Question

statement: Barium sulphate is sparingly soluble in water whereas beryllium sulphate is freely soluble.
Enter 1 if the given statement is true, if false enter 0.

Answer 1

Hint:To answer this question, you must recall the factors on which the solubility of an ionic salt in water depends. The solubility of a salt depends majorly on its lattice energy and hydration energy. Smaller is the size of cation, more will be its hydration energy and thus more will be the solubility. Similarly, greater is the size of the cation, more will be its lattice energy and thus lower will be solubility.

Complete step by step solution:
If the hydration energy of the ions is large enough so as to overcome the lattice energy of the compound, the compound will be soluble in water. If the hydration energy of the ions does not exceed the lattice energy of the compound, it will not be soluble in water.
We know that beryllium is small in size as compared to barium and the sulphate ion. Thus, it has very low lattice energy. Also, due to small size, beryllium has high charge density and thus higher hydration. So it will be freely soluble in water.
Whereas, barium sulphate is not at all soluble in water. It is one of the most insoluble ionic salts. This is because the sizes of barium ion and sulphate ion are almost similar and thus the compound has a very high lattice energy which is significantly greater than its hydration energy.

The given statement is false, enter 0.

Note:
Hydration energy can be termed as the amount of energy which is released when one mole of ions is hydrated.
Lattice energy of a compound is termed as the energy released when ions combine to form a compound. It can also be said as the energy required to break the lattice of a salt in order to dissolve it in water.