Part I: The Dissolving of Solid Sodium Hydroxide in Water
Procedure
1. Measure out approximately 205 mL of distilled water and pour it into the calorimeter. Stir
carefully with a thermometer until a constant temperature is reached. Record the volume of
water and the constant initial temperature of the water on your data table.
2. Place a plastic measuring trough on top of the digital balance, and then zero the balance (press
the tare button) so that the mass of the trough will be "ignored" and will not be added to the
total mass measured by the balance.
3. Measure out approximately three to five scoops of solid sodium hydroxide and record the mass
to your data table.
4. Place the solid sodium hydroxide into the water in the calorimeter and replace the lid
immediately. Stir gently until the solid is completely dissolved and record the highest
temperature reached.
Data and Observations
Distilled Water Volume
Constant Initial Temperature
NaOH Mass
Final Temperature
205.2 mL
24.2°C
2.535g
27.8°C
Calculations
1. Write out a balanced "equation" for the process you investigated in Part I, including phase
symbols.
NaOH(s) + H2O(l) -> Na+(aq) + OH-(aq)
2. Calculate the number of moles of sodium hydroxide dissolved. Show your work.
2.535g NaOH x
1 mol NaOH
= 𝟎. 𝟎𝟔𝟑𝟒𝐦𝐨𝐥 𝐍𝐚𝐎𝐇
40g NaOH
3. Calculate the amount of energy involved in this dissolving process. Show your work.
q = m x c x ∆T
qsurroundings = 207.735g x
3125.996J x
4.18J
x 3.6℃ = 3126J
℃xg
1kJ
= 3.126kJ
1000J
qsurroundings = -qsystem
qsystem = -3.126kJ
4. Determine the enthalpy change, per mole of sodium hydroxide dissolved. Show your work.
−3.126kJ
−𝟒𝟗. 𝟑𝐤𝐉
=
0.0634mol NaOH 𝐦𝐨𝐥 𝐍𝐚𝐎𝐇
Part II: The Reaction of Sodium Hydroxide Solution with Hydrochloric Acid
Procedure
1. Measure out approximately 100 mL of 0.50 M hydrochloric acid solution and 100 mL of 0.50 M
sodium hydroxide solution. Record both volumes on your data table.
2. Pour the hydrochloric acid solution into the calorimeter. Measure and record the initial
temperature of each solution and record on your data table.
3. Add the sodium hydroxide solution to the acid solution in the calorimeter and immediately
replace the lid of the calorimeter. Stir the mixture and record the highest temperature reached.
Data and Observations
HCl Solution
NaOH Solution
Initial HCl Temperature
Mixed Solution Temperature
100.9mL
100.9mL
25.2°C
28.2°C
Calculations
1. Write out a balanced equation for the reaction you investigated in Part II, including phase
symbols.
NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)
2. Determine the enthalpy change of this reaction.
100.9mL + 100.9mL = 201.8mL
1.0g
201.8mL x
= 201.8g
1mL
4.18J
x 3℃ = 2530.6J
℃xg
= 2.531kJ
qsurroundings = 201.8g x
2530.6J x
1kJ
1000J
qsurroundings = -qsystem
qsystem = -2.53kJ
3. Determine the number of moles of NaOH.
100.9mL NaOH x
1g
1mL
x
1 mol NaOH
40g NaOH
= 2.5 mol NaOH
4. Determine enthalpy per mole of NaOH. Show all of your work.
−2.53kJ
−𝟏. 𝟎𝟏𝟐𝐤𝐉
=
2.5 mol NaOH
𝐦𝐨𝐥 𝐍𝐚𝐎𝐇
Conclusion:
1. Determine the enthalpy change for NaOH (s) + HCl (aq) → NaCl (aq) + H2O (l) using Hess's Law.
a) Write the balanced chemical reaction and enthalpy change for Part I
NaOH(s) + H2O(l) -> Na+(aq) + OH-(aq)
∆H = -49.3kJ/mol
b) Write the balanced chemical reaction and enthalpy change for Part II
NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)
∆H = -1.012kJ/mol
c) Calculate the enthalpy change using Hess's Law. Refer to the lesson for an example of
Hess's Law.
NaOH(s) + HCl(aq) -> NaCl(aq) + H2O(l)
∆H = -50.312kJ/mol
2. If the accepted enthalpy change value for the dissolving of sodium hydroxide in water is −44.2
kilojoules per mole, determine the percent error of the experimental value that you calculated
in Part I. Show your work.
[−49.3 − (−44.2)]
−5.1
=
= 0.12 x 100 = 𝟏𝟐%
−44.2
−44.2
3. If the accepted heat of reaction for the neutralization of hydrochloric acid with sodium
hydroxide is −56.0 kilojoules per mole, determine the percent error of the experimental value
that you calculated in Part II. Show your work.
[−56 − (−1.012)]
−54.988
=
= 0.982 x 100 = 𝟗𝟖. 𝟐%
−56
−56
4. Using the accepted values of the processes you've examined, would your estimation of the
enthalpy change for the reaction of solid sodium hydroxide in aqueous hydrochloric acid change
from the prediction you made in question one?
Yes, because somehow my data is way off with a 98.2% error. The enthalpy change should be
almost twice my initial prediction.
5. Give a detailed explanation, using what you know about bonds and forces of attraction, for the
enthalpy changes you observed in parts I and II of this lab.
NaOH is an ionic bond and is stronger than a covalent bond, so naturally it would require a
higher amount of energy to break apart.
6. If the hole for the thermometer in a calorimeter is wider than the diameter of the thermometer,
leaving a gap between the lid and the thermometer itself, how do you think this would this
affect the temperature change observed in the experiment? How would this affect the
calculated enthalpy change?
Heat would escape through the gap and you would observe lower temperatures than what
would actually occur and result in lower enthalpy change.