Experiment 7Class XII · CBSEChemistry 043

To Determine the Enthalpy of Neutralization of a Strong Acid and a Strong Base

Hydrochloric acid and sodium hydroxide poured together warm up by about six degrees, and the figure that comes out is the same whichever strong acid and strong base you choose.

SectionThermochemistry

In the examContent based experiment · 6

01Aim

To determine the enthalpy of neutralization of a strong acid (hydrochloric acid) by a strong base (sodium hydroxide) using a polystyrene cup calorimeter.

02Requirements

No.	Material	Specification / purpose
1	Polystyrene cup with a lid	The calorimeter; it loses very little heat
2	Beaker (250 ml)	To hold the cup steady
3	Thermometer	Reading to 0.1 °C
4	Stirrer	A glass or plastic rod
5	Measuring cylinders	50 ml and 25 ml
6	Chemical balance	Reading to 0.01 g
7	Hydrochloric acid	1 M, 50 ml
8	Sodium hydroxide	1 M, 50 ml

03Principle

The enthalpy of neutralization is the heat evolved when one gram equivalent of an acid is neutralised by one gram equivalent of a base in dilute aqueous solution.

A strong acid and a strong base are both completely ionised in water. The sodium and chloride ions are spectators and take no part. The only change that actually happens is:

H⁺(aq)+OH⁻(aq)⟶H₂O(l)(1)

ΔH=−57.1kJ mol⁻¹(2)

Because that single reaction is all that occurs, the enthalpy of neutralization of any strong acid by any strong base is almost exactly the same, close to −57 kJ mol⁻¹. That constancy is itself strong evidence for the theory of complete ionisation.

With a weak acid or weak base the value is smaller. Some of the heat released has to be spent ionising the weak electrolyte first. Acetic acid and sodium hydroxide give only about −55.9 kJ mol⁻¹, and the 1.2 kJ difference is the enthalpy of ionisation of acetic acid.

As before, the heat is found from the mass, the specific heat and the temperature rise:

q=mcΔTandΔH=−qn(3)

m = total mass of the mixed solutionn = moles of water formed, equal to the moles of acid neutralised

04The calorimeter

Plate I

A polystyrene cup inside a beaker, with a thermometer and a stirrer through the lid. The whole temperature change is read off in one or two minutes.

05Procedure

Measure 50 ml of 1 M hydrochloric acid into the polystyrene cup.
Measure 50 ml of 1 M sodium hydroxide into a separate beaker.
Let both stand side by side for a few minutes and record the temperature of each. They should agree; if not, take the mean as T₁.
Pour the alkali quickly and completely into the acid in the cup.
Replace the lid at once, stir continuously and watch the thermometer.
Record the highest steady temperature reached, T₂.
Repeat the determination and take the mean.
For comparison, repeat the whole thing with 1 M acetic acid in place of the hydrochloric acid.

06Observations

Quantity	Symbol	Reading
Volume of 1 M HCl	V₁	50 ml
Volume of 1 M NaOH	V₂	50 ml
Mass of the mixture	m	100.0 g
Initial temperature of both	T₁	27.0 °C
Highest temperature after mixing	T₂	33.6 °C
Rise in temperature	ΔT	6.6 °C

The comparison run with a weak acid

Acid used	Base	ΔT (°C)	ΔH (kJ mol⁻¹)
1 M HCl, strong	1 M NaOH	6.6	−57.2
1 M CH₃COOH, weak	1 M NaOH	6.4	−55.4

07Calculation

Heat evolved

q=mcΔT=100.0×4.18×6.6=2758.8J

Moles of water formed

n=1×501000=0.050mol

Enthalpy of neutralization

ΔH=−2758.80.050=−55176J mol⁻¹=−55.2kJ mol⁻¹

The accepted value is −57.1 kJ mol⁻¹. The small shortfall is heat lost to the cup, the thermometer and the air; no simple calorimeter escapes it.

08Result

The enthalpy of neutralization of hydrochloric acid by sodium hydroxide was found to be −55.2 kJ mol⁻¹, close to the accepted value of −57.1 kJ mol⁻¹. The negative sign shows the reaction is exothermic. With acetic acid the value was smaller, because part of the heat is used in ionising the weak acid.

09Precautions

Use a polystyrene cup, not glass. Glass absorbs a good deal of heat itself and the answer comes out low.
Keep the cup covered and stir gently and continuously.
Read the thermometer to 0.1 °C, with the bulb fully immersed and not touching the wall.
Note the highest or lowest steady temperature reached, not the first reading after mixing.
Work quickly; the longer the mixture stands the more heat is exchanged with the room.
Both solutions must be at the same starting temperature.
Pour the alkali in all at once, not in portions.

10Viva voce

Q1Define the enthalpy of neutralization.

ANSThe heat evolved when one gram equivalent of an acid is neutralised by one gram equivalent of a base in dilute solution.

Q2Why is it nearly constant for all strong acids and strong bases?

ANSBecause both are completely ionised, so the only reaction taking place is H⁺ + OH⁻ → H₂O, whatever the salt formed.

Q3Why is the value lower for a weak acid?

ANSSome of the heat released by neutralisation is consumed in ionising the weak acid, so less is left to warm the solution.

Q4What is the accepted value for a strong acid and a strong base?

ANSAbout −57.1 kJ per mole of water formed.

Q5Why is the experimental value always a little low?

ANSHeat is lost to the calorimeter, the thermometer, the stirrer and the surrounding air.

Q6What would happen if you used 2 M acid with 1 M alkali?

ANSThe alkali would be the limiting reagent, so the moles of water formed would be set by the alkali; the enthalpy per mole would come out the same.

Class 12 Chemistry Experiment 7 - To Determine the Enthalpy of Neutralization of a Strong Acid and a Strong Base