A solution containing 10.0 ppm (1 ppm = I mg per liter) of a certain colored mat
ID: 880791 • Letter: A
Question
A solution containing 10.0 ppm (1 ppm = I mg per liter) of a certain colored material, measured at 1.00 cm optical path, has, the absorbance and the percent transmittance (%T) shown in the first line of the tabulation below Calculate values to fill in the blanks in the tabulation, as indicated. Assume the system conforms to Beer’s Law.
Conc.,
Ppm
Optical
Path cm
%T
Absorbance
Given:
10.0
1.00
38.0
0.410
(a)
14.2
1.00
--
--
(b)
12.7
1.00
--
--
(c)
8.0
2.00
--
--
(d)
--
1.00
--
0.861
(e)
--
5.00
32.1
--
(f)
10.0
--
19.2
--
Conc.,
Ppm
Optical
Path cm
%T
Absorbance
Given:
10.0
1.00
38.0
0.410
(a)
14.2
1.00
--
--
(b)
12.7
1.00
--
--
(c)
8.0
2.00
--
--
(d)
--
1.00
--
0.861
(e)
--
5.00
32.1
--
(f)
10.0
--
19.2
--
Explanation / Answer
Conc.,
Ppm
Optical
Path cm
%T
Absorbance
Given:
10.0
1.00
38.0
0.410
(a)
14.2
1.00
26.17
0.5822
(b)
12.7
1.00
30.15
0.5207
(c)
8.0
2.00
22.1
0.656
(d)
21.0
1.00
13.8
0.861
(e)
2.40
5.00
32.1
0.493
(f)
10.0
1.75
19.2
0.717
First calculate the molar absorptivity constant by using following expression:
Absorbance = molar absorptivity x path length x concentration
Here; Absorbance = 0.410; path length = 1.00 cm ; concentration = 10.0 ppm
Then;
molar absorptivity = Absorbance / path length x concentration
= 0.410 / 1.00*10=0.0410 ppm-1 cm-1
Now calculate the Absorbance for (b) row:
Absorbance = molar absorptivity x path length x concentration
= 0.0410 ppm-1 cm-1*14.2*1.00
= 0.5822
Relation between %T and Absorbance
A = 2 - log10 %T
0.5822 = 2- log10 %T
log10 %T = 2-0.5822=1.4178
%T =10^1.4178=26.17
Now calculate the Absorbance for (c) row:
Absorbance = molar absorptivity x path length x concentration
= 0.0410 ppm-1 cm-1*12.7*1.00
= 0.5207
Relation between %T and Absorbance
A = 2 - log10 %T
0.5207 = 2- log10 %T
log10 %T = 2-0.5207=1.4793
%T =10^1.4793=30.15
Now calculate the Absorbance for (d) row:
Absorbance = molar absorptivity x path length x concentration
= 0.0410 ppm-1 cm-1*8*2.00
= 0.656
Relation between %T and Absorbance
A = 2 - log10 %T
0.656 = 2- log10 %T
log10 %T = 2-0.656=1.344
%T =10^1.344=22.1
Now calculate the concentration for (e) row:
Absorbance = molar absorptivity x path length x concentration
0.861= 0.0410 ppm-1 cm-1*concentration *1.00 cm
Concentration = 21.0 ppm
Relation between %T and Absorbance
A = 2 - log10 %T
0.861 = 2- log10 %T
log10 %T = 2-0.861=1.139
%T =10^1.139=13.8
Relation between %T and Absorbance
A = 2 - log10 %T
A = 2- log10 32.1
A= 2-1.507 = 0.493
Now calculate the concentration for (e) row:
Absorbance = molar absorptivity x path length x concentration
0.493= 0.0410 ppm-1 cm-1* concentration *5.00
concentration = 2.4 ppm
Relation between %T and Absorbance
A = 2 - log10 %T
A = 2- log10 19.2
A= 2-1.283= 0.717
Now calculate the concentration for (e) row:
Absorbance = molar absorptivity x path length x concentration
0.717= 0.0410 ppm-1 cm-1* path *10.00
Path = 1.75 cm
Conc.,
Ppm
Optical
Path cm
%T
Absorbance
Given:
10.0
1.00
38.0
0.410
(a)
14.2
1.00
26.17
0.5822
(b)
12.7
1.00
30.15
0.5207
(c)
8.0
2.00
22.1
0.656
(d)
21.0
1.00
13.8
0.861
(e)
2.40
5.00
32.1
0.493
(f)
10.0
1.75
19.2
0.717
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