Question

Determine the concentration of the HCl solution from the data for the standardization of HCl with Na2CO3? Calculate the % by mass of aspirin in the tablet? -Calculate moles and mass of acetylsalicylic acid in each tablet, molar mass= 180.16 g/mol -[NaOH]=.101 M -[HCl]=.1M -Trial 1 Mass of Aspirin= .365g -Trial 2 Mass of Aspririn=.375g Trial 1&2 Volumes of NaOH= 50.00 mL Standard HCl: -Mass of Na2CO3=.085g -Volume of HCl= (initial) 26.0 ml (final) 11.3 mL -1st titration of tablet solution: volume of HCl initial= 37.0, final=15.11 mL -2nd titration of tablet solution: volume of HCl= 23.5 mL (initial), 16.02 is final

Answer 1

Titration

standarisation of NaOH

moles of Na2CO3 = 0.085/105.99 = 0.0008 mol

moles of HCl required = 2 x 0.0008 = 0.0016 mol

Volume of HCl used = 26 - 11.3 = 14.7 ml

molarity of HCl solution = 0.0016/0.0147 = 0.109 M

Trial 1

Mass of aspirin = 0.365 g

moles of NaOH added = 0.101 M x 50 ml = 5.05 mmol

moles of HCl used = 0.109 M x (37 - 15.11) ml = 2.38 mmol

moles of NaOH reacted = moles of aspirin = 5.05 - 2.408 = 2.642 mmol

mass of aspirin in the sample = 2.642 mmol x 180.16/1000 = 0.430 g

Trial 2

Mass of aspirin = 0.375 g

moles of NaOH added = 0.101 M x 50 ml = 5.05 mmol

moles of HCl used = 0.109 M x (23.5 - 16.02) ml = 0.815 mmol

moles of NaOH reacted = moles of aspirin = 5.05 - 0.815 = 4.235 mmol

mass of aspirin in the sample = 4.235 mmol x 180.16/1000 = 0.763 g