-
The procedure was the same as HCL except CH3COOH was used.
Results and Calculations:
(I) Titration of HCL
(II) Titration of CH3COOH
Calculations:
3.
Concentration of HCl:
(1)(50) = m (25)
m = 2M
Concentration of CH3COOH:
(1)(50) = m (5)
m = 10M
4.
By E=mcΔT
For HCL:
E = [(100)/1000](4180)[(32) – 25]
E = 2926 J dm-1 K-1
∴ Heat energy released = 2926 J dm-1 K-1
For CH3COOH:
E = [(100)/1000](4180)[(24.5) – 23]
E = 627 J dm-1 K-1
∴ Heat energy released = 627 J dm-1 K-1
5.
Standard enthaply change of neutralization:
ΔHf o = − E/no.of moles of water formed
By the equation:
2NaOH + 2HCl ⇔ 2NaCl + H2O
As NaOH is limited reagent,
∴ no. of mole of H2O = (1)(50/1000) = 0.05 mol
ΔHoneu = 2926/0.05
ΔHoneu = − 58.52 kJ
(Calculations)
CH3COOH + NaOH ⇔ CH3COOH-Na+ + H2O
As NaOH is limited reagent,
∴ no. of mole of H2O = (1)(50/1000) = 0.05 mol
ΔHoneu = 627/0.05
ΔHoneu = − 12.54 kJ
∴The standard enthaply change of neutralization of using HCl and CH3COOH are −58.52kJ mol-1 and −12.54kJ mol-1 respectively.
Questions:
-
[H2O] is a constant when it acts as a solvent. The equilibrium will shifts to the left or shift to the right and [H2O] remanins unchanged. As the [acid] and [base] is very low though it is strong, [H2O] remanins unchanged.
- (I) Some heat is lost to the surrounding.
(II) We assume that the heat capacity of the cup is zero.
- As they are not completely ionized in water, the equilibrium shifts to the right. Hence, less portion of weak acids/bases has reacted with the bases/acid. Thus, it’s heat of neutralization is less negative than the strong ones.
Discussion:
We must stir the solution throughly after each addition of the acid, so as to measure a temperature with higher accuracy. Moreover, we must replace the lid before stirring which can minimize the heat lost to the surrounding. We should stir gently or the plactic lid will break as a result of our volience.
After all, after the experiment we will plot a graph to determine the highest temperature, it is unwise to add the last drop of acid from the burette drop by drop, as it is time consuming and we cannot get the most accurute temperature after adding. We should add the acid with care and about 5.0cm3 each time.
(discussion)
We should try hard to make sure whether the rise in temperature is logical, if the temperature have no willing to increase for a degree after a few times of the addition of the strong acid. We should once report it to the teacher to find if the given reagent has a different concentration from given.
As sodium hydroxide is very corrosive, we must use the pipette filler supplied, and we must wear lab coat and safety spectacles in the laboratory.
The last drop of the pipette should be dipped into the solution.
We should fill the burette with a filter funnel and under eye sight for safety reasons. The filter funnel should be removed after filling.
The burette should be placed upright to the vertical.
Conclusion:
The value from experimential result of neutralizatin using HCl and CH3COOH are −58.52kJ mol-1 and −12.54kJ mol-1 respectively. The one using HCl is more negative than −57.6 kJ mol-1 , as the maximum temperture attained may be higher estimated.
