Chads GC quiz problem

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michaelamae1

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What is the molar solubility of BaF2 (Ksp=3.2 x 10^-8)?

A. 8.0 x 10-9M
B. 1.6 x 10^-8M
C. 2.0 x 10^-3M
D 1.8 x 10^-4M

The answer I got was some of A but mine was 8.0 x 10^-3. The answer is C. Can someone explain how it is C?
 
Ksp = [Ba] [2F]^2 = 3.2 x 10^-8
Ksp = 4x^3 = 3.2 x 10^-8
x^3 = 8 x 10^-9
x = 2 x 10^-3 (choice C)

the 4x^-3 is the solution product when three things can dissolve (e.g. 1 Ba and 2 F)

Hope this helps.
 
lite bear explained it verywell.
BaF2. Ba is alkaline earth metal forming +2 cation and F forming -1anion.
therefore you need two F for each Ba to balance the positive and negative number.
+2=2[-1]
Ba=2F

hope it helps you out.
 
How I did it was
BaF_2===> Ba^2+ + 2F^-
I 0 0
C x 2x
E x 2x
Ksp=(x)(2x)^2
(3.2*10^-8)=(x)(2x)^2
(3.2*10^-8)= (x)(4x^2)
(3.2*10^-8)=4x^3
(3.2*10^-8)/4=x^3
8.0*10^-8=x^3
then cube root it and the cube of 8.0 is 2


Hope this helps 😛
 
How I did it was
BaF_2===> Ba^2+ + 2F^-
I 0 0
C x 2x
E x 2x
Ksp=(x)(2x)^2
(3.2*10^-8)=(x)(2x)^2
(3.2*10^-8)= (x)(4x^2)
(3.2*10^-8)=4x^3
(3.2*10^-8)/4=x^3
8.0*10^-8=x^3
then cube root it and the cube of 8.0 is 2


Hope this helps 😛

Thanks, that is exactly what i needed to know. You cube it!!!! Dahhh, sorry for the dumb question!!
 
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