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Rhiannon Red
Rhiannon Red asked in Science & MathematicsChemistry · 5 years ago

Please please help me with this solubility question, have no idea how to do it?

Lead chloride dissolves in water as described by the following equation:

PbCl2(s) --> Pb2+(aq) + 2 Cl-(aq)

At an unknown temperature, once equilibrium has been reached, the concentration of Pb2+(aq) ions in solution was found to be 0.0157 mol dm-3.

Calculate the solubility product of PbCl2 at this temperature. Hint: the concentration of Cl-(aq) ions must be twice that of the Pb2+(aq) ions

2 Answers

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  • pinkjenna96
    5 years ago

    Ksp= [Pb2+][Cl-]^2 (PbCl2 is ignored because it is a solid)

    You know the concentration of lead ion, and that the concentration of chlorine must be twice that.

  • Fern
    Lv 7
    5 years ago

    PbCl2(s) ==⇒ Pb+2 + 2Cl-

    [Pb+2][Cl-]^2 = Ksp

    [Pb+2] = 0.0157 ; [Cl-] = 2(0.0157) = 0.0314

    0.0157(0.0314)^2 = 1.55 x 10^-5

    Ksp = 1.55 x 10^-5

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