From the information below, calculate: 1) [SCN-]initial 2) [FeSCN2+]eq 3) el (e=
ID: 1034405 • Letter: F
Question
From the information below, calculate: 1) [SCN-]initial 2) [FeSCN2+]eq 3) el (e=constant, l=the diameter of the test tube he sample is in)A (=absorbance) is 0.661 EXPERIMENT 8 histry DETERMINATION OF AN EQUILIBRIUM CONSTANT num when it is at equilibrium. We will study: constants can be determined by finding the concentrations of all the species in the system Fe (ac) + SCN (a)FeSCN for which the equilibrium constant expression is The FeSCN complex is a deep red color, while the other two species are colorless,_ We can find concentration of the EeSCN2 complex with a colorimcter. The complex absorbs light ata r wavelength, in this case 470 nanometers (nm). The red color of the complex absorb blue light well, so the blue LED setting on the colorimeter is used. The colorimeter measures the amount of blue light absorbed by the solutions. Absorbance is directly related to concentration by Beer's law: where ? is a constant, L is the diameter of the test tube the sample is in (the cell), and c is the concentration in molarity
Explanation / Answer
[1] For [SCN-]
From procedure, 1.0 ml of 2.00*10-3 M SCN- was used, and the final volume was 10 ml
Now, using M1V1 = M2V2 where M1, V1 are initial molarity and volume; M2, V2 are final molarity and volume
(2.00*10-3 M) *1.0 ml = M2 (10.0 ml)
or, M2 = (2.00*10-3 M) *1.0 ml / 10.0 ml =2.00*10-4 M
Thus, initial [SCN-] = 2.00*10-4 M (this solution used for el determination)
[3] Calulation of el
[SCN-] = 2.00*10-4 M ,
[Fe2+] = (5.0 ml * 0.200 M) /10 ml = 0.100 M
since, [SCN-] is limiting reactant,[FeSCN2+] = [SCN-] = 2.00*10-4 M
Given, Absorbance, A = 0.661
using Beer's law, A = elc , c is concentration of FeSCN2+
or, el = A/ [FeSCN2+] = (0.661)/2.00*10-4 M = 3305 M-1
Therefore, vaule of el = 3305 M-1
[2] [FeSCN2+]eq has to be obtained experimentally. For this calculation you should be using Beer's law A=ecl
or, [FeSCN2+]eq = c = A/el = (A / 3305) M
As, recorded absorbance data is not attached one can not find [FeSCN2+]eq
And, if you insist [FeSCN2+]eq without providing absorbance data, then it is equal to [SCN-], assuming all of SCN- is used, and no dissocaition of [FeSCN2+] ion occur.
you will need these to complete table 2
For [Fe3+]
Initial molarity, M1 = 2.00*10-3 M
Volume used (same in 5 test tubes), V1 = 5.0 ml
Final Volume, V2 = 10.0 ml
So, [Fe3+] = M2 = M1V1/V2 = (2.00*10-3 M) * 5.0 ml / 10.0 ml = 1.00*10-3 M
Now, [SCN-] can be calculated considering dilution factor (using M1V1 = M2V2)
M1 = 2.00*10-3 M (stock solution concentration of SCN-)
V2 =10 ml (Final volume)
V1 = 1.0, 2.0, 3.0, 4.0, 5.0 ml (five values, so five calculation or simply scale the value obtained for V1=1.0)
In test tube 1 = (2.00*10-3 M) *1.0 ml / 10.0 ml = 2.00*10-4 M
In test tube 2 = 2 (2.00*10-4 M) = 4.00*10-4 M ( as volume of stock SCN- used is twice of test tube 1)
In test tube 3 = 3 (2.00*10-4 M) = 6.00*10-4 M ( as volume of stock SCN- used is three times)
In test tube 4 = 4 (2.00*10-4 M) = 8.00*10-4 M ( as volume of stock SCN- used is four times)
In test tube 5 = 5 (2.00*10-4 M) = 10.00*10-4 M ( as volume of stock SCN- used is five times of test tube 1)
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