Using Henderson–Hasselbalch, a buffer is prepared with 0.100 M acetic acid (HA) and 0.200 M acetate (A-). If pKa = 4.74, what is the pH?

Using Henderson–Hasselbalch, a buffer is prepared with 0.100 M acetic acid (HA) and 0.200 M acetate (A-). If pKa = 4.74, what is the pH?

• A. 4.74
• B. 5.04
• C. 5.50
• D. 3.74

Answer: (B)

This uses Henderson–Hasselbalch to connect pH with the ratio of conjugate base to weak acid in a buffer. The equation is pH = pKa + log([A-]/[HA]). Here [A-] = 0.200 M and [HA] = 0.100 M, so the ratio is 2. log(2) is about 0.301, so pH ≈ 4.74 + 0.301 = 5.041, roughly 5.04. Since there is more conjugate base than acid, the pH shifts above the pKa. This matches the chosen value of about 5.04. If the ratio were 1, pH would equal pKa (4.74); if the ratio is less than 1, pH would be below 4.74.

This uses Henderson–Hasselbalch to connect pH with the ratio of conjugate base to weak acid in a buffer. The equation is pH = pKa + log([A-]/[HA]). Here [A-] = 0.200 M and [HA] = 0.100 M, so the ratio is 2. log(2) is about 0.301, so pH ≈ 4.74 + 0.301 = 5.041, roughly 5.04. Since there is more conjugate base than acid, the pH shifts above the pKa. This matches the chosen value of about 5.04. If the ratio were 1, pH would equal pKa (4.74); if the ratio is less than 1, pH would be below 4.74.