Using the Henderson–Hasselbalch equation, what is the pH of a buffer containing HA 0.100 M and A− 0.100 M with Ka = 1.0×10−5?

Using the Henderson–Hasselbalch equation, what is the pH of a buffer containing HA 0.100 M and A− 0.100 M with Ka = 1.0×10−5?

• A. pH = 4.50
• B. pH = 6.00
• C. pH = 7.00
• D. pH = 5.00

Answer: (D)

Henderson–Hasselbalch links pH to the acid’s pKa and the ratio of conjugate base to acid: pH = pKa + log([A−]/[HA]). The Ka is 1.0×10−5, so pKa = -log(1.0×10−5) = 5.00. The buffer has equal amounts of HA and A−, so [A−]/[HA] = 1 and log(1) = 0. Therefore pH = 5.00 + 0 = 5.00. The buffer’s pH is 5.00.

Henderson–Hasselbalch links pH to the acid’s pKa and the ratio of conjugate base to acid: pH = pKa + log([A−]/[HA]). The Ka is 1.0×10−5, so pKa = -log(1.0×10−5) = 5.00. The buffer has equal amounts of HA and A−, so [A−]/[HA] = 1 and log(1) = 0. Therefore pH = 5.00 + 0 = 5.00. The buffer’s pH is 5.00.