Ethanoic Acid with Bases

Introduction

Ethanoic acid reacts with bases through its carboxyl O–H proton. Sodium hydroxide produces sodium ethanoate and water. The acid is weak in water, but its partial ionisation does not change the one-to-one mole ratio in the balanced neutralisation equation.

Core explanation

Write CH₃COOH + NaOH → CH₃COONa + H₂O. One molecule of acid supplies one transferable proton; one hydroxide ion accepts it to make water. The remaining CH₃COO⁻ pairs with Na⁺ to form sodium ethanoate. The ionic view is CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O; Na⁺ is a spectator ion in solution. Both views conserve atoms and charge when used correctly.

The acid-water equilibrium does not require us to start with all acid molecules ionised. As hydroxide removes acidic protons, more CH₃COOH can react. At the equivalence point of a suitable titration, one mole NaOH has reacted per mole CH₃COOH originally present, assuming no other acid-base-reactive components interfere. “Weak acid” does not mean only a small fraction of its total amount can eventually be neutralised.

Other bases yield other ethanoate salts. With potassium hydroxide, CH₃COOH + KOH → CH₃COOK + H₂O. With a basic metal oxide, a balanced example is 2 CH₃COOH + CuO → Cu(CH₃COO)₂ + H₂O, assuming suitable reaction conditions. Copper(II) oxide accepts two acid protons per oxide ion overall, so two acid molecules are needed to balance the Cu²⁺ ethanoate salt.

Naming salts follows the cation then the carboxylate anion: sodium ethanoate, potassium ethanoate, copper(II) ethanoate. The acid's two-carbon skeleton remains in CH₃COO⁻. Do not write “sodium ethanoic acid” for the salt, and do not leave H attached to the carboxylate after neutralisation.

Ethanoic acid with a carbonate is also an acid-base process, but it produces CO₂ as well as salt and water. Its stoichiometry differs from NaOH because CO₃²⁻ can accept two protons overall. Treating every base as one-to-one with acid would give incorrect coefficients.

Step-by-step reasoning

1. Identify one acidic proton in each CH₃COOH molecule. 2. Identify the base's proton-accepting capacity or hydroxide count. 3. Form ethanoate paired with the base cation. 4. Add water, and CO₂ if carbonate is involved. 5. Balance elements and charges before using mole ratios.

Visual explanation

Draw CH₃COO–H next to Na⁺OH⁻. Move H conceptually to OH⁻, making H₂O, and place Na⁺ beside CH₃COO⁻. Use a second diagram with CuO and two acid molecules to show how charge +2 requires two ethanoate ions.

Real-world analogy

One ticket admits one person; a venue requiring two tickets per group changes the count. NaOH accepts one acid proton per formula unit, while an oxide or carbonate can involve two. The balanced equation is the ticket rule for a particular base.

Real-world example

An acid-base titration can measure ethanoic acid in a vinegar sample. The analyst uses the known NaOH concentration and the one-to-one equation to calculate moles of acid, then divides by sample volume for concentration. The acidity strength affects the pH curve, not the equivalence mole ratio.

Why?

Why does water form? Hydroxide has oxygen and hydrogen; accepting the acid's O–H proton turns OH⁻ into H₂O. The acid residue becomes negatively charged ethanoate, balanced by the metal cation.

Common misconception

“Only the ionised portion of a weak acid reacts with NaOH.” Removing protons shifts the equilibrium, so remaining acid can ionise and react. Stoichiometric neutralisation can consume the full acid amount.

Worked example

Suppose 0.020 mol CH₃COOH reacts fully with NaOH. The balanced equation has coefficient 1 for each, so 0.020 mol NaOH is required and 0.020 mol sodium ethanoate forms. If the NaOH solution is 0.100 mol L⁻¹, volume needed is n/c = 0.020/0.100 = 0.200 L, or 200 mL. This calculation assumes the sample's only titratable acid is ethanoic acid.

Quick check

1. What salt forms from ethanoic acid and KOH? Answer: Potassium ethanoate, CH₃COOK, plus water.

Exam focus

Use the exact balanced equation for the stated base. Name the salt from its cation and ethanoate, and do not confuse acid strength with neutralisation mole ratio. Include units when using titration data.

Advanced insight

The pH at equivalence for a weak acid–strong base titration is commonly above 7 because ethanoate reacts with water to produce some OH⁻. Indicator selection must account for the titration curve, not just the net stoichiometric equation.

Summary

Ethanoic acid transfers its carboxyl proton to a base and forms an ethanoate salt. NaOH reacts in a one-to-one ratio, while divalent bases can require two acid molecules. The weak-acid equilibrium changes pH behaviour but not balanced mole accounting.

Practice questions

1. Balance ethanoic acid with NaOH. Answer: CH₃COOH + NaOH → CH₃COONa + H₂O. 2. What is the net ionic equation with hydroxide? Answer: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. 3. How many moles of NaOH react with 0.5 mol ethanoic acid? Answer: 0.5 mol, from the one-to-one coefficients. 4. Name Cu(CH₃COO)₂. Answer: Copper(II) ethanoate.