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)
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
Lead(II) sulfate is the crust that forms on battery terminals, its solubility in water is 0.00425...
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