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

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:

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:

1. The weak species must ionise (an endothermic step).
2. 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:

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.

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

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

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

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.

Step 2: Moles of water formed.

Step 3: Molar enthalpy.

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).

In one sentence

The enthalpy of neutralisation is the heat released per mole of water formed in an acid-base reaction, equal to about $-57.6$ kJ/mol for any strong acid plus strong base combination (whose only net reaction is $H^+ + OH^- \rightarrow H_2O$) and smaller in magnitude when either reactant is weak (because the endothermic ionisation step reduces the net heat); experimentally, mix known amounts in a polystyrene cup, apply $q = mc\Delta T$ to the solution, and divide by $n(H_2O)$ with a negative sign.