CoreChem:Hydration of Ions
From ChemPRIME
The resolution of this apparent paradox lies in the interactions between ions and the molecules of water or other polar solvents. The negative (oxygen) side of a dipolar water molecule attracts and is attracted by any positive ion in solution. Because of this ion-dipole force, water molecules cluster around positive ions, as shown in Figure 1a. Similarly, the positive (hydrogen) ends of water molecules are attracted to negative ions. This process, in which either a positive or a negative ion attracts water molecules to its immediate vicinity, is called hydration.
When water molecules move closer to ions under the influence of their mutual attraction, there is a net lowering of the potential energy of the microscopic particles. This counteracts the increase in potential energy which occurs when ions are separated from a crystal lattice against their attractions for other ions.
Thus the process of dissolving an ionic solid may be divided into the two hypothetical steps shown in Fig. 2. First, the crystalline salt is separated into gaseous ions. The heat energy absorbed when the ions are separated this way is called the lattice enthalpy (or sometimes the lattice energy). Next, the separate ions are placed in solution; that is, water molecules are permitted to surround the ions. The enthalpy change for this process is called the hydration enthalpy.
| Task: This image should be re-done, according to the corrections for 2nd ed. Jshorb 01:09, 4 September 2009 (UTC) ( John Moore ) |
Since there is a lowering of the potential energy of the ions and water molecules, heat energy is given off and hydration enthalpies are invariably negative.
The heat energy absorbed when a solute dissolves (at a pressure of 1.00 atm) is called the enthalpy of solution. It can be calculated using Hess' law, provided the lattice enthalpy and hydration enthalpy are known.
EXAMPLE 1 Using data given in Fig. 2, calculate the enthalpy of solution for NaCl(s).
Solution According to the figure, the lattice enthalpy is 773 kJ mol–1. The hydration enthalpy is – 769 kJ mol–1. Thus we can write the thermo-chemical equations
NaCl(s) → Na+(g) + Cl–(g) ΔHl = 773 kJ mol–1
Na+(g) + Cl–(g) → Na+(aq) + Cl–(aq) ΔHh = –769 kJ mol–1
NaCl(s) → Na+(aq) + Cl–(aq) ΔHs = ΔHl + ΔHh
ΔHs = (773 – 769) kJ mol–1 = +4 kJ mol–1
When NaCl(s) dissolves, 773 kJ is required to pull apart a mole of Na+ ions from a mole of Cl– ions, but almost all of this requirement is provided by the 769 kJ released when the mole of Na+ and the mole of Cl– becomes surrounded by water dipoles. Only 4 kJ of heat energy is absorbed from the surroundings when a mole of NaCl(s) dissolves. You can verify the small size of this enthalpy change by putting a few grains of salt on your moist tongue. The quantity of heat energy absorbed as the salt dissolves is so small that you will feel no cooling, even though your tongue is quite a sensitive indicator of temperature changes.
Few molecules are both small enough and polar to cluster around positive and negative ions in solution as water does. Consequently water is one of the few liquids which readily dissolves many ionic solids. Hydration of Na+, Cl– and other ions in aqueous solution prevents them from attracting each other into a crystal lattice and precipitating.


