Question

An unknown Fe(II) sample is added to 2.50 mL of a 0.0300 M potassium dichromate solution (in acidic medium) in a 25 mL volumetric flask. 2 M H2SO4 is added to fill the flask to the calibration mark. After the reaction proceeds to completion, the absorbance of the solution at 447 nm is measured to be 0.550 using a 1.00-cm path length. Calculate the following and enter your answers with correct significant figures. Extinction coefficient of Cr2O72-in 2M sulfuric acid at 447 nm is 350. M-1cm-1

Molarity of Cr2O7 2- in the solution after reaction.

Moles of Cr2O7 2− that reacted.

Moles of Fe(II) that reacted.

Answer 1

• Fe2+ reacts with Cr2O72- in acidic medium, according to the following equation,

So, 1 mol of Cr2O72- reacts with 6 moles of Fe2+.

• According to the Lambert-Beer law, we kno,

where, A = absorbance of the solution, = molar extinction coefficient of the absorbing species, b = path length and C = concentration of the absorbing species.

• So, putting the values, from the question we get,

So, absorbing species Cr2O72- which is remaining in the solution has the concentration=1.57 x 10-3 M.

(approximated to three significant figure, as that was lowest among the values involved in the calculation) [Part 1 answered]

• Total volume of the solution = 25 mL. Molarity of the dichromate after reaction = 1.57 x 10-3 M. So, mmoles of dichromate remaining = (1.57 x 10-3 x 25) mLxM = 0.03925 mLx(mol/L) = 0.03925 (mL/L)x mol = 0.03925 x 10-3 mol = 0.03925 mmol.

Before reaction there was 2.50 mL 0.0300 M dichromate solution = (2.50 x 0.0300) mmol = 0.0750 mmol.

So, mmoles of dichromate reacted = (0.0750 - 0.03925) mmol = 0.0358 mmol = 0.0358 x 10-3 mol = 3.58 x 10-5 mol. [Part 2 answered]

• Moles of Fe2+ reacted = (6 x 3.58 x 10-5) mol = 21.48 x 10-5 mol = 2.148 x 10-4 mol. [Part 3 answered]