1. Many analytical methods rely on equilibrium
systems in aqueous solution.
This unit will review
General concepts of aqueous solutions
Chemical equilibrium
Equilibrium calculations
Deviations from ideal behavior

2. acid base conjugate conjugate base acid conjugate conjugate
base of HF acid of H2O acid base
base strength
acid strength
For the general chemical reaction: kinetic equilibrium
region region
aA + bB cC + dD
If A and B are brought together: A
There is an initial reduction in the B
concentrations of A and B. C
Both C and D will increase in concentration. D
We reach a point where the concentrations no
longer change.
time

3. These are dynamic equilibria Equilibrium concentrations are based on:
The specific equilibrium
At equilibrium, the forward and reverse rates of The starting concentrations
reaction are equal.
Other factors such as:
Temperature
Any given species is constantly changing from
Pressure
one form to another.
Reaction specific conditions
Changes in the system will alter the rates and a Altering conditions will stress a system,
new equilibrium will be achieved. resulting in an equilibrium shift.
aCc aDd
Keq = a = activity
aAa aBb
[ C ]c [ D ]d
Keq =
[ A ] a [ B ]b

4. We’ll be using molar concentrations when
working with chemical equilibrium.
This introduces errors that you should be
aware of.
While we will not work with activities, we need
to know what they are.
( )
Except for dilute systems, the effective
concentration of ions is usually less than the
actual concentration.
The term activity is used to denote this effective
concentration.
activity ai = fi [ i ]
where
fi is the activity coefficient for i
[ i ] is the molar concentration

5. For very dilute solutions
As fi -> 1, ai -> [ i ]
For µ up to 0.1
0.5 Zi2 µ ! fi < 1 and ai < [ i ]
1 + 0.33 "i µ!
When u is > 0.1
Results in complicated behavior.
In general, if µ < 0.01, we can safely
use molar concentrations.
[ H3O+ ] [ OH- ]
Keq =
[ H2O ]2
In dilute solutions, [ H3O+] and [ OH- ] is much
smaller than [ H2O ]. [ H2O ] is essentially a
constant of ~55.5 M.

6. KW = 10-14 = [H3O+] [OH-]
[H3O+] = [H3O+]water + [H3O+]HCl
[OH-] = [H3O+]water
Lets set x = [H3O+]water then
[H3O+] = x + 1.0 x 10-8
[OH-] = x
10-14 = (x + 1.0 x 10-8) ( x)
Our equation can be rearranged as x = -10-8 + [ (10-8)2 + 4x10-14]1/2 / 2
x2 + 10-8 x - 10-14 = 0 = 1.9 x 10-7 M
This quadratic expression can be solved by:
pH = 6.72
-b + b2 - 4ac
x= So adding a small amount of HCl to water
2a DOES make it acidic.
Only the positive root is meaningful in While this approach is more time
equilibrium problems. consuming, you’ll find it very useful as our
problems get more complex.

7. KSP expressions are used for ionic materials that At equilibrium, our system is a saturated solution
are not completely soluble in water. of silver and chloride ions.
Their only means of dissolving is by dissociation. The only way to know that it is saturated it to
observe some AgCl at the bottom of the solution.
AgCl(s) Ag+ (aq) + Cl- (aq)
[ Ag+ (aq) ] [ Cl- (aq) ] As such, AgCl is a constant and KSP expressions
Keq = do not include the solid form in the equilibrium
[ AgCl(s) ] expression
Determine the solubility of AgCl in water at 20oC in
grams / 100 ml.
KSP = [Ag+] [Cl-] = 1.0 x 10-10
At equilibrium, [Ag+] = [Cl-] so
1.0 x 10-10 = [Ag+]2
[Ag+] = 10-5 M
g AgCl = 10-5 mol/l * 0.1 l * 143.32 g/mol
Solubility = 1.43x10-4 g / 100 ml
KSP = 1.1 x 10-12 = [ Ag+ ]2 [ CrO42- ] KSP = 1.1 x 10-12 = [ Ag+ ]2 [ CrO42- ]
[CrO42-] = [CrO42-]Ag2CrO4 + [CrO42-]Na2CrO4 [CrO42-] = 0.010 M
[ Ag+ ] = ( KSP / [ CrO42- ] ) 1/2
With such a small value for KSP, we can assume
that is [CrO42-]Ag2CrO4 negligible. = 1.1 x 10-12 / 0.010 M
If we’re wrong, our silver concentration will be = 1.1 x 10-5 M
significant (>1% of the chromate concentration.)
[ Ag+ ] << [ CrO42- ]
Then you’d use the quadratic approach.
so our assumption was valid.

