An acetate buffer is prepared with 0.100 M HOAc and 0.100 M NaOAc. If pKa(HOAc) = 4.74, what is the pH?

An acetate buffer is prepared with 0.100 M HOAc and 0.100 M NaOAc. If pKa(HOAc) = 4.74, what is the pH?

• A. 4.26
• B. 4.74
• C. 4.94
• D. 7.00

Answer: (B)

The pH of a buffer depends on the ratio of its conjugate base to the weak acid, described by the Henderson–Hasselbalch equation: pH = pKa + log([A-]/[HA]). In this acetate buffer, the weak acid is HOAc (HA) and its conjugate base is OAc− (A−). Since both components are present at equal concentrations (0.100 M each), the ratio [A−]/[HA] is 1, and log(1) = 0. Therefore the pH equals the pKa, which is 4.74. This is the key reason the pH is 4.74.

The pH of a buffer depends on the ratio of its conjugate base to the weak acid, described by the Henderson–Hasselbalch equation: pH = pKa + log([A-]/[HA]). In this acetate buffer, the weak acid is HOAc (HA) and its conjugate base is OAc− (A−). Since both components are present at equal concentrations (0.100 M each), the ratio [A−]/[HA] is 1, and log(1) = 0. Therefore the pH equals the pKa, which is 4.74. This is the key reason the pH is 4.74.