Question

A tablet of Pain Be Gone Aspirin, which had a mass of 1.213 g, was pulverized and 1.159 g were dissolved in 10.0 mL of ethyl alcohol and 25.0 mL of DI water. The titration of this solution with 0.1052 M NaOH required 15.62 mL to reach the phenolphthalein endpoint. Answer questions a – d below. a. Determine the moles of NaOH that reacted with the acetylsalicylic acid. b. Determine the mass, in grams, of acetylsalicylic acid in the sample analyzed. c. Determine the mass, in milligrams, of H????9H7????4 in the tablet. d. The manufacturer claims that each tablet contains 325 mg + 10 mg of acetylsalicylic acid. Is the actual amount of acetylsalicylic acid in the tablet acceptable?

Answer 1

a. Moles of
`NaOH` = 0.001643 moles

b. 0.296 g

c. 0.3098 g

d. Not acceptable

Step-by-step explanation:

a.

Considering:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

Or,

`Moles =Molarity * {Volume\ of\ the\ solution}`

Given :

For
`NaOH` :

Molarity = 0.1052 M

Volume = 15.62 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 15.62×10⁻³ L

Thus, moles of
`NaOH` :

`Moles=0.1052 * {15.62* 10^(-3)}\ moles`

Moles of
`NaOH` = 0.001643 moles

b.

The reaction of NaOH with the acetylsalicylic acid is in the ratio of 1:1.

Thus, Moles of NaOH = Moles of acetylsalicylic acid = 0.001643 moles

Molar mass of acetylsalicylic acid = 180.16 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

Mass = Moles * Molar mass = 0.001643 moles * 180.16 g/mol = 0.296 g

c.

1.159 g of sample contains 0.296 g of acetylsalicylic acid

1.213 g of sample contains
`(0.296)/(1.159)* 1.213` g of acetylsalicylic acid

Mass of acetylsalicylic acid = 0.3098 g = 309.8 mg

d. Sample contains = 309.8 mg

Manufacturer claiming = 315 mg to 335 mg

Thus , it is not acceptable.