Assay of Aspirin

Question 1

Place about ___ of Aspirin, accurately weighed, in a flask, add 50.0 mL of 0.5 N sodium hydroxide VS. And boil the mixture gently for 10 minutes. Add phenolphthalein TS, and titrate the excess sodium hydroxide with 0.5 N sulfuric acid VS and perform a blank determination under titrimetry. Each mL of 0.5 N sodium hydroxide is equivalent to 45.04 mg of C9H8O4

Answer 1

1.5 g

Question 2

Place about 1.5 g of Aspirin, accurately weighed, in a flask, add ______. And boil the mixture gently for 10 minutes. Add phenolphthalein TS, and titrate the excess sodium hydroxide with 0.5 N sulfuric acid VS and perform a blank determination under titrimetry. Each mL of 0.5 N sodium hydroxide is equivalent to 45.04 mg of C9H8O4

Answer 2

50.0 mL of 0.5 N sodium hydroxide VS

Question 3

Place about 1.5 g of Aspirin, accurately weighed, in a flask, add 50.0 mL of 0.5 N sodium hydroxide VS. And boil the mixture gently for 10 minutes. Add _______ and titrate the excess sodium hydroxide with 0.5 N sulfuric acid VS and perform a blank determination under titrimetry. Each mL of 0.5 N sodium hydroxide is equivalent to 45.04 mg of C9H8O4

Answer 3

phenolphthalein TS,

Question 4

Place about 1.5 g of Aspirin, accurately weighed, in a flask, add 50.0 mL of 0.5 N sodium hydroxide VS. And boil the mixture gently for 10 minutes. Add phenolphthalein TS, and titrate the excess sodium hydroxide with_________ and perform a blank determination under titrimetry. Each mL of 0.5 N sodium hydroxide is equivalent to 45.04 mg of C9H8O4

Answer 4

0.5 N sulfuric acid VS

Question 5

Place about 1.5 g of Aspirin, accurately weighed, in a flask, add 50.0 mL of 0.5 N sodium hydroxide VS. And boil the mixture gently for 10 minutes. Add phenolphthalein TS, and titrate the excess sodium hydroxide with 0.5 N sulfuric acid VS and perform a blank determination under titrimetry. Each mL of 0.5 N sodium hydroxide is equivalent to _______ of C9H8O4

Answer 5

45.04 mg

Question 6

Official requirement for Aspirin Use

Answer 6

Aspirin contains not less than 99.,5 percent and not more than 100.5 percent of C9H8O4. Calculated on the dried basis (MW 180.16)