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 4,129 M-1cm-1. Next the student measured out 0.29 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 absorbance as 0.343 Calculate the molarity of the diluted solution determined from the absorbency and the slope. Round your answer to 3 significant figures

Answer 1

As we know that

Given;

Molar absorptivity constant = 4129 M^-1cm^-1

Absorbance of the solution measured = 0.343
Assume that path lenght = 1 cm

So,

According to Beer-Lambert Law:
A = ε * C * l

Then,

C=A/ε*I
0.343 = 4129 M^-1cm^-1 * C * 1 cm
Concentration C = 8.307* 10^-5 M

From the experiment, initially, 0.29 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.
= 8.307*10^-5M

Then,

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 = 8.307 *10^-5M

M1V1 = M2V2
M1 = M2V2/V1

= (8.307 *10^-5M * 100 mL) / 5mL = 0.0016M

Then,

Again, this concentration is from 10 mL NaOH solution containing 0.29 g of aspirin.

So,

V1= 10 mL    

M1 =?
V2= 100mL

M2 = 0.0016 M

M1 = M2V2/V1

= (0.0016 M *100 mL) / 10mL = 0.016 M

Therefore,

0.29 g of asprin tablet in 10 mL solution, the concetration is = 0.016M

Hence,0.29 g of aspirin tablet in 10 mL solution gives you a concentration of 0.016 M

When the above solution is dissolved into 100

We get 0.0016 M solution.

Then,

Again, from the above solution we take 5mL and dissolved it into 100 mL

So,

The concentration of final dilution = 8.307* 10^-5M