Let’s see if you can replicate the calculations that you did on your data sheet.
ID: 713072 • Letter: L
Question
Let’s see if you can replicate the calculations that you did on your data sheet. Given in the table below is another student’s data for the same lab experiment for their test tube #3.
Concentration of Fe(NO3)3 in 0.10 M HNO3 solution: 0.002 M
Concentration of NaSCN in 0.1o M HNO3 solution: 0.002 M
Volume of Fe(NO3)3 solution: 5.00 mL
Volume of NaSCN solution: 2.00 mL
Absorbance: 0.346
Equation of the trendline in their graph y = 2278.6x + 0.0121
1) Calculate the initial [Fe3+] in the test tube.
2) Calculate the initial [SCN-] in the test tube.
3) Calculate the equilibrium [Fe3+] in the test tube.
4) Calculate the equilibrium [SCN-] in the test tube.
Explanation / Answer
Solution:
Initial concentration calculations:
[SCN-]o=(Concentration of stock*volume taken)/total volume=0.002M*(2ml/7ml)=0.000571M=5.71*10^-4M
[Fe3+]o=0.002M*(5ml/7ml)=0.00143M=1.43*10^-3M
Equilibrium concentration of SCN=[SCN-]eq
Explanation: Fe3+ (aq) +SCN-(aq)<--->[FeSCN]2+ (aq)
[SCN-]o=initial concentration <<<[Fe3+]o (initial),so it can be assumed that SCN- is completely used up in the reaction and at equilibrium [SCN-]eq=[FeSCN2+]eq
Moreover ,[FeSCN2+]eq is the absorbing species,so its absorbance can be used to calculate its concentration using callibration curve for standard [FeSCN2+] solutions.
[Using Beer's law ,
Absorbance=e*l*C
where e=absorptivity of species
l=path length of light(1 cm)]
The equation for callibration curve ,is y=2278.6x+0.0121
y=Absorbance ,x=[FeSCN2+]
given Abs=0.346
x=(y-0.0121)/2278.6=(0.346-0.0121)/2278.6=1.465*10^-4M
[FeSCN2+]eq=[SCN-]eq=1.465*10^-4M
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