Question

As $N{a_2}S{O_4}$ is water soluble but $BaS{O_4}$ is insoluble explain?
(A) The hydration energy of $N{a_2}S{O_4}$is higher than that of its lattice energy.
(B) The hydration energy of $N{a_2}S{O_4}$is less than that of its lattice energy.
(C) The hydration energy of $BaS{O_4}$ is less than that of its lattice energy.
(D) The hydration energy of $BaS{O_4}$ is higher than that of its lattice energy.

Answer 1

Barium sulphate is an inorganic compound which is an alkaline and divalent metal. It is non toxic in nature. It is white in colour with odourless powder. It has poor solubility in the universal solvent water. When it reacts with aluminium powder it reacts violently. Sodium sulphate is a crystalline powder with an orthorhombic and hexagonal structure.

Complete step-by-step answer: The concept of hydration energy with solubility is that the component is soluble in water if the hydration energy of the solution is more than its lattice energy. Barium sulphate is insoluble in water because it has a high lattice energy. On adding water in barium sulphate its hydration energy decreases more rapidly in comparison to its lattice energy. And to be soluble in water the component lattice energy should be less than its hydration energy. For $N{a_2}S{O_4}$the lattice energy is lower than the hydration energy, so it is soluble in water. This is because the large sized sulphate ion will be more stabilized by large sized cation and sodium size is bigger than barium.
See the substance which has a smaller cation has the higher lattice energy and as we know that the substance which has higher lattice energy tends to decrease the solubility of the substance. The other factor which is important while determining the solubility of substance is the solvation energy of the cation and the solvation energy of the small cation is more. The solvation energy tends to increase the solubility of ionic substances.
Now if you see the factor of hydration energy then on decreasing the hydration energy the solubility of the substance also gets decreased. On dissolving it in water we see that the outermost ions which are present at the edge lattice move away from the lattice and get covered with the neighbouring water molecules. Now if the hydration energy of the substance get equal to greater than the lattice energy then it will be soluble but if the hydration energy is less than that of lattice energy then it will be soluble.
So the $N{a_2}S{O_4}$ is soluble in water because its hydration energy is greater than its lattice energy whereas the $BaS{O_4}$ is insoluble in water because its hydration energy is less than that of its lattice energy.

So the correct answer is option (A) and option (C).

Note: Barium sulphate is used in drilling fluids. It is used as a contrast agent. It is used in diagnosis of certain diseases. Sodium sulphate is used as a fining agent which removes small air bubbles from molten glass. It is also used in the Kraft process of paper pulping.