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CHEMICAL EQUILIBRIUM
the dynamic state in which
rates of the forward and
reverse reactions are identical.
CHEMICAL
EQUILIBRIUM
aA + bB
cC + dD
Equilibrium constant

Cc Dd
K
Aa Bb
K>1 
Forward reaction is
favoured
a,b,c,d – stoichiometry coefficients
[A], [B], [C], [D] –concentrations of A, B, C, D
in standard state:
• For solutes – 1M
• For gases – 1 atm
• For solids and pure liquids: [X] = 1
In these conditions K is dimensionless.
Equilibrium const. for a reverse reaction:
cC + dD
aA + bB
K1 = 1/K

A a Bb
K1 
Cc Dd
Equilibrium const. for two reactions added: K3 = K1 x K2
HA
H+ + C
HA + C
H+ + ACH+
[H ][A  ]
K1 
[HA]
K2 
[CH ]
[C][H ]
CH+ + A- K .K  [H ][A  ]  [CH ]
1 2

[HA]
[C][H ]
[A  ][CH ]

 K3
[HA][C]
THERMODYNAMICS
The equilibrium constant is derived from the
thermodynamics of a chemical reaction.
ENTHALPY
ΔH - enthalpy change is the heat absorbed or released
during reaction
ΔH > 0
ΔH < 0
Heat absorbed
Heat liberated
Endothermic
reaction
Exothermic
reaction
ENTROPY
ΔS – a measure of ‘disorder’ of a substance
Gas  Liquid  Solid
Decrease in disorder
ΔS > 0
ΔS < 0
Products more
disordered than
reactants
Products less
disordered than
reactants
Example:
KCl(s)
K+(aq) + Cl-(aq)
ΔS0 = +76 J/K mol at 250C
FREE ENERGY
Gibbs free energy:
ΔG < 0
The reaction is favoured
K>1
ΔG = ΔH - TΔS
ΔG > 0
The reaction is not favoured
K<1
Free energy and equilibrium:
Ke
 ΔG0 /RT
OR
ΔG  RT ln K
LE CHATELIER’S PRINCIPLE
When a system at equilibrium is disturbed, the direction
in which the system proceeds back to equilibrium is
such that the change is partially offset.
A+B
C+D
If reaction is at equilibrium
and reactants are added
(or products removed),
the reaction goes to the
right.
K
C D
A  B
If reaction is at equilibrium
and products added ( or
reactants are removed),
the reaction goes to the
left.
Recall:
Q = Reaction quotient
has the same form as equilibrium constant (K),
but the solution concentrations do not have to be
equilibrium concentrations.
Thus at equilibrium:
Q=K
According to Le Châtelier:
If Q > K  reaction will proceed in the reverse
direction
If Q < K  reaction will proceed in the forward
direction
A+B
C+D
At equilibrium:
[A] = 0.002M
[B] = 0.025M
[C] = 5.0M
[D] = 1.0M
Say we add double reactant A,
[A] = 0.004M
K
CD
A B

5.01.0
K
0.0020.025
= 1x105 at 250C

5.01.0
Q
0.004 0.025 
= 0.5x105 at 250C
Q<K
To reach equilibrium:
Q=K
and the reaction must go to the right
 spontaneous
THE EFFECT OF TEMPERATURE ON K
Ke
Ke
- G/RT
-( H TS ) /RT
ΔG = ΔH - TΔS
e
 H/RT
.e
S/R
Independent
of T
For endothermic reactions (ΔH > 0):
K increases if T increases.
For exothermic reactions (ΔH < 0):
K decreases if T increases.
K AND SOLUBILITY
K = Ksp (solubility product) when the equilibrium
reaction involves a solid salt dissolving to give its
constituent ions in solution.
Recall: [Solid] = 1
Saturated solution – in equilibrium with undissolved
solid
Thus if an aqueous solution is left in contact with
excess solid, the solid will dissolve until Ksp is
satisfied.
Thereafter the amount of undissolved solid remains
constant.
Example:
Calculate the mass of PbCl2 that dissolves in 100 ml
water. (Ksp = 1.7x10-5 for PbCl2)
PbCl2(s)
Initial:
Final:
Pb2+(aq) + 2Cl-(aq)
(solid)
(solid)
m = 0.45 g
THE COMMON ION EFFECT
Now add 0.03M NaCl to the PbCl2 solution
 We added 0.03M Cl-
PbCl2(s)
Initial:
Final:
Pb2+(aq) + 2Cl-(aq)
(solid)
(solid)
For this system to be at equilibrium when [Cl-] is
added, the [Pb2+] decreases (reverse reaction).
– this is an application of the Le Chatelier’s principle
and is called THE COMMON ION EFFECT
The salt will be less soluble if one of its constituent
ions is already present in the solution.
The Common Ion Effect - experiment
THE NATURE OF WATER AND ITS IONS
H+ does not exist on its own in H2O  forms H3O+
H3O+:
In aqueous solution, H3O+ is tightly associated with 3
molecules of H2O through exceptionally strong
hydrogen bonds.
One H2O is held by weaker ion-dipole attraction
Can also form H5O2+ cation  H+ shared by 2 water
molecules
H3O2- (OH-.H2O) has been observed in solids
AUTOPROTOLYSIS
Water undergoes self-ionisation  autoprotolysis,
since H2O acts as an acid and a base.
H2O + H2O
H3O+ + OH-
The extent of autoprotolysis is very small.
For H2O:
Kw = [H3O+][OH-] = 1.0 x 10-14
(at 25oC)
pH
pH ≈ -log[H+]
Approximate definition of pH
pH + pOH = -log(Kw) = 14.00
at 250C
It is generally assumed that the pH range is 0-14. But
we can get pH values outside this range.
e.g. pH = -1  [H+] = 10 M
This is attainable in a strong concentrated acid.
ACTIVITY
aA + bB
cC + dD
Equilibrium constant
c
d

