NAME________________________ SECTION________________
THE DETERMINATION OF THE ENTHALPY OF
NEUTRALIZATION
HOWARD R. LEO, Ph.D.
When an acid and a base are reacted heat is released. The heat is the result of the formation of
water. The water is produced when the H+ from the acid covalently bonds to a pair of electrons
on the OH‾ from the base. Below are the molecular, ionic and net ionic equations for a typical
neutralization reaction:
HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)
H+(aq) + NO3‾(aq) + K+(aq)+ OH‾(aq) → K+(aq)+ NO3‾(aq) + H2O(l)
H+(aq) + OH‾(aq) → H2O(l)
Since the cation of the base and the anion of the acid start in solution and remain in solution,
they do not contribute to the ΔH of the reaction.
The more acid and base used for the reaction, the greater is the quantity of heat released (Q). By
definition the heat of neutralization (ΔHneut) is the quantity of heat released per mole of water
(n) formed:
ΔHneut = Q/n
(1)
The number of moles of water depends on the number of moles of acid and based used for the
reaction. If the acid and base react in a one-to-one ratio, and if an equal amount of the acid and
base are used, the amount of water formed is equal to both the moles of acid and base. For
example if 0.25 moles of HNO3 and 0.25 moles of KOH are reacted, then 0.25 moles of H2O
will be produced. On the other hand, if they are not mixed in an equal amount, the limiting
reagent will determine the number of moles of water produced.
Both the acid and the base are in aqueous solution. When the neutralization reaction occurs, the
heat released will be absorbed by the solution, and its temperature will increase. ( The Q of the
reaction is negative, so the Q of the solution is positive.) The quantity of heat absorbed can be
calculated from the change in temperature (ΔT) of the solution, its mass (m) and its specific heat
(S.H.):
Q = S.H. x m x ΔT
(2)
To determine the ΔHneut from equation (1), the number of moles of the limiting reagent must be
known. The number of moles of a solute in a solution can be calculated from its volume in
liters (V), and its Molarity (M):
n=VxM
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(3)
Experimental Procedure
Wear your safety glasses!
If you get any acid or base on yourself
immediately wash it off thoroughly.
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1. You will be using two alcohol thermometers during the experiment. The thermometers may
not read exactly the same when they are at the same temperature. To correct for this possible
problem, place your two thermometers in a small beaker of room temperature water. After both
have come to thermal equilibrium with the water, record their readings on line 1) of the second
data page, page 6. Use the lower reading thermometer for the acid. We will assume that the
lower reading is the correct temperature. The difference between the two readings will be the
correction factor for the thermometer that reads high (the base thermometer). For every
measurement with the base thermometer, subtract the correction factor, as indicated on the
second data page.
2. Obtain two Styrofoam cups, and label one acid and the other base. Weigh the acid cup to
0.01 grams, and record its mass on line 1 of the data page.
3. You will use two graduate cylinders, one for the acid and one for the base. Measure out
about 50mL of acid. Record the amount of acid used and its Molarity on lines 2 and 3 of the
first data page, page 5. Be sure to record the volumes to 0.1 mL by estimating between the
lines of the cylinder. Add the acid to its Styrofoam cup. Measure out about 55mL of base.
Record the amount of base used and its Molarity on lines 4 and 5 of the first data page. Insert a
thermometer into each solution as shown in the diagram below. Be sure to use the thermometer
that reads high for the base.
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4. Once both cups are setup, gently swirl the cups, and continue to swirl during the entire
experiment. Begin recording the temperature of the acid and the base. Start with the base, and
alternate between the two at 0.5 minute intervals, up to 4.5 minutes. See the second data page,
page 6.
5. At 5.0 minutes, quickly poor the base into the acid, and continue to swirl. You will not be
able to make a temperature reading during this time. At 5.5 minutes measure the temperature of
the mixture. Continue to measure the temperature at 1.0 minute intervals, up to 20.5 minutes.
Calculations
1. To determine the ΔT, the temperatures must be plotted on a graph. Put time on the x-axis,
and temperature on the y-axis. The initial temperature (Ti) is determined from the acid and base
readings. Extrapolate the best fit line to the 5.0 minute mark (See abbreviated example below.)
Record Ti on line 6 of the first data page.
To determine the final temperature (Tf), extrapolate the mixture’s temperature backwards from
20.5 minutes to 5.0 minutes. Be sure to draw the best fit line from the data points. When you
do this ignore the reading at 5.5 minutes. Record Tf on line 7 of the first data page.
