5.1 - Determination of the Heat of Reaction of Hydrochloric Acid with Sodium Hydroxide Revision Notes for Leaving Cert Chemistry

5.1 - Determination of the Heat of Reaction of Hydrochloric Acid with Sodium Hydroxide

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Experiment Summary

In this experiment, the heat of neutralisation for the reaction between hydrochloric acid ( $HCl$ ) and sodium hydroxide ( $NaOH$ ) is determined.

Equal volumes of 1 M solutions of $HCl$ and $NaOH$ are mixed, and the temperature rise is measured.

The heat liberated (exothermic reaction) is calculated using the formula q = mc∆T

The heat of the reaction is expressed per mole of acid neutralised.

The reaction is:

\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}

Materials and Apparatus Required

Chemicals

1 M hydrochloric acid
1 M sodium hydroxide

Apparatus

Polystyrene cups with lids
Graduated cylinders (100 cm³)
Two thermometers (reading to 0.1°C)
Stopwatch
Stirring rod

Safety Precautions

Wear safety glasses at all times.
Hydrochloric acid and sodium hydroxide are corrosive. Avoid skin contact and use gloves if necessary.
Dispose of the reaction mixture by diluting it with excess water before flushing it down the drain.

Method

Measure 50 cm³ of 1 M hydrochloric acid using a graduated cylinder and pour it into a polystyrene cup.
Measure 50 cm³ of 1 M sodium hydroxide using a second graduated cylinder and pour it into another polystyrene cup.
Record the temperature of both the $HCl$ and $NaOH$ solutions using separate thermometers.
When both solutions are at the same temperature, quickly add the $NaOH$ solution to the $HCl$ solution while stirring continuously. Be careful to avoid splashing.
Cover the cup with a lid and continue stirring.
Record the highest temperature reached after mixing.
Calculate the temperature rise by subtracting the initial temperature from the final temperature.

Results

Measurement	Value
Initial temperature of HCl solution	22°C
Initial temperature of NaOH solution	22°C
Initial temperature of NaOH solution	22°C
Highest temperature after mixing	28°C
Temperature rise	6°C
Number of moles of HCl used	0.05 mol
Number of moles of NaOH used	0.05 mol

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Sample Calculation

The amount of heat liberated is calculated using the formula:

q = mc\Delta T

Where:

$m = :highlight[100 \, \text{g}]$ (mass of the combined solution, assuming 100 cm³ of solution has a mass of 100 g)
$c = :highlight[4.2 \, \text{J/g°C}]$ (specific heat capacity of water)
$\Delta T = :highlight[6 \, \text{°C}]$ (temperature rise)

q = 100 \times 4.2 \times 6 = :highlight[2520 \, \text{J}] = :highlight[2.52 \, \text{kJ}]

The heat of reaction per mole of $HCl$ is:

\frac{2.52 \, \text{kJ}}{0.05 \, \text{mol}} = :highlight[-50.4 \, \text{kJ/mol}]

Example Questions with Answers

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Q1: What precautions are taken to minimise heat loss during the experiment?

A polystyrene cup is used as it is an excellent insulator, and a lid is placed on the cup to reduce heat loss to the surroundings.

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Q2: Why might the experimental value for the heat of reaction differ from the literature value of -57 kJ/mol?

The difference could be due to inaccuracies in measuring volumes with graduated cylinders, heat loss to the surroundings, or assumptions such as the heat capacity of the solution being equal to that of water.

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Q3: Why would the use of nitric acid or potassium hydroxide yield similar results?

Both $HCl$ and $HNO₃$ are strong acids, and both $NaOH$ and $KOH$ are strong bases.

The heat of neutralisation mainly comes from the reaction between hydrogen ions ( $H⁺$ ) and hydroxide ions ( $OH⁻$ ), which is the same for both sets of acids and bases.

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Q4: What is the significance of the negative sign in the heat of reaction?

The negative sign indicates that the reaction is exothermic, meaning it releases heat to the surroundings.

5.1 - Determination of the Heat of Reaction of Hydrochloric Acid with Sodium Hydroxide (Leaving Cert Chemistry): Revision Notes