Question

Which hydroxide will have the lowest value of solubility product at normal temperature$\text{(2}{{\text{5}}^{\text{0}}}\text{C)}$?
A) \[\text{Mg}{{\left( \text{OH} \right)}_{\text{2}}}\]
 B) \[\text{Ca(OH}{{\text{)}}_{\text{2}}}\]
C) \[\text{Ba(OH}{{\text{)}}_{\text{2}}}\]
D) \[\text{Be(OH}{{\text{)}}_{\text{2}}}\]

Answer 1

Hint: The solubility is the property of solute to dissolve in solvent. The solubility of the solute is related to the lattice energy .Lattice energy for a crystal is the amount of energy to separate the solute into its ion.Less the energy required to separate in ions higher is the solubility of solute.

Complete step by step answer:
The hydroxides of alkaline earth metals are less soluble in water as compared to the alkali earth metals.
The solubility of hydroxides of alkaline earth metal depends on the lattice energy of the elements.
The lattice enthalpies or the energies for a crystal is defined as the change in the energy when one mole of the crystal lattice is converted into the gaseous ions. In other words, the amount of energy required to break the crystal into its gaseous ions. Lattice energies are affected by the following factors:
   1) Charge: charge is directly proportional to the lattice enthalpy. Greater the charge on the ion greater is the lattice enthalpy
    2) Size of ion: lattice enthalpy is inversely related to the size of an ion.Smaller the size of the ion greater is the lattice enthalpy required.
Thus lattice energy is given as:
\[\text{Lattice Energy }\propto \text{ }\dfrac{\text{positive ion charge }\times \text{ negative ion charge}}{\text{cation radius + anion radius}}\]
Now, we are trying to get the relation between lattice energy and solubility for alkaline earth metal. As we look towards the group 2 (alkaline earth metal) the ionic radius increases from $\text{ Be to Ba}$.The charge for group 2 elements is always $+2$ .
Thus the lattice energy depends only on the size of the atom. Since the atomic size and radius increases down the group, the lattice energy for the hydroxides alkaline earth metal decreases from the$\text{ Be to Ba}$.
The solubility for hydroxide is inversely related to the lattice energy. Higher is the lattice energy of hydroxide $\text{Be(OH}{{\text{)}}_{\text{2}}}$lower is its solubility. On the other hand, lower is the lattice energy of alkaline hydroxide$\text{Ba(OH}{{\text{)}}_{\text{2}}}$, higher is the solubility.
Thus $\text{Be(OH}{{\text{)}}_{\text{2}}}$is less soluble and least solubility product out of the given alkaline earth metals.

Hence, (D) is the correct option.

Note: The hydration energy also contributes towards the solubility. The effect of lattice energy and hydration energy is given as follows:
\[\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{Solution}}}\text{ = }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{Lattice}}}\text{ - }\!\!\Delta\!\!\text{ }{{\text{H}}_{\text{Hydration}}}\]
This plays an important role in deciding the stability of salt in water and thus there solubility.Hydration enthalpy is the amount of energy released when one mole of the ion undergoes hydration. It is a special case of solvation. The solubility of hydroxides of alkaline earth metal increases down the group. This is because lattice energy decreases down the group with increase in size whereas the hydration energy remains constant. This is because of the small size of ions.