What is the volume of water needed to dissolve one gram of Barium sulphate at 25 degrees is
Dear Student,
The ksp of BaSO4 is 1.1 x 10-10.
On dissociation, BaSO4 produce one mole of Ba2+ and 1 mole of SO42-.
Hence ksp expression would be = s x s = s2.
s = (1.1 x 10-10)1/2 = 1.0488 x 10-5 M
Molecular mass of BaSO4 = 233.38 g/mol
Hence, strength of solution would be = 233.38 x 1.0488 x 10-5 = 2.4476 x 10-3 g/L
Hence, the solubility of BaSO4; 2.4476 x 10-3 g in 1L of water
So 1 g of BaSO4 needed = 1/(2.4476 x 10-3) = 408 L
Regards
The ksp of BaSO4 is 1.1 x 10-10.
On dissociation, BaSO4 produce one mole of Ba2+ and 1 mole of SO42-.
Hence ksp expression would be = s x s = s2.
s = (1.1 x 10-10)1/2 = 1.0488 x 10-5 M
Molecular mass of BaSO4 = 233.38 g/mol
Hence, strength of solution would be = 233.38 x 1.0488 x 10-5 = 2.4476 x 10-3 g/L
Hence, the solubility of BaSO4; 2.4476 x 10-3 g in 1L of water
So 1 g of BaSO4 needed = 1/(2.4476 x 10-3) = 408 L
Regards