Question

In the back titration of aspirin, an excess amount of NaOH was reacted with aspirin and the excess was then back titrated with HCl. Suppose a 0.4567 g sample of aspirin was reacted with 18.12 mL of 0.1345 M NaOH and back titrated with 12.15 mL of 0.1124 M HCl to reach the endpoint. Calculate the weight of aspirin (Asp) in the sample.

Answer 1

Millimole of NaOH taken = 18.12ml×0.1345mol/L = 2.43714mmol

Millimole of HCl used = 12.15ml × 0.1124mol/L = 1.36566mmol

Millimole of NaOH used with aspirin = 2.43714 - 1.36566 = 1.07148mmol

Number of millimole of NaOH reacted with aspirin is equal to the millimole of aspirin present in sample . NaOH would neutralizes the -COOH group in aspirin .

Millimole of aspirin in sample = 1.07148mmol

Molar mass of aspirin = 180.158g/mol

Mass of aspirin in sample = 1.07148×10^-3mol × 180.158g/mol = 0.1930357 g

Weight of aspirin in sample = 0.1930 g. (Answer)