Chad's question on molar solubility

[Cofo]

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[Clancyp]

I had the same question.

 

[bharat008]

I had the same question.

me2..

 

[Clancyp]

Now looking at it, I think it may be because it's molar solubility is different, due to the common ion effect from Ba(NO3)2. So you have to use the expression?



Edit: I have in my notes that, "you can use that short cut of putting (2x)^2 only when its without the common ion effect." I may be wrong but I dont think its comparable with that 0.032M Ba(NO3)2 entered into the equation since Fluoride is present/dissociated in a different amount by its prior molecule.

 

[Double Bonded]

methree


I think it might be from his old (nonHD) lectures.

 

[tRad]

I may be completely wrong lol. Ba(NO3)2 will dissociate similarly to BaF2 ==> Ba^2+ You are given the Molarity for Ba(NO3)2 at 0.032M (which also means 0.032M of Ba^2+) and the Ksp for BaF2. Given this info all you need to do is find the [F^-] and square root it since there are two. You dont use (2x)^2 in this problem because you are already given the concentration of the Ba ion. The concentration of the floride ion will not be the same as the Ba ion, you just have to see that. Hope that helped and again, I may be wrong.

 

[tRad]

Now looking at it, I think it may be because it's molar solubility is different, due to the common ion effect from Ba(NO3)2. So you have to use the expression?



Edit: I have in my notes that, "you can use that short cut of putting (2x)^2 only when its without the common ion effect." I may be wrong but I dont think its comparable with that 0.032M Ba(NO3)2 entered into the equation since Fluoride is present/dissociated in a different amount by its prior molecule.

Good point, the common ion effect plays a part here.

 

[Cofo]

I may be completely wrong lol. Ba(NO3)2 will dissociate similarly to BaF2 ==> Ba^2+ You are given the Molarity for Ba(NO3)2 at 0.032M (which also means 0.032M of Ba^2+) and the Ksp for BaF2. Given this info all you need to do is find the [F^-] and square root it since there are two. You dont use (2x)^2 in this problem because you are already given the concentration of the Ba ion. The concentration of the floride ion will not be the same as the Ba ion, you just have to see that. Hope that helped and again, I may be wrong.

Wow, glad to see so many responses.
I am beginning to think the common ion effect does not play a part in solving this...especially because it doesn't even say the two solutions were mixed. The only reason it gives you the molar solubility for Ba(NO3)2 is to show you that the molar solubility for Ba^2+ is 0.032 M.
Given this info...the molar solubility of Ba^2+ in BaF2 will also be 0.032 M, as will F^2+.
However...if we were to use (2F^-)^2 .... we would be saying that we are trying to solve for 2 fluoride ions...instead of single fluoride ions.
I really think this is why the coefficient 2 is not used. I could be wrong but idk.
Anyone else think something different?

 

[bharat008]

Wow, glad to see so many responses.
I am beginning to think the common ion effect does not play a part in solving this...especially because it doesn't even say the two solutions were mixed. The only reason it gives you the molar solubility for Ba(NO3)2 is to show you that the molar solubility for Ba^2+ is 0.032 M.
Given this info...the molar solubility of Ba^2+ in BaF2 will also be 0.032 M, as will F^2+.
However...if we were to use (2F^-)^2 .... we would be saying that we are trying to solve for 2 fluoride ions...instead of single fluoride ions.
I really think this is why the coefficient 2 is not used. I could be wrong but idk.
Anyone else think something different?

Why don't you post this question on course saveer and see what Chad says. Hopefully, this will allow all of us to clear our doubts...
But please do repost what he says over here..

👍

 

[Cofo]

You dont use (2x)^2 in this problem because you are already given the concentration of the Ba ion. The concentration of the floride ion will not be the same as the Ba ion, you just have to see that. Hope that helped and again, I may be wrong.

Ahh...so you are basically saying that since the concentration of 1 Ba^2+ has a molar solubility of 0.032 M ....then the concentration of 1 F^2+ must also be 0.032 M?

This makes sense, and I think is what I said in my most previous post. Cool, cool...but if this is true, then the common ion effect has nothing to do with trying to solve for the answer. It's as simple as what I said above I'm guessing!

