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
[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
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]
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
O
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 gre
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 u
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
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
contain
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 st
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
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
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
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
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 tit
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 onl
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+
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
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
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

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