Question

One mole of anhydrous salt AB dissolves in water and librates 21.0 J mol of heat. The value of ∆H(hydration) of AB is -29.4 J mol-1. The heat of dissolution of hydrated salt AB.2H2O(s), is?

Answer 1

Answer : The heat of dissolution of hydrated salt is 8.4 J/mol

Explanation :

Heat of hydration : It is defined as the amount of energy liberated when 1 mole of ions undergo hydration. It is a special type of dissolution energy where in the solvent is water.

The reaction of dissolution of anhydrous salt AB  will be,

$AB(aq)+2H_2O\rightarrow AB.2H_2O(aq),\Delta H_1=-21.0J/mol$    .....(1)

The reaction of hydration of AB salt  will be,

$AB+2H_2O\rightarrow AB.2H_2O(s),\Delta H_2=-29.4J/mol$    ....…(2)

Now Subtracting equation (2) from (1) we get,

$AB.2H_2O(s)+(aq)\rightarrow AB.2H_2O(aq),\Delta H_{req}=?$

So,

$\Delta H_{req}=\Delta H_1-\Delta H_2$

$\Delta H_{req}=(-21.0J/mol)-(-29.4J/mol)$

$\Delta H_{req}=8.4J/mol$

Therefore, the heat of dissolution of hydrated salt is 8.4 J/mol

Answer 2

Answer:

The heat of dissolution of hydrated salt AB.2H2O(s), is $-50.4 Jmol^{(-1)}$

Explanation:

Anhydrous salt + water =Hydrate

That is,

$AB+2H_2O>AB.2H_2O\\\\\Delta H_{Lattice}=-21.0Jmol^{-1}$

(- sign used since it says heat is liberated)

The heat energy released when new bonds are made between the ions and water molecules is known as the hydration enthalpy of the ion.

$A^++B^-+2H_2O>AB.2H_2O\\\\\Delta H=-29.4Jmol^{-1}$

$\Delta H_{dissolution}=\Delta H_{hydration(cation)}+\Delta H_{hydration(anion)}+\Delta H_{Lattice}$

$\Delta H_{dissolution}=-29.4+(-21.0)=-50.4 Jmol^{(-1)}$

(Answer)