The active ingredient in aspirin is acetylsalicylic acid (mw = 180.2 g/mol). an aspirin tablet was dissolved in
water and titrated with 0.100 m naoh.
part a
if the titration required 14.85 ml of naoh to reach the phenolphthalein endpoint, how many milligrams of
acetylsalicylic acid were in the tablet?

1214 mg

Explanation:

Data given

Volume of NaOH solution (V) = 14.85 mL

Molarity of NaOH solution (M) = 0.100 M NaOH.

Amount of acetylsalicylic acid = ?

Molecular weight = 180.2 g/mol

Solution:

It is a acid base titration

As equal amount of NaOH will neutralize equal amount of acetylsalicylic acid.

First find the moles of NaOH

As we know

                 M = moles of solute / 1 L x volume of solution

Rearrange and modify the above equation

                 moles of NaOH = M of NaOH x 1 L / volume of NaOH solution

Put values in above equation

                moles of NaOH = 0.100 M x 1 L / 14.85 mL

                moles of NaOH = 0.0067 moles

Now find to find the weight of acetylsalicylic acid

As

equal amount of NaOH will neutralize equal amount of acetylsalicylic acid.

                 moles of acetylsalicylic acid =  moles of NaOH (1)

As we know  

                  no. of moles = mass in g / molar mass

So,

The modified form of equation 1 will be

              mass in g / molar mass (acetylsalicylic acid) =  moles of NaOH

Put values in above equation

                  mass in g / 180.2 g/mol = 0.0067 moles

Rearrange the above equation

                mass of acetylsalicylic acid = 0.0067 moles x 180.2 g/mol

                 mass of acetylsalicylic acid = 1.214 g

Convert the grams to milligrams

1 g = 1000 milligram

So,  

1.214 g = 1214 mg

3.55 × 10n M n = -10