During the titration of a weak acid with a strong base, the pH at the half-equivalence point is equal to which property of the weak acid?

• A. pH at the equivalence point
• B. pKb of the weak base
• C. log Ka
• D. pKa of the weak acid

Answer: (D)

The main idea is to recognize what happens at the half-equivalence point in a weak acid–strong base titration. At this stage, you’ve added exactly enough base to convert half of the weak acid into its conjugate base, so the solution contains equal amounts of the weak acid (HA) and its conjugate base (A−). That pair forms a buffer, and for a buffer the pH is set by the acid’s dissociation constant through the Henderson–Hasselbalch relation: pH = pKa + log([A−]/[HA]).

Since [A−] equals [HA] at half-equivalence, log([A−]/[HA]) is log(1) = 0, giving pH = pKa. So the pH at the half-equivalence point directly reflects the acid’s strength, expressed as pKa.

Why the other ideas don’t fit: the pH at the equivalence point in this titration is governed by the hydrolysis of the conjugate base and is not generally equal to pKa. The pKb of a weak base isn’t the determining factor here. And while pKa is related to Ka through pKa = −log Ka, the value you’re looking for is the pKa itself, not log Ka.