Heat of Neutralization
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flagWhat you'll discover
- arrow_forwardDefine heat (enthalpy) of neutralization and state its value for strong acid-strong base
- arrow_forwardUse a simple calorimeter correctly, recording T₁ before mixing and T₂ at the peak
- arrow_forwardCalculate ΔH from q = mcΔT and the moles of water formed
- arrow_forwardExplain why weak acids give a numerically smaller ΔH
- arrow_forwardIdentify heat-loss errors and how the experiment minimises them
One reaction behind every strong pair
The heat of neutralization is the enthalpy change when one mole of water is formed by an acid neutralising a base in dilute solution. For any strong acid with any strong base the measured value is remarkably constant: about −57.3 kJ/mol. The reason is that strong acids and bases are fully ionised, so the spectator ions (Na⁺, Cl⁻) merely watch while the only real chemistry is identical every time: H⁺ + OH⁻ → H₂O.
That constancy is itself evidence for the ionic theory of acids and bases — HCl + NaOH, HNO₃ + KOH, H₂SO₄ + NaOH (per mole of water) all release the same heat because they are all literally the same reaction. The negative sign declares the reaction exothermic: chemical potential energy leaves the reacting ions and appears as the kinetic energy of the solution, which you observe as a temperature rise.
The calorimeter and the calculation
The apparatus is humble: a polystyrene (thermocol) cup with a lid and a thermometer. Polystyrene is chosen because it is an excellent insulator with negligible heat capacity — nearly all the heat stays in the solution and gets counted. Procedure: measure equal volumes (say 50 mL each) of 1.0 M acid and 1.0 M NaOH, record the steady initial temperature T₁, mix quickly, stir, and record the maximum temperature T₂. The peak matters — after it, heat leaks to the room and the reading only falls.
The arithmetic: total solution mass m = 100 g (taking density ≈ 1 g/mL), specific heat c ≈ 4.18 J/g°C. Heat released q = m·c·(T₂−T₁). Moles of water formed n = 1.0 × 0.050 = 0.050 mol. ΔH = −q/n. With ΔT ≈ 6.8 °C: q ≈ 2840 J, ΔH ≈ −57 kJ/mol. Main error: heat escaping before you read T₂ — which always makes your |ΔH| too small, never too big.
Why weak acids fall short
Repeat the experiment with ethanoic acid and the temperature rise is consistently smaller — ΔH around −55 kJ/mol instead of −57.3. A weak acid is only partly ionised: most CH₃COOH molecules are intact when mixing happens. Before their protons can neutralise OH⁻, those molecules must first ionise, and ionisation costs energy (it is endothermic for most weak acids).
So the measured heat is the full H⁺ + OH⁻ payout minus the ionisation tax: ΔH(neutralization, weak) = −57.3 + ΔH(ionisation). For ethanoic acid the tax is about +2 kJ/mol. The effect is even larger for weaker acids — HCN manages only about −12 kJ/mol because its ionisation is strongly endothermic. Measuring the shortfall is actually a clean indirect way to estimate an acid's enthalpy of ionisation: thermochemistry doing detective work, an application of Hess's law you can verify in this very simulation.