Insoluble salt

[dttom]

We all know that even so called 'insoluble salt' would dissolve a bit in water forming a saturated solution. For example, PbSO4, how could we test for the sulphate ion inside?

I've asked my teacher about this question, he suggested adding enough amount of lead(ll) sulphate to a certain volume of water forming a saturated solution then filter the excess solid and then test for the ions by using the solution.

Now I've got two questions.

As there is an equilibrium between solid lead(ll) sulphate and the soluble mobile ions, PbSO4(s) + (aq) <----> Pb2+(aq) + SO42-(aq), as the solubility of lead(ll) sulphate is quite low, 0.99g per 100 ml water, we may expect that the equilibrium constant for the above equilibrium may be quite small, if we filter the excess solids added to water by filtration, that is, remove the reactant of the above equilibrium, I would expect the equilibrium position to shift leftwards, that is, forming back solid lead(ll) sulphate, is my expectation correct?

Beside, in spite of whether the equilbrium would shift leftwards, as the amount of originally dissolved PbSO4 is very small, if we just add acidified BaCl(aq) to it as usual, could we observe the result (as I think the amount is so small that I can not observe)?

[encipher]

Removing the solid from the solution will not do anything. The equilibrium constant does not contain solids. The Ksp would be as follows: Ksp = [Pb2+][sO42-] Therefore if solid is added or removed, it will not affect the amount of dissolved salt in the solution (given that it is already saturated)

[Tartaglia]

Ksp can actually contain terms for the concentration of solids, its just they are defined as 1 and then usually left out.

 

As for the eqm of solid and aqueous salt, it does actually exist, its just that you only need a infinitesimal amount of precipitate to set up the eqm as the solid automatically attains its defined concentration of 1. Thus there is no observable precipitate

[encipher]

Ksp can actually contain terms for the concentration of solids' date=' its just they are defined as 1 and then usually left out.

 

As for the eqm of solid and aqueous salt, it does actually exist, its just that you only need a infinitesimal amount of precipitate to set up the eqm as the solid automatically attains its defined concentration of 1. Thus there is no observable precipitate[/quote']

 

 

Meaning the solid has no effect on the solubility of the salt. No matter what you do, removing or adding solid will not cause the equilibrium to shift either way since Ksp is directly related to the molar solubility of the salt.

[dttom]

I get this point now, but what about the second question? As there is just very little amount of lead(ll) sulphate dissolved, would the barium sulphate is so little that we can't observe?

[encipher]

Umm.. I don't quite understand your question ?

[John Cuthber]

Make a saturated solution of barium sulphate, filter it, evaporate off nearly all the water. The excess BaSO4 will now be present as a solid.