A 2.11 g sample of acetylsalicylic acid required 23.29 mL of 0.5028 M NaOH for complete reaction. Find the molar mass (in g/mol) of acetylsalicylic acid and its Ka value.

Question

The active ingredient in aspirin is acetylsalicylic acid. A 2.11 g sample of acetylsalicylic acid required 23.29 mL of 0.5028 M NaOH for complete reaction. Addition of 13.08 mL of 0.4476 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 K_a value. Acetylsalicylic acid is a monoprotic acid.

molar mass ________ g/mol

Ka _______

Answer 1

Let acetylsalicylic acid be represented by HA.

HA + NaOH ==> NaA + H₂O ... neutralization reaction

moles NaOH needed for complete neutralization = 23.29 ml x 1 L/1000 ml x 0.5028 mol/L = 0.01171 mol

Since NaOH reacts 1:1 with HA, moles HA = 0.01171

Molar mass HA (acetylsalicylic acid) = 2.11 g /0.01171 moles = 180 g/mol (3 significant figures)

At this point in the neutralization, you have produced the sodium salt, i.e. NaA. When HCl is added, we have..

A^- + H⁺ <==> HA i.e. the sodium acetylsalicylate is converted to acetylsalicylic acid.

Moles H⁺ added = 0.01308 L x 0.4476 mol/L = 0.00585 moles H⁺

A^- + H⁺ ====> HA

0.01171..0.00585....0.....Initial

-.00585...-0.00585....+0.00585...Change

0.00586.....0.............0.00586...Final

pH = pKa + log [A-]/[HA]

pH = pKa + log 1 = pKa + 0

pH = pKa = 3.48

Ka = 1x10^-3.48

Ka = 3.0x10^-4