Question

A sample containing acetaminophen was dissolved in water to make a 50.00-mL solution. The solution was titrated with 0.5065 M NaOH, and the equivalence point was reached when 26.10 mL NaOH solution was added.

The terminal hydrogen (on the OH) is removed during neutralization. The Ka value of acetaminophen is 3.2 × 10−10.

Calculate the pH at the equivalence point.

Answer 1

acetaminophen is a weak acid

we know that

moles = molarity x volume (L)

so

moles of NaOH = 0.5065 x 26.1 x 10-3

moles of NaOH = 13.22 x 10-3

now

at equivalence point

moles of acid = moles of base added

so

moles of acetaminophen = moles of NaOH = 13.22 x 10-3

now

acetaminophen + NaOH ---> sodium salt of acetaminphen + H20

now

this sodium salt of acetamiophen is the weak conjugate base of acetaminophen

and

it undergoes hydrolysis

final volume = 50 + 26.1 = 76.1 ml

now

moles of conjugate base formed= moles of acid reacted = 13.22 x 10-3

conc of conjugate base = 13.22 x 10-3 x 1000 / 76.1 = 0.17372

now

for weak bases

[OH-] = sqrt (Kb x C)

also

Kb = Kw / Ka

so

[OH-] = sqrt ( Kw x C / Ka)

so

[OH-] = sqrt ( 10-14 x 0.17372 / 3.2 x 10-10)

[OH-] = 2.33 x 10-3

now

pOH = -log [OH-]

pOH = -log 2.33 x 10-3

pOH = 2.633

now

pH = 14 - pOH

so

pH = 14 - 2.633

pH = 11.367

so

the pH at equivalence point is 11.367