Question

A student used standard solutions of aspirin in a FeCl3-KCl-HCl mixture to plot a graph of molarity versus absorbance for diluted aspirin solutions of known concentration. The student determined the slope of the graph to be 4,893 M-1cm-1. Next the student measured out 0.209 grams of a headache medicine tablet and dissolved it in 10.0 ml of NaOH and then added enough water to make a 100 ml solution. Five ml of this solution was then added to another 100 ml flask and diluted to the mark with enough FeCl3-KCl-HCl mixture. The solution was then measured for absorbency as 0.341 Calculate the molarity of the diluted solution determined from the absorbency and the slope. Then use the molarity of the original solution, its volume and molar mass to calculate the grams of pure acetylsalicylic acid. Put your answer in 3 significant figures.

Answer 1

The slope of the absorbance vs concentration plot is 4893 M^-1cm^-1 which is equal to the molar absorptivity of aspirin in FeCl₃-KCl-HCl mixture at the wavelength of measurement. Assume the path length of the solution to be 1 cm.

Use Beer’s law to find out the concentration when the absorbance is A = 0.341.

0.341 = (4893 M^-1cm^-1)*C*(1 cm)

====> C = (0.341)/(4893 M^-1) = 6.9691*10^-5 M.

The concentration of the diluted aspirin-FeCl₃-KCl-HCl solution is 6.9691*10^-5 M (ans).

Note that the student performed two dilutions; first he dissolved the tablet in 10 mL NaOH and diluted with water to 100 mL. Then he took 5 mL of the diluted solution and diluted further to 100 mL.

Work backwards; find out the dilution factor for the second dilution. The dilution factor (DF) = (100 mL)/(5 mL) = 20.

Concentration of aspirin in the first dilute solution = (6.9691*10^-5 M)*(DF) = (6.9691*10^-5 M)*(20) = 1.39382*10^-3 M.

The first dilution factor is (100 mL)/(10 mL) = 10; the concentration of aspirin in the original 10 mL sample prepared = (1.39382*10^-3 M)*(DF) = (1.39382*10^-3 M)*(10) = 0.0139382 M.

Find out the moles of aspirin dissolved in 10 mL NaOH as mole of aspirin = (10 mL)*(1 L/1000 mL)*(0.0139382 M)*(1 mol/L/1 M) = 1.39382*10^-4 mole.

Molar mass of aspirin, C₉H₈O₄ = (9*12.01 + 8*1.008 + 4*15.9994) g/mol = 180.1516 g/mol.

Mass of pure aspirin in the tablet = (1.39382*10^-4 mole)*(180.1516 g/mol) = 0.0251 g ≈ 0.025 g (ans).

Percent purity of aspirin = (0.025 g)/(0.209 g)*100 = 11.96% (ans).