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- May 1, 2011
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What is the maximum concentration of fluoride ions that could be present in 0.032M Ba(NO3)2 (Ksp,BaF2 = 3.2x10-8)?
a.)5.0x10-4M
b.) 1.0x10-6M
c.) 2.0x10-3M
d.) 1.0x10-3M
I thought the answer was 5.0x10-4, but that actual answer according to Chad is 1.0x10-3. Here is his explaination:
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
*******I just don't understand why 3.2x10-8 = (0.032)[F-]2 is NOT 3.2x10-8 = (0.032)[2F-]2 ****
If anyone could please explain, it would be greatly appreciated 👍
a.)5.0x10-4M
b.) 1.0x10-6M
c.) 2.0x10-3M
d.) 1.0x10-3M
I thought the answer was 5.0x10-4, but that actual answer according to Chad is 1.0x10-3. Here is his explaination:
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
*******I just don't understand why 3.2x10-8 = (0.032)[F-]2 is NOT 3.2x10-8 = (0.032)[2F-]2 ****
If anyone could please explain, it would be greatly appreciated 👍