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posted by  FishStick on 2/22/2008 3:52:02 PM  |  status: Live  

Please help

Course Textbook Chapter Problem
General Chemistry N/A N/A N/A
Question Details:
A saturated solution of lead(II) fluoride, \rm PbF_2, was prepared by dissolving solid \rm PbF_2 in water. The concentration of \rm Pb^{2+} ion in the solution was found to be 2.08\times 10^{-3}\;M. Calculate K_sp for \rm PbF_2.
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posted by Clark on 2/22/2008 7:46:23 PM  |  status: Live
Asker's Rating: Helpful   
FishStick's comment:
"I figured it out after i posted. Not the correct answer but thanks for trying!"
Response Details:
For the KSP in this question, Ksp = [Pb2+][F1-]2.  The fluorine ion is squared because in PbF2, you have 2 fluorine ions per molecule, as opposed to 1 Pb2+ ion.  If the question was find the KSP for Mg2N3, Mg would be raised to the exponent 2 and N would be raised to the exponent 3.
 
Now all you have to do is plug the concentration of Pb2+ into the equation and solve for KSP:
KSP = [0.00208][0.00208]2
= 8.998 x 10-9
 
I hope this helps!!
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