Determine the concentration of the HCl solution from the data for the standardiz

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

Explanation / Answer

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

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