Question

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 0.550. If a trial's absorbance is measured to be 0.350 and its initial concentration of SCN– was 0.0010 M, the equilibrium concentration of SCN– will be _____?

Can you please explain to me how to work this question out? Thanks

Answer 1

The concentration of SCN- in the first mixture is:
(1/10)(0.0020M) = 0.0020M)
The concentration of Fe+3 was:
(9/10)(0.200M) = 0.18 M
Since the concentration of F+3 was 90 times the concentration of SCN-, we may assume the equilibrium, Fe+3(aq) + SCN-(aq) <=> FeSCN+2(aq), if far enough to the right to assume completion: Most of the SCN- became FeSCN+2.
[SCN-](initial) = [FeSCN+2](final) = 0.0020M
FeSCN+2(aq) is resopnsible for the absorbance.
The absorbance, A, is proportional to the concentration.
A = (const.)C, or
C = A/(const.)
The proprtionality constant is:
0.550/0.0020M= 2750 M^-1
and our formula becomes,
A = 2750*C , where C = [FeSCN+2]
Use A = 0.350 to calculate C, and finally,
[SCN-](eq) = 0.0010 - C

Answer 2

A comprehensive look at this equilibrium system and the experiment for determining the equilibrium constant is given here:
http://www.scienceteacherprogram.org/chemistry/Tehilla98.html

Answer 3

Hi GK,
For the first step, isn't (1/10)(0.0020) = 0.00020 M?
Also, I understand your very clear explanation to the last step: [SCN-] = 0.0010 -C
I understand the minus C part, but how did you get 0.0010?
Thank you!