Question

What is the solubility of `"barium sulfate"` in a `250*mL` volume of `0.01*mol*L^-1` `"sulfuric acid"`?

`K_"sp"("barium sulfate")=1.0842xx10^-10`

Answer 1

We need some data................

Explanation:

This site quotes `K_"sp"` for `BaSO_4` as `1.0842xx10^(−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`, then equilibrium exists between the species in solution, and the solid salt.

If `[Ba^(2+)][SO_4^(2-)]>1.0842xx10^-10`, then precipitation of `BaSO_4` will occur until equality is reached.

So, I think you have specified that `SO_4^(2-)=0.01*mol*L^-1`. And thus, at saturation point, if we add `Ba^(2+)` ion then..........

`[Ba^(2+)]-=(1.0842xx10^-10)/([SO_4^(2-)])`

`-=(1.0842xx10^-10)/(0.01*mol*L^-1)=1.08xx10^-8*mol*L^-1`

And so in a `250*mL` volume, then we have a mass of...........

`1.08xx10^-8*mol*L^-1xx0.250*Lxx208.23*g*mol^-1` with respect to `BaCl_2`, i.e. `=` `"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_4` in `mol*L^-1` with respect to pure water. Good luck.