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#

1 Answer
Apr 20, 2017

Answer:

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.