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9. In the back titration of aspirin, an excess amount of NaOH was reacted with aspirin...


9. In the back titration of aspirin, an excess amount of NaOH was reacted with aspirin and the excess was then back titrated
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
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ReportAnswer #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 use

ReportAnswer #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.

source: Based on equation of aspirin with NaOH
answered by: Eslam Elkaeed, PhD
ReportAnswer #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% 

source: My knowledge
answered by: Eslam B. Elkaeed, PhD
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