The standard solution of FeSCN2+ (prepared by combining 9.00 mL of 0.200 M Fe(NO3)3 w/1.00 mL of 0.0020 M KSCN) has an absorbance of 0.510. If a trial's absorbance is measured to be 0.250 and its initial concentration of SCN− was 0.00050 M, the equilibrium concentration of SCN− will be...
ive tried for a while, but i keep on not getting the correct answer.
you don't need to just give it to me, but could someone show me the steps and reasoning needed to get the proper answer?
Partially answered below.
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To find the equilibrium concentration of SCN−, we can use the formula for Beer-Lambert Law, which relates the concentration of a solution to its absorbance:
A = εbc
Where:
A = Absorbance
ε = Molar absorptivity
b = Path length (in this case, it is not given, so we assume it is constant)
c = Concentration
We know the absorbance of the standard solution (Astd = 0.510) and the absorbance of the trial (Atrial = 0.250).
For the standard solution, we have:
Astd = εstd * b * cstd ---- (1)
For the trial solution, we have:
Atrial = εtrial * b * ctrial ---- (2)
Dividing equation (2) by equation (1), we get:
Atrial / Astd = (εtrial * b * ctrial) / (εstd * b * cstd)
0.250 / 0.510 = (εtrial * ctrial) / (εstd * cstd)
Since b is the same for both solutions, it cancels out. Rearranging the equation, we have:
(ctrial / cstd) = (εtrial / εstd) * (Atrial / Astd)
Now we can substitute the given values:
ctrial = (εtrial / εstd) * (Atrial / Astd) * cstd
ctrial = (0.0020 M / εstd) * (0.250 / 0.510) * 0.00050 M
Note: The molar absorptivity (εstd) is specific to the FeSCN2+ complex and needs to be looked up. Assuming it to be 1.00 M^-1cm^-1,
ctrial = (0.0020 M / 1.00 M^-1cm^-1) * (0.250 / 0.510) * 0.00050 M
Evaluating the expression:
ctrial = 0.000490196 M ≈ 4.90 x 10^-4 M
Therefore, the equilibrium concentration of SCN− in the trial solution is approximately 4.90 x 10^-4 M.