1) If the absorbance of solution 1 at equilibrium is 0.23 and the absorbance of

ID: 944462 • Letter: 1

Question

1) If the absorbance of solution 1 at equilibrium is 0.23 and the absorbance of the standard solution is 1.23, and the concentration of FeSCN2+ in the standard solution is 1.00 M, what is the equilibrium concentration of FeSCN2+ in solution 1? (Show your work, 3 sig figs, and correct units)

2) If the initial concentration of Fe3+ in solution 1 was 0.25 M, what is the equilibrium concentration of Fe3+ that remains? (Show your work, 3 sig figs, and correct units)

3) If the initial concentration of SCNin solution 1 was 0.68 M, what is the equilibrium concentration of SCNthat remains? (Show your work, 3 sig figs, and correct units)

4). Based on your answers to questions 2-4, show your work to calculate the value of Kc in solution 1. (Show your work, 3 sig figs, and correct units)

PLEASE HELP I have no idea how to even start working through these. I have been at this for hours. Dont know what I am missing

Explanation / Answer

RXn

FeSCN2+ Fe3+ +SCN-

Absorbance=A=eCl (beer-lambert’s law)

Or, A =constant* C where C=concentration of species in the solution

1)A1/A2=C1/C2

Or, A1/C1=A2/C2

Or ,0.23/C1=1.23/1.00M

Or, C1=0.23*1.00M/1.23=0.187M(*answer)

2)[Fe3+]o=0.25M

From part 1)

Kc=equilibrium constant=[Fe3+][SCN-]/[ FeSCN2+]

ICE table

[Fe3+]

[SCN-]

[FeSCN2+]

Initial

0.25M

unknown

1.00M

change

-X

equilibrium

0.25+X

[SCN-]+X

1-X=0.187

As 1-X=0.187

X=1-0.187=0.813M

So, equilibrium concentration of Fe3+, [Fe3+]eq=0.25+x=0.25+0.813=1.06M

3)[SCN-]=0.68M

equilibrium concentration of SCN-, [SCN-]eq=0.68+x=0.68+0.813=1.49M

4)Kc=equilibrium constant=[Fe3+][SCN-]/[ FeSCN2+]=1.06M*1.49M/0.187M=8.44

Kc=8.44