HSC · Topic explainer
Finding pKa at the half-equivalence point
For a weak monoprotic acid titrated with a strong base, the half-equivalence point occurs at half the volume needed to reach equivalence. At that point, the weak acid and its conjugate base have equal concentrations. The pH is approximately the pKa under the usual ideal-solution approximation.
Worked example
Suppose the steep rise in the titration curve puts equivalence at 24.0 mL. Read the pH at 12.0 mL. If the curve shows pH 4.60 there, pKa is about 4.60.
Convert using Ka = 10 raised to the power −pKa. Here Ka is about 2.5 × 10⁻⁵. The midpoint to use is halfway along the volume axis to equivalence. It is not halfway between the initial and equivalence pH values.
Official exam practice
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