T
e
m
p
(°C)
time (min)
2. Determine the mass of the mixture and the cup. Report the mass on line 9 of the first data
page. Determine the mass of the solution, line 10.
3. Calculate the quantity of heat, Q, absorbed by the solution using equation (2). Use 3.99J/g°C
for the value of the S.H. of the solution.
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4. Calculate the moles of acid and base used for the reaction using equation (3). Record the
number of moles on lines 12 and 13 of the first data page.
5. Calculate the ΔHneut using equation (1). Remember the number of moles of H2O is
determined by the limiting reagent. Also, ΔHneut = –ΔHsolution. Record your ΔHneut on line 16 in
kilojoules per mole.
6. Repeat the experiment, but this time use about 50 mL 0f base and about 55 mL of acid.
6. Average your two trials, line 17.
7. Using the excepted ΔHneut of –55.9 kJ/mole, calculate your percent error. Report your
percent error on line 18.
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Data Page One
Trial #1
Trial #2
1) Mass of empty acid cup (g)__________
2) Volume of acid (mL)
___________
2) Volume of acid (mL)
___________
3) Molarity of acid (mol/L) ___________
3) Molarity of acid (mol/L)
___________
4) Volume of base (mL)
4) Volume of base (mL)
___________
5) Molarity of base (mol/L) ___________
5) Molarity of base (mol/L)
___________
6) Initial temperature (°C)
___________
6) Initial temperature (°C)
___________
7) Final temperature (°C)
___________
7) Final temperature (°C)
___________
___________
8) Change in temperature(°C) __________
8) Change in temperature(°C) __________
9) Mass of cup plus solution ___________
9) Mass of cup plus solution ___________
10) Mass of solution
___________
10) Mass of solution
___________
11) Quantity of heat (J)
___________
11) Quantity of heat (J)
___________
12) Moles of acid used
___________
12) Moles of acid used
___________
13) Moles of base used
___________
13) Moles of base used
___________
14) Identity of limiting reagent___________
14) Identity of limiting reagent___________
15) Moles of H2O formed
15) Moles of H2O formed
16) ΔHneut (kJ/mol)
___________
____________
16) ΔHneut (kJ/mol)
17)Average ΔHneut(kJ/mol)____________
18) Percent error
_____________
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___________
____________
Data Page Two
1) Temperature of H2O
with acid thermometer
___________
2) Temperature of H2O
with base thermometer
___________
3) Correction factor for
base thermometer
___________
TRIAL #1
Temperature (°C)
Acid
Base Corrected Base
Time
(min)
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.5
7.5
8.5
9.5
10.5
11.5
12.5
13.5
14.5
15.5
16.5
17.5
18.5
19.5
20.5
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TRIAL #2
Temperature (°C)
Acid
Base Corrected Base
Time
(min)
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.5
7.5
8.5
9.5
10.5
11.5
12.5
13.5
14.5
15.5
16.5
17.5
18.5
19.5
20.5
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Questions
1. If the cup acted as a perfect insulator, how would this change the plot of the temperature
versus time? To answer this question, make a sketch of the temperature versus time for an
experiment carried out under this assumption.
2. What does the extrapolating of the temperature of the mixture back to the five minute mark
account for?
3. Styrofoam has a very low density and consequently the cup has a very low mass. So by
equation (2) the heat adsorbed by the cup during the experiment was quite low. Thus we ignored
it. On the other hand, Pyrex has a relatively high density. If you had carried out the experiment
in a Pyrex beaker the same size as the Styrofoam cup, would the measured ΔT have been higher,
lower or unaffected?
4. The heat absorbed by the calorimeter can be determined experimentally. If the calorimeter did
not absorb any heat, when equal volumes of hot and cold water are mixed in it, the ΔTcold =
–ΔThot. In reality some of the heat is adsorbed by the calorimeter. Taking the S.H. of water to be
4.18 J/g°C, use the data below to calculate the amount of heat absorbed by the calorimeter when
the water is mixed in it. Hint: calculate the heat lost by the hot water and the heat gained by the
cold water. The missing heat was absorbed by the calorimeter.
COLD Water: mass = 20.0g, Ti = 20.0°C
HOT Water: : mass = 20.0g, Ti = 40.0°C
Tf after mixing = 29.5°C
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