Vol. Fe Vol. Ligand Absorbance mL 20 18 17 16 Plot the measured absorbance data
ID: 1026462 • Letter: V
Question
Vol. Fe Vol. Ligand Absorbance mL 20 18 17 16 Plot the measured absorbance data (y-axis) versus volume (mL) of Fe" on x-axis. 0L1lines determine the volume of Fe (V ) at which the From the intersection of the two maximum absorbance would occur (it may not be one of your solution values, you will have to fit two linear lines ·1931 and calculate this value as described previously). 07 to calculate the mole ratio of ligand to iron. 14 12 10 Moles Fe-[Fe2+] 10 12 16 20 Moles ligands = [ Ligand] x [(20.00-Yu.) x 10-3] 225 n = molesofligand moles Fe Question: Predict the absorbance of a solution made using 12 mL of0.345 mM [Fe J and s mL of 0.456 mM [Ligand], then diluting to 100 mL. (Note: It is important to take into account the n value determined above, as well as the molar absorptivity constant.) Absorbance Notes: Turn in your datasheets and your graphs. 48Explanation / Answer
Answer:
Predict the absorbance of a solution made using 12 mL of 0.345 M [Fe2+] and 8mL of 0.456 M [Ligand] than diluting to 100 mL( Take into account that n which is moles of ligand/moles Fe2+ is 2.8)
Final conc of [Fe2+]=[0.345M*12mL/100mL]=0.0414M
Final conc of Ligand=0.456M*8mL/100mL=0.03648M
since n=moles of ligand/moles of Fe2+=2.8/1
Fe2+ + 2.8 Ligand ==>product (therefore ligand is a limiting reatant) hence moles or molarity of Fe2+ consumed will be
moles of Fe2+ consumed=0.03648M/2.8=0.013M=Molarity of product
Equations given: A= ELC
A= absorbance, E= epsilon (given as 12500), L= length of cell (given as 1.00cm), and C= concentration (in Molar)
Now you can proceed
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