This site quotes K_"sp"Ksp for BaSO_4BaSO4 as 1.0842xx10^(−10)1.0842×10−10 under standard conditions.......so it is pretty insoluble stuff.
What does this mean? We it means that if the ion product [Ba^(2+)][SO_4^(2-)]=1.0842xx10^-10[Ba2+][SO2−4]=1.0842×10−10, then equilibrium exists between the species in solution, and the solid salt.
If [Ba^(2+)][SO_4^(2-)]>1.0842xx10^-10[Ba2+][SO2−4]>1.0842×10−10, then precipitation of BaSO_4BaSO4 will occur until equality is reached.
So, I think you have specified that SO_4^(2-)=0.01*mol*L^-1SO2−4=0.01⋅mol⋅L−1. And thus, at saturation point, if we add Ba^(2+)Ba2+ ion then..........
[Ba^(2+)]-=(1.0842xx10^-10)/([SO_4^(2-)])[Ba2+]≡1.0842×10−10[SO2−4]
-=(1.0842xx10^-10)/(0.01*mol*L^-1)=1.08xx10^-8*mol*L^-1≡1.0842×10−100.01⋅mol⋅L−1=1.08×10−8⋅mol⋅L−1
And so in a 250*mL250⋅mL volume, then we have a mass of...........
1.08xx10^-8*mol*L^-1xx0.250*Lxx208.23*g*mol^-11.08×10−8⋅mol⋅L−1×0.250⋅L×208.23⋅g⋅mol−1 with respect to BaCl_2BaCl2, i.e. == "something less than a microgram"something less than a microgram
In pure water, would barium sulfate be more or less soluble? In fact, I think it would be a good idea to quote the solubility of BaSO_4BaSO4 in mol*L^-1mol⋅L−1 with respect to pure water. Good luck.
