
amphiprotic compounds that contain both a weak
acid and a weak base functional group.
 When dissolved in water, an amino acid, such as
glycine, undergoes a kind of internal acid/base
reaction to ptoduce a zwitterion.
 A zwitterion is an ion that has both a positive
and a negative charge.

NH 2CH 2COOH  NH 3 CH 2COO 
Glycine
zwitterion
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THE CHEMICAL COMPOSITION OF
AQUEOUS SOLUTIONS
12
Water is the classic example of an amphiprotic
solvent, that is, a solvent that can act either as an
acid or as a base, depending on the solute.
 Other common amphiprotic solvents are methanol,
ethanol, and anhydrous acetic acid.

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THE CHEMICAL COMPOSITION OF
AQUEOUS SOLUTIONS
13
Strengths of acids and bases
 Figure 2 shows the dissociation reactions of a few
common acids in water.
Figure 2: Dissociation reactions & relative strengths
of some common acids & their conjugate base.
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THE CHEMICAL COMPOSITION OF
AQUEOUS SOLUTIONS
14
The common strong bases include NaOH, KOH,
Ba(OH)2, and the quaternary ammonium hydroxide
R4NOH (R = an alkyl group: CH3 or C2H5).
 The common strong acids include HCl, HBr, HI,
HClO4, HNO3, the first proton in H2SO4, and the
organic sulfonic acid RSO3H.
 Perchloric and hydrochloric acids are strong acids
in water.
 If anhydrous acid, a weaker proton acceptor that
water, is substituted as the solvent, neither of these
acids undergoes complete dissociation.
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
THE CHEMICAL COMPOSITION OF
AQUEOUS SOLUTIONS
15

Equilibria below is established:

2

2
CH 3COOH  HClO4  CH 3COOH  ClO
Base1
Acid 2
Acid 2
Base 2
In a differentiating solvent, various acids
dissociate to different degrees and have different
strengths.
 In a leveling solvent, several acids are completely
dissociated and show the same strength.

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CHEMICAL EQUILIBRIUM
16
Many reactions used in analytical chemistry never
result in complete conversion of reactants to
products.
 Instead, they proceed to a state of chemical
equilibrium in which the ratio of concentrations
of reactants and products is constant.
 Equilibrium-constant expressions are algebraic
equations that describe the concentration
relationship among reactants and products at
equilibrium.
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
CHEMICAL EQUILIBRIUM
17
The Equilibrium State:
 Consider the chemical reaction:



3
H 3 AsO4  3I  2 H  HAsO3  I  H 2O
The rate of this reaction and the extent to which it
proceeds to the right by monitoring the
appearance of the orange-red color of the
triiodide ion I3-.
 The final position of a chemical equilibrium is
independent of the route to the equilibrium state.

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CHEMICAL EQUILIBRIUM
18
The equilibrium state is altered by applying stress
(temperature, pressure, concentration of reactant
or product) to the system.
 These effects can be predicted qualitatively from
the Le Chatelier’s principle.
 The Le Chatelier’s principle:
 It states that the position of an equilibrium
always shifts in such a direction as to relieve a
stress that is applied to the system.

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CHEMICAL EQUILIBRIUM
An increase in temperature of a system
alters the concentration relationship in the
direction that tends to absorb heat, and an
increase in pressure favors those
participants that occupy a smaller total
volume.
 The mass-action effect is a shift in the
position of an equilibrium caused by
adding one of the reactants or products to
a system.
19

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CHEMICAL EQUILIBRIUM
Equilibrium is a dynamic process.
 Although chemical reactions appear to stop at
equilibrium, in fact, the amounts of reactants
and products are constant because the rates of
the forward and reverse processes are exactly
the same.
 Chemical thermodynamics is a branch of
chemistry that concerns the flow of heat & energy
in chemical reactions.
 The position of a chemical equilibrium is related
to these energy changes.

20
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CHEMICAL EQUILIBRIUM
21

Equilibrium-constant expressions
 The influence of concentration or pressure on
the position of a chemical equilibrium is
conveniently described in quantitative terms by
means of an equilibrium-constant expression.
 They allow us to predict the direction and
completeness of chemical reactions.
 However, yields no information concerning the
rate of reaction.
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CHEMICAL EQUILIBRIUM
22
Consider a generalized equation for a chemical
equilibrium: wW + xX ↔ yY + zZ
y
z
 The equilibrium-constant expression:

Y  Z 
K
w
x
Molar concentration or partial
W  X 
pressure in atmospheres if they are
gas phase reactants or products.


If a reactant or product is a pure liquid, a pure
solid, or the solvent present in excess, no term for
this species appears in the equilibrium-constant
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expression.
CHEMICAL EQUILIBRIUM
23

Y
K
w
x
W  X 
y

A temperature dependent equilibrium constant:
y
Y
w
W
a a
K
a a

z
Z
x
X
The activities of
species Y, Z, W, X
Table 1 summarizes the types of chemical
equilibria and equilibrium constant that are of
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importance in analytical chemistry.
CHEMICAL EQUILIBRIUM
Table 1: Equilibria & Equilibrium constants important
in analytical chemistry
24
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CHEMICAL EQUILIBRIUM
25
Applying the ion-product constant for water
 Aqueous solutions contain small concentrations of
hydronium and hydroxide ions as a consquence of
the dissolution reaction.


