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Chemical Oceanography, 4th Edition
The heat involved in hydrating the ions Na + and Cl – is also quite large:
Na + (g) + Cl – (g) → Na + (aq) + Cl – (aq)
(4.7)
This can be clearly demonstrated by examining the Born–Haber cycle for the
hydration of NaCl (ΔH h is the heat of hydration of the ions). This heat of hydration
for the salt (MX) is given by
M + (g) + X – (g) → M + (aq) + X – (aq)
(4.8)
It is related to the heats of the reactions of the formation of MX in the solid state:
M + (g) + X – (g) → MX(s)
(4.9)
which is called the lattice heat and the heat of solution of the salt.
MX(g) → M + (aq) + X – (aq)
(4.10)
ΔH h = ΔH LATTICE + ΔH SOLN
(4.11)
For NaCl,
ΔH h = 181 + 0.6 = 182 kcal mol –1
(4.12)
The fact that ΔH h is the same order of magnitude as ΔH LATTICE indicates that the
energies involved in solvating the ions are the same order of magnitude as those
forming ionic bonds in a crystal.
2. The solution formed after the addition of NaCl no longer freezes at 0°C, but does
at approximately 2.3°C. The freezing point depression is given approximately by
ΔT f = 1.86νm
(4.13)
where ν is the number of ions formed when the salt completely dissociates. This
equation gives ΔT f = 2.3°C, T f = –2.3°C. This indicates that the interaction of Na +
and Cl – with water breaks down the structure of water.
3. The solution also boils at a higher temperature. The boiling point elevation is
given by
ΔT h = 0.52νm
(4.14)
This equation gives ΔT h = 0.63 or T b = 100.63°C. This indicates that the hydration
interactions tend to keep the water molecules in the liquid state.
4. The vapor pressure of the NaCl solution is lower than for pure water. The ratio
of the vapor pressure over the solution P/ P H2O = 0.98. The 2% depression in vapor
pressure also indicates that the hydration “ties up” the water molecules, making it
harder for them to go into the gaseous state.