EK vs. Princeton on molar solubility

[nr6unhH]

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on one of the princeton FL, it asked to find molar solubility of F- when given Ksp of barium flouride = 1.7*10^-6 and M for Ba is 4.3*10^-3

Their explanation for answer goes...
AbCd <-b, d are subscripts
[A]^b x [C]^d

Ksp=[Ba2+]^1 [F-]^2= 1.7*10^-6= (4.3*10^-3)(x^2)

; x^2= 4.0x10^-4
x=2.0*10^-2



BUT THEN EK HAS BEEN EXPLAINING THIS solubility product to be..

Ba(OH)2-> Ba2+ + 2OH-
Ksp= [Ba2+][OH-]^2
2.4*10^-5= (x)(2x)^2
x=1.8*10^-2


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for Barium flouride i thought it also dissociates to Ba2+ and 2F- so wouldn't it follow the same procedure as EK? what's with this discrepancy

thanks!

 

[Frazier]

on one of the princeton FL, it asked to find molar solubility of F- when given Ksp of barium flouride = 1.7*10^-6 and M for Ba is 4.3*10^-3

Their explanation for answer goes...
AbCd <-b, d are subscripts
[A]^b x [C]^d

Ksp=[Ba2+]^1 [F-]^2= 1.7*10^-6= (4.3*10^-3)(x^2)

; x^2= 4.0x10^-4
x=2.0*10^-2



BUT THEN EK HAS BEEN EXPLAINING THIS solubility product to be..

Ba(OH)2-> Ba2+ + 2OH-
Ksp= [Ba2+][OH-]^2
2.4*10^-5= (x)(2x)^2
x=1.8*10^-2


---

for Barium flouride i thought it also dissociates to Ba2+ and 2F- so wouldn't it follow the same procedure as EK? what's with this discrepancy

thanks!

Both of those answers are correct with an ounce of rounding...

What is the trouble?

 

[Frazier]

Ksp BaF2 = 1.7*10^-6

1.7*10^-6 = [(4.3*10^-3)][(x)^2]
3.95*10^-4 = [(x)^2]
3.95*10^-4 = (x^2)
1.99*10^-2 = x (molar solubility of F-)

Ksp Ba(OH)2 = 2.4*10^-5
2.4*10^-5 = [(x)][(2x)^2]
2.4*10^-5 = [(x)][4x^2]
2.4*10^-5 = [4x^3]
6.0*10^-6 = [x^3]
1.8*10^-2 = x (molar solubility)

 

[nr6unhH]

Just didnt understand why it would be x^2 for the first one and (2x)^2 for the second one.. when both of anions dissociate into 2 molecules.

 

[fizzgig]

molecule AB^2 dissociates into A and 2B

Ksp = A^1 B^2

one of the solns GAVE you the concentration of A. plug it in. now you're solving directly for B. that is why it's just Ksp = GIVENVALUE * X^2

the other does not. but you know B is 2*A.
so Ksp = X*(2X)^2 = 4X^3
here you are solving for X, the solubility of A. the solubility of B is 2X, so double it and that's the solubility of B.