Question

The maximum allowable concentration of pb2+ ions in drinking water is 0.05 ppm (i.e., 0.05 g of pb2+ in 1 million grams of water). (a) what is the pb2+ concentration of an underground water supply at equilibrium with the mineral anglesite (pbso4) (ksp = 1.6 × 10−8) ?

Answer 1

PbSO₄ partially dissociates in water. the balanced equation is;
                    
                       PbSO₄(s) ⇄  Pb²⁺(aq) + SO₄²⁻(aq)
Initial                                     -                -
Change             -X               +X           +X
Equilibrium                           X              X

Ksp           =    [Pb²⁺(aq)] [SO₄²⁻(aq)]
1.6 x 10⁻⁸  =    X * X
1.6 x 10⁻⁸  =    X²
          X    =   1.3 x 10⁻⁴ M
      
Hence the Pb²⁺ concentration in underground water is 1.3 x 10⁻⁴ M. 
[Pb²⁺]  = 1.3 x 10⁻⁴ M.
           = 1.3 x 10⁻⁴ mol / L x 207 g / mol 
           = 26.91 ppm