Question

Lead(II) sulfate is the crust that forms on battery terminals, its solubility in water is 0.00425 g/ 100 mL. If 20.0 kg of lead(II) sulfate has been stored of in a 100 L tank filled with water what is the concentration of Pb2+ in the water? (KSP = 6.3 x 10-7, MW = 303.26 g/mol)

Answer 1

mass of PbSO4 = 20.0 kg = 20 x 10^3 g

moles of PbSO4 = 20 x 10^3 / 303.26 = 65.95 mol

volume = 100 L

Molarity = 65.95 mol /100

             = 0.660 M

            = 0.00200 g / mL

solubility = 0.00425 g / 100 mL

              = 0.0425 g / L

              = 1.40 x 10^-4 M

concentration of Pb2+ = 0.660 M

                                     = 0.00200 g / mL