[ThatGuy]

I'm having difficulty understanding what Ksp is. I know how to calculate it and whatnot, but I don't quite understand how it works.

For example:

BaSO_{4 (s)} <--> Ba ²⁺ _(aq) + SO₄ ^2- _(aq)

Ksp = 1.0 x 10^-10

I can figure out that the solubility is the square root of 1.0 x 10^-10 which is 1.0 x 10^-5 .

But, what does this mean? Suppose I were to add an amount of  BaSO₄ less than 1.0 x 10^-5 to a beaker . Would there be some solid deposited at the bottom of the beaker... to me it seems that there would be some deposit at the bottom and it would be in dyanmic equilibrium.

What if the amount of BaSO₄ added were to exceed the solubility?

[Corribus]

It means that you can add barium sulfate to a solution until the product of the concentrations of barium ion and sulfate ions is equal to the value of Ksp, and the barium sulfate will completely dissolve. At that point, any additional barium sulfate added to the solution will remain undissolved, at least until you remove some of the barium or sulfate ions.

[ThatGuy]

It means that you can add barium sulfate to a solution until the product of the concentrations of barium ion and sulfate ions is equal to the value of Ksp, and the barium sulfate will completely dissolve. At that point, any additional barium sulfate added to the solution will remain undissolved, at least until you remove some of the barium or sulfate ions.

Ok. So if I understand correctly, any barium sulfate added up to this point will dissolve completely, it will not deposit at the bottom of the beaker. If I were to add any more barium sulfate past this point, it would begin to settle at the bottom?

[Borek]

Ok. So if I understand correctly, any barium sulfate added up to this point will dissolve completely, it will not deposit at the bottom of the beaker.

Yes.

If I were to add any more barium sulfate past this point, it would begin to settle at the bottom?

Yes.

Note that it can be enough to add only one of the ions to precipitate the solid. All that is required is that the product of the concentrations is higher than the Ksp, concentrations don't have to be identical (which is the case in the saturated solution prepared by dissolving solid barium sulfate).

[ThatGuy]

Note that it can be enough to add only one of the ions to precipitate the solid. All that is required is that the product of the concentrations is higher than the Ksp, concentrations don't have to be identical (which is the case in the saturated solution prepared by dissolving solid barium sulfate).

Yes, I understand this part.

When you say that it is enough to 'precipitate the solid' - you mean that it is enough to cause some of the solid to settle or all of the solid?

Also, when you say 'saturated solution', you mean when the products of the ion concentrations equal the Ksp, correct?

(Sorry if these are 'dumb' questions, I'm trying to make sure I understand what I'm doing... thank you for the help)

[Borek]

When you say that it is enough to 'precipitate the solid' - you mean that it is enough to cause some of the solid to settle or all of the solid?

Some. Just enough for the product of concentrations to get down to the Ksp.

Also, when you say 'saturated solution', you mean when the products of the ion concentrations equal the Ksp, correct?

Yes.

(Sorry if these are 'dumb' questions, I'm trying to make sure I understand what I'm doing... thank you for the help)

Nothing dumb about them. At least so far