Method 2:
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25.00 cm3 of the given alkali solution was pipette into a dry polystyrene cup.
- The content in the polystyrene cup was equilibrated for a few minutes and the temperature of the solution was measured and recorded.
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The acid was poured into burette to the mark’ 0 cm3’ through a filter funnel.
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2.00 cm3 of the acid was added from the burette into the polystyrene cup containing 25.00cm3 alkali solution. The mixture was swirled gently and the maximum temperature was recoded.
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Another 2.00 cm3 acid was added into the polystyrene cup after the maximum temperature had been reached, and the maximum temperature was recorded again.
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Step 5 was repeated until a total volume of 40 cm3 of acid was added.
- A graph was plotted and (i) the concentration of alkali solution and (ii) the enthalpy change of neutralization were determined.
Results:
Date of experiment: 23rd October, 2004
Temperature of the laboratory: 23.6°C
Alkali used in the experiment: NH3(aq)
Acid used in the experiment: HCl(aq)
Method 1
Table1: The temperature change of mixing HCl and NH3
From the results obtained, the maximum change in temperature of neutralization is 5.30°C by mixing 15.00cm3 of HCl(aq) and 15.00cm3 of NH3(aq).
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According to the above data, the change in temperature versus the volume of NH3 (aq) added was plotted on Graph(I).
Graph (I).
- Calculate the molarity of the alkali solution from the data in Method 1.
No. of moles of hydrochloric acid in 15cm3 of 1.0 mol/dm3 solution
= the molarity of acid x the volume of acid added
= 1.0 moldm-3 x (15/ 1000) cm3
= 0.015 mol
Since
NH3 + H2O → NH4+ + OH-
OH- + HCl + NH4+ → NH4Cl + H2O
NH3+ + HCl → NH4Cl
No. of mole of HCl = No. of mole of NH3
=0.015 mol
, and when 15.2 cm3 of NH3 was added, it gave the greatest change of temperature
Therefore, the molarity of the alkaline solution is
= 0.015 mol / (15.2 cm3/ 1000)
= 0.987 mol/dm3
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Determine the enthalpy change of neutralization (KJmol-1) in this case.
(Given: Density of water = 1g/cm3, specific heat capacity of water = 4.184 Jg-1K-1)
Total volume of solution = (15.20 + 14.8) cm3.
Heat produced = specific heat capacity x mass of solution x temperature
= - 4.184 Jg-1K-1 x (30 cm3) x 5.4K
= -0.6771KJ
Enthalpy change of neutralization (ΔH) = -0.6771KJ / 0.015 mol
= - 45.19 KJmol-1
Therefore, the reaction is exothermic and the enthalpy change of neutralization is - 45.19 KJmol-1.
Method 2
- Using the above data in Method 2, the temperature against volume of acid (aq) added was plotted on Graph (II).
Graph (II)
- Calculated the molarity of the alkali solution from the data in Method 2.
No. of moles of hydrochloric acid in 25.5cm3 of 1.0 mol/dm3 solution
= the molarity of acid x the volume of acid added
= 1.0 moldm-3 x (25.5/ 1000) cm3
= 0.0255 mol
Since
NH3++ HCl→ NH4Cl
No. of mole of HCl = the no. of mole of NH3
= 0.0255 mol
, and when 25cm3 of NH3 was added, it gave the greatest change of temperature
Therefore, the molarity of the alkaline solution is
= 0.0255mol / (25/ 1000) cm3
= 1.02mol/dm3
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Determine the enthalpy change of neutralization (KJmol-1) in this case.
(Given: Density of water = 1g/cm3, specific heat capacity of water = 4.184 Jg-1K-1)
Total volume of solution = (25+ 25.5) cm3 = 50.5cm3, and
The temperature changed = (28.0 – 22.5) °C = 5.5°C
Heat produced = specific heat capacity x mass of solution x temperature
= - 4.184 Jg-1K-1 x (50.5 cm3) x5.5K
= -1.161KJ
Enthalpy change of neutralization (ΔH) = -1.161KJ / 0.0255mol
= - 45.53 KJmol-1
Therefore, the reaction is exothermic and the enthalpy change of neutralization is –45.53KJmol-1.
