Measuring the Enthalpy of Neutralisation

When hydrochloric acid and sodium hydroxide are mixed, the solution warms up noticeably. That rise in temperature is energy being released as hydrogen ions and hydroxide ions combine to form water. Measuring it is one of the classic practicals in chemistry: simple equipment, clear results, and a number you can compare with the accepted value.

What you’re measuring

The standard enthalpy change of neutralisation, ΔH(neut), is the enthalpy change when an acid and an alkali react to form one mole of water, under standard conditions.

For any strong acid with any strong alkali, the reaction that actually happens is:

H⁺(aq) + OH⁻(aq) → H₂O(l)

The accepted value is about −57 kJ/mol (values between −55.8 and −57.6 kJ/mol are quoted, depending on the source and concentrations). The negative sign means the reaction is exothermic: energy is released to the surroundings, so the temperature rises.

Aim

To measure the enthalpy change of neutralisation of hydrochloric acid by sodium hydroxide, and compare it with the accepted value.

Equipment

• Expanded polystyrene cup with lid (with a hole for the thermometer), in a beaker for stability
• Thermometer reading to ±0.1 °C or a digital temperature probe
• Two 25 cm³ measuring cylinders or, better, pipettes
• 1.00 mol/dm³ hydrochloric acid
• 1.00 mol/dm³ sodium hydroxide solution
• Stopwatch
• Eye protection

Why polystyrene? It’s a good thermal insulator and has a very low heat capacity, so it absorbs very little of the heat released. That makes it an effective, cheap calorimeter.

Method

1. Put the polystyrene cup inside a beaker to stop it tipping over.
2. Measure 25.0 cm³ of 1.00 mol/dm³ HCl into the cup.
3. Measure 25.0 cm³ of 1.00 mol/dm³ NaOH into a separate clean container.
4. Measure the temperature of each solution. Ideally they should be the same (let them stand in the room for a while). Record the starting temperature, or the average if they differ slightly.
5. For better accuracy (see the extrapolation method below), record the temperature of the acid in the cup every 30 seconds for 2–3 minutes.
6. At a known time (say 3 minutes), add the sodium hydroxide all at once, put the lid on, and stir gently with the thermometer.
7. Record the temperature every 30 seconds for a further 5–6 minutes.
8. Repeat the experiment to check reproducibility.

Simple method: highest temperature reached

In the simplest version, record the starting temperature and the highest temperature reached after mixing.

Sample results:

Measurement	Value
Starting temperature of acid and alkali	21.2 °C
Highest temperature after mixing	27.9 °C
Temperature rise, ΔT	6.7 °C

Calculation

Step 1. Energy transferred to the solution

q = m × c × ΔT

• m = mass of solution. Assume both solutions have density 1.00 g/cm³, so 50.0 cm³ has a mass of 50.0 g.
• c = specific heat capacity. Assume it’s the same as water: 4.18 J/g/°C.
• ΔT = 6.7 °C

q = 50.0 × 4.18 × 6.7 = 1400 J = 1.40 kJ

Step 2. Moles of water formed

HCl + NaOH → NaCl + H₂O (1 : 1 : 1 : 1)

Moles HCl = 1.00 × (25.0 ÷ 1000) = 0.0250 mol Moles NaOH = 0.0250 mol Moles H₂O formed = 0.0250 mol

Step 3. Enthalpy change per mole of water

ΔH = −q ÷ moles of water = −1.40 ÷ 0.0250 = −56.0 kJ/mol

The negative sign is added because the reaction released energy (the temperature went up).

This is within about 2% of the accepted value. For more on why the temperature rises and what the sign of ΔH means, see exothermic vs endothermic.

Better method: extrapolation

Heat starts leaking out of the cup as soon as the reaction happens. By the time the thermometer reaches its maximum, some energy has already been lost, so the simple method underestimates ΔT.

To correct for this:

1. Plot temperature (y-axis) against time (x-axis) for all your readings.
2. Draw a line of best fit through the readings before mixing (usually a flat line).
3. Draw a line of best fit through the readings after the maximum, as the solution slowly cools.
4. Extrapolate the cooling line back to the time of mixing.
5. The difference between the two lines at the time of mixing is the corrected ΔT.