C D
K
A a Bb
[A], [B], [C], [D] –concentrations of A, B, C, D
BUT in a real solution all charged ions are:
surrounded by ions with opposite charge – ionic
atmosphere
hydrated - surrounded by tightly held water dipoles
Adding an “inert” salt to a sparingly soluble salt
increases the solubility of the sparingly soluble salt.
WHY?
“inert” salt = a salt whose ions do not
react with the compound of interest
Consider:
BaSO4 (Ksp = 1.1x10-10) as the sparingly soluble salt and
KNO3 as the “inert” salt. In solution:
The cation (Ba2+) is surrounded by anions (SO42-, NO3-)
net positive charge is reduced
The anion (SO42-) is surrounded by cations (Ba2+, K+)
net negative charge is reduced
 attraction between oppositely charged ions is
decreased.
The net charge in
the ionic
atmosphere is
less than the
charge of the ion
at the center.
The ionic
atmosphere
decrease the
attraction
between ions.
The greater the ionic strength of a solution, the higher the
charge in the ionic atmosphere.
Each ion-plus-atmosphere contains less charge and there
is less attraction between any particular cation and anion.
Activity of the
ion in a solution
depends on its
hydrated radius
not the size of
the bare ion.
IONIC STRENGTH, µ
A measure of the total concentration of ions in solution.
The more highly charged an ion, the more it is counted.
1
μ   ci zi2
2 i
Where ci = concentration of the ith species
zi = charge for all ions in solution
Example:
Find the ionic strength of 0.010 M Na2SO4 solution.
Effect of ionic strength on solubility
Explain all 4 cases
ACTIVITY COEFFICIENTS
To account for the effect of ionic strength,
concentrations are replaced by activities.
AC  CγC
Activity of C
And general form of equilibrium constant is:
c
d
c
d
A C AD
[C] γ C [D] γD
K

a
b
a
b
a
b
A A AB
[A] γ A [B] γB
c
d
Activity coefficient:
• Measure of deviation of behaviour from ideality
(ideal   = 1)
• Allows for the effect of ionic strength
At low ionic strength:
activity coefficients  1
and K  concentration equilibrium
Thus for the sparingly soluble salt BaSO4, dissolving in
the presence of the “inert” salt KNO3:
Ksp = aBa aSO4 = [Ba2+]Ba [SO42-]SO4
If more BaSO4 dissolves in the presence of KNO3,
[Ba2+] and [SO42-] increases and Ba and SO4 decreases
ACTIVITY COEFFICIENTS
OF IONS
Extended Debye-Hűckel equation relates
activity coefficients to ionic strength:
 0.51z 2 μ
log  
 μ 

1  

305


at 250C
 = effective hydrated radius of the ion
Effect of Ionic Strength, Ion charge and Ion Size on the
Activity Coefficient
(Over the range of ionic strength
from 0 to 0.1M)
1. As ionic strength increases, the
activity coefficient decreases.
  1 as   0
2. As the charge of the ion increases, the
departure of its activity coefficient
from unity increases.
Activity corrections are much more
important for an ion with a charge of
3 than one with the charge 1.
3. The smaller the hydrated radius of
the ion, the more important activity
effects become.
Activity coefficients for
differently charged ions
with a constant hydrated
radius of 500pm.
Obtain values for  from the table:
Use interpolation to find values of  for ionic strengths not listed
How to interpolate - SELF STUDY!!
Linear
interpolation:
Unknown y interval Known x  interval

y
x
At high ionic strengths:
activity coefficients of
most ions increase
Concentrated salt
solutions are not the
same as dilute aqueous
solutions

