Question

what is number 3 and 4? thank you.

TT 3. A solution of [FeSCN2] is found to have 23% transmittance at 447 nm. If the molar absorption coefficient (e) is 4400 ( M cm) at y 447 nm, what is the concentration of [FeSCN] in the solution? Assume a 1 cm path length. (Show all work.) 4. A student mixes 5.00 mL of 0.0020 M Fe(NO.), with 3.00 mL of 0.0020 M KSCN Solution and 2.00 mL water at a certain temperature. He places a sample of the resulting Solution in a spectrometer, and measures the transmittance as 23% at 447 nm. a. What are the initial concentrations of [Fe] and [SCN)? b. What are the final concentrations of [Fe], [SCN), and [FeSCN*]? c. What is the measured value of K? 4 SPECTROSCOPIC DETERMINATION OF AN EQUILIBRIUM CONSTANT: PRE-LAB 80

Answer 1

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* transmittance of [Fes (N 2+] = 23% Molar absorption coefficient k = 4400 Mtemel concentration of [FE SCN 2+ ] in solution c=2 Fraction of light transmitted through solution at 447 nm. is called transmittance. Now according to Beer-hambart low transmittance C = 10 kex K = molan absorption coefficient = 4400 transmittance = 23% = 23 = 0.23 path length x = 1cm. = 0.23 = 10-4400 x CX Icm. or log 0.23 = log 10-4400 X CX lan) -0.638 = – 4400X0 x 1cm & log to concentration c = 0.638 4400 x 1cm - C = 1.458 10-4 M. >1