Confusing Solubility Question

[lazypremed]

Barium fluoride is a slightly soluble salt with Ksp=1.7*10^-6. If the concentration of barium ions in a saturated aqueous solution is equal to 4.3*10^-3 M, what is the concentration of fluoride ions in the solution?
a. 4.0*10^-4 M
b. 2.0*10^-2 M
c. 4.0*10^-2 M
d. 2.0*10^-1 M

Okay, so I thought it's possible to calculate the molarity of each ion in a saturated solution based only on the Ksp. So why do we need the concentration of barium ions? Furthermore, the concentration of barium ions they give us does not result in the given Ksp. And the answer I would normally look for would just be a concentration of fluoride that is twice that of barium, but that's not an answer and even if it were, it wouldn't combine with the given barium concentration to equal the given Ksp. What am I missing here?

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[ashtonjam]

If you set up the solubility product expression with BaF2:

Ksp = [Ba][F]^2

Plug in values for Ksp and [Ba]

1.7e-6 = (4.3e-3) * [F]^2

Solve for [F]^2.

(1.7/4.3) * 10^-3 = [F]^2

1.7/4.3 is around 1/2.5, which is 2/5. 1.7/4.3 is therefore 0.4.

0.4 * 10^-3 = [F]^2

Now, you have to turn the left side into something that is easily square rooted. Multiply 0.4 by 10 and divide 10^-3 by 10. This trick is pretty nice.

4 * 10^-4 = [F]^2
[F] = 2 * 10^-2

Main idea: Use the values they give you and solve for [F]. Part of it is setting up the right Ksp expression in the first place and remembering that Barium Fluoride is really BaF2. Then, the hardest part is the number manipulation to figure out 1.7/4.3, which is kind of a leap in logic but still doable, then finding out a good way to make your number square-rootable.

 

[lazypremed]

Thanks for the reply! The thing is, I understand how to do that, but what I don't understand is how the ratio of moles of barium to fluoride can be anything other than 1:2! How would you explain that?

 

[ashtonjam]

Not actually sure why but it just seems to be a consequence of the math >.> Maybe someone else can help shed light on that.

 

[gettheleadout]

This question brings to light an interesting nuance. The saturated aqueous solution isn't necessarily only water and BaF2. The solution is saturated with BaF2 such that the ion-product is equal to the Ksp. As ashtonjam showed us, that's how you have to solve it.

But why is the concentration of fluoride not twice that of barium ions? Maybe the solid BaF2 was dissolved in a solution of another fluoride salt, where the common ion puts the value of [F-] at greater than 2*[Ba2+] when the BaF2 is maximally dissolved, since the common ion effect fluoride exerts decreases the apparent solubility of the solid BaF2.