NSWChemistrySyllabus dot point

Inquiry Question 2: What happens when acids react?

Investigate the enthalpy of neutralisation, including the calorimetric determination of the heat released when strong and weak acid-base combinations react

A focused answer to the HSC Chemistry Module 6 dot point on the enthalpy of neutralisation. The standard value for strong acid plus strong base, why weak acid neutralisations release less heat, calorimetric procedure with q = mcDeltaT, and worked HSC past exam questions.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-22

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to define enthalpy of neutralisation, perform and analyse a calorimetric experiment (using $q = mc\Delta T$ ), and explain why the magnitude of $\Delta H_{neut}$ is essentially constant for any strong acid plus strong base combination but smaller for combinations involving a weak acid or weak base. The chemistry builds on reactions of acids and strong vs weak acids and bases, and underpins the analysis of titration curves.

The answer

What is the enthalpy of neutralisation?

The enthalpy of neutralisation ( $\Delta H_{neut}$ ) is the heat released per mole of water formed when an acid neutralises a base under standard conditions. The sign is always negative (exothermic), and HSC quotes the accepted value for strong acid plus strong base as about $-57.6$ kJ/mol.

The reason this value is essentially the same for every strong-strong combination is that the only chemistry happening is the net ionic equation:

H^+_{(aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)} \quad \Delta H = -57.6 \text{ kJ/mol}

The spectator ions ( $Na^+$ , $K^+$ , $Cl^-$ , $NO_3^-$ , ...) do not participate energetically. Whether you mix $HCl$ with $NaOH$ , $HNO_3$ with $KOH$ , or $HCl$ with $Ca(OH)_2$ , the heat released per mole of water formed is the same.

Why weak combinations release less heat

If either the acid or the base (or both) is weak, two things happen in sequence:

The weak species must ionise (an endothermic step).
The resulting $H^+$ and $OH^-$ combine to form water (the standard $-57.6$ kJ/mol step).

The measured enthalpy is the sum of these two contributions, so the magnitude is smaller than $-57.6$ kJ/mol. Typical values:

System	$\Delta H_{neut}$ (kJ/mol)
HCl + NaOH (strong + strong)	-57.6
HNO3 + KOH (strong + strong)	-57.6
CH3COOH + NaOH (weak acid + strong base)	-55.2
HCl + NH3 (strong acid + weak base)	-52.2
CH3COOH + NH3 (weak + weak)	-50.4

Read these values as estimates; exact magnitudes vary slightly with concentration and temperature.

Calorimetric procedure

A polystyrene cup (or insulated foam cup) is an excellent simple calorimeter because polystyrene has very low thermal conductivity.

Measure a known volume (for example 50.0 mL) of acid into the cup and record the initial temperature $T_1$ .
Measure an equal volume of base at the same temperature.
Pour the base rapidly into the acid, stir gently with a thermometer, and record the maximum temperature reached, $T_2$ .
Compute $\Delta T = T_2 - T_1$ and apply:

q_{soln} = m \cdot c \cdot \Delta T

assuming the combined solution has the density (1.00 g/mL) and specific heat (4.18 J/g/K) of water.

Divide by the moles of water formed (limiting reagent based on the stoichiometric net ionic equation) and apply the sign:

\Delta H_{neut} = -\frac{q_{soln}}{n(H_2O)}

The negative sign converts "heat gained by the solution" into "heat released by the reaction".

Sources of error

Heat lost to surroundings. The cup is not perfectly insulating; heat is lost to the air and the cup material.
Heat absorbed by the cup and thermometer. A more rigorous calculation includes a calorimeter constant.
Approximating $c$ and $\rho$ as water. Dilute solutions are close, but concentrated solutions differ.
Slow mixing. If the maximum temperature is reached gradually, heat is lost before the peak is recorded. Plot $T$ vs time and extrapolate back to mixing if accuracy matters.
Reading error on the thermometer. A digital probe to 0.1 K improves precision considerably.

To reduce these errors, use a stirred, insulated calorimeter, a graphed temperature-time correction, and equal initial temperatures for the two reactants.

