Lab Report. Objectives 1. To determine the enthalpy of neutralization of strong acid and strong base. 2. To determine the quantity and direction of the heat transfer in the dilution of a salt.

4. A graph for vs time (s) and the maximum temperature is determined for the neutralization reaction.

5. The experiment is repeated for more accurate data.

6. By using the same procedure, the experiment procedure is repeated for reaction of NaOH and HNO3. The for the reaction is calculated and the result is compared with the first the neutralization reaction (NaOH + HCl).

PART B : Heat of Salt Solution

1. 5.0 g of prepared NaCl salt is weighed.

2. The mass of a dry Styrofoam cup is measured and the data is recorded.

3. The cup is filled with 20 ml of distilled water and its initial temperature is recorded.

4. The NaCl salt is added into the Styrofoam cup and closed tightly. The cup is swirled to dissolve completely the NaCl salt. The temperature of dilution is measured for every 15 seconds. The data is recorded.

5. Then, a graph is plotted for the preparing of salt solution vs time (s) and the maximum temperature is determined for the reaction.

6. The experiment is repeated twice to get more accurate data.

3.6 Results and Data

Section A : Table of Data to Plot Graph

Enthalpy (heat) of neutralization acid-base

Enthalpy (heat) of the salt solution

HCl - NaOH

HNO3 – NaOH

Trial 1

Trial 2

Trial 1

Trial 2

Trial 1

Trial 2

0

28

0

27.5

0

27

0

27.5

0

27

0

27

15

28

15

27.5

15

27

15

27.5

15

27

15

27

30

28

30

27.5

30

27

30

27.5

30

27

30

27

45

28

45

27.5

45

27

45

27

45

27

45

27

60

28

60

27.5

60

27

60

27

60

27

60

27

75

27.5

75

27

75

27

75

27

75

27

75

27

90

27.5

90

27

90

27

90

27

90

27

90

27

105

27.5

105

27

105

27

105

27

105

27

105

27

120

27.5

120

27

120

27

120

27

120

27

120

27

135

27.5

135

27

135

27

135

27

135

27

135

27

Section B : Heat of Neutralization

Item

HCl + NaOH

HNO3 + NaOH

Trial 1

Trial 2

Trial 1

Trial 2

1.Acid volume (mL)

50.0

50.0

50.0

50.0

2.Acid temperature ()

22

22

22

22

3.NaOH volume (mL)

50.0

50.0

50.0

50.0

4.NaOH temperature ()

21.5

21.5

21.5

21.5

5.NaOH concentration (molL-1)

1

6.Maximum temperature from graph ()

28

27.5

27

27

Section C : Calculation of the Heat of Neutralization

Item

HCl + NaOH

HNO3 + NaOH

Trial 1

Trial 2

Trial 1

Trial 2

1.Average initial temperature of acid and base

21.8

21.8

21.8

21.8

2.Temperature change

6.2

5.7

5.2

5.2

3.Volume of final mixture (mL)

100

100

100

100

4.Mass of final mixture (g)

100

100

100

100

5.Specific heat of the mixture

4.18 Jg-1

4.18 Jg-1

6.Yielded heat (J)

2591.6

2382.6

2173.6

2173.6

7.Amount of OH- reacted (g)

1.95

1.95

1.95

1.95

8.Amount of H2O produced (g)

0.90

0.90

0.90

0.90

9.Yielded heat per mole H2O ()

-51.83

-47.65

-43.47

-43.47

10.Average (kJmol-1 H2O)

-49.74

-43.47

* Density of the solution is 1.0 g/mL

PART A : Heat of Neutralization of Acid-Base

Section D : Heat of Salt Solution

Salt : NaCl

Item

Trial 1

Trial 2

1.Mass of salt (g)

5.002

5.005

2.Mole of salt (mol)

0.0856

0.0856

3.Mass of cup and water (g)

21.500

21.054

4.Mass of Styrofoam cup (g)

2.659

2.280

5.Mass of water (g)

18.841

18.774

6.Initial temperature of water

22

22

7.Initial temperature of water from graph

22

22

From Trial 1 :

Mole of salt = Mass Atomic mass

= 5.002 (22.990 + 35.453)

= 5.002 58.443

= 0.0856 mole

Mass of cup and water = 21.500 g

Mass of Styrofoam cup = 2.659 g

Mass of water = 21.500 – 2.659 = 18.841 g

Section E : Calculation for Heat of Salt Solution

Item

Trial 1

Trial 2

1. Temperature change

5

5

2. Water heat released (J)

393.78

392.38

3. Salt heat released (J)

104.54

104.60

4. Total enthalpy change ()

498.32 J

496.98 J

5. Amount of OH- reacted (g)

1.55

1.55

6. Amount of H2O produced (g)

1.55

1.55

7. , NaCl salt

+ 5.79

+ 5.78

8. Average

+ 5.785

PART B : Heat of Salt Solution

3.7 Questions

. Show your calculations for Trial 1 (in Section C)

HCl and NaOH

Volume of HCl = 50 ml

Temperature of HCl = 22

Volume of NaOH = 50 ml

Temperature of NaOH = 21.5

Maximum temperature from graph = 28

Calculations :

Average initial temperature of acid and base = = 21.8

Temperature change = 28 – 21.8 = 6.2

Volume of final mixture = 50 ml HCl + 50 ml NaOH = 100 ml

Mass of final mixture = Density Volume = (1.00 g/ml)(100 ml) = 100g

Specific heat of the mixture (Given) = 4.18 Jg-1

mol NaOH = MV = = 0.050 mol

Molar mass of NaOH = 22.990 + 15.999 + 1.008 = 38.997

Amount of NaOH is used = 0.050 mol 38.997 g/mol = 1.95 g NaOH

Amount of NaOH is used = Amount of OH- reacted = 1.95 g

Yielded heat = Specific heat H2O Total mass of acid and base

= (4.18) (100 g) (6.2 )

= 2591.6 J or 2.592 kJ

mol NaOH reacted will produce 1 mol H2O.

