 Chemistry Help Forum Back Titration (or Indirect titration) to determine concentration of acetylsalicylic Feb 28th 2019, 11:11 PM #1 MW2121 Junior Member   Join Date: Feb 2019 Location: Adelaide Posts: 1 Back Titration (or Indirect titration) to determine concentration of acetylsalicylic MW21 New Members 0 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 )﻿  Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode 