Barium selanate has the molecular formula of BaSeO4
and the molecular weight of 280.2748 g/mol. It can be
prepared by reacting sodium selenate with barium chloride
in solution:
BaCl2(aq)+ Na2SeO4(aq) ? BaSeO4+ 2NaCl(aq)
Barium selenate is slightly soluble in water at
0.0052 g/100 ml at 20°C. Its CAS number is 7787-41-9.
It is a white rhombic crystal which decomposes when
heated beyond about 425°C. It is slightly soluble in
water at 0.0152 g/100 ml at 20°C. Its solubility product
is 3.40×10-8.
The heat of formation of barium selenate is:
△H298 BaSeO4 = -249.0 kcal/mol (-1041.8 kJ/mol).
As a result of the measurements, it has been found
that for the reaction:
Basolid+ Sesolid+ 2O2 = BaSeO4
△H298 = -279.2 cal/mol
△F298 = -249.1 cal/mol
