The type and degree of splitting depends on the
ligand and geometry of the complex that is formed.
This ‘splitting’ of the d orbitals results in a d>d
transition that is in the UV/Vis range.
Chromium(III) examples
octahedral
tetrahedral
square planar
Lig
[A]
[HA]
Initially, each solution is at pH 7.00.
After adding 10 ml of 1.0 M HCl we have:
Pure water
(10 ml)(1.0 M)
[H3O+] =
(110 ml)
pH
= 1.04
This is a pretty big jump!
= 0.091
Determine the concentration of silver ion when
excess AgCl is added to 0.1M NaBr.
AgCl(s)
Ag+ + Cl
We are adding a relative insoluble material to a
solution that contains an anion that also
produces a precipitate with silver.
AgBr(s)
Ag+ + Br
KSP AgBr =
H5Y+
1
H 3 YY4
H4Y
H6Y2+
HY3
H2Y2
Under normal conditions, the H6Y2+ and H5Y+
forms are not present at significant levels.
The effect of the hydrogen ion can then be
calculated using !Y.
! .5
[Y4]
!Y = [H Y] + [H Y] + [H Y2] + [HY3] + [Y4]
4
3
2
0
If the goal is to oxidize your analyte to a single
form, no material is available that can be used
as a column.
You must have a method to remove any
excess oxidizing reagents prior to titration.
HCl
Often, the results in the preparation step being
more co
mol A
L solution
mmol A
mL solution
dissolve
in water
grams A
Formula Weight A
liters of solution
molA =
M=
1 M H2SO4  analytical M
[H2SO4]
]
= 0.00 M
[HSO4
+ SO4
2
~1%
= 1.01 M
O+]
This is all great but lets consider what to do when
we actually conduc
In order to get sharp endpoints, solutions are
typically buffered at basic conditions.
This also insures that:
Y4 form is available.
EDTA will be in a soluble form.
In addition, only one !Y must be used in
calculating equivalence and over titration
condi
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
The best way to see what is going on is to give an
overview of the approach.
We can then introduce the relative simple
calculations with each step.
The first premise is that have a sample that
contains an analyte of interest.
Your goal is to determine the
A large portion of this course deals with methods
that are used to determine how much of a
material is present.
We need to review the general steps that are
taken for any quantitative method.
These steps are taken to insure an accurate
and reliable answer
Equivalent weight
Formula Weight
Eq Wt = # equivalents/mole
Determining the number of equivalents in a
mole requires that you know the type of
reaction and how the species involved actually
combine
In other words  you already know R
If the normality of y
Ag+ + Cl
AgCl(s) + Excess Ag+
+
Fe(SCN)2+
(red)
endpoint
SCNFe3+
AgSCN (s)
The endpoint is not very sharp but gives good results.



+ + + +
+
+
+
+
+
+
AgCl
+
+
+
+
+
+ + + + +
After the equivalence point,
silver ion is in excess.
It becomes our prim
With UV, Vis, IR  absorption occurs over a
range. By scanning and measuring absorption
over a range of wavelengths, we can produce
a spectrum.
Molecular absorption
spectra example
UV
Visible
Near
IR
IR
Because of differences in equipment, we
typically ob
So the E for electrolysis is actually:
Eapp = (Eanode + !ac + !aa)  (Ecathode + !cc + !ca)
!ac = concentration overpotential at anode
!aa = activation overpotential at anode
!cc = concentration overpotential at cathode
!ca = activation overpotential at c
PA
The rate can be expressed mathematically
based on the disappearance of A with respect
to time as:
d [A]

where
A
time
dt
=kt
k is the rate constant expressed
in units of time1 (sec1, min1, .)
A firstorder reaction depends only on the
value of k an
carrier
gas
sample
introduction
heated
zones
column
detector
b
a
c
d
a  compressed gas cylinder
b  pressure regulator
c  valve
d  filter
data
system
As with other types of reactions, the formation of
a precipitate can be used as the basis of a
titration.
analyte + titrant
precipitate
The approach assumes that under the
experimental conditions used, the product is
virtually insoluble.
panalyte
equivale
titration
analyte of
unknown
concentration
By accurately measuring
the volume of titrant that
is added, the amount of
sample can be determined.
A buret is used to control
and measure the amount
of titrant that is added.
titrant  AgNO3
 a standard soluti
Assumes that you have some idea as to what the
standard deviation should be or can calculate
it.
M=
 suspect  mean
S
If M > 4 then you can reject the point.
This is simply a crude t test. It is only useful for
discarding obviously bad data.
Use ratio '
Both oxidation and reduction occur during a
titration.
The equivalence point is based on the
concentration of the oxidized and reduced form
of all species involved.
Aox + Bred = Ared + Box
Ce4+ + Fe2+ = Ce3+ + Fe3+
For a REDOX titration, the equivalence p
E1/2 values are listed verses the SCE.
Not the same as Eo values
Are also dependent on the supporting
electrolyte used for the analysis.
Typical range is 1.9 to +0.2 V vs SCE.
Supporting
E1/2 vs SCE electrolyte
Reaction
Cu2+
Sn4+
Cu
Sn2+
0.04
0.25
0.1 M
Rs =
v1
v2
w1
v2  v1
(w2+ w1)/2
w2
since w = 4 v / N1/2
Rs =
N v2  v1
2 v2 + v1
Note: times or any
representative unit can be
used in place of v and w terms.
Since
v = Vm (1  k)
Rs =
resolution in terms of k is:
N
k2  k1
2 2 + k 2 + k1
Rs =
N
2
!1
!
The average response is then used for
subsequent data analysis.
Two types of
information are
obtained.
id
id
diffusion current
E1/2
halfwave potential
E1/2
id
STD
id
UNK
=
CSTD
CUNK
KA1 / KA2
= 7.5x103 / 6.0x108
= 1.25 x 105
KA2 / KA3
= 6.0x108 / 4.8x1013
= 1.25 x 105
This indicates that it is possible to
treat each step separately during
a titration.
Just because you have a material where KA1/KA2
is not greater than 1000 does no

A quantitative method based on weight gain. It is
also referred to as electrodeposition.
+
A
!A very old method.
!When it works, it works well.
!Unfortunately, it only works for a limited
number of materials.
V
R
R  potentiometer
A  ammeter
V  Voltme