Why Na2SO4 soluble

The dissolving of Na2SO4 can be divided into two stages:

In stage 1, energy is absorbed to break the lattice.
Na2SO4(s) → 2Na+(g) + SO42-(g) .. ΔH1 = -(lattice enthalpy)
ΔH1 depends on two factors: Firstly, the greater the interionic distance, the smaller the magnitude of lattice enthalpy, and thus the less energy is needed to break the lattice. Secondly, the greater the charges on the ions, the greater is the magnitude of lattice enthalpy, and thus the more energy is needed to break the lattice. Na+ is large in size and carries only one positive charge. Besides, SO42- is bulky to cause a long interionic distance. Therefore, the amount of energy is needed to break the lattice of Na2­SO4 is low.

In stage 2, energy is released when the ions are hydrated.
2Na+(g) + SO42-(g) + aq → 2Na+(aq) + SO42-(aq) .. ΔH2 = hydration enthalpy
ΔH2 depends on two factors: Firstly, the greater the ionic sizes, the smaller the magnitude of hydration enthalpy. Secondary, the greater the charges on the ions, the greater is the magnitude of hydration enthalpy. Na+ is large in size and carries only one positive charge. Besides, SO42- is a bulky ion. Therefore, the amount of energy released in hydration is low.

Both the amount of energy absorbed in breaking of lattice and that released in hydration are low, and they are comparable. On the other hand, the entropy (degree of randomness) is increased during dissolving. Therefore, dissolving of Na2SO4 is a feasible process when combining the effect of enthalpy and that of entropy (i.e. a negative free energy change).

when Na2SO4 in water
it have 4 kinds of ions
Na+,SO4 2-, H+, OH-

Na is more want to form compound with OH- then SO4 2-
so OH- will attract Na+
and H+ will attract SO4 2-
this is why Na2SO4 is soluble

all Na+ compound is soluble in water