Solubility for a Sparingly Soluble Salt

Chemistry
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The solubility product of lead (2) fluoride, PbF2, is 3.7x10^-8. Estimate the solubility of the salt.

Mar 10th, 2015

The dissociation of lead (ii) fluoride is given by…

           PbF2 (s)             <-->            Pb2+       +       2 F-

I..........................................................0……………….0

C......................................................+ S…..............+ 2 S

E.......................................................S………….…..2S

The equilibrium expression is given by…

Ksp = [Pb2+] * [F-]^2

3.7E-8 = (S)(2S)^2

Rearranging to solve for S…

S = (3.7E-8 / 4) ^ (1/3)

S = 2.099168E-3 M

  = 2.1 E-3 M or 0.0021 M (with sigfigs in mind)


Mar 10th, 2015

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