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 : pKₐ (CH₃COOH) = 4.76)
log 2 = 0.30
log 3 = 0.48
log 5 = 0.69
log 7 = 0.84
log 11 = 1.04

Solution

Chemistry ma’am · 4 steps

Half the acid is neutralised here, so acid and salt are equal and pH is just the pKₐ.

1. Count millimoles. Acid taken: 50 × 0.1 = 5 mmol CH₃COOH. Base added: 25 × 0.1 = 2.5 mmol NaOH.
2. After reaction. CH₃COOH + NaOH → CH₃COONa + H₂O uses up all 2.5 mmol of NaOH. That leaves 2.5 mmol of acid and forms 2.5 mmol of CH₃COO⁻, so the mixture is an acidic buffer.

   pH = pKₐ + log[CH₃COO⁻]/[CH₃COOH] = 4.76 + log 2.5/2.5

3. Use the log. Both species are in the same 75 mL, so the volume cancels in the ratio. log 1 = 0, which gives pH = 4.76.
4. So. pH = 4.76 = 476 × 10⁻², so the answer is 476.

The common slip: 446. Putting the starting 5 mmol of acid in the ratio instead of the 2.5 mmol left gives 4.76 + log 0.5 = 4.46, which reads as 446, not 476.

The idea to keep. At the half-equivalence point of a weak acid titration, the remaining acid equals the salt formed, so pH = pKₐ. Always compare the leftover acid with the salt, not the starting acid with the salt.

Answer from the NTA final answer key. Solution by PrepWiser.