Problem: What concentration of the lead ion, Pb 2+, must be exceeded to precipitate PbF2 from a solution that is 1.00x10-2 M in the fluoride ion, F- =  Ksp for lead(II) fluoride is 3.3x10-8.

FREE Expert Solution

For this problem, we´re being asked to calculate the concentration of the lead ion Pb2+ that must be exceeded to precipitate PbF2 from a solution of fluoride ion F-.

In water, the dissociation of PbF2 is the following:

PbF2(s)  Pb2+(aq) + 2 F-(aq)


The Ksp expression for PbF2 is:

Where solids are ignored (reactants), so:

Ksp=products=Pb2+F-2

Note that each concentration is raised by the stoichiometric coefficient: [Pb2+] is raised to 1 and [F-] is raised to 2.


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Problem Details

What concentration of the lead ion, Pb 2+, must be exceeded to precipitate PbF2 from a solution that is 1.00x10-2 M in the fluoride ion, F- =  Ksp for lead(II) fluoride is 3.3x10-8.

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Our tutors have indicated that to solve this problem you will need to apply the Ksp concept. You can view video lessons to learn Ksp. Or if you need more Ksp practice, you can also practice Ksp practice problems.