What may happen when \[N{a_2}S{O_4}\] is hydrolysis?

Hint: The inorganic compound sodium sulfate, as well as other similar hydrates, has the formula \[N{a_2}S{O_4}\]. Most of those forms are white solids that are extremely water soluble. It is mostly used in the making of detergents and the pulping of kraft paper.

Complete answer:
A molecule of water is attached to a product in a chemical reaction known as hydrolysis. When this happens, both the substance and the water molecule will break into two. One fragment of the target or parent molecule gains a hydrogen ion in such reactions.
First, we have to write the equation as per the given question.
\[N{a_2}S{O_4}\left( s \right) + {H_2}O\left( l \right)\] . That is sodium sulfate \[ + \] water.
So, when we look at \[N{a_2}S{O_4}\], sodium is a metal, and then \[S{O_4}\], \[S\] and \[{O_{}}\] are non-metals. The sulfate ion is called a polyatomic ion.
We look up sodium on the periodic table. It is in Group \[1\] that means it’s ionic charge is going to be a \[ + 1\] , and then \[S{O_4}\], the sulfate ion has \[2\] minus charges.
We take solid sodium sulfate and liquid water, and solid dissolves, and it dissociates into its ions
So, we got
\[N{a_2}S{O_4}\left( s \right) + {H_2}O\left( l \right) \to 2N{a^ + }\left( {aq} \right) + SO_4^{2 - }\left( {aq} \right)\].
The solid dissolved in water, and we got aqueous \[N{a^ + }\] and aqueous \[SO_4^{2 - }\]. Because we mentioned aqueous in the equation, we don't need to write \[{H_2}O\] on the right side of the equation.
Therefore, the answer is \[N{a_2}S{O_4}\left( s \right) + {H_2}O\left( l \right) \to 2N{a^ + }\left( {aq} \right) + SO_4^{2 - }\left( {aq} \right)\].

Note:
In water, sodium sulfate has unusual solubility properties. Between \[0^\circ C\] and \[32.384^\circ C\], its solubility in water increases by more than tenfold, reaching a limit of \[49.7g/100mL\].