While performing a thermodynamics experiment, a student made the following observations.
HCl + NaOH → NaCl + H₂O ΔH = -57.3 kJ mol⁻¹
CH₃COOH + NaOH → CH₃COONa + H₂O ΔH = -55.3 kJ mol⁻¹
The enthalpy of ionization of CH₃COOH as calculated by the student is _____________ kJ mol⁻¹. (nearest integer)

Solution

Chemistry ma’am · 4 steps

The weak acid gives out 2 kJ less heat because it spends that energy ionising first.

1. Strong acid baseline. HCl and NaOH are fully ionised, so their neutralisation is only H⁺ + OH⁻ → H₂O. That step releases 57.3 kJ mol⁻¹.
2. Weak acid route. CH₃COOH is only partly ionised. It must first split into CH₃COO⁻ and H⁺, which absorbs heat, and then the same H⁺ + OH⁻ step releases 57.3 kJ mol⁻¹.

   Δ Hₙₑᵤₜ(CH₃COOH) = Δ Hᵢₒₙ + Δ Hₙₑᵤₜ(strong)

3. Put in numbers. -55.3 = Δ Hᵢₒₙ + (-57.3), so Δ Hᵢₒₙ = -55.3 + 57.3 = +2.0 kJ mol⁻¹.
4. So. The enthalpy of ionization of CH₃COOH is +2 kJ mol⁻¹, so the answer is 2.

The common slip: -2. Subtracting the wrong way round, -57.3 - (-55.3), gives -2. Ionization absorbs heat, so the value must be positive.

The idea to keep. For a weak acid or base, enthalpy of ionization = (its heat of neutralisation) minus (−57.3 kJ). It always comes out positive, because ionisation absorbs heat.

Answer from the NTA final answer key. Solution by PrepWiser.