Standard enthalpy of neutralisation ΔH^θ_{neutralisation} (NaOH/HCl) is the heat energy released when 1 mole of hydrochloric acid solution and 1 mole of sodium hydroxide solution completely neutralise each other at 298K/101 kPa (though pressure effects are insignificant for reactions only involving liquids, solutions or solids)

HCl(aq)  +  NaOH(aq)  ===>  NaCl(aq)  +  H₂O(l)

Ionic equation: H⁺(aq)  +  OH^-(aq)  ===>  H₂O(l)

Remember chloride and sodium are spectator ions in the context of a homogeneous neutralisation.

 

Method (for experimental details see method 1.3a1)

You can do this experiment by mixing equal volumes of equimolar concentrations of dilute hydrochloric acid and dilute sodium hydroxide. e.g. 25 cm³ of each in the polystyrene calorimeter as previously described.

Make sure the acid and alkali solution are at the same laboratory temperature.

Read the temperature every 30 seconds of one of the solutions for a few minutes before adding the 2nd solution.

After the mixing, continue to read the temperature every 30 seconds for a few more minutes.

 

Data processing - graph

Typical temperature-time graph shape for an exothermic experiment

A graph is drawn of temperature versus time and the true temperature rise estimated by extrapolation as shown by the purple lines on the graph.

Suppose after mixing, via accurate pipettes, 25.0 cm³ of 1.0 mol dm^–3 hydrochloric acid and 25.0 of 1.0 mol dm^–3, sodium hydroxide solutions the extrapolated temperature rise from a graph like the one above, is 7.1^oC, calculate the enthalpy of neutralisation for the reaction.

This is obtained using the extrapolated purple lines to correct as much as possible for heat energy loss as the calorimeter and contents start cooling down as soon as the hydrochloric acid reacts with the sodium hydroxide!

 

Calculation.

(i) Using the SHC for pure water and the total mass is effectively 50 g (actually 50 cm³ of NaCl solution).

q = m c ΔT  (m in g, c = 4.18 Jg^-1K^-1, ΔT temperature rise in ^oC)

=  50 x 4.18 x 7.1 = 1483.9 J, 1.4839 kJ

From the equation: mol HCl = mol NaOH = 1.0 x 25/1000 = 0.025 mol

Therefore scaling up to 1 mol gives a numerical enthalpy change of 1.4839 x 1/0.025 = 59.4 kJ

Since the temperature rose indicating an exothermic reaction, the enthalpy of neutralisation is ..

ΔH_{neutralisation}(HCl + NaOH) = –59.4 kJ mol^–1  (only accurate to 3 sf)

 

(ii) Using the SHC of 0.50 molar sodium chloride solution, which is 4.03 J g^{–1 o}C^–1

(see the discussion on the specific heat capacity of salt solutions)

By mixing 1 molar solutions of HCl and NaOH, you actually produce a 0.5 molar solution of sodium chloride, for which, we actually have the true specific heat capacity!

q = m c ΔT

=  50 x 4.03 x 7.1 = 1430.65 J, 1.43065 kJ

From the equation: mol HCl = mol NaOH = 1.0 x 25/1000 = 0.025 mol

Therefore scaling up to 1 mol gives a numerical enthalpy change of 1.43065 x 1/0.025 = 57.2 kJ

Since the temperature rose indicating an exothermic reaction, the enthalpy of neutralisation is ..

ΔH_{neutralisation}(HCl + NaOH) = –57.2 kJ mol^–1  (only accurate to 3 sf)

 

Comments

The data book value is –57.1 kJ mol^–1

% error = 100 x  (experimental value – accepted value) / accepted value

Calculation (i) gives an error of 4.0% and calculation (ii) gives an error of 0.2%.

So you can see that by using the wrong specific heat capacity you incur a 4.0% error due to using the higher specific heat capacity of pure water.

thermochemistry of determining the enthalpy of neutralisation of hydrochloric acid & sodium hydroxide solution, method, apparatus, data, calculation, plastic cup calorimeter, thermochemistry for AQA, Edexcel, OCR, Salters, CIE, WJEC Eduqas & CCEA A-level chemistry exam students, US grades 11-12 K12 AP Honors chemistry courses