This gives the temperature rise that would have occurred if the reaction had been instantaneous and no heat had been lost. Corrected values of ΔT are typically a few tenths of a degree larger than the simple maximum.

Why do different strong acids give the same value?

Try the experiment with nitric acid instead of hydrochloric, or potassium hydroxide instead of sodium hydroxide. The result is almost the same, about −57 kJ/mol.

That’s because strong acids and strong alkalis are completely ionised. Whatever the acid and alkali, the only reaction taking place is H⁺ + OH⁻ → H₂O. The other ions (Na⁺, K⁺, Cl⁻, NO₃⁻) are spectator ions that don’t take part. This was one of the early pieces of evidence for the Arrhenius theory of ions; see Arrhenius acids and bases.

Extension: weak acids

Repeat the experiment with 1.00 mol/dm³ ethanoic acid in place of hydrochloric acid. The result is typically a little less exothermic, around −55 to −56 kJ/mol.

Why? Ethanoic acid is weak, so most of it is un-ionised. As hydroxide removes H⁺ ions, more ethanoic acid molecules ionise to replace them:

CH₃COOH → CH₃COO⁻ + H⁺

Breaking the O–H bond and separating the ions requires a small amount of energy, which slightly reduces the overall energy released. For some weak acids and bases, the difference is larger. For hydrofluoric acid, oddly, neutralisation is more exothermic than for strong acids (around −68 kJ/mol), because the fluoride ion is so strongly hydrated in water.

Extension: sulfuric acid

With 25.0 cm³ of 1.00 mol/dm³ sulfuric acid and 25.0 cm³ of 1.00 mol/dm³ NaOH, the NaOH is the limiting reagent (sulfuric acid is diprotic and in excess). Moles of water formed = 0.0250 mol, so the calculation is the same, but make sure you divide by moles of water formed, not moles of acid. Using 50.0 cm³ of NaOH to fully neutralise 25.0 cm³ of the sulfuric acid would form 0.0500 mol of water.

Sources of error

Error	Effect	Improvement
Heat lost to the surroundings	ΔT too small, ΔH less negative	lid, insulation, extrapolation
Heat absorbed by the cup and thermometer	ΔT slightly too small	use polystyrene; ignore or calculate heat capacity of the calorimeter
Assuming density and specific heat capacity equal to water	small error	measure the density of the mixture
Solutions at different starting temperatures	inaccurate ΔT	use the average starting temperature, or let both equilibrate
Slow mixing	heat loss before maximum reached	add quickly and stir
Thermometer resolution	reading uncertainty	use a ±0.1 °C thermometer or digital probe

Percentage uncertainty

With a thermometer reading to ±0.1 °C, each reading has an uncertainty of ±0.1 °C, so ΔT (from two readings) has ±0.2 °C. For ΔT = 6.7 °C:

Percentage uncertainty = 0.2 ÷ 6.7 × 100 = 3.0%

That uncertainty alone is bigger than the difference between our result (−56.0) and the accepted value (−57), so the experiment agrees with the accepted value within its uncertainty. Using more concentrated solutions gives a larger ΔT and a smaller percentage uncertainty.

Safety

• Wear eye protection throughout. 1 mol/dm³ sodium hydroxide is corrosive to eyes and irritates skin; 1 mol/dm³ hydrochloric acid is an irritant.
• Handle glass thermometers carefully; many schools use digital probes instead.
• The final solution is dilute sodium chloride and can be poured away with plenty of water.

Key takeaways

• The enthalpy of neutralisation is the enthalpy change for forming 1 mol of water from an acid and an alkali.
• For strong acid + strong alkali it’s about −57 kJ/mol, because the reaction is always H⁺ + OH⁻ → H₂O.
• Calculate q = mcΔT, divide by moles of water formed, and add a negative sign for an exothermic reaction.
• Extrapolating the cooling curve corrects for heat loss.
• Weak acids usually give slightly less exothermic values because some energy is used in ionisation.