Question

What may happen when $N{a_2}S{O_4}$ is hydrolysis?

Answer 1

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 $49.7g/100mL$.