pH = pKa when:

pH = pKa when:

• A. [A-] = [H3O+]
• B. [A-] = [HA]
• C. [A-] >> [HA]
• D. [HA] >> [A-]

Answer: (B)

The key idea is Henderson–Hasselbalch: pH = pKa + log([A-]/[HA]). This pH equals pKa exactly when the ratio [A-]/[HA] is 1, meaning the conjugate base and the weak acid are present in equal amounts. That situation occurs at the midpoint of the buffer’s titration, where the acid is half dissociated, so [HA] = [A-]. Therefore pH = pKa. If [A-] were equal to [H3O+], there’s no direct reason for pH to match pKa; and if either form dominates ([A-] >> [HA] or [HA] >> [A-]), the log term becomes positive or negative, pushing the pH away from pKa.

The key idea is Henderson–Hasselbalch: pH = pKa + log([A-]/[HA]). This pH equals pKa exactly when the ratio [A-]/[HA] is 1, meaning the conjugate base and the weak acid are present in equal amounts. That situation occurs at the midpoint of the buffer’s titration, where the acid is half dissociated, so [HA] = [A-]. Therefore pH = pKa.

If [A-] were equal to [H3O+], there’s no direct reason for pH to match pKa; and if either form dominates ([A-] >> [HA] or [HA] >> [A-]), the log term becomes positive or negative, pushing the pH away from pKa.