STANSW Science Education News Journal 2019 2019 SEN Vol 68 Issue 3 | Seite 37

ARTICLES
Understanding and Calculating the Heat Capacity of Solutions
By Malcolm Hooper (Normanhurst Boys HS)
OH – (aq):
Thermodynamics is a cornerstone of chemistry, physics and
engineering. The senior chemistry texts describe experiments
to measure the heat of dissolution of a substance, but usually
approximate the heat capacity of the solution as the heat capacity
of the water alone. This is invalid, hence also inaccurate, since it
introduces a small systematic error. The SI Chemical Data Book
usually contains enough data to estimate the heat capacity of
the solution directly and so better estimate the molar heat of
dissolution of a substance. Hepler and Hovey reviewed heat
capacities of solutions and their data tables list more values.
TOTAL = 418 + 4.6 – 14.9 = 407.7 J/K, ∴ true temperature rise
= 4500 J /(407.7 J/K) = 11.04 K
This difference is small but interesting. With more concentrated
solutions the effect is larger, because more ions are present and
a greater fraction of the water molecules are closely interacting
with them. Good thermometers reading to one decimal place
should detect the effect even with 1 molal solutions. The most
interesting part is the sign of the change. The heat capacity of the
solution is less than the heat capacity of the water, even though
the solution contains more atoms. This is usually attributed to
the strong bonding between the dissolved ions and their several
surrounding water molecules 'freezing out' many of the vibrational
modes of the water molecules. Liquid water is a weird substance;
its heat capacity is slightly greater than 9R, considered to be the
classical limit expected from the rule of Dulong and Petit if all
three atoms vibrate in all three directions. Einstein’s 1907 model
explaining the heat capacity of solids at low temperatures was
an early triumph of quantum theory, and Debye’s refinement of
the model to treat low frequency vibrational modes is a good fit
across the complete temperature range for solids.
The usual substances used in these experiments are sodium
hydroxide (exothermic dissolution) and ammonium nitrate
(endothermic dissolution). I prefer to use potassium nitrate, which
gives a larger temperature change. All of these substances are
highly soluble, so they can form highly concentrated solutions
and change the temperature by several degrees.
Dissolution of sodium hydroxide
We use two polystyrene foam cups as our calorimeter. The heat
capacity of the foam is negligible and the system is close to
adiabatic because the foam is such a good insulator. We use
about 100.0 g of water and make a solution that is close to 1.00
molal by adding 4.00 gram of sodium hydroxide. Please note the
spelling 'molal'; this is 1.00 mole of NaOH per kilogram of solvent,
not 'molar' which is 1.00 mole of NaOH per litre of solution.
The heat capacity of the H + (aq) ion is zero. This value is arbitrary.
The solution has no net charge so for every univalent cation in
solution, there is a matching univalent anion in solution. The sum
of the heat capacities of the anions and cations is correct, even
though the individual values given are not 'real'.
The amount of heat released is calculated using standard heats
of formation and the equation:
NaOH(s) → Na + (aq) + OH – (aq)

−425
−240
0.100 × –149 J.K –1 = –14.9 J/K
Dissolution of potassium nitrate
Δ H = –45 kJ
We use about 100.0 g of water and make a solution that is close
to 1.00 molal by adding 10.11 gram of potassium nitrate.
−230
Therefore the heat released while dissolving 0.100 mol of
NaOH is 4.5 kJ
The amount of heat released is calculated:
KNO 3 (s) → K + (aq) + NO 3 – (aq)
Δ H = +36 kJ
Usually the heat capacity of the solution is treated as the heat
capacity of the water,
Heat capacity = m.Cp = 100 g × 4.18 J.K –1 .g –1 = 418 J/K Therefore the heat absorbed dissolving 0.100 mol is 3.6 kJ
Hence, the temperature rise would be
However, a more precise estimate of the heat capacity of the
solution should include the amounts of dissolved ions:
Water:
+
Na (aq):
–1
100 g × 4.18 J.K .g
–1
0.100 mol × 46 J.K .g
−207
Heat capacity = m.Cp = 418 J/K
Temperature fall = 3600 J /(418 J/K) = 8.61 K
= 418 J/K
–1
−252
When the heat capacity of the solution is treating as being the
heat capacity of the water,
4500 J /(418 J/K) = 10.77 K
–1
−495
However again when the heat capacity of the solution is better
estimated by including the amounts of dissolved ions:
= 4.6 J/K
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SCIENCE EDUCATIONAL NEWS VOL 68 NO 3