Write the equation that expresses K_eq for the reaction between Fe^3+ and Sal^-,

Question

Write the equation that expresses K_eq for the reaction between Fe^3+ and Sal^-, K_eq = _________ The concentration of NaSal is 4.4 times 10^-4 M. If 15.0 mL of this solution is added to 10.0 mL of an Fe(NO_3)_3 solution with a concentration of 4.0 times 10^-4 M. What is the concentration of your diluted NaSal solution? what is the concentration of your diluted Fe(NO_3)_3 solution? Given an absorbance of (FeSal)^2+ of 0.190 what is the concentration of (FeSal)^2+ Use the beer's law-equation froth the procedure. Using your answers to questions 1 -3 find the value of the equilibrium constant for the reaction of NaSal and Fe(NO_3)_3. Disclaimer: these are made up values to give you practice and will not necessarily match the values that you find in your lab.

Explanation / Answer

1) Fe3+ +Sal- <-->[FeSal]2+

keq=[FeSal]2+/[Fe3+][SaI-]

2) concentration of NaSal

M1V1=M2V2

M1=4.4*10^-4M

V1=15ml

after mixing total volume=10+15=25ml

V2=25ml

moles of NaSal=M1V1=4.4*10^-4M*0.015L=6.6*10^-6 moles

So final [NaSal]=M2=M1V1/V2=6.6*10^-6 moles/0.025L=2.64*10^-4M

initial moles of Fe(NO3)3=M3V3=4.0*10^-4M *10ml=4.0*10^-4M *0.010L=4.0*10^-6 moles

Similarly,final [Fe(NO3)3]=M3V3/V2=4.0*10^-6 moles/0.025L=1.6*10-4M

4)keq=[FeSal]2+/[Fe3+][SaI-]

initial concentrations

[Fe3+]o=1.6*10-4M

[Sal-]o=2.64*10^-4M

and [Fesal2+]o=0

So at equilibrium,

[Fe3+]eq=1.6*10-4M-x

[Sal-]eq=2.64*10^-4M-x

[Fesal2+]eq=9.74*10^-5M=X say

or,[Fe3+]eq=1.6*10-4M-9.74*10^-5M=0.626*10^-4M

[Sal-]eq=2.64*10^-4M-0.974*10^-4M=1.666*10^-4M

keq=9.74*10^-5M/(1.666*10^-4M)(0.626*10^-4M)=9.34*10^3

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