Henderson-Hasselbalch: For a buffer with pKa = 4.75, what [A-]/[HA] ratio yields pH = 4.00?

Henderson-Hasselbalch: For a buffer with pKa = 4.75, what [A-]/[HA] ratio yields pH = 4.00?

• A. 0.178
• B. 1.78
• C. 4.75
• D. 0.021

Answer: (A)

The Henderson-Hasselbalch equation links pH to pKa and the base/acid ratio: pH = pKa + log([A-]/[HA]). To get pH = 4.00 with pKa = 4.75, the log term must be -0.75. So [A-]/[HA] = 10^(-0.75) ≈ 0.178, i.e., about 0.18. This ratio being less than 1 means the acid form HA is more prevalent than its conjugate base A-, which is consistent with a pH below the pKa. The other given numbers would yield pH values further from 4.00 because they produce different log ratios (e.g., a larger ratio pushes pH higher; a much smaller ratio would push pH lower).

The Henderson-Hasselbalch equation links pH to pKa and the base/acid ratio: pH = pKa + log([A-]/[HA]). To get pH = 4.00 with pKa = 4.75, the log term must be -0.75. So [A-]/[HA] = 10^(-0.75) ≈ 0.178, i.e., about 0.18. This ratio being less than 1 means the acid form HA is more prevalent than its conjugate base A-, which is consistent with a pH below the pKa. The other given numbers would yield pH values further from 4.00 because they produce different log ratios (e.g., a larger ratio pushes pH higher; a much smaller ratio would push pH lower).