Wat are the equilibrium concentrations of pb2+ and f- in a saturated solution

So the question goes

"What are the equilibrium concentrations of Pb2+ and F- in a saturated solution of lead fluoride if the Ksp of PbF2 is 3.2*10^-8"

So I used the formula

(x)(2x)^2 = 3.2 *10^-8

4x^3 = 3.2 *10^-8

I solved for .002. Which IS the correct conc. for Pb2+.

However I am unable to find the conc. for F-. Help?

1 Answer

  • If 0.002 mole/L of PbF2 dissolves, the concentration of Pb^+2 will be 0.002 mole/L and the concentration of F^-1 must be 0.004 mole/L (2 x 0.002 = 0.004)

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