the dynamic state in which rates of the forward and reverse reactions are identical.
aA + bB cC + dD
Equilibrium constant K

   
    d 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: K
1
= 1/K cC + dD aA + bB
K
1

   
    b d
Equilibrium const. for two reactions added: K
3
= K
1 x K
2
H +
HA H +
+ C CH
+ A
+
K
1

[H

][A

]
[HA]
K
2

[CH

[C][H

]
]
HA + C CH + + A -
K
1
.K
2

[H

][A

]
[HA]

[CH

[C][H

]
]

[A

][CH

]
[HA][C]

K
3

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
Heat absorbed
Endothermic reaction
ΔH < 0
Heat liberated
Exothermic reaction

ENTROPY
ΔS – a measure of ‘disorder’ of a substance
Gas

Liquid

Solid
Decrease in disorder
ΔS > 0
Products more disordered than reactants
ΔS < 0
Products less disordered than reactants
Example:
KCl(s) K + (aq) + Cl (aq)
ΔS 0 = +76 J/K mol at 25 0 C

FREE ENERGY
Gibbs free energy: ΔG = ΔH - TΔS
ΔG < 0
The reaction is favoured
K>1
ΔG > 0
The reaction is not favoured
K<1
Free energy and equilibrium:
K
 e

0
OR
 

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 K

   
   
If reaction is at equilibrium and reactants are added
(or products removed), the reaction goes to the right.
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 Ch âtelier:
If Q > K
 reaction will proceed in the reverse direction
If Q < K
 reaction will proceed in the forward direction

A + B C + D K

  
  
At equilibrium:
[A] = 0.002M [B] = 0.025M
[C] = 5.0M [D] = 1.0M
K


5.0

0.002


1.0

0.025

= 1x10 5 at 25 0 C
Say we add double reactant A,
[A] = 0.004M
Q


5.0

0.004


1.0

0.025

= 0.5x10
5 at 25 0 C
Q < K
To reach equilibrium:
Q = K and the reaction must go to the right
 spontaneous

K
 e -

G/RT
ΔG = ΔH - TΔS
K
 e -(

H

T

S ) /RT  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 = K sp
(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 K satisfied. sp is
Thereafter the amount of undissolved solid remains constant.

Example:
Calculate the mass of PbCl
2 water. (K sp that dissolves in 100 ml
= 1.7x10
-5 for PbCl
2
)
Initial:
Final:
PbCl
2
(s) Pb 2+ (aq) + 2Cl (aq)
(solid)
(solid) m = 0.45 g

Now add 0.03M NaCl to the PbCl
2

We added 0.03M Cl solution
Initial:
Final:
PbCl
2
(s) Pb 2+ (aq) + 2Cl (aq)
(solid)
(solid)
For this system to be at equilibrium when [Cl ] is added, the [Pb 2+ ] 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

H + does not exist on its own in H
2
O
 forms H
3
O +
H
3
O + :

In aqueous solution, H
3
O + is tightly associated with 3 molecules of H
2
O through exceptionally strong hydrogen bonds.
One H
2
O is held by weaker ion-dipole attraction

Can also form H
5
O
2
+ molecules cation

H + shared by 2 water
H
3
O
2
(OH .H
2
O) has been observed in solids

Water undergoes self-ionisation
 autoprotolysis , since H
2
O acts as an acid and a base.
H
2
O + H
2
O H
3
O + + OH -
The extent of autoprotolysis is very small.
For H
2
O: K w
= [H
3
O + ][OH ] = 1.0 x 10 -14 (at 25 o C)

pH ≈ -log[H + ]
Approximate definition of pH pH + pOH = -log(K w
) = 14.00 at 25 0 C

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.

aA + bB cC + dD
Equilibrium constant K

   
    d 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:
BaSO
4
KNO
3
(K sp
= 1.1x10
-10 ) as the sparingly soluble salt and as the “inert” salt. In solution:
The cation (Ba 2+ ) is surrounded by anions (SO
4
2, NO
3
)
 net positive charge is reduced
The anion (SO
4
2) is surrounded by cations (Ba 2+ , 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.

A measure of the total concentration of ions in solution.
The more highly charged an ion, the more it is counted.
μ 
2
1  i c i z i
2
Where c i z i
= concentration of the i th species
= charge for all ions in solution

Example:
Find the ionic strength of 0.010 M Na
2
SO
4 solution.

