Back Titration (or Indirect titration) to determine concentration of acetylsalicylic

Mar 2019
New Members
1 post
Report post
Posted 1 hour ago
For school, we performed a back titration involving one brand of aspirin and were told to determine the quantity of acetylsalicylic acid in a single tablet. sodium hydroxide was added in excess then the solution was heated. the excess unreacted NaOH was then reacted with HCl. my calculations are as follows:

V(NaOH) = 0.04085L

c(NaOH) = 0.111M hence n(NaOH) total = 0.00435 moles

v(HCl) = 0.01778L

c(HCl) = 0.0582M hence n(HCl) = 0.001035 moles

number of moles excess = number of moles titrated of HCl

hence 0.001035 moles didn't react with acetylsalicylic acid

the number of moles of NaOH that reacted with acetylsalicylic acid is 0.00435-0.001035 = 0.003495

therefore 0.003495 moles of NaOH reacted with the acetylsalicylic acid

the mole ration between NaOH and acetylsalicylic acid is 1:2

meaning the number of moles of acetylsalicylic acid is 1/2 x 0.003495

which is = 0.0017475 moles

Molar mass acetylsalicylic acid = 180.158g/mol

0.0017475 x 180.158 = 0.314g or 314mg

this does not make sense considering I only added 300mg of crushed aspirin powder in the first place, meaning that I would have a percentage purity of 105%

if anyone can see any obvious errors on my part or could suggest any reason for this ridiculous result, I would be extremely appreciative :))