solution containing the complex formed between bi( iii) and thiourea has a molar absorptivity of 9.32 x 103 l mol-1 cm at 470 nm_ (a) what is the absorbance of a 6.24xl0 5 m solution of the complex at 470 nm in a 1.00 cm cell? (b) what is the percent transmittance of the solution described in (a)? (c) what is the molar concentration of the complex in solution that has the absorbance described in
C
=
2
×
10
−
4
M
A
=
0.123
l
=
2
c
m
?
T
, ?
%
T
, ?
ε
The relationship between transmittance T
and absorbance is:
A
=
−
log
T
⇒
T
=
10
−
A
Since the length of the cell was increased to l
=
2
c
m
, the absorbance will double and therefore, A
2
=
2
×
0.123
=
0.246
⇒
T
=
10
−
0.246
=
0.568
Therefore, %
T
=
100
×
T
=
100
×
0.568
=
56.8
%
In order to solve for the molar absorptivity ε
, we will use the Beer-Lambert law:
A
=
ε
×
l
×
C
⇒
ε
=
A
l
×
C
ε
=
0.123
1
c
m
×
2
×
10
−
4
M
=
615
M
−
1
⋅
c
m
−
1
The basic idea here is to use a graph plotting Absorbance vs. Concentration of known solutions. Once you have that you can compare the absorbance value of an unknown sample to figure out its concentration.
http://intro.chem.okstate.edu/ChemSource/Instrument/inst4.htm
You will be applying Beer's law to calculate the concentration.
The equation for Beer's law is: A = εmCl
(A=absorbance, εm = molar extinction coefficient, C = concentration, l=path length of 1 cm)
You should have a data set which was used to create a standard curve. The graph should plot concentration (independent variable) on the x-axis and absorption (dependent variable) on the y axis.
You'll need to add a line of best fit to the data points and determine the equation for the line. The equation should be in y=mx + b form.
y = absorbance (A)
Note: no unit for absorbance
x = concentration (C)
Note: unit is M or mol/L
m = (εm) = slope or the molar extinction coefficient in beers law which has units of M
−
1
c
m
−
1
So A = εmC +b
If you solve for C you should get
C = (A-b)/εm
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