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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Tom on July 01, 2004, 04:31:13 PM

Title: Solubility
Post by: Tom 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!
   

Title: Re:Solubility
Post by: AWK on July 02, 2004, 08:27:34 AM

K_sp=[Ca²⁺ ]x[F^- ]²
[F^- ]=2[Ca²⁺ ]
Hence molar solubility= [Ca²⁺ ]=SQRT(K_sp/4)=7.5x10^-3

Title: Re:Solubility
Post by: Tom 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.

Title: Re:Solubility
Post by: Donaldson Tan on July 03, 2004, 10:17:02 PM

Quote from: AWK on July 02, 2004, 08:27:34 AM

K_sp=[Ca²⁺ ]x[F^- ]²
[F^- ]=2[Ca²⁺ ]
Hence molar solubility= [Ca²⁺ ]=SQRT(K_sp/4)=7.5x10^-3

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

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

Hence, molar solubility is ( K_sp/4 ) ^1/3

Title: Re:Solubility
Post by: AWK on July 05, 2004, 07:54:49 AM

Thanks, Geodome
I put wrong cation, but calculations ar OK.