Aspirin tablet, mass of tablet= 0.4149 g concentration of iron III salicylate complex = 3.997E-4, What is the concentration of ASA in the original solution? What is the mass of ASA in the original solution?

The procedure was as follows: accurately weigh by difference a tablet and transfer it into a 125ml erlnemyer flask, add 20 ml of 0.5 M NaOH and gently heat the mixture to boiling, Transfer the solution to a 100 ml volumetric flask and dilute the solution with water to the 100ml mark, pipet 2ml of the solution into a 100 ml volumetric flask and dilute to 100ml mark with 0.02 M iron III chloride Solution.

To find the concentration of ASA (acetylsalicylic acid) in the original solution, we need to calculate the moles of iron III salicylate complex formed in the reaction and use the stoichiometric relationship between ASA and iron III salicylate complex.

Step 1: Calculate the moles of iron III salicylate complex:
Mass of tablet = 0.4149 g
Concentration of iron III salicylate complex (FeSali) = 3.997E-4 M
Volume of FeSali solution used = 2 ml = 0.002 L

Moles of FeSali = Concentration × Volume
Moles of FeSali = 3.997E-4 M × 0.002 L

Step 2: Calculate the moles of ASA:
Using a balanced equation for the reaction between ASA and iron III salicylate complex, we know that 1 mole of ASA reacts with 1 mole of FeSali.

Moles of ASA = Moles of FeSali

Step 3: Calculate the concentration of ASA in the original solution:
Volume of original solution = 100 ml = 0.1 L

Concentration of ASA = Moles of ASA / Volume of original solution

Step 4: Calculate the mass of ASA in the original solution:
Mass of ASA = Moles of ASA × Molar mass of ASA

You will need to provide the molar mass of ASA to calculate the final answer.

To find the concentration of ASA (acetylsalicylic acid) in the original solution, we need to understand the reaction that occurs between the ASA in the tablet and the iron III chloride solution.

In this reaction, ASA forms a complex with iron III chloride, known as iron III salicylate complex. The concentration of this complex is given as 3.997E-4. Now, we can use this information to determine the concentration of ASA.

First, we need to calculate the concentration of the iron III chloride solution. We know that 2 mL of the original solution was diluted to a final volume of 100 mL with the 0.02 M iron III chloride solution.

Using the dilution formula:

C1V1 = C2V2

Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

We have:
C1 = 0.02 M
V1 = 2 mL = 0.002 L
C2 = ?
V2 = 100 mL = 0.1 L

Plugging these values into the formula, we get:
(0.02 M)(0.002 L) = C2(0.1 L)

Rearranging the equation, we can find C2:
C2 = (0.02 M)(0.002 L) / (0.1 L) = 0.0004 M

Now that we have the concentration of the iron III chloride solution, we can calculate the concentration of ASA in the original solution using the given concentration of the iron III salicylate complex.

The iron III salicylate complex is formed through a 1:1 reaction with ASA. This means that the concentration of ASA is equal to the concentration of the iron III salicylate complex.

Therefore, the concentration of ASA in the original solution is 3.997E-4 M.

To find the mass of ASA in the original solution, we need to consider the dilution process. We initially had a tablet with a mass of 0.4149 g, and we transferred it into a 125 mL flask.

Then, 2 mL of the solution was diluted to 100 mL with the iron III chloride solution. This means that the final volume of the solution containing ASA is 100 mL.

Now, we can calculate the mass of ASA using the concentration and volume.

Mass = Concentration x Volume

Mass = (3.997E-4 M)(0.1 L)

Mass = 3.997E-5 g

Therefore, the mass of ASA in the original solution is 3.997E-5 g.