Question

Acetaminophen, shown below, is a weak acid used pharmaceutically as an aspirin
alternative. The molecular formula of acetaminophen is HOC6H4NHCOCH3. The
terminal hydrogen (on the OH) is removed during neutralization. The Ka value of
acetaminophen is 3.2 × 10−10
.
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.

Calculate the pH of the sample solution before any NaOH was added.

Answer 1

Moles of NaOH used= 0.5065*26.1/1000=0.01322 moles

Acetaminophen]OH + NaOH → [Acetaminophen]O- Na+

moles of Acetaninophen = 0.01322

Molarity= 0.01322*1000/50=0.2643M

Let Acetaminophen be HA

HA----> H+ + A-

A- is conjugate base

Ka= [H+] [A-]/[HA]

let x= drop in concentration of HA to reach equilibrium

3.2*10-10= x2/(0.2643-x)

asuming that x is smaller than 0.2643

hence

3.2*10^-10 =x²/0.2643

x²= 8.5*10-11

x= 9.2*10-6 (<<<0.2643)

pH= -log(9.2*10-6)=5.03