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What is the molar solubility of lead(II) chromate (ksp=1.8x10^-14) in 0.13 M potassium chromate?

  • chemistry -

    Let X = solubility Pb(CrO4)2.
    ....Pb(CrO4)2 ==> Pb^2+ + 2CrO4^2-
    .......X...........X.......2X

    K2CrO4 is 100% dissociated.
    .........K2CrO4 ==> 2K^+ + CrO4^2-
    initial..0.13......0.......0
    change..-0.13......+0.13....0.13
    equil......0.......0.13......0.13

    Ksp = (Pb^2+)(CrO4^2-)
    For (Pb^2+) substitute X
    For (CrO4^2-) substitute 2X for that from Pb(CrO4)2 + 0.13 from K2CrO4 for total of 0.13+2X.
    Solve for X.

  • chemistry -

    4.46x10^-11 g/l

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