1. A student determined the absorption spectrum of a 0.1049M standard solution a
ID: 917465 • Letter: 1
Question
1. A student determined the absorption spectrum of a 0.1049M standard solution and then repeated the process for a solution of the same solute but with an unknown concentration (unknown #1). A graph of the two absorption spectra is show below. A 2.00cm wide cuvette was used in the spectrophotometer.
A. What is the concentration of the unknown solution? Explain your method, identifying any wavelength(s) used and any data taken from the graph.
B. The student tried to determine the absorbance of another solution at lmax, but its absorbance was too high to be measured accurately. The student then set the spectrophotometer to 560nm and found that unknown solution #2 had an absorbance of 0.765. Calculate the concentration of unknown solution #2, showing your work. What are the benefits and dangers of working at 560nm, compared to working at lambda max and compared to working at some other wavelength?
C. To determine the concentration of unknown solution #2 using the spectrophotometer set at lmax, what can we do in preparing the sample to lower the absorbance? What other changes could we make without changing the sample preparation to lower the absorbance?
Absorption Spectra 0.5 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 standard unknown #1 400 450 550 600 wavesiongh (nm)Explanation / Answer
A.)
We know that.
A=ebc
Where A is absorbance (no units, since A = log10 P0 / P )
e is the molar absorbtivity with units of L mol-1 cm-1
b is the path length of the sample - that is, the path length of the cuvette in which the sample is contained. We will express this measurement in centimetres.
c is the concentration of the compound in solution, expressed in mol L-1
Since, molar absoptivity is constant.
Hence,
A1/b1c1 = A2/b2c2
Here, grom the graph and information.
At lamda max, 420 nm,
A1 = 0.47
A2 = 0.19
b1 = 2 cm
b2 = 2cm
c1 = 0.1049 M
c2 = ?
So,
0.47/(2*0.1049) = 0.19/(2*x)
x =0.095/2.24 = 0.0424M
So, concentration of unknown solution is 0.0424 M.
and method, identifying any wavelength(s) used and any data taken from the graph has been mentioned
B.)
At wavelength, 560 nm
A=ebc
Since, molar absoptivityi s constant.
Hence,
A1/b1c1 = A2/b2c2
Here, from the graph and information. standard and unknown solution
A1 = 0.02
A2 = 0.765
b1 = 2 cm
b2 = 2cm
c1 = 0.1049 M
c2 = ?
So,
0.02/(2*0.1049) = 0.765/(2*x)
x =0.3825/0.09532 = 4.0127 M
Hence, concentration of unknown solution #2, = 4.0127 M
The benefits and dangers of working at 560nm, compared to working at lambda max and compared to working at some other wavelength. Working at lamda max makes it easy for the calculation and morever the result comes precisely at lamda max as the wavelength of maximum absorbance is a characteristic value, designated as lamda max so it further makes a calculation easy and presied.
C.)
Since, A=ebc
Where A is absorbance (no units, since A = log10 P0 / P )
e is the molar absorbtivity with units of L mol-1 cm-1
b is the path length of the sample - that is, the path length of the cuvette in which the sample is contained. We will express this measurement in centimetres.
Here, A is directly proportional to c and directly proportional to b
Therefore, for preparing the sample to lower the absorbance we can reduce the concentration of the sample i.e. we can dilute the concentration by adding more and more solvent.
Other changes could we make without changing the sample preparation to lower the absorbance is decreasing the width of the cuvette of spectrophotometry.
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