 

[Cool Beans]

For BaF2, Ksp=[Ba2+][F-]^2. For finding the molar solubility (x) in water we would substitute into the expression giving Ksp=(x)(2x)^2. Note that [F-]=2x.

Note that setting [Ba2+]=x and [F-]=2x assumes that these ions are being produced solely by the dissolving of an 'x' molar concentration of BaF2(s).

But in the actual problem that you're trying to solve this is not the case. The Ba2+ ions are not coming from BaF2(s) at all but from Ba(NO3)2, a strong electrolyte which dissociates completely. And likewise the F- could just simply be coming from the addition of some other strong electrolyte (like NaF or something). So doing any substitution involving x is really not appropriate in this problem.

For the problem you're working on, we're still going to start with the same expression
Ksp=[Ba2+][F-]^2
You're given the Ksp and [Ba2+] so you can just substitute them in and solve for [F-].

Note again that the question didn't ask you to solve for the molar solubility (x), it asked you to solve for [F-] and so there's no reason to substitute 2x in for [F-]. But assuming we did it anyway. If you did then when you solved for x you'd have to multiply by 2 as [F-]=2x according to your substitution.

Hope this helps.

 

[Cofo]

For BaF2, Ksp=[Ba2+][F-]^2. For finding the molar solubility (x) in water we would substitute into the expression giving Ksp=(x)(2x)^2. Note that [F-]=2x.

Note that setting [Ba2+]=x and [F-]=2x assumes that these ions are being produced solely by the dissolving of an 'x' molar concentration of BaF2(s).

But in the actual problem that you're trying to solve this is not the case. The Ba2+ ions are not coming from BaF2(s) at all but from Ba(NO3)2, a strong electrolyte which dissociates completely. And likewise the F- could just simply be coming from the addition of some other strong electrolyte (like NaF or something). So doing any substitution involving x is really not appropriate in this problem.

For the problem you're working on, we're still going to start with the same expression
Ksp=[Ba2+][F-]^2
You're given the Ksp and [Ba2+] so you can just substitute them in and solve for [F-].

Note again that the question didn't ask you to solve for the molar solubility (x), it asked you to solve for [F-] and so there's no reason to substitute 2x in for [F-]. But assuming we did it anyway. If you did then when you solved for x you'd have to multiply by 2 as [F-]=2x according to your substitution.

Hope this helps.

Ahh...I understand now. Thank you very much Chad. You rock 👍
I'm sending you a check after I become a dentist.

 

[Double Bonded]

For BaF2, Ksp=[Ba2+][F-]^2. For finding the molar solubility (x) in water we would substitute into the expression giving Ksp=(x)(2x)^2. Note that [F-]=2x.

Note that setting [Ba2+]=x and [F-]=2x assumes that these ions are being produced solely by the dissolving of an 'x' molar concentration of BaF2(s).

But in the actual problem that you're trying to solve this is not the case. The Ba2+ ions are not coming from BaF2(s) at all but from Ba(NO3)2, a strong electrolyte which dissociates completely. And likewise the F- could just simply be coming from the addition of some other strong electrolyte (like NaF or something). So doing any substitution involving x is really not appropriate in this problem.

For the problem you're working on, we're still going to start with the same expression
Ksp=[Ba2+][F-]^2
You're given the Ksp and [Ba2+] so you can just substitute them in and solve for [F-].

Note again that the question didn't ask you to solve for the molar solubility (x), it asked you to solve for [F-] and so there's no reason to substitute 2x in for [F-]. But assuming we did it anyway. If you did then when you solved for x you'd have to multiply by 2 as [F-]=2x according to your substitution.

Hope this helps.

Ahh...I understand now. Thank you very much Chad. You rock 👍
I'm sending you a check after I become a dentist.

Is this chad?

 

[tRad]

Thats the explanation we all needed!!

 

[nimman1126]

Hey I tried it the other way to see and I got x= 2 X 10^-3... So to get the F- concentration from here I divide by 2 right? Or did I make an error in calculating?