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 4,838 M^-1cm^-1. Next the student measured out 0.22 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 0.34

Calculate the original molarity (Mo) of the aspirin solution using the dilution equation

Answer 1

Molar absorptivity constant = 4838 M-1cm-1

Absorbance of the solution measured = 0.34
Assume that path lenght = 1 cm

According to Beer-Lambert Law:
A = ε * C * l
0.34 = 4838 M-1cm-1 x C x 1 cm
Concenreation C = 7.028 x 10^-5 M

From the experiment, initially, 0.22 g of an aspirin tablet is dissolved in 10mL NaOH and diluted to 100 mL.
and then 5 mL of the solution is taken and diluted again with 100 mL.

Therefore, the final aspirin solution concentration = the concentration determined from molar absorptivity constant.
                           = 7.028 x 10^-5M

From this, we go backward to find the concentration of aspirin in the sample from the dilution.

Initial V1 = 5 mL    M1 = ?
final V2 = 100 mL    M2 = 7.028 x 10^-5M

M1V1 = M2V2
M1 = M2V2/V1 = (7.028 x 10^-5M x 100 mL) / 5mL = 0.0014 M

Again, this concentration is from 10 mL NaOH solution containing 0.22 g of aspirin.

So, V1 = 10 mL     M1 =?
V2= 100mL        M2 = 0.0014 M

M1 = M2V2/V1 = (0.0014 M x 100 mL) / 10mL = 0.014 M

Therefore, 0.22 g of asprin tablet in 10 mL solution, the concetration is = 0.014 M

Hence, 0.22 g of aspirin tablet in 10 mL solution gives you a concentration of 0.014 M

When the above solution is dissolved into 100....we get 0.0014 M solution.

Again, from the above solution we take 5mL and dissolved it into 100 mL ....so, the concentration of final dilution = 7.028 x 10^-5M