Beer’s Law can be used to determine the concentration of two substances, A and B
ID: 695260 • Letter: B
Question
Beer’s Law can be used to determine the concentration of two substances, A and B in solution, provided that they do not react or interact, so they absorb radiation independently. The following data were obtained for A and B in three different solutions in a cuvette of pathlength 1.00 cm.
Using the data below calculate the molar absorptivity of A and B at both 400 and 500 nm.
What are the concentrations of A and B in solution 3.
[A] in mol/L
[B] in mol/L
Absorbance at 400 nm
Absorbance at 500 nm
Solution 1
0.00100
0
0.242
0.121
Solution 2
0
0.00200
0.115
0.690
Solution 3
Unknown
Unknown
0.398
0.301
[A] in mol/L
[B] in mol/L
Absorbance at 400 nm
Absorbance at 500 nm
Solution 1
0.00100
0
0.242
0.121
Solution 2
0
0.00200
0.115
0.690
Solution 3
Unknown
Unknown
0.398
0.301
Explanation / Answer
solution -1 is pure A, Since concentration of B in this solution is zero.
Since Absorbance(A)= e( molar absorptivity)*b(path length)*C( concentration) at 400nm
0.242= e*1* 0.001, e= 0.242/(0.1*0.001)=2420/cm.M, at 500nm, e= 0.121/(0.1*0.001)= 1210/cm.M
solution-2 is pure B. So concentration of A in this solution is zero.
hence at 400 nm, 0.115 =0.002*e*1, e= 0.115/0.002 =57.5/M.cm, at 500 nm, e= 0.301/0.002= 150.5/M.cm
in case of solution-3, let CA concentration of A and CB is concentration of B.
hence at 400nm, 2420*1*CA+57.5*1CB=0.398 (1) at 500nm, 1210*1CA+150.5CB=0.301 (2)
Eq.1* 150.5 gives 2420*150.5 CA+CB*57.5*150.5= 0.398*150.5 (1A)
Eq.2* 57.5 gives 1210*57.5*CA+150.5*57.5= 0.301*57.5 (2A)
Eq.1A- Eq.2A 294635CA= 42.5915, CA= 0.000145 moles/L and hence From Eq.2, 1210*0.000145+150.5CB=0.301, CB=0.000835 moles/L
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