A 0.300 g sample of an aspirin tablet is used to prepare 100.0 mL of a stock sol

Question

A 0.300 g sample of an aspirin tablet is used to prepare 100.0 mL of a stock solution by dissolving the tablet in 1 M NaOH then diluting with distilled water to the 100 mL mark of a volumetric flask. A 0.30 mL sample of the stock solution is then transferred to a volumetric flask and diluted with 0.02 M iron(III) chloride to the 10.00 mL mark. The tetraaquasalicylatoiron(III) complex concentration in this 10.00 mL sample is determined to be 4.3x10-4M. What is the percentage of acetylsalicylic acid in the original aspirin tablet? Do not include the % sign in your answer.

Explanation / Answer

Mass of aspirin tablet (g) = 0.43 Volume of hydrochloric acid required in back titration (ml) = 2.0 Concentration of NaOH = 0.179 M Concentration of HCl = 0.1000 M 1) Moles of sodium hydroxide = 0.0100 L x 0.179 mol•L-1 = 8.95 x 10-3 mol NaOH 2) Moles of hydrochloric acid = 0.002 L x 0.100 mol•L-1 = 2.00 x 10-4 mol HCl 3) Moles of sodium hydroxide= 0.002 L x 0.100 mol•L-1 = 2.00 x 10-4 mol NaOH 4) Moles of acetylsalicylic acid= 2.00 x 10-4 mol / 2 = 1.00 x 10-4 mol 5) Mass of acetylsalicylic acid= 1.00 x 10-4 mol x 180.17 g/mol = 0.018 g 6) % of acetylsalicylic acid= 0.018 g / 0.43 g = 4.19 %

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