IB IA: Determination of Heat of Neutralization

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NAME: Nur Amira Rozali

SUBJECT: Chemistry HL

ASSESSMENT CRITERIA: DC, DP, EV

INTERNAL ASSESSMENT

CHEMISTRY HIGHER LEVEL

PRACTICAL 16: DETERMINATION OF HEAT OF NEUTRALISATION

Data collection:

TABLE 1

Volume of acid used for each experiment = 50cm3

Volume of base used for each experiment = 50cm3

Specific heat capacity of water = 4.18 kg-1K-1

Concentration of each reactant = 1 M

Data processing:

TABLE 2

Assumption: The acid and base solutions, each has the same density and specific heat capacity as water, so the mass of “water”, m in this case = the total amount of solutions used = 50cm3 + 50cm3 = 100cm3

Specific heat capacity of water, c = 4.18 kJ kg-1 K-1

Temperature change, ө = Temperature difference, T3

Since the concentration of the each of the acids and bases is 1 M, which is 1 moldm-3.

Therefore, number of moles of acids or bases used for each experiment

                 = 1 x volume of each acid or base used

Experiment 1:

Equation for the experiment:

HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)

Heat evolved = mcӨ

                      = 0.1 x 4.18 x 5.5

                  = 2.299 kJ

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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

                                                       = -45.98kJmol-1

                                  ...

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