Worked example

A student adds 100.0 mL of 0.500 mol/L $CH_3COOH$ at 21.5 degrees C to 100.0 mL of 0.500 mol/L $NaOH$ at 21.5 degrees C in a polystyrene cup. The peak temperature is 24.8 degrees C. Calculate $\Delta H_{neut}$ and compare it with the strong-strong value.

Step 1: $q$ absorbed by solution.

m = 200.0 \text{ g}, \quad \Delta T = 24.8 - 21.5 = 3.3 \text{ K}

q = (200.0)(4.18)(3.3) = 2759 \text{ J} = 2.76 \text{ kJ}

Step 2: Moles of water formed.

n = 0.1000 \times 0.500 = 0.0500 \text{ mol}

Step 3: Molar enthalpy.

\Delta H_{neut} = -\frac{2.76}{0.0500} = -55.2 \text{ kJ/mol}

Step 4: Comparison. The strong-strong value is $-57.6$ kJ/mol. The weak-acid case is about 2 kJ/mol less exothermic because the ionisation of $CH_3COOH$ absorbs roughly that amount, consistent with the trend.

Common traps

Forgetting the negative sign: $\Delta H$ for an exothermic reaction is negative. The solution gains heat (positive $q$ for the solution), so the reaction releases heat ( $\Delta H < 0$ ).
Using only one volume in $m$: The mass in $q = mc\Delta T$ is the total mass of the combined solution, not just the acid or the base.
Mixing up moles of acid and moles of water: When a diprotic acid like $H_2SO_4$ neutralises a monoprotic base, each mole of acid produces 2 mol of water. Always express $\Delta H$ per mole of water formed.
Quoting $-57.6$ kJ/mol for a weak combination: Weak combinations release less heat. Use the experimental value, not the textbook strong-strong value.
Calling heat loss the only error: Heat loss is the dominant systematic error, but markers reward students who also mention the cup absorbing heat, $c$ approximation, and slow-mixing effects.
Using degrees C in $\Delta T$ vs K: A temperature change in degrees C equals the same change in K. Either unit is fine in $\Delta T$ ; just keep $c$ in matching units (J/g/K).

Examples in context

Example 1. Heat management at Orica Botany Bay sulfuric acid plant. Orica's Botany Bay sulfuric acid facility neutralises waste alkaline streams from adjacent manufacturing with concentrated $H_2SO_4$ . Each mole of water formed liberates 57.6 kJ, so neutralising a 5000 L tank of 1.0 mol L $^{-1}$ NaOH releases approximately $5000 \times 57.6 = 2.88 \times 10^{5}$ kJ. Process engineers must size cooling coils to remove this heat at a rate exceeding the dosing rate, otherwise the mixing vessel temperature exceeds the design limit of 60 degrees C. The same $q = mc\Delta T$ calculation an HSC student does in a polystyrene cup with 100 mL of acid scales by five orders of magnitude to determine industrial cooling requirements.

Example 2. NSW HSC depth study calorimetry of acetic acid plus sodium hydroxide. Stage 6 students mix 50.0 mL of 1.00 mol L $^{-1}$ acetic acid with 50.0 mL of 1.00 mol L $^{-1}$ NaOH in a polystyrene cup, measuring an initial temperature of 21.0 degrees C and a maximum of 27.5 degrees C. Applying $q = mc\Delta T$ with $m = 100$ g and $c = 4.18$ J g $^{-1}$ K $^{-1}$ gives $q = 100 \times 4.18 \times 6.5 = 2717$ J. Dividing by 0.050 mol of water formed yields $\Delta H = -54.3$ kJ mol $^{-1}$ . The value is less exothermic than the strong-strong figure of $-57.6$ kJ mol $^{-1}$ because the endothermic ionisation of acetic acid consumes some heat.

Try this

Q1. Define the enthalpy of neutralisation and state its standard value for a strong acid and strong base. [2 marks]

Cue. Heat released per mole of water formed in an acid-base reaction; value approximately $-57.6$ kJ mol $^{-1}$ .

