For a buffer containing 0.50 M acetic acid (Ka = 1.8×10^-5) and 0.50 M sodium acetate, what is the pH? (use Henderson-Hasselbalch)

For a buffer containing 0.50 M acetic acid (Ka = 1.8×10^-5) and 0.50 M sodium acetate, what is the pH? (use Henderson-Hasselbalch)

• A. 4.74
• B. 3.74
• C. 5.74
• D. 4.00

Answer: (A)

This uses Henderson-Hasselbalch for a buffer. The pH of a buffer is pH = pKa + log([A-]/[HA]). First find pKa from Ka: pKa = -log(1.8×10^-5) ≈ 4.74. In the given buffer, [A-] (acetate) and [HA] (acetic acid) are both 0.50 M, so the ratio [A-]/[HA] = 1 and log(1) = 0. Therefore pH = 4.74 + 0 = 4.74. When acid and conjugate base are present in equal amounts, the pH equals the pKa.

This uses Henderson-Hasselbalch for a buffer. The pH of a buffer is pH = pKa + log([A-]/[HA]). First find pKa from Ka: pKa = -log(1.8×10^-5) ≈ 4.74. In the given buffer, [A-] (acetate) and [HA] (acetic acid) are both 0.50 M, so the ratio [A-]/[HA] = 1 and log(1) = 0. Therefore pH = 4.74 + 0 = 4.74. When acid and conjugate base are present in equal amounts, the pH equals the pKa.