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1. 2.0 mL of a stock solution of 0.010M Iron(III) Nitrate was added to a 100...

1. 2.0 mL of a stock solution of 0.010M Iron(III) Nitrate was added to a 100 mL volumetric flask, and DI water was added to the fill line. A solution was prepared by adding 10.0 mL of that diluted solution to 15.0 mL of 1M KSCN. If the molar absorptivity of Fe(SCN)2+ at 430 nm is 6000 M-1cm-1, what should the absorbance of the resulting solution be?

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Answer #1

Given;

2mL solution of 0.01 M ferric nitrate solution diluted to 100mL with water then the resulting molarity (M) of diluted solution =( initial molarity x initial volume)/final volume = 0.01 x 2/ 100 = 0.0002 M.

10 mL of 0.0002 M ferric nitrate reacted with 15mL solution of KSCN (1M)

No. of moles of Ferric nitrate in 10 mL of reactant = 0.0002 x10/ 1000 = 2 x 10-6 moles so it produces 2 x 10-6 moles of ferrous thiocyanate in final volume of 25 mL (10ml + 15 mL).

Concentration of ferrous isothiocyanate = (2 x 10-6 x 1000)/ 25 = 8 x 10-5 M

By beer lambert law;

Absorbance (A) = molar absorptivity (ε) x concentration (c) x path length (l)

A = 6000 M-1 cm-1 x 8 x 10-5 x l (cm)

Let path length be 1cm (since data not provided)

Absorbance = 0.48 (Answer)

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