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.
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