Molar solubility questions

[pilotdentist]

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

Solution: 3.2X10^-8 = .32 * X^2
2-What is the molar solubility of a salt of the formula AB3 (Ksp=3.0x10-19)?

Solution: 3.0X10^-19 = X * (3X)^3
Why does problem 1 use X^2 and not 2X^2 similar to question 2? Does it have to do with the question asking for IONS present vs molar solubility? Does anyone have a good set of rules for evaluating these problems?

Thank you!

 

[1Overdrive]

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

Solution: 3.2X10^-8 = .32 * X^2
2-What is the molar solubility of a salt of the formula AB3 (Ksp=3.0x10-19)?

Solution: 3.0X10^-19 = X * (3X)^3
Why does problem 1 use X^2 and not 2X^2 similar to question 2? Does it have to do with the question asking for IONS present vs molar solubility? Does anyone have a good set of rules for evaluating these problems?

Thank you!

Where did you get these solutions? I am pretty sure it is definitely supposed to be (2x)^2 which would come out to 4*x^2.

Let's work through it:

BaF2 --> Ba(2+) + 2F(-)
I _______.032___ 0
C _______+s____ +2s
E ______.032____ 2s

As a general rule, s is negligible when Ksp is < 1x10^-5. Usually always negligible for the purposes of the DAT.

Equilibrium expression: [Ba(2+)][F(-)]^2 = Ksp

Now calculations: .032(2s)^2 = 3.2E-8
.032(4s^2) = 3.2E-8
s is the molar solubility

Something is wrong with the answer they gave in the first problem. In the second problem, the solution seems to work out, not sure why they didnt do it that way for the first problem.

 

[pilotdentist]

It's from Chad's video's. I did some more digging and found the quoted solution below:

My basic understand is that it asks for the concentration of fluoride ions NOT the molar solubility, you do NOT use the coefficient. I am still struggling with the WHY. Molar solubility is moles of solute per L of solution before saturation. I realize F- has a 2:1 ratio with Ba2+ so I guess it is 2 * the molar solubility.... i dunno.



"This is the single most common question I get. I'll try to make a video solution to it soon to help with the confusion.

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.

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 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-]. If you did then when you solved for x you'd have to multiply by 2 as [F-]=2x according to your substitution.
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"

 

[Cool Beans]

If you wanted to solve for the molar solubility then you could indeed substitute in 2x for [F-]. If you did this you would get x = 5.0x10^-4.

But the question asked for the concentration of F- so you'd have to go back to the fact that [F-] = 2x.

So multiplying 5.0x10^-4 by 2 gets us 1.0x10^-3, the same answer as above.

The key is that this is the long way to do this problem whereas the solution in the above post is the short way. If the question is asking us to solve for [F-] and we already have it in an equation then why would we want to substitute in 2x for it. The only reason is that this is what we'd normally do solve for the molar solubility and we're in the habit of solving similar solubility problems by doing so. It really is the long way around but you will still arrive at the correct answer.

Hope this helps!