A 42.36 mL aliquot of 0.09410 M NaOH was added to a 0.3952 g sample of crushed a

Question

A 42.36 mL aliquot of 0.09410 M NaOH was added to a 0.3952 g sample of crushed aspirin tablet dissolved in ethanol. The acetylsalicylic acid in the aspirin tablet was neutralized by NaOH, leaving excess OH in solution. The excess OH was back-titrated with 17.77 mL of 0.08140 M HCI The molecular mass of ASA is 180.157 g/mol. For scientific notation, use the format 2.4E-4, the "E" must be capitalized (a) How many total moles of NaOH were added to the sample mol NaOH total (b) How many total moles of NaOH were present in excess? mol NaOH in excess (c) What mass of ASA was present in the 0.3952 g sample of crushed aspirin tablet? g ASA in the sample (d) If the average mass of a tablet is 1.197 g, determine the mass of ASA (in mg) present in a single tablet. mg ASA per tablet

Explanation / Answer

a) no of mol of naOH added = 42.36*0.0941/1000 = 0.004 mol

b) no of mol of HCl reacted = 17.77*0.08140/1000 = 0.00145 mol

   no of mol of naOH were present excess = 0.00145 mol

c) no of mol of ASA reacted = no of mol of naOH reacted = 0.004-0.00145 = 0.0025 mol

   mass of ASA present in tablet = 0.00255*180.157 = 0.46 g

d) mass of ASA present in single tablet = 0.46*1.197/0.3952 = 1.39 g

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