**Heat of Hydration Formula**

Heat of hydration is defined as the amount of energy released when one mole of ions undergo hydration. It is a special type of dissolution energy where in the solvent is water.

The enthalpy of a hydrated salt is the change in heat, when 1 mole of an anhydrous substance combines with requisite number of water molecules to form the hydrate. The heat of hydration can be determined if the heat of the solutions of anhydrous salt and the hydrated forms are known.

Anhydrous salts readily combine with water to form hydrates and dissolve with the evolution of heat. The only difference between hydrate and anhydrous salt is the heat is evolved as heat of hydration in the formation of hydrates.

**The heat of hydration formula is given by:**

**Heat of hydration = **Δ**H **solution** – **Δ**H **lattice energy

Where,

Δ**H **solution** = Heat of the solution**

Δ**H **lattice energy = Lattice energy of the solution

**Example 1**

The heat of solution of anhydrous and hydrated copper sulphate are – 65 and lattice energy is 11 KJ respectively. Determine the heat of hydration.

**Solution:**

Given parameters are

Heat of solution = – 65

Lattice energy = 11 KJ

Substitute the values in the given formula,

Heat of hydration = ΔHsolution – ΔHlattice energy

= – 65 – 11

Heat of hydration = -56

**Example 2**

The lattice enthalpy of sodium chloride ΔH for **NaCl **→→** Na****+**** + Cl****–** is 700 kJ/mol. The heat of solution in making up 1 M NaCl is +5.0kJ/mol. Determine the heat of hydration of Na+ and Cl–, where the heat of hydration of Cl– is -300kJ/mol.

**Solution:**

Given data,

Lattice energy = 700 kJ/mol

Heat of solution = 5.0kJ/mol

Substitute the values in the given formula,

Heat of hydration = ΔHsolution – ΔHlattice energy

= 5 – 700

Therefore, Heat of hydration = -695

Heat of hydration of Na+ + Cl– = -695

Heat of hydration of Na+ = -695 – (-300)

Therefore, Heat of hydration of Na+ = -395