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Topic: Solubility  (Read 6457 times)

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Tom

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Solubility
« on: July 01, 2004, 04:31:13 PM »
If the solubility product for BaF2 is 1.7 x 10-6 at 25 º C.  Acording to my calculations the molar solubility is   1.3 x 10-3 M. Can anyone verify this for me?

P.S. I still have not been able to utilize the subscript, can anyone indicate the procedure?   The -3 and -6 are suppose to be subscripted!
   
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Offline AWK

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Re:Solubility
« Reply #1 on: July 02, 2004, 08:27:34 AM »
Ksp=[Ca2+ ]x[F- ]2
[F- ]=2[Ca2+ ]
Hence molar solubility= [Ca2+ ]=SQRT(Ksp/4)=7.5x10-3
« Last Edit: July 02, 2004, 08:49:50 AM by AWK »
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Tom

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Re:Solubility
« Reply #2 on: July 02, 2004, 11:02:51 AM »
Thank you I see were I made my mistake I had calculated the wrong numerals. Thanks again.
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Offline Donaldson Tan

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Re:Solubility
« Reply #3 on: July 03, 2004, 10:17:02 PM »
Quote from: AWK on July 02, 2004, 08:27:34 AM
Ksp=[Ca2+ ]x[F- ]2
[F- ]=2[Ca2+ ]
Hence molar solubility= [Ca2+ ]=SQRT(Ksp/4)=7.5x10-3

I think you meant [Ba2+ ], not [Ca2+ ]

However, since
[F- ]=2[Ba2+ ]
then
Ksp=[Ba2+ ]x[F- ]2 = 4 [Ba]2+ ] 3

Hence, molar solubility is ( Ksp/4 ) 1/3

« Last Edit: July 03, 2004, 10:17:55 PM by geodome »
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Offline AWK

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Re:Solubility
« Reply #4 on: July 05, 2004, 07:54:49 AM »
Thanks, Geodome
I put wrong cation, but calculations ar OK.
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