For the first table you are correct, moles of SCN:
<-----Vf final volume 25mL
<---- this is true assuming the following reaction:
Transmittance is calcualted using
Blank | 1 | 2 | 3 | 4 | 5 | |
Vol NaSCN | 0 | 1 | 2 | 3 | 4 | 5 |
moles SCN | 0 | 0.000001 | 0.000002 | 0.000003 | 0.000004 | 0.000005 |
[SCN] | 0 | 0.00004 | 0.00008 | 0.00012 | 0.00016 | 0.0002 |
FeNCS | 0 | 0.00004 | 0.00008 | 0.00012 | 0.00016 | 0.0002 |
%T | 100.000 | 77.625 | 57.016 | 42.364 | 33.963 | 29.040 |
Abs | 0 | 0.11 | 0.244 | 0.373 | 0.469 | 0.537 |
Using this calibribation date you have to make a plot of Abs Vs [FeNCS]
And make a linear regression (straight line in the graph) to calculate the equation to calculate the concentration of FeNCS, thus the equation is:
For the second table:
The moles for Fe3+ are the same for all the solutions (you are adding the same volume)
The moles of SCN :
<----the volume added depending on the solution (1mL, 2mL, 3mL)
Trasnmittance the same as above.
6 | 7 | 8 | 9 | 10 | |
Vol Fe(NO3)3 | 5 | 5 | 5 | 5 | 5 |
moles Fe3+ (initial) | 0.00001 | 0.00001 | 0.00001 | 0.00001 | 0.00001 |
Vol NaSCN | 1 | 2 | 3 | 4 | 5 |
moles SCN | 0.000002 | 0.000004 | 0.000006 | 0.000008 | 0.00001 |
%T | 73.282 | 52.360 | 37.670 | 23.988 | 15.812 |
Abs | 0.135 | 0.281 | 0.424 | 0.62 | 0.801 |
And using the equation from the first part you can calculate the equilibrium concentration of FeNCS each solution:
For example in 6:
And the moles of FeNCS, you need to multiply the concentration by the volume of the solution (10mL)
6 | 7 | 8 | 9 | 10 | |
[FeNCS] eq | 0.00006 | 0.00012 | 0.00017 | 0.00025 | 0.00032 |
mols FeNCS | 6.0E-07 | 1.2E-06 | 1.7E-06 | 2.5E-06 | 3.2E-06 |
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