When a strong acid is added to a buffer, how can you estimate the new pH using Henderson–Hasselbalch?

When a strong acid is added to a buffer, how can you estimate the new pH using Henderson–Hasselbalch?

• A. pH = pKa - log([A-]/[HA]); H+ converts HA to A-
• B. pH = pKa; buffer resists change
• C. pH = pKa + log([HA]/[A-]); H+ converts HA to A-
• D. pH = pKa + log([A-]/[HA]); H+ converts A- to HA

Answer: (D)

The Henderson–Hasselbalch idea is that pH is set by the acid’s pKa and the ratio of conjugate base to weak acid in the buffer. When a strong acid is added, the extra H+ reacts with the conjugate base (A−), converting it into the weak acid (HA). This shifts the ratio [A−]/[HA] downward. Because pH = pKa + log([A−]/[HA]), a smaller ratio makes the log term more negative, so the pH falls. Thus you estimate the new pH with pH = pKa + log([A−]/[HA]), reflecting that H+ converts A− to HA. Using the reciprocal ratio would misrepresent the direction of the pH change.

The Henderson–Hasselbalch idea is that pH is set by the acid’s pKa and the ratio of conjugate base to weak acid in the buffer. When a strong acid is added, the extra H+ reacts with the conjugate base (A−), converting it into the weak acid (HA). This shifts the ratio [A−]/[HA] downward. Because pH = pKa + log([A−]/[HA]), a smaller ratio makes the log term more negative, so the pH falls. Thus you estimate the new pH with pH = pKa + log([A−]/[HA]), reflecting that H+ converts A− to HA. Using the reciprocal ratio would misrepresent the direction of the pH change.