Molar mass of H2O = 2(1.008) + 15.999 = 18.015

Amount of H2O produced = 0.050 mol 18.015 = 0.90 g H2O

Yielded heat per mole H2O () = - 2.592 kJ / 0.05 mol OH = - 51.83 kJ/mol

HNO3 and NaOH

Volume of HNO3 = 50 ml

Temperature of HNO3 = 22

Volume of NaOH = 50 ml

Temperature of NaOH = 21.5

Maximum temperature from graph = 27

Calculations :

Average initial temperature of acid and base = = 21.8

Temperature change = 27 – 21.8 = 5.2

Volume of final mixture = 50 ml HNO3 + 50 ml NaOH = 100 ml

Mass of final mixture = Density Volume = (1.00 g/ml)(100 ml) = 100g

Specific heat of the mixture (Given) = 4.18 Jg-1

mol NaOH = MV = = 0.050 mol

Molar mass of NaOH = 22.990 + 15.999 + 1.008 = 38.997

Amount of NaOH is used = 0.050 mol 38.997 g/mol = 1.95 g NaOH

Amount of NaOH is used = Amount of OH- reacted = 1.95 g

Yielded heat = Specific heat H2O Total mass of acid and base

= (4.18) (100 g) (5.2 )

= 2173.6 J or 2.1736 kJ

mol NaOH reacted will produce 1 mol H2O.

Molar mass of H2O = 2(1.008) + 15.999 = 18.015

Amount of H2O produced = 0.050 mol 18.015 = 0.90 g H2O

Yielded heat per mole H2O () = - 2.1736 kJ / 0.05 mol OH = - 43.47 kJ/mol

2. Compare both values and give your opinion.

Answer :

Based on the experiment, we got the average ΔHn between HCl and NaOH is 49.74 kJ/mol and the average ΔHn between HNO3 and NaOH is 43.47 kJ/mol.

Actually, they should have the same ∆Hn value because HCl and HNO3 are strong acids. Strong acid and strong base will ionize completely in water and the actual ΔHn value for strong acid and strong base is -57 kJ/mol. In my opinion, 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.

3. Write down the balance thermochemistry equation for the neutralization.

Answer :

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ΔH = -57 kJ/mol

H+(aq) + OH-(aq) → H2O(l) ∆H = -57 kJ/mol

HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l) ∆H= -57 kJ/mol

H+(aq) + OH-(aq) → H2O(l) ∆H =-57 kJ/mol

4. Show your calculation for Trial 1 (in Section E)

Calculations :

Temperature change = 27 – 22 = 5

Water heat released = mass × specific heat capacity × change in temperature

= m × s ×

= (18.841) (4.18) (5)

= 393.78 J

Salt heat released = m × s ×

= (5.002) (4.18) (5)

= 104.54 J

Total mass = mass of water + mass of salt

= 18.841 + 5.002

= 23.843 g

Total enthalpy change (q) = Total mass × s ×

= 23.843 g × 4.18 × 5

= 498.32 J / 1000

= + 0.498 kJ

mol NaCl = mass NaCl ÷ Molar mass

= 5.002 g ÷ 58.44 g/mol

= 0.086 mol

Molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol

mol NaCl reacted will produce 1 mol H2O.

Molar mass of H2O = 2(1.008) + 15.999 = 18.015

Amount of H2O produced = 0.086 mol 18.015 g/mol = 1.55 g H2O

Amount of H2O produced = Amount of OH- reacted = 1.55 g

, NaCl = q mol NaCl

= 0.498 kJ ÷ 0.086 mol

= + 5.79 kJ/mol

Average = (5.79 + 5.78) ÷ 2

= + 5.785 kJ/mol

5. Instead of using specific heat, what other formula can you use to calculate the heat of reaction? Explain in what condition that formula can be applied.

Answer : Instead of using specific heat, I suppose we can use Hess’s Law which is used when you know equations and their ∆H values and the equations can be added to get the equation you desire to know the ∆H for.

6. What is the appropriate name should be given to the Styrofoam cup as an apparatus to measure heat of reaction?

Answer : Calorimeter

3.8 Discussion

PART A : Heat of Neutralization of Acid-Base

The value obtain are experimental data that must be compared to the true value in order to gauge the accuracy of this experiment. Both 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 Styrofoam cup is not included in the calculation.

A few suggestions to improve the accuracy of the experiments are the thermometer had to be calibrated, which improves accuracy but is itself an imprecise technique. The thermometer and the calorimeter should wipe till it completely dry to prevent the errors in reading. The calorimeter also should run in closed area to prevent the heat easily lost to the surrounding.

PART B : Heat of Salt Solution

The experimental value for is different from the value of ). If the system is not well insulated, energy from the surroundings will be absorbed as well as energy from the water so that the fall of water temperature will not be as great as expected.

3.9 Conclusions

From this experiment, we get the average ΔHn between HCl and NaOH is -49.74 kJ/mol and the average ΔHn between HNO3 and NaOH is -43.47 kJ/mol. 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 ΔHn from the theoretical value (ΔHn = -57.1 kJ/mol). must be positive (+) because energy has been absorbed, that is, the process is endothermic.

3.10 References

Brown, T.L.(2014). Chemistry: The Central Science (3rd ed.). Australia : Pearson Group Pty Ltd.

Chang, R. (2010). Chemistry (10th ed.). New York : McGraw-Hill Companies, Inc.

SKU3023 : Chemistry II, Laboratory Manual.