Question

50 mL of 0.1 M CH₃COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be ____ × 10^–2. (Nearest integer)
(Given : pK_a (CH₃COOH) = 4.76)
log 2 = 0.30
log 3 = 0.48
log 5 = 0.69
log 7 = 0.84
log 11 = 1.04

Answer 1

Correct Answer: 476

We need to determine the pH of a solution when 25 mL of 0.1 M NaOH is added to 50 mL of 0.1 M CH₃COOH during a titration. The reaction is CH₃COOH + NaOH → CH₃COONa + H₂O.
Total mmol of CH₃COOH = 0.1 M × 50 mL = 5 mmol.
Total mmol of NaOH = 0.1 M × 25 mL = 2.5 mmol.
After adding NaOH, CH₃COOH left = 5 mmol - 2.5 mmol = 2.5 mmol, CH₃COONa produced = 2.5 mmol.
Concentration of CH₃COOH = 2.5 mmol / 75 mL = 0.0333 M.
Concentration of CH₃COONa = 2.5 mmol / 75 mL = 0.0333 M.
Using the Henderson-Hasselbalch equation: pH = pK_a + log([A^-]/[HA]) = 4.76 + log(0.0333/0.0333) = 4.76.
pH = 4.76 × 10⁰ = 476 × 10^-2. The computed value 476 is within the specified range of 476,476.