# CHEMISTRY

posted by Anonymous

The solubility of barium sulfate varies with the composition of the solvent. In which one of the solvent mixtures below would BaSO4 have the lowest solubility?
i)1.0 M (NH4)2SO4(aq), ii) 0.10 M Na2SO4(aq), iii) pure water, iv) 1.0 M HCl(aq)
v) 0.5 M Ba(NO3)2(aq)

I have no idea how to do this question

1. DrBob222

So you understand the theory behind this.
BaSO4 ==> Ba^2+ + SO4^2-
Le Chatelier's Principle says that when we do something to a reaction at equilibrium the reaction will shift so as to undo what we did. SO, if we add (NH4)2SO4 to the reaction, that's adding SO4^2-. Adding that common ion make the reaction move to the left because it wants to LOWER the sulfate so making itself less soluble it can do that
To calculate the solubility of BaSO4 in 1.0 M (NH4)2SO4, we do the following:
........BaSO4 ==> Ba^+2 + SO4^2-
I.......solid.....0........0
C.......solid.....x........x
E.......solid.....x........x
Ksp = 1E-10 = (Ba^2+)(SO4^2-)
in pure water, then,
(Ba^2+) = x
(SO4^2-) = x
So 1E-10 = x*x and x = 1E-5 = solubility of BaSO4 in pure water.

i, ii, and v are done the same way lik this for i.
(NH4)2SO4 is 100% ionized like this.
..........(NH4)2SO4 ==> 2NH4^+ + SO4^-
I..........1.0M.........0.........0
C.........-1.0.........2.0........1.0
E...........0..........2.0........1.0

Ksp is the same but the concentrations are not.
(Ba^2+) = x
(SO4^2-) = x from BaSO4 and 1.0 from the (NH4)2SO4. Solve for x - solubility and you get 1E-10 which is far less than the solubility in pure water which I did above. You can work the others, ii is done the same way. v is the same way EXCEPT Ba is the common ion and not the SO4^2-.

You need not worry about the 1.0M HCl since that INCREASES the solubility. WHY? For this reason.
BaSO4 ==> Ba^2+ + SO4^2- BUT
HCl ==> H^+ + Cl^- (no common ion here)
Note that SO4^2- + H^+ ==> HSO4^- and this is k2 for H2SO4. Note the H^+ is readcting with the SO4 which removes the SO4^2-, remember if we remove SO4^2- the reaction will shift to ADD more so it moves to the right which INCREASES the solubility. Hope all of this helps.

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