In the titration of a weak acid with a strong base, at what point does pH = pKa?

Question

BlackTom AI · Accepted Answer

Henderson-Hasselbalch equation shows pH = pKa when the concentrations of the conjugate base and the weak acid are equal.

1) Relevant concepts
- Titration of a weak acid HA with strong base OH− converts HA to its conjugate base A−: HA + OH− → A− + H2O.
- The buffer region exists before the equivalence point; in a buffer the Henderson-Hasselbalch relation applies: pH = pKa + log([A−]/[HA]).
- The half-equivalence (midpoint) is when exactly half the initial acid has been neutralized.

2) Step-by-step reasoning
- Start with n0 moles of HA. At the half-equivalence point, n0/2 moles of OH− have been added.
- Moles remaining HA = n0 − n0/2 = n0/2. Moles of A− produced = n0/2.
- Therefore [A−] = [HA] (assuming same solution volume or accounting for equal dilution).

3) Intermediate calculation using Henderson-Hasselbalch
- pH = pKa + log([A−]/[HA]) = pKa + log(1) = pKa + 0 = pKa.

4) Why the correct answer is correct
- Because at the half-equivalence point the acid and its conjugate base are present in equal amounts, the logarithmic term in Henderson-Hasselbalch becomes zero, giving pH = pKa. This is the defining experimental way to determine pKa from a titration curve (the midpoint of the buffer region).

5) Why the other options are incorrect
- b. At the equivalence point: incorrect because all HA has been converted to A−; the solution is basic (pH > 7) due to hydrolysis of A−, so pH ≠ pKa.
- c. At the end-point: the end-point is the indicator color change and may not exactly coincide with any specific stoichiometric ratio; it is not generally the point where pH = pKa.
- d. Never: incorrect because the half-equivalence is a clear case where pH equals pKa.

6) Conclusion
- Therefore the correct answer is (a) At the half-equivalence point, confirming pH = pKa.

In the titration of a weak acid with a strong base, at what point does pH = pKa?单项选择题

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