The half-equivalence point is a key feature of weak acid–strong base and weak base–strong acid titrations. It links the titration curve directly to $pK_a$ (or $pK_b$) with minimal computation.

Understanding the half-equivalence point

What “half-equivalence” means chemically

For a weak acid titrated with a strong base, the reaction converts HA to A−. At the half-equivalence point:

• Exactly half of the initial HA has been converted to A−

• The solution contains a buffer made of the conjugate pair HA/A−

• The conjugate pair concentrations are equal: [HA] = [A−] (as stated in the syllabus focus)

For a weak base titrated with a strong acid, the analogous conjugate pair is B/HB+, and at half-equivalence [B] = [HB+].

Practical identification on a titration curve

This titration curve shows pH versus volume of titrant for a weak acid–strong base titration, highlighting the buffer region before equivalence and the rapid pH change near the equivalence point. On this kind of graph, the half-equivalence point occurs at half the equivalence-point volume, and the pH read there equals $pK_a$ for the weak acid. Source

To locate half-equivalence on a graph of pH vs volume of titrant:

• Find the equivalence-point volume $V_\text{eq}$ (the inflection/steepest-rise centre for a monoprotic titration curve)

• Compute (conceptually) half that volume: $V_\text{half} = \frac{1}{2}V_\text{eq}$

• Read the pH at $V_\text{half}$; this pH equals $pK_a$ (weak acid case)

This method is primarily graphical/interpretive: you are extracting $pK_a$ from the curve rather than from tabulated equilibrium data.

Why pH equals pKa at half-equivalence

The key relationship is the Henderson–Hasselbalch equation, which applies to a buffer containing a conjugate acid–base pair.

At the half-equivalence point, the syllabus condition [HA] = [A−] makes the ratio $\frac{[A^-]}{[HA]} = 1$, so:

• $\log(1) = 0$

• Therefore $pH = pK_a$

This is why the half-equivalence point is so valuable: the curve gives $pK_a$ directly.

Weak base version (how to extract pKb)

These paired titration curves compare (a) a weak acid titrated with a strong base and (b) a weak base titrated with a strong acid, emphasizing how curve shape and equivalence-point pH differ from strong/strong cases. In both cases, the half-equivalence (midpoint) occurs halfway to the equivalence volume, where the conjugate pair concentrations are equal and the buffer equation simplifies to $pH=pK_a$ (or $pOH=pK_b$). Source

For a weak base titrated with strong acid, the buffer pair is B/HB+ and:

• At half-equivalence, [B] = [HB+]

• The analogous buffer form gives $pOH = pK_b$

• You can then convert to pH using $pH = 14 - pOH$ at 25 °C (if needed)

Using half-equivalence to determine pKa from data

What you must read or infer

To determine $pK_a$ from a weak acid titration curve:

• Identify that the titration is weak acid + strong base (buffer region present; equivalence pH > 7, but the half-equivalence method itself relies only on the buffer logic)

• Locate $V_\text{eq}$ and then $V_\text{half}$

• Read the pH at $V_\text{half}$; report that value as $pK_a$

Common pitfalls to avoid

• Half-equivalence logic requires a conjugate pair buffer; it does not apply to strong acid–strong base titrations (no meaningful $pK_a$ to extract).

• Ensure you use the volume corresponding to half of the equivalence volume, not “halfway up the steep rise.”

• For systems with more than one protonatable site, multiple buffer regions can appear; each half-equivalence corresponds to a different acid step (qualitatively).