✔ 最佳答案
Method 1 :
The full equation for the neutralization with NaOH:
CH₃COOH(aq) + NaOH(aq) → H₂O(l) + CH₃COONa(aq)
CH₃COOH is a weak acid which exists mainly as molecules in aqueous solution. NaOH is a strong base which completely dissociates in water to give Na⁺ and OH⁻. H₂O is molecular. CH₃COONa is a solution salt which completely dissociates in water to give CH₃COO⁻ and Na⁺ ions. Hence, the complete ionic equation is:
CH₃COOH(aq) + Na⁺(aq) + OH⁻(aq) → H₂O(l) + CH₃COO⁻(aq) + Na⁺(aq)
Cancel the dummy ion, i.e. Na⁺(aq), on the both sides. The ionic equation:
CH₃COOH(aq) + OH⁻(aq) → H₂O(l) + CH₃COO⁻(aq)
The answer: b) CH₃COOH(aq) + OH⁻(aq) → H₂O(l) + CH₃COO⁻(aq)
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Method 2 :
In all neutralization, the reaction is between H⁺(aq) and OH⁻(aq) ions, i.e.
H⁺(aq) + OH⁻(aq) → H₂O(l)
However, CH₃COOH is a weak acid which incompletely in water. As the consumption of H⁺(aq) ions, CH₃COOH(aq) will then continuously dissociates in water to give H⁺(aq) ions until the reaction completes.
CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq)
Add up the above two equations, and cancel H⁺(aq) on the both sides. The net ionic equation is:
CH₃COOH(aq) + OH⁻(aq) → H₂O(l) + CH₃COO⁻(aq)
The answer: b) CH₃COOH(aq) + OH⁻(aq) → H₂O(l) + CH₃COO⁻(aq)
