Sketch a titration curve for a weak acid-strong base titration. Select the FALSE statement.

Question

BlackTom AI · Accepted Answer

Henderson–Hasselbalch: pH = pKa + log([A–]/[HA]). This central relation governs a weak acid–strong base titration and shows how the ratio [A–]/[HA] controls pH.

1) Relevant concepts
- Weak acid HA and its conjugate base A– form a buffer; buffer capacity depends on the relative amounts of HA and A–.
- At the half-equivalence point [A–] = [HA], so pH = pKa and buffer capacity is maximal.
- Before much base is added, HA predominates; after equivalence, excess OH– dominates and pH rises sharply (equivalence pH > 7 because A– hydrolyzes).

2) Step-by-step reasoning
- Initially (no or little base): [A–]/[HA] is small (<<1) because most material is HA. Use HH: if [A–]/[HA] << 1, log([A–]/[HA]) is negative, so pH < pKa (acidic). Thus early pH is low and increases gradually as base converts HA → A–.
- Approaching half-equivalence: [A–]/[HA] → 1 so pH → pKa; buffer capacity is highest here and the curve is relatively flat.
- Near equivalence: rapid pH change (steep inflection) because buffer is exhausted.
- After equivalence: excess OH– controls pH; pH rises and then asymptotes to that of the added strong base.

3) Calculational intermediates (illustrative)
- HH at start: if [A–]/[HA] = 0.01, pH = pKa + log(0.01) = pKa − 2 → clearly acidic.
- At half-equivalence: [A–]/[HA] = 1 → pH = pKa.
- After equivalence with large excess OH–, pOH ≈ −log[excess OH–] → pH ≈ 14 − pOH.

4) Why the correct (false) statement is false
- Statement 3 claims “Early in the titration, [A–]/[HA] is very high…” This contradicts the chemistry: early [A–]/[HA] is very low because HA has not yet been converted. Therefore statement 3 is false.

5) Briefly on other options
- 1 True: the pre-inflection curve is sigmoidal (gradual then flatter near buffer region).
- 2 True: the sigmoidal shape arises from buffer formation and changing species ratios.
- 4 True: buffer capacity highest near half-equivalence ([A–]/[HA]=1).
- 5 True: later the buffer’s effectiveness declines as one species dominates, so pH changes more.
- 6 Essentially true: after equivalence excess OH– dominates and pH rises toward that of the strong base.

6) Conclusion
Statement 3 is false; this matches the provided correct answer.

Sketch a titration curve for a weak acid-strong base titration. Select the FALSE statement.单项选择题

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