A buffer solution contains 0.40 M acetic acid and 0.20 M acetate. With Ka 1.8×10^-5, calculate pH using Henderson-Hasselbalch.

A buffer solution contains 0.40 M acetic acid and 0.20 M acetate. With Ka 1.8×10^-5, calculate pH using Henderson-Hasselbalch.

• A. pH ≈ 4.74
• B. pH ≈ 4.44
• C. pH ≈ 5.00
• D. pH ≈ 3.44

Answer: (B)

A buffer's pH is set by the balance between the weak acid and its conjugate base, described by Henderson-Hasselbalch: pH = pKa + log([A-]/[HA]). For acetic acid, Ka = 1.8×10^-5, so pKa ≈ 4.74. The buffer has [A-] = 0.20 M and [HA] = 0.40 M, giving a ratio [A-]/[HA] = 0.20/0.40 = 0.50. log(0.50) ≈ -0.301. Plugging in, pH ≈ 4.74 + (-0.301) ≈ 4.44. Since there is more acid than base, the pH is below the pKa, yielding about 4.44.

A buffer's pH is set by the balance between the weak acid and its conjugate base, described by Henderson-Hasselbalch: pH = pKa + log([A-]/[HA]). For acetic acid, Ka = 1.8×10^-5, so pKa ≈ 4.74. The buffer has [A-] = 0.20 M and [HA] = 0.40 M, giving a ratio [A-]/[HA] = 0.20/0.40 = 0.50. log(0.50) ≈ -0.301. Plugging in, pH ≈ 4.74 + (-0.301) ≈ 4.44. Since there is more acid than base, the pH is below the pKa, yielding about 4.44.