“different solvents”
H+ in NaClO4 solution of
varying ionic strengths
pH AND ACTIVITY COEFFICIENTS
pH ≈ -log[H+]
Approximate definition of pH
The real definition of pH is:
pH  log AH  log[H ]γH
NOTE:
A pH electrode measures activity of H+ and NOT
concentration
SYSTEMATIC TREATMENT OF EQUILIBRIUM
Chemical equilibrium provides a basis for most
techniques in analytical chemistry and application of
chemistry to other disciplines such as like biology,
geology etc.
The systematic treatment of equilibrium gives us the tool
to deal with all types of complicated chemical equilibria.
The systematic procedure is to write as many
independent algebraic equations as there are unknowns
(species) in the problem. This includes all chemical
equilibrium conditions + two balances: charge and mass
balances.
CHARGE BALANCE
The sum of positive charges in solution equals the sum
of negative charges.
Charge neutrality
E.g.
An aqueous solution of KH2PO4 and KOH contains
the following ionic species:
H+, OH-, K+, H2PO4-, HPO42-, PO43-
The charge balance is:
[H+] + [K+] = [OH-] + [H2PO4-] + 2[HPO42-] + 3[PO43-]
The coefficient in front of each species
= the magnitude of the charge on the ion
MASS BALANCE
Conservation of matter.
Quantity of all species in a solution containing a
particular atom must equal the amount of that atom
delivered to the solution.
E.g.
Mass balance for 0.02 M phosphoric acid in water:
0.02 M = [H3PO4] + [H2PO4-] + [HPO42-] + [PO43-]
SYSTEMATIC TREATMENT OF EQUILIBRIUM:
Step 1. Write the pertinent reactions.
Step 2. Write the charge balance equation.
Step 3. Write the mass balance equations.
Step 4. Write the equilibrium constant for each
chemical reaction.
Step 5. Count the equations and unknowns.
Step 6. Solve for all the unknowns.
E.g.: The ionization of water
H2O H+ + OHKw = 1.0x10-14 at 250C
Find the concentrations of H+ and OH- in pure water
Step 1. Pertinent reaction
– only one above.
Step 2. Charge balance:
(1)
Step 3. Mass balance:
(2)
Step 4. Equilibrium constants – the only one
(3)
Step 5. Count equations and
unknowns
Step 6. Solve.
– 2 eq. and 2 unknowns
For pure water the ionic strength approaches 0 and we can write eq.3 as:
E.g.: The solubility of Hg2Cl2
Find the concentration of Hg22+ in a saturated solution of
Hg2Cl2
Step 1. Pertinent reactions
Hg2Cl2 Hg22+ + 2Cl- Ksp
H2O H+ + OH-
Step 2. Charge balance:
Kw
(1)
Step 3. Mass balance:
(2)
Step 4. Equilibrium constants:
(3)
(4)
Step 5. Count equations and unknowns – 4 eqs. and 4 unknowns
Step 6. Solve. Using eqn 2 we can write eqn 3 as:
THE DEPENDENCE OF SOLUBILITY ON pH
Coupled equilibria – the product of one reaction is reactant
in the next reaction
Problem:
The mineral fluorite, CaF2,
has a cubic crystal structure
and often cleaves to form
nearly perfect octahedra.
Find the solubility of CaF2 in
water.
Step 1. Pertinent reaction
CaF2 dissolves:
CaF2(s)
Ca2+ + 2F- Ksp = 3.9x10-11
The F- ions reacts with water to give HF:
F- + H2O
HF + OH- Kb = 1.5x10-11
For every aqueous solution:
H 2O
H+ + OHKw = 1x10-14
Step 2. Charge balance
(1)
Step 3. Mass balance
Some fluoride ions react to give HF.
(2)
Also
CaF2(s)
Ca2+ + 2F-
Ksp
F- + H2O
HF + OH-
Kb
H 2O
H+ + OH-
Kw
Step 4. Equilibrium constants
(3)
(4)
(5)
Step 5. Count equations and unknowns
5 eqs. and 5 unknowns:
Step 6. Solve
To simplify the problem let us solve it for a fixed pH = 3
That means:
[H+] =
and [OH-] =
Then from eqn 4:
Kb = [HF][OH-]/[F-]
[HF]/[F-] = Kb/[OH-] = 1.5x10-11/1.0x10-11 = 1.5
Thus
[HF] = 1.5[F-]
Substitute [HF] in the eqn 2:
[F-] + [HF] = 2[Ca2+]
[F-] + 1.5[F-] = 2[Ca2+]
And
[F-] = 0.80[Ca2+]
Using this expression in eqn 3:
Ksp = [Ca2+][F-]2 = [Ca2+](0.80[Ca2+])2
Thus
[Ca2+]
= (Ksp
/0.802)1/3
=
3.9x10-4M
Now find:
[F-] = 3.1x10-4 M
[HF] = 4.7x10-4 M
NOTE: To fix the pH of a solution an ionic compound is
added. Thus the charge balance equation as
written not longer holds.
pH dependence of the
conc. of Ca2+, F- and
HF in a saturated
solution.
Also [OH-] = [H+] + [HF]
No longer holds
Applications of coupled equilibria in the modeling of
environmental problems
Found [Ca] in acid rain that has
washed off marble stone
(largely CaCO3) increases as the
[H+] of acid rain increases.
CaCO3(s) + 2H+(aq) 
Ca2+(aq) + CO2(g) + H2O(l)
SO2(g) + H2O(l)  H2SO3(aq)
oxidation
H2SO4(aq)
Al is usually “locked” into insoluble minerals e.g. kaolinite and
bauxite. But due to acid rain, soluble forms of Al are introduced into
the environment. (Similarly with other minerals containing Hg, Pb etc.)
Total [Al] as a function of pH in
1000 Norwegian lakes.