question_answer

\[N{{a}_{2}}S{{O}_{4}}\] is water soluble while \[BaS{{O}_{4}}\] not.

Answer:

                Lattice enthalpy of\[N{{a}_{2}}S{{O}_{4}}\]  is less compared to that of \[BaS{{O}_{4}}\] because of smaller charge density on\[N{{a}^{+}}\]ion as compared to\[B{{a}^{2+}}\] ions. Therefore, hydration enthalpy released easily overcomes the lattice enthalpy when\[N{{a}_{2}}S{{O}_{4}}\] dissolves in water. But this is not possible in case of \[BaS{{O}_{4}}\] when tried to be dissolved in water. Therefore, \[N{{a}_{2}}S{{O}_{4}}\] is readily soluble in water while \[BaS{{O}_{4}}\]  is almost insoluble.