A series of solutions is prepared in which the amount of iron(II) is held consta
ID: 1072953 • Letter: A
Question
A series of solutions is prepared in which the amount of iron(II) is held constant at 2,00 ML of 7.12 times 10^-4, while the volume of 7.12 times 10^-4 M 1, 10-phenantholine is varied. After dilution to 25 mL absorbance data these solutions in 1 .00-cm cuvettes at 510 nm are shown in the following table, Evaluate the composition of the complex, Estimate the value of the formation constant of the complex. 1, 10-Phenanthline (mL):) Absorbance: 2.00 0.240 3 00 4.00 5.00 6.00 8.00 10.00 12.00 0.360 0.480 0.593 0.700 0.720 0.720 0.7201Explanation / Answer
Calculation
Fe(II)i = 7.14 x 10^-4 M x 2 ml/25 ml = 5.712 x 10^-5 M
From absorbance,
[Fe-phenanthroline]eq = absorbance/molar absorptivity
molar absorptivity = 11,100 M-1.cm-1
[Fe(II)eq = [Fe(II)i - [Fe-phenanthroline]eq
So,
phenanthroline (ml) [Fe-Phenanthroline]eq (M) [Fe(II)]eq (M)
2 0240/11,100 = 2.16 x 10^-5 5.712 x 10^-5 - 2.16 x 10^-5 = 3.552 x 10^-5
3 3.24 x 10^-5 2.472 x 10^-5
4 4.32 x 10^-5 1.392 x 10^-5
5 5.34 x 10^-5 3.72 x 10^-6
Similarly rest can be done
[Phenanthroline]i [Phenanthroline]eq
7.12 x 10^-4 x 2/25 = 5.712 x 10^-5 M 5.712 x 10^-5 - 2.16 x 10^-5 = 3.552 x 10^-5
8.544 x 10^-5 5.30 x 10^-5
1.14 x 10^-4 7.08 x 10^-5
1.42 x 10^-4 8.86 x 10^-5
Similarly the others can be calculated
So,
Kf = [Fe-pheanthroline]eq/[Fe(II)]eq.[Phenanthroline]eq
From first row data,
Kf = 2.16 x 10^-5/(3.552 x 10^-5)(3.552 x 10^-5) = 17120
similarly other values from above can be fed into this equation to get other Kf values for the complex formation.
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