A buffer solution contains 0.20 M acetic acid (CH3COOH, Ka = 1.8 * 10^-5) and 0.20 M sodium acetate (CH3COONa). What is the pH of this buffer? (log10(1.8) approx 0.26)

Answer and explanation

Correct answer: A. 4.74

  1. Step 1: Apply the Henderson-Hasselbalch equation: pH = pKa + log10([Conjugate Base] / [Acid]).
  2. Step 2: Compute pKa: pKa = -log10(Ka) = -log10(1.8 * 10^-5) = 5 - log10(1.8) = 5 - 0.26 = 4.74.
  3. Step 3: Since [Base] = [Acid] = 0.20 M: [Base]/[Acid] = 1, and log10(1) = 0.
  4. Step 4: Therefore: pH = 4.74 + 0 = 4.74.
Answer
4.74.
Why not the others: '5.04' adds 0.30; '4.44' subtracts 0.30; '7.00' assumes neutral pH for any salt/acid mixture.

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