A 39.44-mL aliquot of 0.0960 M NaOH was added to a 0.3480-g sample of crushed as
ID: 960620 • Letter: A
Question
A 39.44-mL aliquot of 0.0960 M NaOH was added to a 0.3480-g sample of crushed aspirin tablet dissolved in ethanol. The acetylsalicylic acid in the aspirin tablet was neutralized by the NaOH, leaving excess OH in solution. This excess was back-titrated with 17.70 mL of 0.0860 M HCl. The molecular weight of ASA is 180.157 g/mol. How many total moles of NaOH were added to the sample? How many moles of NaOH were presented in excess? How many grams of ASA were presented in the 0.3480-g sample? The average mass of a tablet is 0.3429 g. Based on this information and the results of the analysis of the 0.3480-g sample, determine how many mg of ASA are presented in a single tablet.Explanation / Answer
ANSWER
(a) Moles of NaOH added to the sample = Molarity X Volume in liters
Moles of NaOH added to the sample = 0.0960 X 39.44 = 3.78624 mmole,
(b) Similarly moles of NaOH in excess = moles of HCl used for neutraization
moles of NaOH in excess = Molarity X 17.7 = Q mmole
NOTE: The quality of image is poor, the molarity of HCl is not clear. So please insert the correct value
(c) Moles of ASA present = moles of NaOH consumed.
Moles of NaOH consumed = Total moles - excess moles =3.78624 mmole,moles - Qmoles =R mmole
Hence moles of ASA present in 0.3480g of sample =R mmole
mass of ASA 0.3480g of the sample = No. of moles X Molar mass = R X 180.157 = S g
(d) Sg = 10-3 X S = T mg