Question

Which of the following ions forms a hydroxide that is highly soluble in water?
A. \[{\rm{N}}{{\rm{i}}^{{\rm{2 + }}}}\]
B. \[{{\rm{K}}^{\rm{ + }}}\]
C. \[{\rm{Z}}{{\rm{n}}^{{\rm{2 + }}}}\]
D. \[{\rm{A}}{{\rm{l}}^{{\rm{3 + }}}}\]

Answer 1

Hint: Solubility depends on the difference between lattice energy and hydration energy. Lattice energy is the energy needed to change one mole of a compound into gaseous ions. Hydration energy is the energy released when a mole of ions is hydrated.

Complete Step by Step Solution:
Hydration is the process by which the dissociated ions interact with water dipoles to form hydrated ions. It is an exothermic process as energy is released in this process. This released energy is hydration energy. Hydration energy is the energy released when the ions acquired by the dissolution of the compound in water react with water dipoles forming hydrated ions. So, a compound with increased lattice energy possesses a greater intermolecular force which makes the breaking down of the compound into cations and anions in water harder. Hence, for the solvation process, hydration energy must be more than lattice energy.

We know that the higher the charge on the cation or anion, the greater the extent of lattice energy. Here the anion i.e.hydroxide ion is the same in all the compounds. So, we have to compare only cations. Out of the given options,\[{{\rm{K}}^{\rm{ + }}}\] has the lowest charge i.e., +1 thus the hydroxide of this cation i.e. KOH will have the lowest lattice energy.
KOH will dissociate readily in water as compared to the other given hydroxides. The ions formed will react with water to form hydrated ions and release hydration energy. This hydration energy is greater than the lattice energy.
So, option B is correct.

Note: The greater the lattice energy, the more energy is required for the breaking of bonds between the ions. So, it has less tendency to dissociate into ions for interaction with water dipoles to form hydrated ions i.e., less is the solubility of the compound.