The maximum allowable concentration of Pb2+ ions indrinking water is 0.05 ppm (that is, 0.05 g of Pb2+ in 1million g of water). Is this guideline exceeded if an undergroundwater supply is at equilibrium wiht the mineral anglesite,PbSO4 (Ksp=1.6 x10-8)?

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  1. ...........PbSO4 ==> Pb^2+ + SO4^2-

    Ksp = 1.6E-6 = (Pb^2+)(SO4^2-)
    Substitute the E line into the Ksp expression and solve for x = (PbSO4) in mols/L. Convert to grams Pb/L which will essentially be grams Pb in 1000 g. Compare with grams/10^6 g H2O

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  2. this aint it chief

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