Part Il Prepare and Test Equilibrium Systems the solutions according to the char

Question

Part Il Prepare and Test Equilibrium Systems the solutions according to the chart below, tie 10.0 mL graduated cylinders to measure the solutions. Mix each soluti Note: You are using 0,0020 M Fe(NOs)s in this test. 9. Label three new small beakers A-C. on thoroughly H20 (mL) 4.00 3.00 2.00 0.0020 M Fe(NO)s 0.0020 M SCN- (mL) 3.00 4.00 5.00 Beaker (mL) 3.00 3.00 3.00 10. Collect absorbance-concentration data for the three beakers of equilibrium mixtures a. Using the solution in Beaker A, rinse the cuvette twice with ~1 mL amounts and then fill it 3/4 full. Wipe the outside with a tissue and place the cuvette in theSspectrometer. b. Write down, in your data table, the absorbance of the sample in Beaker A. c. Trace along the linear regression equation to find the FeSCN2+ concentration for the sample in Beaker A. Wnite down the concentration in your data table. Discard the cuvette contents as directed. Rinse and fill the cuvette with the solution in Beaker B and place it in the device. After the reading stabilizes, write down the absorbance in your data table and use the Interpolate function to determine the concentration of the sample. Repeat Step d for the mixtures in Beaker C. d. e.

Explanation / Answer

Using Beer's law ,

Absorbance=e*l*C

where e=absorptivity of species(FeSCN2+)

l=path length of light(1 cm)

A callibration curve between Abs and concentration is linear of the form y=mx+c,

y=Abs ,x=concentration ,m=slope=el

Linear regression Equation : y=3345x+0.045136

Here el=3345 M^-1 cm^-1,,for l=1cm,e=3345 M^-1

y=Absorbance, c=0.045136 =abs of blank solution

Abs of the species=Abs -abs of blank solution =Abs-

data analysis

part ii)

Beaker A)

[Fe3+]initial=0.002M*(3ml/10ml)=0.0006M

[SCN-]initial=0.002M*(3ml/10ml)=0.0006M

Beaker B)

[Fe3+]initial=0.002M*(3ml/10ml)=0.0006M

[SCN-]initial=0.002M*(4ml/10ml)=0.0006M

Beaker C)

[Fe3+]initial=0.002M*(3ml/10ml)=0.0006M

[SCN-]initial=0.002M*(5ml/10ml)=0.001M

[Fe3+]eq=[Fe3+]initial-[FeSCN2+]eq

[SCN-]eqm=[SCN-]initial-[FeSCN2+]eq

0.0006-6.308*10^-5

=5.369*10^-4M

0.0006-7.593*10^-5

=5.241*10^-4M

0.0008-6.308*10^-5

=7.369*10^-4M

=0.0010-7.593*10^-5

=9.241*10^-4M

Keq=[FeSCN2+]eq/[Fe3+]eq[SCN-]eq

beaker 1)Keq=(6.517*10^-5M)/(5.348*10^-4M)(5.348*10^-4M)=227.858

beaker 2)Keq=(6.308*10^-5M)/(5.369*10^-4M)(7.369*10^-4M)=159.437

beaker 3)Keq=(7.593*10^-5M)/(5.241*10^-4M)(9.241*10^-4M)=156.776

Keq (average)=(227.858+159.437+156.776)/3=181.357

Absorbance Actual Abs=abs-0.045136 [FeSCN2+]eqm=actual abs/e beaker A 0.263 0.218 0.218/3345 M^-1 =6.517*10^-5M beaker B 0.256 0.211 0.211/3345 M^-1 =6.308*10^-5M beaker C 0.299 0.254 0.254/3345 M^-1 =7.593*10^-5M

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