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.
Determine the concentration of silver ion when
excess AgCl is added to 0.1M NaBr.
Ag+ + Cl-
We are adding a relative insoluble material to a
solution that contains an anion that also
produces a precipitate with silver.
Ag+ + Br-
KSP AgBr =
H 3 YY4-
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.
!Y = [H Y] + [H Y-] + [H Y2-] + [HY3-] + [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.
Often, the results in the preparation step being
Formula Weight A
liters of solution
1 M H2SO4 - analytical M
= 0.00 M
= 1.01 M
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
Many analytical methods rely on equilibrium
systems in aqueous solution.
This unit will review
General concepts of aqueous solutions
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
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
In other words - you already know R
If the normality of y
Ag+ + Cl-
AgCl(s) + Excess Ag+
The endpoint is not very sharp but gives good results.
+ + + +
+ + + + +
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
Because of differences in equipment, we
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
The rate can be expressed mathematically
based on the disappearance of A with respect
to time as:
k is the rate constant expressed
in units of time-1 (sec-1, min-1, .)
A first-order reaction depends only on the
value of k an
As with other types of reactions, the formation of
a precipitate can be used as the basis of a
analyte + titrant
The approach assumes that under the
experimental conditions used, the product is
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
| suspect - mean|
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
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.
E1/2 vs SCE electrolyte
v2 - v1
since w = 4 v / N1/2
N v2 - v1
2 v2 + v1
Note: times or any
representative unit can be
used in place of v and w terms.
v = Vm (1 - k)
resolution in terms of k is:
k2 - k1
2 2 + k 2 + k1
KA1 / KA2
= 7.5x10-3 / 6.0x10-8
= 1.25 x 105
KA2 / KA3
= 6.0x10-8 / 4.8x10-13
= 1.25 x 105
This indicates that it is possible to
treat each step separately during
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 very old method.
!When it works, it works well.
!Unfortunately, it only works for a limited
number of materials.
R - potentiometer
A - ammeter
V - Voltme