2 H 2O  H 3O  OH


An equilibrium constant for this reaction can be
formulated:


[ H 3O ][ OH ]
K
2
[ H 2O ]
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CHEMICAL EQUILIBRIUM

26
When compared with the concentration of
hydrogen and hydroxide ions:


K [ H 2O ]  KW  [ H 3O ][ OH ]
2
Ion-product constant for water
At 25oC, Kw = 1.008 x 10-14.
 Table 2:
Variation
of KW with
temperature

bblee@unimap
CHEMICAL EQUILIBRIUM
27
Example 1:
 Calculate the hydronium and hydroxide ion
concentrations of pure water at 25oC and
100oC.
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CHEMICAL EQUILIBRIUM
28
Using solubility-product constants:
 Sparingly soluble salts are essentially completely
dissociated in saturated aqueous solution.
 When an excess of barium iodate is equilibrated
with water, the dissociation process is adequately
described by:

Ba IO3 2 s   Ba
2
aq   2 IO aq 

3
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CHEMICAL EQUILIBRIUM
2
29

2
[ Ba ][ IO3 ]
K
[ Ba IO3 2 s  ]
K [ Ba( IO3 )2 s ]  K sp  [ Ba ][ IO3 ]
2

2
Solubility-product constant or
Solubility product
bblee@unimap
CHEMICAL EQUILIBRIUM
30
Example 2:
 What mass (in grams) of Ba(IO3)2 (487
g/mol) can be dissolved in 500 ml of water
at 25oC?
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CHEMICAL EQUILIBRIUM
31
The effect of a common ion in the solubility of a
precipitate:
 The common-ion effect is a mass-action effect
predicted from Le Chatelier principle.
 The solubility of an ionic precipitate decreases
when a soluble compound containing one of the
ions of the precipitate is added to the solution.

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CHEMICAL EQUILIBRIUM
32
Example 3:
 Calculate the molar solubility of Ba(NO3)2 in
a solution that is 0.0200 M in Ba(NO3)2.
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LE CHATELIER’S PRINCIPLE
33

When a change is imposed on a system at
equilibrium the position of the equilibrium shifts in
a direction that tends to reduce the effect of that
change.
It is like the “undo” button on your
computer!
Henri Le Chatelier
 1850-1936
 Studied mining engineering.
 Interested in glass and ceramics.
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
LE CHATELIER’S PRINCIPLE
34
Factors that Affect Equilibrium:
i.
Concentration
ii. Temperature
iii. Pressure (For gaseous systems only!)
iv. The presence of a catalyst
(i) Effect of a Change in Concentration
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LE CHATELIER’S PRINCIPLE
35
• Add more reactant  Shift to products
• Remove reactants  Shift to reactants
• Add more product  Shift to reactants
• Remove products  Shift to products

The reaction quotient for an equilibrium system is
the same as the equilibrium expression, but the
concentrations are NOT at equilibrium!
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LE CHATELIER’S PRINCIPLE
36
N2O4(g)  2NO2(g)

Keq at
equilibrium
Q = [NO2]2
[N2O4]
Changes in concentration are best understood in
terms of what would happen to “Q” if the
concentrations were changed.
If Q< K then there are too many reactants, the
reaction will shift in the forward direction (the
products).

If Q>K then there are too many products, the
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reaction will shift to the reactants.

LE CHATELIER’S PRINCIPLE
37
(ii) Effect of a Change in Pressure (Volume)
The system is
initially at
equilibrium.
The piston is pushed in,
decreasing the volume and
increasing the pressure. The
system shifts in the direction
that consumes CO2 molecules,
lowering the pressure again.
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LE CHATELIER’S PRINCIPLE

Decreasing volume:
38
bblee@unimap
LE CHATELIER’S PRINCIPLE
39
It only affects equilibrium systems with unequal
moles of gaseous reactants and products.
 Increase Pressure
 Stress of pressure is reduced by reducing the
number of gas molecules in the container.

N2(g) + 3H2(g) = 2NH3(g)
There are 4 molecules of reactants vs. 2
molecules of products.
 Thus, the reaction shifts to the product ammonia.

bblee@unimap
LE CHATELIER’S PRINCIPLE
Increasing the volume:
 The system shifts in the direction that increases its
pressure.
 Decrease Pressure:
 Stress of decreased pressure is reduced by
increasing the number of gas molecules in the
container.

