Ksp on chad's notes quiz

[Molarcule]

---

Members do not see this ad.

What is the maximum concentration of fluoride ions that could be present in 0.032M Ba(NO3)2 (Ksp,BaF2 = 3.2x10-8)?

This is how you do it.
BaF2(s) → Ba2+(aq) + 2F-(aq)
Ksp = [Ba2+][F-]2
3.2x10-8 = (0.032)[F-]2

(1.0x10-6)1/2=[F-]
[F-]=1.0x10-3M

But my question is. Since it is BaF2 it would yield 2F. When I do the I.C.E table I would get [x][2x]^2. Why is it not 2x and instead x?

 

[HannibalLecter]

What is the maximum concentration of fluoride ions that could be present in 0.032M Ba(NO3)2 (Ksp,BaF2 = 3.2x10-8)?

This is how you do it.
BaF2(s) → Ba2+(aq) + 2F-(aq)
Ksp = [Ba2+][F-]2
3.2x10-8 = (0.032)[F-]2

(1.0x10-6)1/2=[F-]
[F-]=1.0x10-3M

But my question is. Since it is BaF2 it would yield 2F. When I do the I.C.E table I would get [x][2x]^2. Why is it not 2x and instead x?

I think you are having difficulty with understanding the basics of this particular Ksp problem. The question is asking for the total possible number of fluoride ions that will be present in the solution if a common ion solution of Ba(NO3)2 is added to a solution of BaF2. If you set up the Ksp problem for this yes it will look like you have stated. But since the problem asks for the total number of fluoride ions in the solution figuring out the x would just be half the fluoride ions that will be present in the solution, get it? Let me know if you need any further explanation.

 

[FROGGBUSTER]

i'm pretty sure it should be 2x like you're saying.

 

[Molarcule]

I guess if you look at per flouride ion then it would just be x. But as flouride ions all together it would be times 2.

 

[brainache]

The question's asking for CONCENTRATION of fluoride, not molar solubility. Molar solubility would be the (2x)^2 you were referring to. Concentration = (x*molar solubility)^x, where x is the coefficient of the ion.

 

[Golfguy]

What is the maximum concentration of fluoride ions that could be present in 0.032M Ba(NO3)2 (Ksp,BaF2 = 3.2x10-8)?

This is how you do it.
BaF2(s) → Ba2+(aq) + 2F-(aq)
Ksp = [Ba2+][F-]2
3.2x10-8 = (0.032)[F-]2

(1.0x10-6)1/2=[F-]
[F-]=1.0x10-3M

But my question is. Since it is BaF2 it would yield 2F. When I do the I.C.E table I would get [x][2x]^2. Why is it not 2x and instead x?

[F-] = the concentration from the equation:
Ksp = [Ba2+][F-]2

Plugging in "2x" would simply give you the molar solubility. Then you would have to multiply it by 2 to get the concentration back.