Effect of ionic strength on solubility
Explain all 4 cases

To account for the effect of ionic strength, concentrations are replaced by activities.
Activity of C
A
C

 
C
And general form of equilibrium constant is:
K

A
C c
A
D d
A
A a
A
B b

[C] c γ
C c
[D] d γ
D d
[A] a γ
A a
[B] b γ
B b

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 BaSO
4
, dissolving in the presence of the “inert” salt KNO
3
:
K sp
= a
Ba a
SO4
= [Ba 2+ ]

Ba
[SO
4
2]

SO4

If more BaSO
4 dissolves in the presence of KNO
3
,
[Ba 2+ ] and [SO
4
2] increases and

Ba and

SO4 decreases

Extended Debye-H űckel equation relates activity coefficients to ionic strength:
 

2
 


 




at 25 0 C

= 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

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 NaClO
4 solution of varying ionic strengths

pH ≈ -log[H + ]
Approximate definition of pH
The real definition of pH is: pH
  log A
H

  log [H

]
H

NOTE:
A pH electrode measures activity of H + and NOT concentration

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 KH
2
PO
4 the following ionic species: and KOH contains
H + , OH , K + , H
2
PO
4
, HPO
4
2, PO
4
3-
The charge balance is:
[H + ] + [K + ] = [OH ] + [H
2
PO
4
] + 2[HPO
4
2] + 3[PO
4
3]
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 = [H
3
PO
4
] + [H
2
PO
4
] + [HPO
4
2] + [PO
4
3]

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
H
2
O H + + OH K w
= 1.0x10
-14 at 25 0 C
Find the concentrations of H + and OH in pure water
Step 1. Pertinent reaction
Step 2. Charge balance:
– only one above.
Step 3. Mass balance:
Step 4. Equilibrium constants – the only one
(1)
(2)
(3)
Step 5. Count equations and unknowns – 2 eq. and 2 unknowns
Step 6. Solve.
For pure water the ionic strength approaches 0 and we can write eq.3 as:

E.g.: The solubility of Hg
2
Cl
2
Find the concentration of Hg
2
2+
Hg
2
Cl
2 in a saturated solution of
Step 1. Pertinent reactions Hg
H
2
2
Cl
2
O H +
Hg
2
2+
+ OH -
+ 2Cl K sp
K w
Step 2. Charge balance:
Step 3. Mass balance:
(1)
Step 4. Equilibrium constants:
(2)
(3)
Step 5. Count equations and unknowns – 4 eqs. and 4 unknowns
(4)
Step 6. Solve. Using eqn 2 we can write eqn 3 as:

Coupled equilibria – the product of one reaction is reactant in the next reaction
Problem:
The mineral fluorite, CaF
2
, has a cubic crystal structure and often cleaves to form nearly perfect octahedra.
Find the solubility of CaF
2 water.
in

Step 1.
Pertinent reaction
CaF
2 dissolves:
The F -
CaF
2
(s) Ca 2+ + 2F K sp
= 3.9x10
ions reacts with water to give HF:
-11
F + H
2
O HF + OH -
For every aqueous solution:
K b
= 1.5x10
-11
H
2
O H + + OH K w
= 1x10 -14
Step 2.
Charge balance
(1)
Step 3.
Mass balance
Some fluoride ions react to give HF.
(2)
Also

CaF
2
(s) Ca 2+ + 2F -
F + H
2
O HF + OH -
H
2
O H + + OH -
Step 4.
Equilibrium constants
K sp
K b
K w
(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:
K b
= [HF] [OH ] /[F ]
[HF]/[F ] = K b
/ [OH ] = 1.5x10
-11 / 1.0x10
-11 = 1.5
Thus [HF] = 1.5[F ]
Substitute [HF] in the eqn 2:
[F ] + [HF] = 2[Ca 2+ ]
[F ] + 1.5[F ] = 2[Ca 2+ ]
And [F ] = 0.80[Ca 2+ ]
Using this expression in eqn 3:
K sp
= [Ca 2+ ] [F ] 2 = [Ca 2+ ]( 0.80[Ca 2+ ] ) 2
Thus [Ca 2+ ] = (K sp
/0.80
2 ) 1/3 = 3.9x10
-4 M
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 Ca 2+ , 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 CaCO
3
) increases as the
[H + ] of acid rain increases.
CaCO
3
(s) + 2H + (aq)

Ca 2+ (aq) + CO
2
(g) + H
2
O(l)
SO
2
(g) + H
2
O(l)

H
2
SO
3
(aq) oxidation
H
2
SO
4
(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.