Further question:
- What assumptions have been made when calculating the enthalpy change of neutralization in this experiment?
Ans: -no heat loss to the surroundings
-the specific heat capacity of the solution involved is equal to that of water
-the density of the solutions is equal to that of water
-the specific of the polystyrene cup is neglected
-the temperature and the pressure in the lab are kept constant as the standard condition (25°C, 1atm)
- Which method, 1 or 2, provides a more accurate result in the determination of (a) the concentrations of alkali solution; and (b) the enthalpy change of neutralization? Explain.
Ans: As for (a), method 2 is more accurate rather than method 1. Since we obtained more data and it has a larger variation in the experiment, so more data are used to calculate the concentration of alkaline and the results become more accurate.
As for (b), method 1 is more accurate than method 2. Because the experiment we carried out in method 1 is taken a shorter time than method 2. Also, the titration in method 1 is done by separately instead of continuously. So, the heat loss in method 1 is less than method 2. Therefore, the more accurate results can be obtained.
- Combinations of acid-alkali neutralization result from other classmates.
The combination of acid- alkali neutralization
- According to the results in the above table, comment on the relationship between the enthalpy change of neutralization and the combination of acid and alkali used.
Ans: From the results obtained, it shows that the value of the enthalpy change of neutralization become less exothermic when it goes from upward to downward in the table. It is because when a strong acid and a strong base react together, the ions are completely ionized, there is no energy consumed for bond breaking (ionized the molecules). Similarly, when a weak base and weak acid react together, it consumes a large amount of energy that is used for bond breaking (ionized the molecules), and this process is endothermic causing less energy released to the system. Therefore, the values of the enthalpy change of neutralization become less exothermic (smaller value) from strong acid-strong base to weak acid-weak base.
Discussion:
From the results obtained, it shows that when an acid and alkaline mix together, an exothermic reaction will be involved.
Yet, the value of standard enthalpy change depends on the strength of the acid and alkaline. As a strong acid reacts with the strong base, the enthalpy value is quite large. Since the molecules are completely ionized and it does not require too much energy for ionize the molecules. So, more energy will be released.
On the other hand, as a strong acid react with weak base or a weak acid react with strong base, the reaction consume more energy than “strong acid-strong base” reaction. Take ammonia as an example, because ammonia only partially ionized and it provides only some of the free OH- ions for neutralization, and so more energy is required for ionized the molecules, so less energy is released to the system.
Similarly, the weak acid with a weak base requires a largest amount of energy for ionization. Since both of the weak acid and base is required energy to carry out ionization.
Therefore, the value we obtained is smaller than that of the combination of strong acid and strong base but larger than that of combination of weak acid and weak base. As some of the energy is used in dissociating the non-ionized weak base molecules so that neutralization can proceed.
There are some possible errors in the experiment. First, there are heat loss to the surroundings due to the evaporation and convection. Second, the heat capacity of vacuum flask or expanded polystyrene cup is neglected. Third, the specific heat capacity of the product solution is assumed to be the same as that of water. Forth, the ordinary thermometer is not precise enough. Fifth, human errors were involved during in the investigation of temperature value.
There are number of improvement for the experiment. For instance, we can find the heat capacity of the vacuum flask, expanded polystyrene cup or the solution. We can also replace a thermometer with a more precise one. Taking the reading by the same person to achieve the results in order to minimize the errors
Conclusion:
From the results obtained, the value of enthalpy change of neutralization with hydrochloric acid and ammonia is in a medium range when compare to the combination with another strong acid-strong base and weak acid-weak base. In method1, the molarity of ammonia is 0.987mol/dm3 and the value of enthalpy change of neutralization is - 45.19 KJmol-1. In method 2, the molarity of ammonia is 1.02mol/dm3and the value of enthalpy change of neutralization is - 45.53 KJmol-1
References:
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W.H.Skoog, (1996) Fundamentals of Analytical Chemistry, 7th ed., P.17
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P.W.Atkins (1996) The Elements of Physical Chemistry, 2nd ed., P.49-70
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