Experiment 11

Objective

To determine the enthalpy of neutralization for the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH), both strong electrolytes.

Introduction

In an acid-base neutralization reaction, 1 mole of hydrogen ions (H+) from the acid combines with hydroxide ions (OH−) from the base to form water. This reaction releases heat, known as the enthalpy of neutralization, and is exergonic (releases energy). For strong acid-strong base reactions, the enthalpy of neutralization is consistent across different combinations.

The neutralization reaction:

HCl + NaOH → NaCl + H2​O

The enthalpy of neutralization is calculated using:

ΔH = 4.2 × ΔT × 10KJ / mol

Materials

Chemicals

●      2N Hydrochloric acid (HCl)

●      2N Sodium hydroxide (NaOH)

Apparatus

●      250 mL beaker

●      Glass rod

●      Thermometer

●      Stand

●      Balance

Reagent Preparation

●      2N solutions of HCl and NaOH

Procedure

1.    Preparation

●      Take 50 mL of 2N HCl and pour it into a 250 mL beaker.

2.    Temperature Measurement Setup

●      Place the thermometer in the beaker using the stand to suspend it, ensuring the bulb is fully immersed in the solution.

3.    Addition of NaOH

●      Add 50 mL of 2N NaOH to the beaker containing HCl.

4.    Temperature Observation

●      Record the initial temperature before mixing (24°C).

5.    Calculation of Enthalpy of Neutralization

●      Calculate the change in temperature, ΔT = 37 – 24 = 13∘C.

●      Using the formula:

ΔH = 4.2 × ΔT × 10 = 4.2 × 13 × 10 = −54.6kJ / mol

Observations

●      Initial Temperature: 24°C

●      Final Temperature: 37°C

●      Calculated Enthalpy of Neutralization: -54.6 kJ/mol

Questions

1.    What is an acid-base neutralization?

2.    What is enthalpy of neutralization?

3.    How is enthalpy of neutralization determined?