8. [H3O+] [A-]
[HA] Water is not
included in the
expressions
because it is a
[OH-] [BH+] constant
[B]
KA = [ H3O+ ] [ A- ] = 2.24 x 10-5 KA = 2.24 x 10-5 = X2 / (0.1 - X)
[ HA ]
Rearranging give us:
Since both a H3O+ and a A- is produced for X2 + 2.24x10-5X - 2.24x10-6 = 0
each HA that dissociates:
[ H3O+ ] = [ A- ] We can solve this quadratic or possibly
assume that the amount of acid that
[ HA ] = 0.1 M - [ H3O+ ]
dissociates is insignificant (compared
to the undissociated form)
Lets set X = [ H3O+ ]
KA = 2.24 x 10-5
[ H3O+ ] = [ A- ]
[ HA ] = 0.1 M - [ H3O+ ]
KA / 0.1 < 10-3 - can assume that [HA] = 0.1 M
2.24x10-5 = X2 / 0.1M
X = (2.24x10-6)1/2
= 0.00150
pH = 2.82

9. -2.24x10-5 + [(2.24x10-5)2 +4x2.24x10-6]1/2
Now, lets go for the exact solution X=
2
Earlier, we found that
X2 + 2.24x10-5X - 2.24x10-6 = 0 X = 0.00149
pH = 2.82
-b + b2 - 4ac
X=
2a No significant difference between our two
answers.
If you are starting with an acid, acidic conditions or
the conjugate acid of a base
Do your calculations using KA
If starting with a base, basic conditions or the
conjugate base of an acid
Do your calculations using KB
You can readily convert pH to pOH and KA to KB
values.
With complex formation, two or more species will
join, forming a single, new species. If we are evaluating the decomposition of a
complex, we can use a K decomposition.
Mn+ + xL M(L)xn+
M(L)xn+ Mn+ + xL
[ Mn+ ] [ L ]m
[ M(L)x ]
n+ KD = [ M(L)xn+ ]
KF =
[ Mn+ ] [ L ]m
Since water is not a portion of these
equilibria, KD = 1 / KF

10. Equilibrium expressions for REDOX systems are
derived from standard electrode potentials.
6Fe2+ + Cr2O72- + 14 H3O+ 6Fe3+ + 2Cr3+ + 21H2O
[Fe3+]6 [Cr3+]2
KREDOX =
[Fe2+]6 [Cr2O72-][H3O+]14
We’ll review how to determine and work with
KREDOX when we cover the units on
electrochemistry.
[H3O+] [H2PO4-]
K A1 = Note:
[H3PO4] [H3O+] is the same for
each expression.
[H3O+] [HPO42-] The relative amounts of
K A1 K A2 K A3 K A2 = each species can be
[H2PO4-] found if the pH is known.
The actual amounts can
be found if pH and total
[H3O+] [PO43-] H3PO4 is known.
K A3 =
[HPO42-]
One possible ‘first step’ would be to determine the
relative amounts of each species.
K A1 [H2PO4-]
=
[H3 O+] [H3PO4]
K A2 [HPO42-]
=
[H3 O+] [H2PO4-]
K A3 [PO43-]
=
[H3O+] [HPO42-]

11. 7.5 x 10-2 [H2PO4-] Total phosphate = 1.0 M so
10-7
=
[H3PO4]
= 750000
1.0 = [H3PO4] + [H2PO4-] + [HPO42-] + [PO43-]
6.2 x 10-8 [HPO42-]
= = 0.62 This is referred to as a mass balance.
10-7 [H2PO4-]
Based on the ratio, we know that [H3PO4] and
4.8 x 10-13 [PO43-] [PO43-] are not present at significant levels for
= = 4.8x10-6 this mass balance so:
10-7 [HPO42-]
These ratios show that only H2PO4- and HPO42- 1.0 = [H2PO4-] + [HPO42-]
are present at significant levels at pH 7.
[HPO42-] Since you now know the concentration of H2PO4-,
= 0.62 you can now sequentially solve the other
[H2PO4-] equilibrium expressions.
[HPO42-] = 0.62 [H2PO4-]
This approach is very useful when dealing with
complex equilibria.
1.0 = [H2PO4-] + 0.62 [H2PO4-]
= 1.62 [H2PO4-]
The mass balance is on means of eliminating
‘insignificant’ species based on addition or
[H2PO4-] = 1.0 / 1.62 = 0.617 M subtraction