chemistry

A solution is prepared by adding some solid PbF2 to water and allowing the solid to come to equilibrium with the solution. The equilibrium constant for the reaction below is Kc = 3.6x10−8.

PbF2(s) ⇌ Pb2+(aq) + 2 F−(aq)

What is the equilibrium concentration of Pb2+ in the solution? Assume that some of the solid remains undissolved.

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  1. ......................PbF2(s) ⇌ Pb2+(aq) + 2 F−(aq)
    I.....................solid.............0......................0
    C....................solid.............x.....................2x
    E.....................solid.............x.....................2x
    K = (Pb^2+)(F^-)^2
    Plug the E line into the Kc expression and solve for x = (Pb^2+) = ?

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    DrBob222
  2. what do you mean plug it into the kc expression? do i just use x and 2x to plug into K = (Pb^2+)(F^-)^2?

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  3. 3.6x10−8 = x(2x)^2
    x^3=9.0x10^-9
    x=0.002

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