The active ingredient in aspirin is acetylsalicylic acid. A 2.08 g sample of acetylsalicylic acid required 23.06 mL of 0.5007 M NaOH for complete reaction. Addition of 13.02 mL of 0.4434 M HCl to the flask containing the aspirin and the sodium hydroxide produced a mixture with pH = 3.48. Find the molar mass (in g/mol) of acetylsalicylic acid and its Ka value. Acetylsalicylic acid is a monoprotic acid.
Number of moles of acetylsalicylic acid and NaOH will be equal since one carboxylic acid group is present in acetylsalicylic acid.
No.of moles Molarity= Volume(L) No.of moles of NaOH=Molarity×Volume(L) =0.5007M×0.02306L =0.01155mol No.of moles of acetylsalicylic acid = 0.01155mol mass Molar mass=. No.of moles 2.08g 0.01155mol
Number of moles of HCl can be determin...
No.of moles Molarity=- Volume(L) No.of moles of HC1=Molarity×Volume(L) =0.4434M×0.01302L =0.00577mol No.of moles of salt of resulting solution=0.01155-0.00577 =0.00578 mol
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