Concentration of diluted bench HCl
Moles of NaOH in 10cm³ = V × C = 10 ×0.2 = 0.002 moles
1000 1000
Equation NaOH + HCl → NaCl + H2O
Moles 1 1
Vol/cm³ 10 10.4 ← This figure is the average titre gained from Stage One.
Conc. g/dm³ 8 x
From the equation 1 mole of NaOH reacts with 1 mole of HCl to give 1 mole of NaCl and H2O. However, there are only 0.002 moles of NaOH and therefore there must be 0.002 moles of HCl. The concentration of HCl can be worked out the following equation.
Concentration = Moles × 1000 = 0.002 ×1000 = 0.19 mol/dm³
Volume 10.4
Concentration of the bench HCl
The dilute bench HCl is diluted by a factor of ten and its concentration was found to be 0.19mol/dm³. The bench HCl should be ten times more concentrated.
Bench HCl (dilute) = 0.19 mol/dm³
Bench HCl = 0.19 × 10 = 1.9 mol/dm³
Therefore bench HCl has a concentration of 1.9 mol/dm³.
Calculations for Stage Two
In this stage the heat of neutralisation needs to be worked. Firstly, a graph needs to be plotted with the results from stage two in order to work out the maximum temperature rise. Refer to graph 1.
Heat of Neutralisation
Heat of neutralisation for this experiment can be represented by the following ionic equation:
H3O+ (aq) + OH- (aq) → 2H2O (l) (Na+ (aq) and Cl- (aq) are spectator ions)
The equation for heat of neutralisation is as follows:
Q = M × SHC × ΔT
For this experiment an assumption is made that the specific heat capacities of NaOH (aq) and HCl (aq) are the same as that of water, which is 4.2 J/g ºC.
The temperature of NaOH and HCl was 16ºC at room temperature. When HCl was added neutralisation took place – this is an exothermic reaction which produced a maximum temperature of 29.3ºC (which is shown on graph 1).
Heat of neutralisation is worked out by adding the heat received by the solution to the heat received by the polystyrene cup. To simplify the calculations I am assuming that polystyrene is an insulator and it only takes a very small amount of the heat of neutralisation.
Calculations to work out heat of neutralisation
Heat from neutralisation = Heat received by water
Q = M × SHC × ΔT
Q = 20 × 4.2 × 13.3
Q = 1111.88 J/mole
Q = 1.11 kJ/mole
The moles of water formed is 0.02 which can be worked out by referring to the word equation. For stage two the bench acid was used. It was worked out that 10cm3 of NaOH consists of 0.002 moles for the diluted bench HCl (this is diluted by a factor of ten). Therefore the bench acid on its own must consist of 0.02 moles. By ratio the equation shows that 0.02 moles of NaOH reacts with 0.02 moles of HCl to form 0.02 moles of water.
Therefore the heat of neutralisation per mole
= Q = 1117.2 = 55860 = -55.86 kJ/mol
moles 0.02
Evaluation
The experiment went according to plan and there were no anomalous readings in stage two. The aim of the experiment was completed successfully. The heat of neutralisation (exothermic) in my experiment was -55.86 kJ/mol which was very close to the actual reading of -57.3 kJ/mol. My result was within an accuracy of 2.51%. This loss in accuracy may have been due to heat losses through convection, conduction and radiation. This can be minimised by using a vacuum flask which is shown below:
The experiment was also simplified because the heat received by the polystyrene beaker was assumed to be negligible. The experiment could have been modified so that the heat received by the polystyrene beaker was also taken account of. This would have produced an accurate result for the heat of neutralisation.
In order to investigate this experiment further I would try different acids (sulphuric acid) and alkalis (sodium chloride) in order to prove that heat of neutralisation works for any strong acid or alkali.