40
PCl5(g) = PCl3(g) + Cl2(g)
 There
are two product gas molecules vs. one
reactant gas molecule.
 Thus, the reaction shifts to the products. bblee@unimap
LE CHATELIER’S PRINCIPLE
(iii) Effect of a Change in Temperature
 The value of K changes with temperature (We can
use this to predict the direction of this change).
 Exothermic reaction – produces heat (heat is a
product)
 Adding energy shifts the equilibrium to the left
(away from the heat term).
 Endothermic reaction – absorbs energy (heat is a
reactant)
Adding energy shifts the equilibrium to the
right (away from the heat term).
41
bblee@unimap
LE CHATELIER’S PRINCIPLE
42
The meaning of K:
 K>1  the equilibrium position is far to the right
 K<1  the equilibrium position is far to the left

(iv) Presence of a Catalyst
 A catalyst lowers the activation energy and
increases the reaction rate.
 It will lower the forward and reverse reaction
rates.
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LE CHATELIER’S PRINCIPLE
43

Therefore, a catalyst has NO EFFECT on a
system at equilibrium!
 It just gets you to equilibrium faster!
Presence of an Inert Substance
 An inert substance is a substance that is notreactive with any species in the equilibrium
system.
 These will not affect the equilibrium system.
 If the substance does react with a species at
equilibrium, then there will be a shift!
bblee@unimap
LE CHATELIER’S PRINCIPLE
44
Example 4
 Given:
S8(g) + 12O2(g)  8 SO3(g) + 808 kcals
What will happen when:
 Oxygen gas is added?
Shifts to products?
??

The reaction vessel is cooled?
Shifts to Products?–?to
? replace heat
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LE CHATELIER’S PRINCIPLE
45

The size of the container is increased?
V increases, Pressure decreases,
? ? ? – to reactants !
shifts to more particles

Sulfur trioxide is removed?
Shift to products to? replace
it !
??

A catalyst is added to make it faster?
No change !
???
bblee@unimap
EXAMPLE 1
 Since
OH- and H3O+ are formed only
from the dissolution of water, their
concentration must be equal:

 Then,

[ H 3 0 ]  [ OH ]


[ H 3O ]  [ OH ]  KW
2

2

[ H 3O ]  [ OH ]  KW
 At
25C,


[ H 3O ]  [ OH ]  1.00 x10
14
= 1.00 x 10-7 M
46
bblee@unimap
EXAMPLE 1
 At
100C,


[ H 3O ]  [ OH ]  49 x10
14
= 7.0 x 10-7 M
47
bblee@unimap
EXAMPLE 2
 The
solubility-product constant for
Ba(IO3)2 is 1.57 x 10-9 (Appendix 2).
 Equilibrium between the solid and its
ions in solution is described:
Ba IO3 2 ( s )  Ba

K sp  Ba
2
2
 2IO
IO   1.57x10
 2
3

3
9
 The
equilibrium reveals that 1 mol of
Ba2+ is formed for each mole of
Ba(IO3)2 that dissolves.
48
bblee@unimap
EXAMPLE 2
 Molar
solubility of Ba(IO3)2 = [Ba2+]
 Since
two moles of iodate are produced
for each mole of barium ion, the iodate
concentration is twice the barium ion

2
concentration:
IO   2Ba 
3
 The
equilibrium constant expression:
Ba 2Ba 
2
2
2

 4 Ba

2 3
= 1.57 x 10-9
bblee@unimap
49
EXAMPLE 2
Ba 
2
 1.57 x10
 
4

9
1/ 3



= 7.32 x 10-4 M
 Since
1 mol of Ba2+ is produced for
every mole of Ba(IO3)2,
Solubility = 7.32 x 10-4 M
 The number of milimoles of Ba(IO3)2
dissolved in 500 ml of solution,
 4 mmol Ba IO3 2
No.mmol Ba IO3 2  7.32 x10
x 500ml
mL
50
bblee@unimap
EXAMPLE 2
 The
mass of Ba(IO3)2 in 500 ml is given
by:
Mass Ba(IO3)2 =
g Ba IO3 2
7.32 x10 x500 mmol Ba IO3 2 x0.487
mmol Ba(IO 3 ) 2

4

= 0.178 g
51
bblee@unimap
EXAMPLE 3
 Molar
solubility of Ba(IO3)2 = 0.5 [IO3-]
 There are two sources of barium ions:
Ba(NO3)2 and Ba(IO3)2.
 The contribution from the nitrate is
0.0200 M, and that from the iodate is
equal to the molar solubility, or 0.5
[IO3-].
Ba 
2
 
1

 0.0200  IO3
2
52
bblee@unimap
EXAMPLE 3
 The
solubility-product expression,
  
1

 
 2
9
 0.0200  IO3  IO3  1.57 x10
2


 Assumption:

 

1

Ba  0.0200  IO3  0.0200M
2
 Simplified equation:
2
 
 2
3
0.0200 IO
IO  

3
1.57 x10
9
 1.57 x10
9
0.0200  7.85 x10
=2.80 x 10-4 M
8
53
bblee@unimap
EXAMPLE 3
 Solubility
of Ba(IO3)2 =0.5 [IO3-]
=0.5 x 2.80 x 10-4
= 1.40 x 10-4 M
54
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