Question

9. 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 percent of aspirin (Asp) in the sample. 8.457% 37.90% 21.14% 42.28% none of the above

Answer 1

0.1345 m 1000 18112 mL of Naou = - 18.12 X 0.1345 nol Nhow 2.437x10-3 moy NAOH Back tilnetien done by AICI. 12.15 80 1124 used for barn hifrahe = mo. of miles of the 7 11 366 x10-3 mol - HCl Amount of aspirin in the scriple = (2-4377(0-3 _ 1.366x103) s 1071810-3 mol 1071810-3 x 1801 189 like! = 0.193g % of Aspirin in the sample = 0, 190 474 X 10% → 42 28% Ans. o 42-28 Y.

Answer 2

The answer is c) 21.14%

because 1 mole of aspirin reacts with 2 moles of NaOH. For example, if the amount of consumed NaOH is 100 moles, the aspirin should be 50 moles.

Answer 3

• Amount of consumed NaOH = 2.437*10^-3  -̶ 1.366*10^-3  =  1.071*10^-3 mol

• From equation, the amount of consumed NaOH is double the amount of aspirin (2 moles NaOH = 1 mole aspirin)

• So the amount of aspirin = 1.071*10^-3 / 2 = 0.535*10^-3

• gm aspirin = mole*mwt = 0.535*10^-3 * 180.16 = 0.096 gm

• % aspirin = 0.096/0.4567 *100 = 21.12%