Question

b) In this experiment an aspirin tablet mark with deionized water. Using a pipette into a 100-mL volumetric flask and diluted to t make a 'diluted aspirin solution'. an aspirin tablet was placed in a 250-ml volumetric flask o dissolved in 10 mL of 1.0M NaOH solution, the flask was then 1.0M NaOH solution, the flask was then filled to the nized water. Using a pipette, 5.0 mL of this solution was added mL volumetric flask and diluted to the mark with FeCl3 solution to A = absorbance C = concentration y = 1442.7x + 0.0149 0.0002 0.0008 0.001 0.0004 0.0006 C(mol/L) Using the standard curve above determine the concentration of acetylsalicylic acid in the 'diluted aspirin solution'[Fe3+ SAJ if the absorbance reading was 0.667. c) Calculate the concentration of the initial solution prepared by the dissolving the aspirin tablet in the 250 mL volumetric flask. If the mass of ASA is 180 g/mole, how many milligrams were in the tablet?

Answer 1

Answer to the first part labelled as (b)

The line equation for the straight line standard graph between concentration (mol/L) of Acetyl Salicylic Acid (ASA) and absorbance is given. It is:

y = 1442.7 x + 0.0149 (Here x refers to the concentration while y refers to the absorbance)

Now, the concentration of ASA when the absorbance reading was 0.667 is required.

Putting y= 0.667 in the given line equation and x can be found

So, 0.667 = 1442.7 x + 0.0149

=> x = (0.667 -0.0149)/1442.7 = 4.520 $\times$ 10^-4

$\therefore$ concentration of ASA in diluted aspirin solution = 4.520 $\times$ 10^-4 mol/L or 0.0004520 mol/L (Answer)

Answer to the first part labelled as (c)

From the concentration of ASA calculated above, the amount of ASA in moles can be calculated in 100 mL of the solution which in turn is equal to the amount of ASA in 5mL of solution that was pipetted out from 250 mL volumetric flask.

$\therefore$ Amount of ASA in 5 mL of ASA solution that was pipetted out from 250 mL flask =( 4.520 $\times$ 10^-4 mol/1000mL) $\times$ 100 mL

[Note that the units of volume must be changed at one place above so that they cancel out]

=> Amount of ASA in 5mL solution = 4.520 $\times$ 10^-5 mol

The above 5 mL of the solution came from 250 mL flask and all the amount of ASA in it came from the aspirin tablet that was dissolved in 10 mL of NaOH solution

$\therefore$ Amount of ASA in 10 mL solution = [(4.520 $\times$ 10^-5 mol)/5mL]$\times$ 250mL = 2.260 $\times$ 10^-3 mol

The above amount of ASA was originally present in the aspirin tablet.

Now, converting the above amount in moles to grams and then milligrams with the help of molar mass of ASA (given to be 180g/mol):

Amount of ASA present in mg in an aspirin tablet = 2.260 $\times$ 10^-3 mol $\times$ 180 g/mol $\times$ 1000mg/1g = 406.8mg (Answer)