[sillykid31]

Hi, I completed this question but my teacher gave me the answer 1.1x10^-10. Not sure where my mistake was, any help at all would be greatly appreciated

The solubility of barium sulphate, BaSO4, is 2.4 x 10-4 g/100 mL of solution. Calculate the Ksp of this solution.

2. The dissociation equation is:
BaSO4 = Ba2+  +  SO42-
Ksp= [Ba2+] [SO42-]
[Ba2+]2 = x
[Ba]=[SO4]=x2
√xmolBasO4/ 1L x 233.43gBaSO4/ 1mol = 2.4 x 10-3gBaSO4
√X x 233.43= 2.4 x 10-3 / 233.43
√ x = 0.000010281
x= 3.2 x 10-3 (3.2 x 10-3)
Therefore, the Ksp is 1.024 x 10-5

[Corribus]

Because of your formatting, it's hard to follow your work. Your answer is the square root of the given answer, which should give you some clue.

If the solubility is given, how does this relate to the concentration at equilibrium of each ion?

[sillykid31]

Thank you for your reply. Sorry about the formatting

I see that \i'm off by the square root and I noticed a typo in my explanation:

[Ba]=[SO4]=x2  should be [Ba]=[SO4]=root

• . After redoing the math, I realized that I actually did use the square root to get the answer, but mistyped it here. The problem is that with that corrected, the math is sound, so I still don't know where I went wrong. It would appear that I should not have taken the square root there, but then the math is inconsistent.

Did I go about this entirely the wrong way?

Thanks again
 

[Corribus]

In my experience, solubility equilibria is one of the more confusing types of equilibrium topics encountered in general chemistry because it involves two phases, one of which does not show up in the equilibrium expression.

The solubility of a substance is the maximum amount of a substance that will dissolve in a solvent. Any more substance added will not dissolve. In the case of a salt like barium sulfate, the dissolved material actually dissociates into two ions. For every x moles of barium sulfate that dissolves, x moles of barium ion and sulfate ion appears in solution, and no barium sulfate remains, so long as x is less than the solubility of barium sulfate. The solubility product is the product of the concentrations of all the dissolved products of the dissolution when an equilibrium is formed between the solid and its dissolved products, which only occurs when more barium sulfate is added than the solution can hold. (In a way, up until the solubility limit has been reached, the substance behaves a little like a concentrated acid - complete dissociation.)

SO!

To find the solubility product, you need to know how much of each of barium and sulfate are in the solution, on a molar basis, at the point where the solution can dissolve no more of either of them - i.e., when the solubility point has been reached.

Most graciously, this last piece of information has been given to you, although you'll have to convert it to a molarity.

Call this value Q. If Q moles of barium sulfate are added to 1L of water, and this is the most moles that 1L of water can hold, how much of barium and sulfate ion are in solution? Call this value... er... Z1 and Z2 for barium and sulfate, respectively. You can then take the product of these two values, and this is your solubility product.

So, let's start with: what is Q, Z1, and Z2?