In Henderson-Hasselbalch calculations, if [A-] equals [HA], what is the pH relative to the pKa?

• A. pH equals pKa.
• B. pH equals pKb.
• C. pH equals pKa + 1.
• D. pH equals log([A-]/[HA]).

Answer: (A)

The key idea is that a buffer’s pH is set by the ratio of its conjugate base to its acid, via pH = pKa + log([A-]/[HA]). If the two forms are present in equal amounts, the ratio [A-]/[HA] is 1, and log(1) equals 0, so pH = pKa. This is the buffer’s midpoint, where the acid and its conjugate base are equal in concentration and the pH matches the pKa regardless of how much total acid or base is present. The other options don’t fit: pH = pKb would not describe this relationship for a conjugate-acid/base buffer, pH = pKa + 1 would require a ratio of 10, and pH = log([A-]/[HA]) omits the pKa term from the Henderson–Hasselbalch equation.