Q2. 50.0 mL of 0.500 mol L $^{-1}$ HCl is mixed with 50.0 mL of 0.500 mol L $^{-1}$ NaOH. The temperature rises by 3.40 degrees C. Calculate $\Delta H$ of neutralisation. [3 marks]

Cue. $q = 100 \times 4.18 \times 3.40 = 1421$ J; moles of water formed = 0.025; $\Delta H = -1421 / 0.025 = -56.8$ kJ mol $^{-1}$ .

Q3. Account for the following observations: (a) the enthalpy of neutralisation for HF + NaOH is $-68$ kJ mol $^{-1}$ , more exothermic than the strong-strong value. (b) The enthalpy for $CH_3COOH$ + NaOH is $-55$ kJ mol $^{-1}$ , less exothermic than the strong-strong value. (c) State one source of error in a polystyrene-cup calorimeter. [2+2+1 marks]

Cue. (a) HF ionisation is exothermic due to favourable hydration of $F^-$ . (b) Acetic acid ionisation is endothermic, reducing net heat release. (c) Heat loss to surroundings, or neglecting cup heat capacity.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 HSC5 marksA student mixes 50.0 mL of 1.00 mol/L HCl at 22.0 degrees C with 50.0 mL of 1.00 mol/L NaOH at 22.0 degrees C in a polystyrene cup calorimeter. The maximum temperature reached is 28.7 degrees C. Calculate the molar enthalpy of neutralisation, stating any assumptions. Compare the result to the accepted value of -57.6 kJ/mol and explain any discrepancy.

Show worked answer →

A 5 mark answer needs the heat calculation, the moles of water formed, the molar enthalpy with sign, the accepted comparison, and a stated source of error.

Step 1: Heat absorbed by the solution. Assume the combined solution has the density and specific heat of water ( $\rho = 1.00$ g/mL, $c = 4.18$ J/g/K). Total mass $m = 100.0$ g. Temperature change $\Delta T = 28.7 - 22.0 = 6.7$ K.

q = mc\Delta T = (100.0)(4.18)(6.7) = 2801 \text{ J} = 2.80 \text{ kJ}

Step 2: Moles of water formed. $n(HCl) = n(NaOH) = 0.0500 \times 1.00 = 0.0500$ mol. Limiting reagent: stoichiometric, so $n(H_2O) = 0.0500$ mol.

Step 3: Molar enthalpy.

\Delta H_{neut} = -\frac{q}{n} = -\frac{2.80}{0.0500} = -56.0 \text{ kJ/mol}

The sign is negative because the reaction is exothermic (the solution gained heat from the reaction).

Step 4: Comparison. The measured value (-56.0 kJ/mol) is slightly less exothermic than the accepted -57.6 kJ/mol. The discrepancy (about 3 percent) is due to heat loss from the polystyrene cup to the surroundings and to the cup itself, which a perfect calorimeter would prevent.

Markers reward (1) correct $q = mc\Delta T$ with stated assumptions, (2) correct moles and ratio, (3) the negative sign on $\Delta H$ , (4) numerical comparison, (5) a sensible source of error.

2019 HSC3 marksExplain why the enthalpy of neutralisation of ethanoic acid with sodium hydroxide is less exothermic (about -55 kJ/mol) than the enthalpy of neutralisation of hydrochloric acid with sodium hydroxide (-57.6 kJ/mol).

Show worked answer →

The strong acid plus strong base reaction has the simple net ionic equation:

H^+_{(aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)}

All of the energy released goes into forming water from already-ionised reactants.

With ethanoic acid (weak), the actual reaction is:

CH_3COOH_{(aq)} + OH^-_{(aq)} \rightarrow CH_3COO^-_{(aq)} + H_2O_{(l)}

Before the proton can combine with $OH^-$ , the $CH_3COOH$ must first ionise (break the $O-H$ bond), which is endothermic. The net release is therefore the strong-strong value minus the ionisation enthalpy of the weak acid, giving a smaller magnitude.

Markers reward (1) the strong-strong net ionic equation, (2) recognising that weak acid ionisation costs energy, (3) the conclusion that the magnitude of $\Delta H_{neut}$ is smaller for weak combinations.