- 2075 Baisakh · 2 marks
pH of acetic acid and NaOH mixture
Short answer
Moles of CH₃COOH initially 30 mmol, NaOH 10 mmol; reaction leaves 20 mmol acid and 10 mmol salt, forming a buffer. pH = pKa + log(10/20) with pKa = 4.733 gives 4.43.
60 mL of 0.5 M acetic acid is mixed with 40 mL of 0.25 M sodium hydroxide solution. What will be the pH of the mixture? (Given Ka = 1.85×10⁻⁵)
Acetic acid is a weak acid and NaOH is a strong base. NaOH reacts with part of the acid, so the mixture contains the acid and its salt: an acidic buffer.
Moles CH₃COOH = 0.5 × 60/1000 = 0.030 mol (30 mmol)
Moles NaOH = 0.25 × 40/1000 = 0.010 mol (10 mmol)
CH₃COOH + NaOH → CH₃COONa + H₂O
Acid left = 30 − 10 = 20 mmol
Salt formed = 10 mmol
(The total volume is common to both, so mmol can be used directly in the ratio.)
pKa = −log(1.85×10⁻⁵) = 4.733
pH = pKa + log([salt]/[acid])
= 4.733 + log(10/20)
= 4.733 − 0.301
= 4.43
Answer: pH ≈ 4.43
✓ Answer checked by Claude Opus 5.5 (high effort)
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