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.

Question
Asked Dec 3, 2019
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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.

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Expert Answer

Step 1

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
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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

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Step 2

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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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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