Using the Henderson–Hasselbalch equation, estimate the pH of a solution containing 0.10 M acetic acid and 0.05 M acetate, given pKa = 4.76.

Using the Henderson–Hasselbalch equation, estimate the pH of a solution containing 0.10 M acetic acid and 0.05 M acetate, given pKa = 4.76.

• A. 4.46
• B. 4.76
• C. 3.46
• D. 5.00

Answer: (A)

Henderson–Hasselbalch shows that a buffer’s pH depends on the ratio of its conjugate base to the weak acid, not on their absolute amounts. For acetic acid and acetate, pH = pKa + log([acetate]/[acetic acid]). Here the acetate concentration is 0.05 M and the acetic acid is 0.10 M, so the ratio is 0.05/0.10 = 0.5. The common log of 0.5 is about -0.301, giving pH ≈ 4.76 + (-0.301) = 4.46. Since there is more weak acid than conjugate base, the pH sits below the pKa. If the ratio were 1, pH would equal the pKa; changing concentrations while preserving the ratio leaves pH unchanged.

Henderson–Hasselbalch shows that a buffer’s pH depends on the ratio of its conjugate base to the weak acid, not on their absolute amounts. For acetic acid and acetate, pH = pKa + log([acetate]/[acetic acid]). Here the acetate concentration is 0.05 M and the acetic acid is 0.10 M, so the ratio is 0.05/0.10 = 0.5. The common log of 0.5 is about -0.301, giving pH ≈ 4.76 + (-0.301) = 4.46. Since there is more weak acid than conjugate base, the pH sits below the pKa. If the ratio were 1, pH would equal the pKa; changing concentrations while preserving the ratio leaves pH unchanged.