Number of moles of acids used = 1 x 0.05 = 0.05 moles
From the equation, 1 mole of HNO3 reacts with 1 mole of NaOH and gives 1 mole of water.
Therefore, 0.05 moles of HNO3 reacts with 0.05 moles of NaOH and gives 0.05 moles of water.
Therefore, heat of neutralization = 2.299 ÷ 0.05
= 45.98kJmol1
= 46.0kJmol1 (3 s.f.)
Experiment 2:
Equation for the experiment:
HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)
Heat evolved = mcӨ
= 0.1 x 4.18 x 5.75
= 2.4035 kJ
Number of moles of acids used = 1 x 0.05 = 0.05 moles
From the equation, 1 mole of HNO3 reacts with 1 mole of KOH and gives 1 mole of water.
Therefore, 0.05 moles of HNO3 reacts with 0.05 moles of KOH and gives 0.05 moles of water.
Therefore, heat of neutralization = 2.4035 ÷ 0.05
= 48.07kJmol1
= 48.1kJmol1 (3 s.f.)
Experiment 3:
Equation for the experiment:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Heat evolved = mcӨ
= 0.1 x 4.18 x 6.00
= 2.508 kJ
Number of moles of acids used = 1 x 0.05 = 0.05 moles
From the equation, 1 mole of HCl reacts with 1 mole of NaOH and gives 1 mole of water.
Therefore, 0.05 moles of HCl reacts with 0.05 moles of NaOH and gives 0.05 moles of water.
Therefore, heat of neutralization = 2.508 ÷ 0.05
= 50.16kJmol1
= 50.2kJmol1 (3 s.f.)
Experiment 4:
Equation for the experiment:
HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)
Heat evolved = mcӨ
= 0.1 x 4.18 x 6.00
= 2.508 kJ
Number of moles of acids used = 1 x 0.05 = 0.05 moles
From the equation, 1 mole of HCl reacts with 1 mole of KOH and gives 1 mole of water.
Therefore, 0.05 moles of HCl reacts with 0.05 moles of KOH and gives 0.05 moles of water.
Therefore, heat of neutralization = 2.508 ÷ 0.05
= 50.16kJmol1
= 50.2kJmol1 (3 s.f.)
Experiment 5:
Equation for the experiment:
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
Heat evolved = mcӨ
= 0.1 x 4.18 x 12.00
= 5.016 kJ
Number of moles of acids used = 1 x 0.05 = 0.05 moles
From the equation, 1 mole of H2SO4 reacts with 2 moles of NaOH and gives 2 moles of water.
Therefore, 0.05 moles of H2SO4 reacts with 0.1 moles of NaOH and gives 0.1 moles of water.
Therefore, heat of neutralization = 5.016 ÷ 0.1
= 50.16kJmol1
= 50.2kJmol1 (3 s.f.)
Experiment 6:
Equation for the experiment:
H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + H2O(l)
Heat evolved = mcӨ
= 0.1 x 4.18 x 12.50
= 5.225 kJ
Number of moles of acids used = 1 x 0.05 = 0.05 moles
From the equation, 1 mole of H2SO4 reacts with 2 moles of KOH and gives 2 moles of water.
Therefore, 0.05 moles of H2SO4 reacts with 0.1 moles of KOH and gives 0.1 moles of water.
Therefore, heat of neutralization = 5.225 ÷ 0.1
= 52.25kJmol1
= 52.3kJmol1 (3 s.f.)
Evaluation:

Standard enthalpy change of neutralization, ∆Hfө, is the energy evolved when 1 mole of water is formed when 1 mole of hydrogen ions, H+ reacts with 1 mole of hydroxide ions, OH under standard conditions, which are 298K and 1 atm.
 The table below shows examples of some neutralization reactions and the theoretical value of each reaction’s standard enthalpy change of neutralization.

From the above table, we can conclude that standard enthalpy change of neutralization for reactions between strong acids and strong bases is theoretically 57kJmol1.
 However, all the 5 experiments have shown a significant deviation in terms of the value for the standard enthalpy change for neutralization. The experimental value for the enthalpy change for neutralization is less than the theoretical value of standard enthalpy change for neutralization.
 This is due to the fact that some heat is lost to the surrounding and the heat absorbed by the polystyrene cup is not included in the calculation.

Sulphuric acid is a dibasic acid that produces 2 moles of hydrogen ions upon dissociation in water. Thus, when the ions are reacted with sodium hydroxide or potassium hydroxide, 2 moles of water molecules are formed. Thus, heat released in such neutralization reactions is twice as much as the heat released by neutralization reactions between monobasic acid, such as NHO3 and HCl with KOH or NaOH.

However, enthalpy of neutralization is still approximately 57kJmol1 because enthalpy change of neutralization is defined in term of the formation of 1 mole of water molecules, not 2 moles of water molecules.

The major weakness of this experiment is the lost of significant amount of heat to the surrounding. This causes deviation of the experimental value of enthalpy change of neutralization from the theoretical value.
 This is unavoidable due to the opening of the polystyrene cup, the pouring of acid into the base solution and the absorption of heat by the polystyrene.
 Besides, the mixture of acid and base solutions may not be well stirred.

Solutions to improve the accuracy of the experiment include:
 Initial temperatures of the solutions are to be taken after several minutes to make sure that the solutions have achieved a consistent temperature.
 Acids should be poured quickly and carefully into the bases solutions to reduce the heat lost.
 Any spilling of the chemical substances should be avoided.
 The mixture of solutions should be stirred throughout the experiment to make sure that the temperature is always consistent.
 The reading of the temperature should be observed from time to time so that a maximum temperature can be obtained.