Ethanoic acid deprotonation yields ethanoate ions

Steps:

• Ethanoic acid (CH₃COOH) forms CH₃COO⁻ ions when deprotonated by bases or in aqueous dissociation.
• Reagent 1 (aqueous Na₂CO₃): Neutralization reaction CH₃COOH + Na₂CO₃ → CH₃COONa + CO₂ + H₂O produces sodium ethanoate, yielding CH₃COO⁻ in solution.
• Reagent 2 (LiAlH₄): Reduces CH₃COOH to CH₃CH₂OH (ethanol), no CH₃COO⁻ formed.
• Reagent 3 (water): Partial dissociation CH₃COOH ⇌ CH₃COO⁻ + H⁺ in aqueous solution produces CH₃COO⁻ ions.

Why B is correct:

• B selects 1 and 3, matching acid-base neutralization (reagent 1) and equilibrium dissociation (reagent 3) per Brønsted-Lowry theory.

Why the others are wrong:

• A includes 2, but LiAlH₄ reduction yields alcohol, not ions.
• C excludes 3, ignoring aqueous dissociation of the weak acid.
• D includes 2 and excludes 1, missing neutralization while adding irrelevant reduction.

Final answer: B